a-math-360-sol-29-sept-2014.pdf

Additional Maths 360 solutions (Unofficial) [29 Sept 2014] Visit sleightofmath.com for up-to-date pdf and video solutions. Authors: Daniel and Samuel from Sleight of Math Disclaimer: Sleight of Math has no affiliations with the publisher of Additional Maths 360 i.e. Marshall Cavendish

Table of Contents Ex 1.1 ......................................................................................................... 3

Ex 6.1 ................................................................................................... 143

Ex 1.2 ......................................................................................................... 9

Ex 6.2 ................................................................................................... 149

Ex 1.3 .......................................................................................................20

Ex 6.3 ................................................................................................... 153

Ex 1.4 .......................................................................................................29

Ex 6.4 ................................................................................................... 162

Rev Ex 1..................................................................................................35

Rev Ex 6 .............................................................................................. 167

Ex 2.1 .......................................................................................................42

Ex 7.1 ................................................................................................... 174

Ex 2.2 .......................................................................................................51

Ex 7.2 ................................................................................................... 180

Ex 2.3 .......................................................................................................55

Ex 7.3 ................................................................................................... 184

Ex 2.4 .......................................................................................................60

Ex 7.4 ................................................................................................... 189

Rev Ex 2..................................................................................................63

Ex 7.5 ................................................................................................... 193 Rev Ex 7 .............................................................................................. 196

Ex 3.1 .......................................................................................................67 Ex 3.2 .......................................................................................................72

Ex 8.1 ................................................................................................... 202

Ex 3.3 .......................................................................................................78

Ex 8.2 ................................................................................................... 209

Ex 3.4 .......................................................................................................81

Rev Ex 8 .............................................................................................. 224

Ex 3.5 .......................................................................................................86 Ex 3.6 .......................................................................................................91

Ex 9.1 ................................................................................................... 233

Rev Ex 3..................................................................................................98

Ex 9.2 ................................................................................................... 242 Rev Ex 9 .............................................................................................. 252

Ex 4.1 .................................................................................................... 103 Ex 4.2 .................................................................................................... 108

Ex 10.1 ................................................................................................. 259

Rev Ex 4............................................................................................... 114

Ex 10.2 ................................................................................................. 266 Ex 10.3 ................................................................................................. 272

Ex 5.1 .................................................................................................... 118

Rev Ex 10............................................................................................ 280

Ex 5.2 .................................................................................................... 127 Rev Ex 5............................................................................................... 138

Ex 11.1 ................................................................................................. 288 Ex 11.2 ................................................................................................. 291 Ex 11.3 ................................................................................................. 298 Rev Ex 11............................................................................................ 319

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1

Additional Maths 360 solutions (Unofficial) [29 Sept 2014] Visit sleightofmath.com for up-to-date pdf and video solutions. Authors: Daniel and Samuel from Sleight of Math Disclaimer: Sleight of Math has no affiliations with the publisher of Additional Maths 360 i.e. Marshall Cavendish Ex 12.1 ................................................................................................. 326

Ex 18.1 ................................................................................................. 495

Ex 12.2 ................................................................................................. 330

Ex 18.2 ................................................................................................. 503

Rev Ex 12 ............................................................................................ 339

Ex 18.3 ................................................................................................. 509 Ex 18.4 ................................................................................................. 512

Ex 13.1 ................................................................................................. 347

Rev Ex 18............................................................................................ 517

Ex 13.2 ................................................................................................. 356 Ex 13.3 ................................................................................................. 366

Ex 19.1 ................................................................................................. 523

Rev Ex 13 ............................................................................................ 376

Ex 19.2 ................................................................................................. 530 Rev Ex 19............................................................................................ 535

Ex 14.1 ................................................................................................. 383 Ex 14.2 ................................................................................................. 390

Ex 20.1 ................................................................................................. 539

Ex 14.3 ................................................................................................. 396

Rev Ex 20............................................................................................ 548

Ex 14.4 ................................................................................................. 400 Rev Ex 14 ............................................................................................ 407 Ex 15.1 ................................................................................................. 410 Ex 15.2 ................................................................................................. 418 Ex 15.3 ................................................................................................. 421 Ex 15.4 ................................................................................................. 423 Rev Ex 15 ............................................................................................ 430 Ex 16.1 ................................................................................................. 436 Ex 16.2 ................................................................................................. 445 Rev Ex 16 ............................................................................................ 454 Ex 17.1 ................................................................................................. 460 Ex 17.2 ................................................................................................. 470 Ex 17.3 ................................................................................................. 479 Rev Ex 17 ............................................................................................ 488


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2

A math 360 sol (unofficial)

Ex 1.1 2(a)

Ex 1.1 1(a)

y = 2x + 1 −(1) 2 y = x + 2x − 3 −(2) sub (1) into (2): 2x + 1 = x 2 + 2x − 3 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 ✓ or x = 2 ✓ y|x=−2 = 2(−2) + 1 y|x=2 = 2(2) + 1 = −3 ✓ =5✓

1(b)

y=2+x y = 2x 2 − 5x − 6

y=2−x 2x 2 + xy + 1 = 0 sub (1) into (2): 2x 2 + x(2 − x) + 1 2x 2 + 2x − x 2 + 1 x 2 + 2x + 1 (x + 1)2 x = −1 y|x=−1 = 2 − (−1) ⇒ (−1,3) ✓

2(b)

−(1) −(2)

y = 1 − 3x x2 + y2 = 5

−(1) −(2)

=5 =5 =0 =0 =0 or

5

x=1

2

y|x=−2 = 1 − 3 (− )

y|x=1 = 1 − 3(1)

5

5

11

= −2

5 2 11

2x + y = 4 y = 4 − 2x −(1)

⇒ (− , 5

2(c)

5

)✓

⇒ (1, −2) ✓

3x + 2y = 1 2y = 1 − 3x y

=

1−3x 2

−(1)

3x 2 + 2y 2 = 11 −(2) sub (1) into (2): 3x 2 + 2 ( 3x 2 + 2 (

1−3x 2

)

2 1−6x+9x2 4

= 11 )

= 11

6x 2 + (1 − 6x + 9x 2 ) = 22 15x 2 − 6x − 21 =0 5x 2 − 2x − 7 =0 (5x − 7)(x + 1) =0 x=

7

or x = −1

5

y|x=7 =

7 5

1−3( )

5

=− 7

8

5

5

2 8 5

⇒ ( ,− ) ✓


=3

2

x=−

=

y 2 − 4x = 0 −(2) sub (1) into (2): (4 − 2x)2 − 4x =0 2 (16 − 16x + 4x ) − 4x = 0 4x 2 − 20x + 16 =0 2 x − 5x + 4 =0 (x − 1)(x − 4) =0 x=1 or x = 4 ✓ y|x=1 = 4 − 2(1) y|x=4 = 4 − 2(4) =2✓ = −4 ✓

=0 =0 =0 =0

sub (1) into (2): x 2 + (1 − 3x)2 x 2 + (1 − 6x + 9x 2 ) 10x 2 − 6x − 4 5x 2 − 3x − 2 (5x + 2)(x − 1)

sub (1) into (2): 2+x = 2x 2 − 5x − 6 2 2x − 6x − 8 = 0 x 2 − 3x − 4 =0 (x + 1)(x − 4) = 0 x = −1 ✓ or x = 4 ✓ y|x=−1 = 2 + (−1) y|x=4 = 2 + (4) =1✓ =6✓ 1(c)

−(1) −(2)

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y|x=−1 =

1−3(−1) 2

=2 ⇒ (−1,2) ✓

3

A math 360 sol (unofficial) 3

x2

y

Ex 1.1

+ =3

−(1)

x + y= 8 y =8−x

−(2)

4

3

5(ii)

Perimeter 2x + 2y x+y y

sub (2) into (1): x2 4

+

(8−x)

(3x 2 ) + (32 − 4x) = 36 3x 2 − 4x − 4 =0 (3x + 2)(x − 2) =0 2 3

4(i) 𝑦

4x + 4y = 32 x+y =8 y =8−x ✓

−(1)

x 2 + y 2 = 34 ✓

−(2)

4(iii) sub (1) into (2): x 2 + (8 − x)2 = 34 2 2 x + (64 − 16x + x ) = 34 2x 2 − 16x + 30 =0 2 x − 8x + 15 =0 (x − 3)(x − 5) =0 x=3 or x=5 y|x=3 = 5✓ y|x=5 = 3✓ ∴ the length of the sides are 3cm & 5cm ✓ 5(i)

= 60 = 60 = 30 = 30 − x −(2)

sub (2) into (1): x(30 − x) = 216 2 (30x − x ) = 216 2 x − 30x + 216 = 0 (x − 12)(x − 18) = 0 x = 12 or y|x=12 = 30 − (12) = 18 ∴ 12 m by 18 m ✓

x = − or x = 2 ✓

4(ii)

−(1)

=3

3

𝑥

Area = 216 xy = 216

𝑦 𝑥 Area = xy ✓ Perimeter = 2x + 2y ✓


6(a)

x = 18 y|x=18 = 30 − (18) = 12

3x − 2y = 1 2y = 3x − 1 y

=

3x−1 2

−(1)

(x − 2)2 + (2y + 3)2 = 26 sub (1) into (2): (x − 2)2

+ [2 (

3x−1 2

) + 3]

−(2) 2

= 26

x 2 − 4x + 4 + (3x − 1 + 3)2 = 26 x 2 − 4x + 4 + (3x + 2)2 = 26 2 2 (9x x − 4x + 4 + + 12x + 4) = 26 10x 2 + 8x + 8 = 26 10x 2 + 8x − 18 = 0 5x 2 + 4x − 9 =0 (5x + 9)(x − 1) = 0 x=−

9

or

5

y|x=−9 =

9 3(− )−1 5

5

=−

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2 16 5

✓

x=1✓ y|x=1 =

3(1)−1 2

=1✓

4

A math 360 sol (unofficial) 6(b)

Ex 1.1

x 2 − 2xy + y 2 = 1 x − 2y −2y

=2 = −x + 2

y

= x−1

7(a)

−(1)

xy + 20 = 5x

−(1)

x − 2y − 3 = 0 −2y = −x + 3

1

y

−(2)

2

1

3

2

2

= x−

−(2)

sub (1) into (2): 1

1

x 2 − 2x ( x − 1) + ( x − 1) 2

2

2

x − x + 2x 2x 1 2 x 4 1 2 x 4 2

sub (2) into (1):

=1

+x

=0

1

y|x=0 = (0) − 1

2 1 2 x 2 2

2

= −1 ✓

−

2 13 2

x + 20

1

3

2

2

y|x=5 = (5) −

=0

2

= −3 ✓ 7(b)

1

= x + 1 −(1)

=

⇒ (5,1) ✓

1

3

2 5

2

y|x=8 = (8) −

=1

1

= (−4) − 1

3y − x = 3 3y =x+3 y

= 5x

x − 13x + 40 =0 (x − 5)(x − 8) =0 x=5 or x = 8

x = −4 ✓ y|x=−4

2 3

1

=0 =0 or

3

2

( x 2 − x) +20 = 5x

=1 =1

1

x ( x − ) + 20

=1

+x+1

x + 4x x(x + 4) x=0✓

6(c)

2

2 1 + ( x 2 − x + 1) 4 1 2 +( x − x + 1) 4

2 5

⇒ (8, ) ✓ 2

2x − y = 4 −y = −2x + 4 y = 2x − 4

−(1)

2x 2 + 4xy − 3y = 0

−(2)

3

2

1

3y

− =2

−(2)

x

sub (1) into (2): 2

−

1 3

3( x+1) 2

−

x+3

1 1

=2

x

2x −(x + 3) 2x −x − 3 x−3 0 2x 2 + 5x + 3 (2x + 3)(x + 1) 3

x=− ✓

= 2x(x + 3) = 2x 2 + 6x = 2x 2 + 6x = 2x 2 + 5x + 3 =0 =0 or

2

1

3

3 1

2

y|x=−3 = (− ) + 1 2

sub (1) into (2): 2x 2 + 4x(2x − 4) − 3(2x − 4) 2x 2 + (8x 2 − 16x) −6x + 12 (10x 2 − 16x) −6x + 12 10x 2 − 22x + 12 5x 2 − 11x + 6 (5x − 6)(x − 1)

=2

x

= ✓ 2


x=

6

or

5 6

y|x=6 = 2 ( ) − 4

x = −1 ✓

5

=−

1

y|x=−1 = (−1) + 1 3 2

6

8

5

5

5 8 5

⇒ ( ,− ) ✓

= ✓

=0 =0 =0 =0 =0 =0

x=1 y|x=1 = 2(1) − 4 = −2 ⇒ (1, −2) ✓

3

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5

A math 360 sol (unofficial) 7(c)

Ex 1.1

3x + y = 1 y = 1 − 3x

9 4𝑥

By Pythagoras Theorem, (4x)2 + (3x)2 = 152 16x 2 + 9x 2 = 225 25x 2 = 225 2 x =9 x = 3 or x = −3 (rej)

(x + y)(x + 2y) = 3 −(2) (1) (2): sub into [x + (1 − 3x)][x + 2(1 − 3x)] = 3 (1 − 2x)(x + 2 − 6x) =3 (1 − 2x)(2 − 5x) =3 (5x − 2)(2x − 1) =3 2 10x − 9x + 2 =3 2 10x − 9x − 1 =0 (10x + 1)(x − 1) =0 x=−

1

y|x=− 1 = 1 − 3 (− 10

= ⇒ (− 8

2x y

,

1

)

10

y|x=1 = 1 − 3(1)

13 10 13

)✓

⇒ (1, −2) ✓

10 10

y

−(1)

x

3x − y = 2 3x − 2 = y y = 3x − 2 −(2) sub (2) into (1): 2x

+

10(i) 𝑦

= −2

+ =3

3x−2 2

Width = 4(3) = 12 inch ✓ Height = 3(3) = 9 inch ✓

or x = 1

10

1

3𝑥

−(1)

3x−2

=3

x 2

2x + (3x − 2) = 3(3x 2 − 2x) 2x 2 + (9x 2 − 12x + 4) = 9x 2 − 6x 11x 2 − 12x + 4 = 9x 2 − 6x 2 2x − 6x + 4 =0 2 x − 3x + 2 =0 (x − 1)(x − 2) =0 x=1 or x = 2 y|x=1 = 3(1) − 2 y|x=2 = 3(2) − 2 =1 =4 ⇒ A(1,1) ✓ ⇒ B(2,4) ✓


5 𝑥

By Pythagoras’ Theorem, x 2 + y 2 = 52 x 2 + y 2 = 25 ✓ −(1) 10(ii) y − x = 1 y =x+1 −(2) sub (1) into (2): x 2 + (x + 1)2 = 25 2 2 x + (x + 2x + 1) = 25 2x 2 + 2x + 1 = 25 2x 2 + 2x − 24 =0 2 x + x − 12 =0 (x + 4)(x − 3) =0 x = −4 ✓ or x=3✓ y|x=−4 = (−4) + 1 y|x=3 = (3) + 1 = −3 ✓ =4✓ 10(iii) ∵ length cannot be negative, x = 3, y = 4

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6


Ex 1.1

11(i) x 2 + xy + ay = b at {x = 2, y = 1} (2)2 + (2)(1) + a(1) = b 6+a =b b =a+6 −(1) 2ax + 3y = b at {x = 2, y = 1} 2a(2) + 3(1) = b 4a + 3 =b sub (1) into (2): 4a + 3 = 6 + a 3a =3 a =1✓ b|a=1 = 7 ✓

12(i) 1st eqn 12x 2 − 5y 2 = 7 At (1, p), 12(1)2 − 5(p)2 = 7 12 − 5p2 =7 2 −5p = −5 p2 =1 p = ±1

−(2)

2nd eqn 2p2 x − 5y = 7 At (1, p), 2p2 (1) − 5(p) = 7 2p2 − 5p − 7 = 0 (2p − 7)(p + 1) = 0

11(ii) Put a − 1, b = 7 into both equations, x 2 + xy + y = 7 −(1)

2

7

3

3

= − x + −(2)

2

7

3

3 7

x + x (− x + ) 2 x + (− x 2 + x) 3 3 1 2 7 x + x 3 3 1 2 5 7 2

x + x+

3 1 2 x 3 2

3 5

3 14

3

3

+ x−

or p = −1

2

∴ p = −1 (common sol) ✓

−(1)

2x − 5y = 7 −5y = −2x + 7

sub (2) into (1): 2

7

12(ii) At p = −1, 12x 2 − 5y 2 = 7

2x + 3y = 7 3y = −2x + 7 y

p=

y 2

7

3 2

3 7

3 2

3 7

3

3

2

7

5

5

= x−

+ (− x + ) = 7 + (− x + ) = 7

sub (2) into (1):

+ (− x + ) = 7

12x 2 − 5 ( x − )

=7

12x 2 − 5 (

=0

x + 5x − 14 = 0 (x + 7)(x − 2) = 0 x = −7 or x = 2 (taken) 2

7

3

3

y|x=−7 = − (−7) + =7 {x = −7, y = 7} ✓

2

7 2

5

5

4

x2 −

25

12x 2 − x 2 +

4

28

5

5

56 2 x 5

+

49

56 2 x 5

+

28 5 28 5

−(2)

x− x−

=7

28 25

x−

x+ 49 5

5

=0

(2x + 3)(x − 1)

=0

3

or

2

x = 1 (taken)

2

3

7

5

2

5

y|x=−3 = (− ) − 2

=7

=0

2x 2 + x − 3

x=−

) =7

=7

5 84

49

25

= −2 3

⇒ (− , −2) ✓ 2


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7


Ex 1.1

13(i)

14(ii) If k = 1, 𝑦 h

(7𝑥 − 20)2 + (6𝑦 + 10)2 = 200

r Total surface area = 32π 2πr 2 + 2πrh = 32π 2 r + hr = 16 [shown] ✓ −(1) 13(ii) h = 4 + r sub (2) into (1): r 2 + (4 + r)r = 16 2r 2 + 4r = 16 2 2r + 4r − 16 = 0 r 2 + 2r − 8 = 0 (r + 4)(r − 2) = 0 r = −4 or (rej ∵ r > 0)

=

y r=2✓ h|r=2 = 4 + (2) =6✓

6

2

) + 10]

(7x − 20) + (20 − 21x)

2

2

(x−2.86)2 (2.02)2

=0

49x 2 − 112x + 60

=0

−(−112)±√(−112)2 −4(49)(60) 2(49)

=

112±√784 98

6

or x = ✓ 7

✓


6

+

+

5 2 62 (y+ ) 3 5 2 (y+ ) 3 200 62

=1 2

5 3

(y−(− ))

=1

2 200 (√ 2 ) 6 2

5 3

(y−(− )) (

+

=1

200

10√2 ) 6

(y−(−1.67))

=1

2

2

=1

(2.36)2 20 7

5

,− ) 3

⇒ horizontal radius = √

200 72

200

490x 2 − 1120x + 600

3

+

⇒ centre (

= 200 = 200

=−

1

2

5 2

+

20 2 7 2 10√2 ( ) 7

= 200

490x − 1120x + 800

6 10

7

3

(x− )

= 200

2

10 10−21( ) 7

7

20 2 7 2 200 (√ 2 ) 7

−(2)

(49x 2 − 280x + 400) +(400 − 840x + 441x 2 )

7

6

2

(x− )

(7x − 20)2 + [(10 − 21x) + 10]2 = 200

y|x=10 =

1

2

) + 62 (y + ) = 200

20 2 (x− ) 7 200 72

−(1)

10−21x

7

7

20 2

72 (x −

sub (1) into (2):

x=

3

Ellipse (7x − 20)2 +(6y + 10)2 = 200

200

6

10

,− )

=− x+

20 72 (x− ) 7

Ellipse (7x − 20)2 + (6y + 10)2 = 200

x=

7

5

Line 21x + 6y = 1 6y = −21x + 1

10−21x

(7x − 20)2 + [6 (

(

20

𝑦=− 𝑥+

−(2)

14(i) If k = 10, Line 21x + 6y = 10 6y = 10 − 21x y

𝑥

𝑂

y|x=6 =

⇒ vertical radius = √

62

Using the graphing calculator to plot curves, the line and ellipse don’t intersect. Hence, there are no solutions ✓ =

112±28 98

14(iii) Geometrically, a line and ellipse can only intersect twice at most. ✓

6 7

10−21( ) 6

7

=−

4 3

✓

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8


Ex 1.2 4(i)

Ex 1.2 1(a)

3x 2 + 9x =1 2 3x + 9x − 1 = 0 i.e. a = 3, b = 9, c = −1 Roots: α & β Sum of roots

1(b)

Sum of roots

b a

c

=

=

a

(−1) (3)

=− ✓ 3

1 2

=( )

α

2(ii)

3

b a

=−

(−4)

=4✓

(1)

(2)

c

2

2β+2α

a

1

= (4 )

β

αβ

+ =

=

αβ

=

5(i) b a

= (1)

a

2(−3) (1)

(3)

= − (1) = −3 (1)

c

= =

=1

= −6 ✓

√33 2

✓

x 2 − 4x + c = 0 i.e. a = 1, b = −4, c = c Roots: α & (α + 2)

=−

b a (−4) (1)

2α + 2 =4 2α =2 α =1✓ α+2 =3✓ 5(ii)

Sum of roots = α + β = −

=

4

=±

⇒ 2α + 2

40x − 138x + 119 = 0 i.e. a = 40, b = −138, c = 119 Roots: α & β are heights of two men

2

= −2

Sum of roots

2

Average height =

2

33

= α + (α + 2) = −

(2α − 1)(2β − 1) = 4αβ −2α − 2β +1 = 4αβ −2(α + β) +1 = 4(1) −2(−3) +1 = 11 ✓

α+β

1

−2(−2)

4

α−β

=α+β =−

2(α+β)

(2)

=

(α − β)2 = α2 − 2αβ + β2 = (α2 + β2 ) −2αβ

=2✓

= (1)

x + 3x + 1 = 0 i.e. a = 1, b = 3, c = 1 Roots: α & β

2

(2) (−4)

4

=

Product of roots = αβ

=

(−1)

=4 ✓

2

Sum of roots

a

=−

−2(−2)

2 1

=α+β =−


=

a

(α + β)2 = α2 + 2αβ + β2 (α + β)2 − 2αβ = α2 + β2 α2 + β2 = (α + β)2 −2αβ

1

4x + 2x 2 = 3x 2 + 2 x 2 − 4x + 2 = 0 i.e. a = 1, b = −4, c = 2 Roots: α & β Sum of roots

c

b

(9)

= − (3) = −3 ✓

4(ii)

2(i)

=α+β =−


=α+β =−


2x 2 − x − 4 = 0 i.e. a = 2, b = −1, c = −4 Roots: α & β

69 20

2

b a

=

=− 69 40

Product of roots c = α(α + 2) = a

⇒ α(α + 2) = c 1(1 + 2) =c ∵α=1 c =3✓

(−138) (40)

=

69 20

6(a)

✓

Roots: α = 2 & β = 5 Sum of roots = α + β = (2) + (5) Product of roots = αβ = (2)(5)

=7 = (10)

x 2 − (SOR)x + (POR) = 0 x 2 − 7x + 10 =0✓


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9


Ex 1.2

Roots: α = −1 & β = 3 Sum of roots = α + β = (−1) + (3) = 2 Product of roots = αβ = (−1)(3) = −3

8(i)

x 2 − (SOR)x + (POR) = 0 x 2 − 2x − 3 =0✓ 7(i)

Sum of roots

1st equation 2x 2 − 4x + 5 = 0 i.e. a = 2, b = −4, c = 5 Roots: α & β

=−

= 5α

= −p ✓ =

= 4α2

=α+β =−


=

c a

b a

=−

(−4)

(5)

= (2)

(2)

=2 =

From (1): 5α = −p p α =−

a

−(1)

c a

=q✓

−(2)

−(3)

5

5

b

= α + 4α

Product of roots = α(4α)

8(ii) Sum of roots

x 2 + px + q = 0 i.e. a = 1, b = p, c = q Roots: α & 4α

2

sub (3) into (2): 2nd equation Roots: (α − 1) & (β − 1) Sum of roots Product of roots = (α − 1) + (β − 1) = (α − 1)(β − 1) = (α + β) −2 = αβ −α − β +1 (2) = −2 = αβ −(α + β) +1 5 =0 = ( ) −(2) +1

5

4p2

4p 9(i)

3 2

x − (SOR)x + (POR) = 0 3

2x 2 + 3 7(ii)

2x 2 − x − 2 = 0 i.e. a = 2, b = −1, c = −2 Roots: α & β =α+β =− =

b a

c

=− =

a

(−1)

(2) (−2) (2)

=

1 2

= −1

=0✓ α2 + β2 = (α + β)2 −2(αβ)

3rd equation Recall α + β = 2,

= 25q [shown] ✓


=0

2

2

Sum of roots

2

x 2 − 0x +

=q

25

2

=

2

1

4 (− p) = q

αβ =

5

1 2

=( )

2

−2(−1)

2 1

Roots: 2α & 2β

=2 ✓ 4

Sum of roots = 2α + 2β = 2(α + β) = 2(2) = 4

9(ii)

β α

+

α β

=

β2 +α2 αβ

1

(2 )

4 = (−1) = −2

1 4

✓

Product of roots = (2α)(2β) = 4(αβ)

5

= 4 ( ) = 10

9(iii)

2

α4 + β4

= (α2 + β2 )2 −2α2 β2 = (α2 + β2 )2 −2(αβ)2

x 2 − (SOR)x + (POR) = 0 x 2 − 4x + 10 =0 ✓

1 2

= (2 ) 4

=


sleightofmath.com

49 16

−2(−1)2

✓

10

A math 360 sol (unofficial) 10(i)

Ex 1.2

x2 = (k − 1)x + k 2 (1 x + − k)x − k = 0 i.e. a = 1, b = (1 − k), c = −k Roots: α & (α + 3)

12(i)

Sum of roots = α + (α + 3) = −

Sum of roots

b

=−

2α + 3 k

=k−1 = 2α + 4

=α+β =−


a (1−k)

⇒ 2α + 3

=

b

=−

a

c

=

a

−3 2

6

=

3 2

=3

2

(1)

2nd equation −(1)

Roots:

2α β

&

2β α

Sum of roots =

Product of roots c = α(α + 3) = ⇒ α(α + 3) =

1st equation 2x 2 − 3x + 6 = 0 i.e. a = 2, b = −3, c = 6 Roots: α & β

a (−k)

=

2α β

+

2β α

=

2α2 +2β2 αβ

2[(α+β)2 −2αβ]

(1)

αβ 2α

2β

β

α

=

=

2(α2 +β2 ) αβ

3 2 2[( ) −2(3)] 2

3

=−

5 2

Product of roots = ( ) ( ) = 4

= −k −(2) sub (1) into (2): α(α + 3) = −(2α + 4) 2 α + 3α = −2α − 4 2 α + 5α + 4 =0 (α + 1)(α + 4) = 0 α = −1 or α = −4 ✓ α+3=2 α + 3 = −1 ✓ (rej ∵ negative roots)

x 2 − (SOR)x + (POR) = 0 5

x 2 − (− ) x +4

=0

2

5

x2 + x 2

2

2x +5x

+4

=0

+8

=0✓

12(ii) 3rd equation 3

Recall α + β = , αβ = 3 2

10(ii) Put α = −4 into (1): k|α=−4 = 2(−4) + 4 = −4 ✓ 11

Roots: (3α + β) & (α + 3β) Sum of roots = (3α + β) + (α + 3β) = 4α + 4β = 4(α + β)

3x 2 − 3kx + k − 6 = 0 i.e. a = 3, b = −3k, c = k − 6 Roots: α & β

3

= 4( ) 2

Sum of roots

=α+β =−

Product of roots = αβ α2 + β2

=

2

(α + β) − 2αβ k

2

−2 (

k−6 3

)

= =

=

c a

b a

=− =

(−3k)

(3) (k−6) (3)

=6 =k =

k−6 3

20 3 20 3 20

3 2

= 3 [( ) − 2(3)] 2

3

3k 2 − 2k + 12 = 20 2 3k − 2k − 8 =0 (3k + 4)(k − 2) = 0 k=−

4 3

Product of roots = (3α + β)(α + 3β) = 3α2 + 10αβ + 3β2 = 3(α2 + β2 ) +10αβ = 3[(α + β)2 − 2αβ] +10αβ

=

75 4

x 2 − (SOR)x + (POR) = 0

or k = 2 ✓

x 2 −(6)x +

75 4

4x 2 −24x +75


+10(3)

sleightofmath.com

=0 =0✓

11


Ex 1.2

1st equation 2x 2 = 8x + 3 2x 2 − 8x − 3 = 0 i.e. a = 2, b = −8, c = −3 Roots: α & β

13(iii) 4th equation Recall α + β = 4, αβ = −

=α+β =−

Product of roots

= αβ

=

b a

c

=− =−

a

(−8) (2) 3

Sum of roots = (α − β) + (β − α) =0

=4

Product of roots = (α − β)(β − α) = −(a − β)2 = −[(α + β)2 − 4αβ]

2

2nd equation 1 α2

&

1

= − [(4)2

β2

Sum of roots = =

1

+

α2

1

=

β2

(α+β)2 −2(αβ) (αβ)2

1

=

α2 +β2 α2 β2 3 2

(4)2 −2(− )

1

3

− 4 (− )] 2

= −22

3 2 (− ) 2

1

Product of roots = ( 2 ) ( 2 ) = (αβ)2 = α

2

Roots: (α − β) & (β − α)

Sum of roots

Roots:

3

β

=

1 3 2 (− ) 2

x 2 − (SOR)x + (POR) = 0 x 2 − (0)x + (−22) = 0 x 2 − 22 =0✓

76 9

=

4

14(i)

9

2x 2 − 71x + 615 = 0 i.e. a = 2, b = −71, c = 615 Roots: α & β are base and height

x 2 − (SOR)x + (POR) = 0 76

4

9

9

x2 − ( ) x + ( )

=0

Sum of roots = α + β = −

b

=−

a

2

9x − 76x +4 = 0 ✓


=

c a

−71

=

2 615 2

71

=

2

= 307.5

13(ii) 3rd equation Recall

α + β = 4, αβ = −

Roots:

α2 β & αβ2

Area = αβ = 307.5 ✓

3 2

71

Perimeter = 2(α + β) = 2 ( ) = 71 ✓ 2

Sum of roots = α2 β + αβ2 = αβ(α + β)

14(ii)

3

= (− ) (4) = −6

Base

2

Product of roots 3 3

27

2

8

= (α2 β)(αβ2 ) = (αβ)3 = (− ) = −

) =0

8x 2 + 48x − 27


Base

Perimeter is not valid because the height is not the side of the quadrilateral ✓

27 8

Height

Area is valid because area = base × height

x 2 − (SOR)x + (POR) = 0 x 2 − (−6)x + (−

Height

=0✓

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12


Ex 1.2

kx 2 + (k − 1)x + 2k + 3 = 0 i.e. a = k, b = (k − 1), c = (2k + 3) Roots: α & 2α Sum of roots = α + (2α) = − ⇒ 3α

=−

3α

=

α

=

1st equation x 2 − 3x − 2 = 0 i.e. a = 1, b = −3, c = −2 Roots: α & β

b a k−1

Sum of roots

k


1−k k 1−k 3k

−(1)

2nd equation x 2 − 6x + p = 0 i.e. a2 = 1, b2 = −6, c2 = p

−(2)

Roots:

Product of roots c = α(2α) = 2

⇒ 2α

=

a 2k+3 k

2( 2(

)

3k k2 −2k+1

9k2 2k2 −4k+2 9k2 2

)

= = =

k

2k+3

= +

k 2k+3

⇒

1 16

k

α β kβ+kα

αβ k(α+β)

k 2k+3

&

c a

a

=− =

−3

−2 1

1

=3 = −2

k β

=−

b2 a2 −6 1

=6 = −4 ✓

k

or k = −2 ✓

=−

=6

αβ k(3) (−2)

k

2k − 4k + 2 = 18k 2 + 27k 16k 2 + 31k − 2 = 0 (16k − 1)(k + 2) = 0 k=

k α

=

b

Sum of roots

sub (1) into (2): 1−k 2

=α+β =−

Product of roots

rej ∵ k is a ( ) non − zero integer

k

k

c2

α

β

a2 (p)

= ( )( ) = ⇒

k2 αβ

k2 αβ

= (1) =p

k2 (−2)

=p

p

=− =−

k2

∵ k = −4

2 (−4)2 2

= −8 ✓


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13


Ex 1.2

17(a) 3x 2 + kx + 96 = 0 i.e. a = 3, b = k, c = 96 Roots: α & 2α

18

5x 2 − 102x + 432 = 0 i.e. a = 5, b = −102, c = 432 Roots: α & β are lengths of the two shorter sides of a ⊿

Sum of roots = α + 2α = −

b

Sum of roots

a k

⇒ 3α

=−

α

=−

−9α k

=k = −9α


3 k

⇒ 2α

=

a

a

=− =

(−102) (5)

=

102 5

432 5

√α2 + β2

α β

By Pythagoras Theorem, Hypotenuse = √α2 + β2

a 96 3

2

2α = 32 2 α = 16 α = 4 or α = −4 (rej ∵ positive roots) k|α=4 = −9(4) = −36 ✓ 2

=

c

b

9

Product of roots c = α(2α) = 2

=α+β =−

1 432

2

2

5

)= 43.2cm2 ✓

Perimeter = α + β + √α2 + β2 = α + β + √(α + β)2 − 2αβ =(

2

17(b) p + q = 13 pq = 6 Roots: p2 & q2 Sum of roots = p2 + q2 = 13 Product of roots = p2 q2 = (pq)2 = 62

1

Area = (αβ) = (

102 5

) + √(

102 2 5

) − 2(

432 5

)

= 36cm ✓

= 36

x 2 − (SOR)x + (POR) = 0 x 2 − 13x + 36 =0 (x − 4)(x − 9) =0 x=4 or x = 9 p2 = 4 p2 = 9 p = ±2 p = ±3 ✓


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14


Ex 1.2

19(a) 2x 2 + px + 24 = 0 i.e. a = 2, b = p, c = 24 Roots: α & β

19(b) 1st equation 3x 2 − x + 2 = 0 i.e. a = 3, b = −1, c = 2 Roots: α & β

α−β =4 α = β + 4 −(1)

Sum of roots

=α+β =−


Sum of roots =α+β

=−

⇒α+β

=−

(β + 4) + β = −

=

=−

a

c

=

a

(−1)

=

(3)

1 3

2 3

b a p

2nd equation Roots: α2 & β2

2 p 2 p

2β + 4

=−

p

= −4β − 8

Sum of roots = α2 + β2 = (α + β)2 −2(αβ)

2

1 2

2

=( )

=

−2 ( )

3

Product of roots c = αβ = ⇒ αβ

b

=−

a 24

3

11 9

2

(β + 4)β = 12 β2 + 4β = 12 2 β + 4β − 12 = 0 (β + 6)(β − 2) = 0 β = −6 or p|β=−6 = −4(−6) − 8 = 16 (rej ∵ p < 0)

2 2

4

3

9

Product of roots = α2 β2 = (αβ)2 = ( ) = x 2 − (SOR)x + (POR) = 0 x 2 − (−

11 9

)x +

4

=0

9

9x 2 + 11x + 4 β=2 p|β=2 = −4(2) − 8

20(i)

= −16 ✓

=0✓

1st equation 4x 2 − x + 36 = 0 i.e. a = 4, b = −1, c = 36 Roots: α2 & β2 = α2 + β2 = −

Sum of roots

2 2

Product of roots = α β

=

c a

b a

=− =

(−1)

(4) (36) (4)

=

1 4

=9

2nd equation Roots:

1 α2

&

1 β2

Sum of roots

=

1

1

α

1

β2 1

α

β

2 +

=

Product of roots = ( 2 ) ( 2 ) =

α2 +β2 α2 β2 1 α2 β2

= =

1 4

( ) (9) 1

=

1 36

9

x 2 − (SOR)x + (POR) = 0 x2 − 2

1 36

x+

1 9

36x − x + 4


sleightofmath.com

=0 =0✓

15


Ex 1.2

20(ii) 3rd equation: 2

1

2

Recall α + β = , 4

21(ii) 2nd equation Roots: (α + 2) & (β + 2)

2 2

α β =9

Roots: α & β Sum of roots =α+β


= ±√(α +

=

±√α2

β2

= ±√9 = ±3

+

+ 2αβ

1

= ±√( ) + 2αβ

=−

±√α2 β2

β)2

=

Sum of roots = (α + 2) + (β + 2) = (α + β) + 4 b

1

= [a(α + 2)(β + 2)] a 1

+4

a

= (4a − 2b + c) a

x 2 − (SOR)x

4

Product of roots = (α + 2)(β + 2)

+ (POR)

b

1

a

a 1

=0

x 2 − (− + 4) x + [ (4a − 2b + c)] = 0

For αβ = 3:

2

1

25

4

4

SOR = ±√ + 2(3) = ±√

=±

b

x + ( − 4) x

5

a

+ (4a − 2b + c) a

ax 2 + (b − 4a)x + (4a − 2b + c)

2

=0 =0✓

For αβ = −3: 1

23

4 5

24

SOR = ±√ + 2(−3) = ±√−

(rej)

∴ SOR = ± , POR = 3 2

x 2 − (SOR)x + (POR) = 0 5

x 2 − (± ) x + 3

=0

2

2

=0✓

2x ± 5x + 6 21(i)

1st equation ax 2 + bx + c = 0 Roots: α & β Sum of roots

=α+β =−


=

c

b a

a

a(α + 2)(β + 2) = a(αβ + 2α + 2β + 4) = a[αβ + 2(α + β) + 4] c

b

a

a

= a [( ) + 2 (− ) + 4] = c − 2b + 4a = 4a − 2b + c [shown] ✓


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16


Ex 1.2

1st equation 4x 2 + kx =1 4x 2 + kx − 1 = 0 i.e. a = 4, b = k, c = −1

22(ii) 3rd equation k

Recall α + β = − , Roots:

=α+β =−


=

b a

c

=− =−

a

k 4 1

=

4

Sum of roots

= 5(α + β)

9 9 = 1 αβ + 2(α + β) + 4 (− ) + 2 (− k) + 4 4 4 36 = 15 − 2k

= 5 (− ) 4

5

=− k 4

Product of roots

x2

= (2α + 3β)(3α + 2β) 2

= 6α + 13αβ + 6β

23(i)

+13αβ

1

1

4

4

1

k2 + )

−

2

8

Sum of roots

4

5

x 2 + kx 4

2

8x + 10kx

=α+β =−


x − (SOR)x +(POR) x − (− k) x

) =0 = 0✓

1st equation 2x 2 + 4x + 5 = 0 i.e. a = 2, b = 4, c = 5 Roots: α & β

4

4

5

15−2k

13

1

2 2

)x +(

+13αβ

4

= k −

15−2k

=0 36

(15 − 2k)x +(3k − 48)x +36

2)

k 2

2

−(

48−3k

2

= 6 [(− ) − 2 (− )] +13 (− )

3

−(SOR)x +(POR)

x2

2

= 6[(α + β)2 − 2αβ]

1

3β + 6 + 3α + 6 3(α + β) + 12 = αβ + 2(α + β) + 4 αβ + 2(α + β) + 4

=

k

16

β+2

3 3 9 )( )= =( (α + 2)(β + 2) α+2 β+2

= 5α + 5β

= 6(

4

3

Product of roots

= (2α + 3β) + (3α + 2β)

= 6(α + β

α+2

&

k 3 (− ) + 12 48 − 3k 4 = = 1 k (− ) + 2 (− ) + 4 15 − 2k 4 4

2nd equation Roots: (2α + 3β) & (3α + 2β)

2

3

1

Sum of roots 3 3 3(β + 2) + 3(α + 2) = + = (α + 2)(β + 2) α+2 β+2

Roots: α & β Sum of roots

αβ = −

4

3 + ( k2 8

=0 1

1

8

4

2

Roots:

4

3

+3k − 2

a

c

=−

4 2

= −2 ✓

5

= ✓

a

2

23(ii) 2nd equation

− )=0

+ k2 −

=

b

1 α

&

1 β

=0 Sum of roots

=0✓

1

1

α

β

= +

=

α+β αβ

1

1

1

α

β

αβ

Product of roots = ( ) ( ) =

= =

−2 5 2

1 5 2

=− =

4 5

2 5

x 2 − (SOR)x + (POR) = 0 4

2

5

5

x 2 − (− ) + 2

5x + 4 + 2


sleightofmath.com

=0 =0✓

17


Ex 1.2

23(iii) 1st equation ax 2 + bx + c = 0 Roots: α & β Sum of roots

25

=α+β =−


=

c

b a

Sum of roots

a

=α+β =−

2nd equation Roots:

1

&

α

2x 2 − kx + k =2 2 2x − kx + k − 2 = 0 i.e. a = 2, b = −k, c = (k − 2) Roots: α & β

1

⇒α+β =−

β

Sum of roots

1

1

α

β

= +

=

α+β αβ

1

1

1

α

β

αβ


= =

b − a c a

1

=− =

c a

b

b a −k 2

k

α+β

=

k

= 2α + 2β

2

a c

Product of roots c = αβ = ⇒ αβ =

x 2 − (SOR)x + (POR) = 0

a k−2 2

x 2 − (− ) x + ( )

=0

sub (1) into (2):

cx 2 − bx

=0✓

αβ =

b

a

c

c

+a

−(1)

c

−(2)

(2α+2β)−2 2

αβ = α + β − 1 24(i)

x(2 − x) =3 2 x − 2x + 3 = 0 α is root ⇒ α2 − 2α + 3 = 0 α2 = 2α − 3

If α < 0, β < 0 ⇒ LHS = αβ > 0 ⇒ RHS = α + β − 1 < 0 ⇒ LHS ≠ RHS ∴ both roots cannot be negative ✓

−(1)

(1) × α: α3 = 2α2 − 3α −(2) (1) (2): sub into α3 = 2(2α − 3) − 3α α3 = (4α − 6) − 3α α3 = α − 6 [shown] ✓ 24(ii) Roots: α & β Sum of roots

=α+β =−


=

c a

b a

=− =

(−2)

(3) (1)

(1)

=2 =3

Following the same manipulation in (i) ⇒ β3 = β − 6 α3 + β3 = (α − 6) + (β − 6) = (α + β) −12 = (2) −12 = −10 ✓


sleightofmath.com

18


Ex 1.2

x 2 + 4(c + 2) = (c + 4)x 2 (c x − + 4)x + 4(c + 2) = 0 i.e. A = 1, B = −(c + 4), C = 4(c + 2) Roots: a & b Sum of roots =a+b

=−

⇒a+b

=−

B A −(c+4) 1

a+b =c+4 2 (a + b) =c+4 2 2 a + 2ab + b = c 2 + 8c + 16

−(1)

Product of roots = ab = ⇒ ab = ab

C A 4(c+2) 1

= 4c + 8

−(2)

sub (2) into (1): a2 + 2(4c + 8) + b2 = c 2 + 8c + 16 a2 + (8c + 16) + b2 = c 2 + 8c + 16 a2 + b2 = c2 ∵ sides are related by pythagoras theorem, it is a right angle triangle & 90° is the largest angle ✓


sleightofmath.com

19


Ex 1.3 2(c)

Ex 1.3 1(a)

y = 3(x − 1)2 − 1 ⇒ turning pt (1, −1) ⇒ ∪ shape

𝑦 𝑦 = 3(𝑥 − 1)2 − 1 2 𝑂

𝑥 (1, −1)

y|x=0 = 2 ⇒ y − intercept = 2 1(b)

y = −2(x + 1)2 − 3 ⇒ turning pt (−1, −3) ⇒ ∩ shape y|x=0 = −5 ⇒ y − intercept = −5

1(c)

𝑂

1

(−2,1) 𝑂

y = −3(x − 2)2 ⇒ turning pt (2,0) ⇒ ∩ shape

𝑥

3(a) (2,0)

𝑂

𝑥

−12 y = −3(x − 2)2

Quadratic equation px 2 − 6x + p = 0 i.e. a = p, b = −6, c = p Discriminant For equal real roots: b2 − 4ac =0 (−6)2 − 4(p)(p) = 0 36 − 4p2 =0 2 p −9 =0 (p + 3)(p − 3) = 0 p = −3 or p = 3 ✓

✓

Quadratic equation 5x 2 − x − 2 = 0 i.e. a = 5, b = −1, c = −2

3(b)

Quadratic equation 3x 2 + 2x − p = 0 i.e. a = 3, b = 2, c = −p Discriminant For two distinct real roots: b2 − 4ac >0 (2)2 − 4(3)(−p) > 0 4 + 12p >0 12p > −4

Quadratic equation 9x 2 + 6x + 1 = 0 i.e. a = 9, b = 6, c = 1 Discriminant b2 − 4ac = (6)2 − 4(9)(1) =0✓ ⇒ 2 Real Roots ✓


Quadratic equation (x − 2)2 = −6 2 x − 4x + 4 = −6 x 2 − 4x + 10 = 0 i.e. a = 1, b = −4, c = 10 Discriminant b2 − 4ac = (−4)2 − 4(1)(10) = −24 ✓ <0 ⇒ 0 Real Root ✓

✓

𝑦

Discriminant b2 − 4ac = (−1)2 − 4(5)(−2) = 41 ✓ >0 ⇒ 2 Real roots ✓ 2(b)

2(d)

𝑦 1 𝑦 = (𝑥 + 2)2 + 1 4 2

4

y|x=0 = 2 ⇒ y − intercept = 2

2(a)

𝑥

✓

y = (x + 2)2 + 1

y|x=0 = −12 ⇒ y − intercept = −12

✓

−5 𝑦 = −2(𝑥 + 1)2 − 3

⇒ turning pt (−2,1) ⇒ ∪ shape

1(d)

Discriminant b2 − 4ac = (1)2 − 4(3)(1) = −11 ✓ <0 ⇒ 0 Real Root ✓

𝑦 (−1, −3)

Quadratic equation x 2 + x + 1 = −2x 2 3x 2 + x + 1 = 0 i.e. a = 3, b = 1, c = 1

p

sleightofmath.com

1

>− ✓ 3

20


Ex 1.3

Quadratic equation 2x 2 + 3x + 2p = 0 i.e. a = 2, b = 3, c = 2p

4(c)

Discriminant For real roots: b2 − 4ac ≥0 2 (3) − 4(2)(2p) ≥ 0 9 − 16p ≥0 16p ≤9 p 3(d)

≤

9 16

Discriminant For no x − intercepts: b2 − 4ac <0 2 (−2) − 4(p)(3) < 0 4 − 12p <0 12p >4 ✓ 5(a)

4(a)

1 16

✓

Quadratic curve y = 4x 2 − 4x − p i.e. a = 4, b = −4, c = −p

k 5(b)

1

> ✓ 2

Quadratic Inequality −3x 2 + 6x + k − 1 is always negative i.e. a = −3, b = 6, c = k − 1 2 conditions (i) a < 0 −3 < 0 (ii) b2 − 4ac (6)2 − 4(−3)(k − 1) 36 + 12(k − 1) 36 + 12k − 12 12k k

Quadratic curve y = 9x 2 − px + 1 i.e. a = 9, b = −p, c = 1 Discriminant For one x − intercept: b2 − 4ac =0 (−p)2 − 4(9)(1) = 0 p2 − 36 =0 (p + 6)(p − 6) = 0 p = −6 or p = 6 ✓


Quadratic Inequality 2x 2 + 2x + k is always positive i.e. a = 2, b = 2, c = k

(ii) b2 − 4ac <0 (2)2 − 4(2)k < 0 8k >4

Discriminant For two x − intercepts: b2 − 4ac >0 2 (−4) − 4(4)(−p) > 0 16 + 16p >0 p > −1 ✓ 4(b)

3

2 conditions (i) a > 0 2>0

Discriminant For no real roots: b2 − 4ac <0 2 (−1) − 4(p)(−4) < 0 1 + 16p <0 <−

1

> ✓

p

Quadratic equation px 2 − x − 4 = 0 i.e. a = p, b = −1, c = −4

p

Quadratic curve y = px 2 − 2x + 3 i.e. a = p, b = −2, c = 3

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<0 <0 <0 <0 < −24 < −2 ✓

21


Ex 1.3

Quadratic Inequality y = 2x 2 + x − 2k lies entirely above x − axis i.e. a = 2, b = 1, c = −2k

7(b)

2 conditions (i) a > 0 2>0

6

<−

1 16

Discriminant For curve to meet x − axis: B 2 − 4AC ≥0 2 (−2) − 4(3)(c − 1) ≥ 0 4 − 12(c − 1) ≥0 4 − 12c + 12 ≥0 16 ≥ 12c

13 (2,1)

✓ does not cut x − axis ⇒ discriminant < 0 ⇒ negative ✓ 7(c)

2 1

= (x − 4)2 2

⇒ turning pt (4,0) ⇒ ∪ shape f(0) = 8 ⇒ y − intercept = 8 𝑦 𝑦 = 𝑓(𝑥)

+3 +3

= −2[(x − 1)2 − 12 ] +3 = −2(x − 1)2 ⇒ turning pt (1,5) ⇒ ∩ shape

+8

= [(x − 4)2 − 42 ] +8

3

= −2(x 2 − 2x)

+8

2 1 2 1

4

f(x) = −2x 2 + 4x

1

f(x) = x 2 − 4x = [x 2 − 8x]

≤ ✓

c

𝑥

𝑂

≥c

3

7(a)

✓

Quadratic equation y = 3x 2 − 2x + c − 1 i.e. A = 3, B = −2, C = c − 1

4

+13 +13 +13 +1

f(0) = 13 ⇒ y − intercept = 13 𝑦 𝑦 = 𝑓(𝑥)

(ii) b2 − 4ac <0 (1)2 − 4(2)(−2k) < 0 1 + 16k <0 k

f(x) = 3x 2 − 12x = 3(x 2 − 4x) = 3[(x − 2)2 − 22 ] = 3(x − 2)2 ⇒ turning pt (2,1) ⇒ ∪ shape

+5

8 𝑂

(4,0)

𝑥 ✓

cuts x − axis once ⇒ discriminant = 0 ⇒ zero ✓

f(0) = 3 ⇒ y − intercept = 3 𝑦 (1,5) 3 𝑂

𝑦 = 𝑓(𝑥) 𝑥 ✓

cuts x − axis twice ⇒ discriminant > 0 ⇒ positive ✓


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22

A math 360 sol (unofficial) 7(d)

Ex 1.3

f(x) = −4x 2 − 8x = −4[x 2 + 2x] = −4[(x + 1)2 − 12 ] = −4(x + 1)2 ⇒ turning pt (−1, −1) ⇒ ∩ shape

−5 −5 −5 −1

8(c)

Discriminant For two unequal real roots: b2 − 4ac >0 (1)2 − 4(2)(1 − p) > 0 1 − 8(1 − p) >0 1 − 8 + 8p >0 8p >7

f(0) = −5 ⇒ y − intercept = −5 𝑦 𝑂

𝑥

(−1, −1) −5 𝑦 = 𝑓(𝑥)

p ✓

8(d)

does not cut x − axis ⇒ b2 − 4ac < 0 ⇒ negative ✓ 8(a)

Quadratic equation 3x 2 = 2x + p − 1 2 3x − 2x + 1 − p = 0 i.e. a = 3, b = −2, c = 1 − p Discriminant For distinct real roots: b2 − 4ac >0 2 (−2) − 4(3)(1 − p) > 0 4 − 12(1 − p) >0 4 − 12 + 12p >0 12p >8 p

2

8

Quadratic equation p(x + 1)(x − 3) = x − 4p − 2 2 p(x − 2x − 3) = x − 4p − 2 px 2 − 2px − 3p = x − 4p − 2 2 px − (2p + 1)x + p + 2 = 0 i.e. a = p, b = −(2p + 1), c = p + 2 Discriminant For no real roots: b2 − 4ac [−(2p + 1)]2 − 4(p)(p + 2) (4p2 + 4p + 1) − 4p(p + 2) (4p2 + 4p + 1) − 4p2 − 8p −4p + 1 −4p 4p

<0 <0 <0 <0 <0 < −1 >1

p

> ✓

1 4

3

Quadratic equation x 2 + p2 = 3px − 5 2 2 x − 3px + p + 5= 0 i.e. a = 1, b = −3p, c = p2 + 5

Quadratic equation x 2 − 2kx + k 2 =3+x 2 2 (2k x − + 1)x + k − 3 = 0 i.e. a = 1, b = −(2k + 1), c = −3 Discriminant For real roots: b2 − 4ac ≥0 2 2 [−(2k + 1)] − 4(1)(k − 3) ≥ 0 (4k 2 + 4k + 1) − 4(k 2 − 3) ≥ 0 4k 2 + 4k + 1 − 4k 2 + 12 ≥0 4k + 13 ≥0

Discriminant For equal real roots: b2 − 4ac =0 2 2 (−3p) − 4(1)(p + 5) = 0 9p2 − 4p2 − 20 =0 2 5p − 20 =0 p2 − 4 =0 (p + 2)(p − 2) =0 p = −2 or p = 2 ✓


7

> ✓

> ✓ 9

8(b)

Quadratic equation (x + 1)(2x − 1) = p − 2 2x 2 + x − 1 =p−2 2 2x + x + 1 − p = 0 i.e. a = 2, b = 1, c = 1 − p

k ⇒ least value of k is −

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≥− 13 4

13 4

✓

23


11

Ex 1.3

Quadratic equation 2x 2 + p = 2(x − 1) 2x 2 + p = 2x − 2 2 2x − 2x + 2 + p = 0 i.e. a = 2, b = −2, c = 2 + p

12(b) Quadratic inequality kx 2 + 1 > 2kx − k for all real values of x kx 2 − 2kx + k + 1> 0 i.e. a = k, b = −2k, c = k + 1 2 conditions (i) a > 0 k>0

Discriminant For no real roots: b2 − 4ac (−2)2 − 4(2)(2 + p) 4 − 8(2 + p) 1 − 2(2 + p) 1 − 4 − 2p −2p

<0 <0 <0 <0 <0 <3

p

> − [shown] ✓

(ii) b2 − 4ac <0 2 (−2k) − 4(k)(k + 1) < 0 4k 2 − 4k(k + 1) <0 2 2 4k − 4k − 4k <0 −4k <0 4k >0 k >0✓

3 2

Quadratic equation px 2 + 3px + p + q = 0 i.e. a = p, b = 3p, c = p + q

13(a) Line & curve y = kx − 5 x 2 = 2y + 1

Discriminant For two equal real roots: b2 − 4ac =0 2 (3p) − 4(p)(p + q) = 0 9p2 − 4p2 − 4pq =0 2 5p − 4pq =0 p(5p − 4q) =0

−(1) −(2)

sub (1) into (2): x2 = 2(kx − 5) + 1 2 x = 2kx − 10 + 1 2 x − 2kx + 9 = 0 i.e. a = 1, b = −2k, c = 9 Discriminant For line to be tangent to curve: b2 − 4ac =0 2 (2k) − 4(1)(9) = 0 4k 2 − 36 =0 2 k −9 =0 (k + 3)(k − 3) =0 k = −3 or k = 3 ✓

4

p = 0 (rej ∵ p ≠ 0) or p = q [shown] ✓ 5

12(a) Quadratic inequality 3x 2 − 3x > x + k for all real values of x 3x 2 − 4x − k > 0 i.e. a = 3, b = −4, c = −k 2 conditions (i) a > 0 3>0 (ii) b2 − 4ac <0 (−4)2 − 4(3)(−k) < 0 16 − 12(−k) <0 12k < −16 k


4

<− ✓ 3

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24


Ex 1.3

13(b) Line & curve x + 3y = k − 1 3y =k−1−x y

=

k−1−x

13(d) Line & curve y=x+k−1

−(1)

(y − 1)2 = 4x

−(2)

−(1)

3

y 2 = 2x + 5

sub (1) into (2): (x + k − 1 − 1)2 = 4x 2 (x + k − 2) = 4x 2 [x + (k − 2)] = 4x x 2 + 2(k − 2)x + (k − 2)2 = 4x 2 2 x + (2k − 4)x +(k − 4k + 4) = 4x x 2 +(2k − 8)x +(k 2 − 4k + 4) = 0 i.e. a = 1, b = (2k − 8), c = (k 2 − 4k + 4)

−(2)

sub (1) into (2): (

k−1−x 2

)

3 (k−1−x)2 9

= 2x + 5 = 2x + 5

(k − 1 − x)2 = 18x + 45 (−1)2 (x + 1 − k)2 = 18x + 45 [x + (1 − k)]2 = 18x + 45 2 x + 2(1 − k)x + (1 − k)2 = 18x + 45 2 2 x + (2 − 2k)x + (1 − 2k + k ) = 18x + 45 2 2 x + (−16 − 2k)x +(k − 2k − 44) = 0 i.e. a = 1, b = (−16 − 2k), c = (k 2 − 2k − 44) Discriminant For line to meet curve: b2 − 4ac ≥0 (−16 − 2k)2 − 4(1)(k 2 − 2k − 44) ≥0 2 2 (4k + 64k + 256) − 4k + 8k + 176 ≥ 0 72k + 432 ≥0 k ≥ −6 ✓ 13(c) Line & curve y = kx + 2 y 2 = 8x − x 2

Discriminant For line to not meet curve: b2 − 4ac (2k − 8)2 −4(1)(k 2 − 4k + 4) 22 (k − 4)2 −4(k 2 − 4k + 4) (k 2 − 8k + 16) −k 2 + 4k − 4 −4k + 12 −4k k 14(i)

−(1) −(2)

−(1)

y=x−k

−(2)

sub (1) into (2): kx(x + 2) =x−k 2 kx + 2kx =x−k 2 kx + (2k − 1)x + k = 0 i.e. a = k, b = (2k − 1), c = k

sub (1) into (2): (kx + 2)2 = 8x − x 2 k 2 x 2 +4kx +4 = 8x − x 2 (k 2 + 1)x 2 +(4k − 8)x +4 = 0 i.e. a = (k 2 + 1), b = (4k − 8), c = 4 Discriminant For line to intersect curve at two distinct points: ⇒ b2 − 4ac >0 2 2 (4k − 8) −4(k + 1)(4) > 0 42 (k − 2)2 −42 (k 2 + 1) >0 2 2 (k − 2) −(k + 1) >0 (k 2 − 4k + 4) −k 2 − 1 >0 −4k + 3 >0 4k <3 k

Line & curve y = kx(x + 2)

<0 <0 <0 <0 <0 < −12 >3✓

Discriminant For curve to meet the line: b2 − 4ac ≥0 2 (2k − 1) −4(k)(k) ≥ 0 4k 2 − 4k + 1 −4k 2 ≥ 0 −4k + 1 ≥0 4k ≤1 1

≤ ✓

k

4

14(ii) k = 1 ✓ 4

3

< ✓ 4

14(iii) ∵1 is greater than 1 and it does not satisfy the 4

inequality to have any intersections, 0 intersections✓ © Daniel & Samuel A-math tuition 📞9133 9982

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25


Ex 1.3

Quadratic inequality −5x 2 + 20x + c < 50 for all real values of x −5x 2 + 20x + c − 50 < 0 i. e. A = −5, B = 20, C = c − 50

17(i)

2 conditions (i) A < 0 −5 < 0 (ii) B 2 − 4AC < 0 (20)2 − 4(−5)(c − 50) 400 + 20(c − 50) 400 + 20c − 1000 20c c 15(ii) y = −5x 2 + 20x = −5(x 2 − 4x) = −5[(x − 2)2 − 22 ] = −5(x − 2)2 ⇒ turning pt (2,40) ⇒ ∩ −shape

Discriminant b2 − 4ac = (4)2 − 4(1)(−1) = 20 >0 ⇒ intersects x − axis twice ✓

<0 <0 <0 < 600 < 30 ✓

17(ii) Quadratic curve When p = −2: y = x 2 + (−2)x − (−2) + 3 = x 2 − 2x + 5 i.e. a = 1, b = −2, c = 5

+20 +20 +20 +40

Discriminant b2 − 4ac = (−2)2 − 4(1)(5) = −16 <0 ⇒ y is always positive ✓

y|x=0 = 20 ⇒ y − intercept = 20 𝑦

17(iii) Quadratic curve y = x 2 + px − p + 3 i.e. a = 1, b = p, c = −p + 3

(2,40) 20 𝑂

𝑦 = −5(𝑥 − 2)2 + 40 𝑥 ✓

15(iii) Indeed as c = 20 < 30, the baseball does not reach the height of 50m (at most 40m). ✓ 16(i)

k 2 − 4k + 12 = (k − 2)2 − 22 + 12 = (k − 2)2 +8 [shown] ✓

16(ii) Quadratic equation x 2 + kx =3−k 2 x + kx + k − 3 = 0 i.e. a = 1, b = k, c = k − 3

Discriminant For curve to be tangent to x-axis, b2 − 4ac =0 (p)2 − 4(−p + 3) = 0 p2 + 4p − 12 =0 (p + 6)(p − 2) = 0 p = −6 or p = 2 At p = −6, At p = 2, 2 y = x − 6x + 6 + 3 y = x 2 + 2x − 2 + 3 = x 2 − 6x + 9 = x 2 + 2x + 1 At y = 0, x 2 − 6x + 9 = 0 (x − 3)2 =0 x =3 ⇒ (3,0) ✓

Discriminant b2 − 4ac = k 2 − 4(1)(k − 3) = k 2 − 4k + 12 = (k − 2)2 + 8 >0 ∵ (k − 2)2 ≥ 0 ⇒ real for all real values of k ✓ © Daniel & Samuel A-math tuition 📞9133 9982

Quadratic curve When p = 4: y = x 2 + (4)x − (4) + 3 = x 2 + 4x − 1 i.e. a = 1, b = 4, c = −1

sleightofmath.com

At y = 0, x 2 + 2x + 1 = 0 (x + 1)2 =0 x = −1 ⇒ (−1,0)✓

26


Ex 1.3

Quadratic equation f(x) = g(x) 2 x + 6x − 5 = 8x + c 2 x − 2x − 5 − c = 0 i.e. A = 1, B = −2, C = −5 − c

19(ii) Quadratic function At C = 80: 1.2n2 − 14.4n + 53.7 = 80 1.2n2 − 14.4n − 26.3 = 0 i.e. a = 1.2, b = −14.4, c = −26.3

Discriminant For y = f(x) to intersect y = g(x): B 2 − 4AC ≥0 (−2)2 − 4(1)(−5 − c) ≥ 0 4 + 4(5 + c) ≥0 1+5+c ≥0 c ≥ −6 ✓

Discriminant b2 − 4ac = (−14.4)2 − 4(1.2)(−26.3) = 333.6 >0 ∴ Possible to reach 80 thousand dollars ✓

18(ii) Quadratic inequality f(x) > g(x) for all real values of x 2 x + 6x − 5 > 8x + c 2 x − 2x − 5 − c > 0 i.e. A = 1, B = −2, C = −5 − c

19(iii) Quadratic inequality C >x 1.2n2 − 14.4n + 53.7 >x 2 1.2n − 14.4n + 53.7 − x > 0 i.e. a = 1.2, b = −14.4, c = 53.7 − x 2 conditions (i) A > 0 1.2 > 0

(i) A > 0 1>0 (ii) B 2 − 4AC (−2)2 − 4(1)(−5 − c) 4 + 4(5 + c) 1 + (5 + c) 6+c c 19(i)

(ii) b2 − 4ac < 0 (−14.4)2 −4(1.2)(53.7 − x) 207.36 −4.8(53.7 − x) 43.2 −53.7 + x −10.5 + x x

<0 <0 <0 <0 <0 < −6 ✓

<0 <0 <0 <0 < 10.5

Combine inequalities: x < 10.5 and x > 0 ⇒ 0 < x < 10.5 ✓

Quadratic function At C = 10: 1.2n2 − 14.4n + 53.7 = 10 1.2n2 − 14.4n + 43.7 = 0 i.e. a = 1.2, b = −14.4, c = 43.7

20(i)

Discriminant b2 − 4ac = (−14.4)2 − 4(1.2)(43.7) = −2.7 <0 ∴ not possible to have a cost of production of 10 thousand dollars ✓

y = −0.5t 2 + 7t + k Height is 26m after 10s, y|t=10 = 26 2 −0.5(10) + 7(10) + k = 26 −50 + 70 + k = 26 k =6✓

20(ii) y = −0.5t 2 + 7t + 6 y|t=0 = 6 ✓


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27


Ex 1.3

20(iii) y > 30 2 −0.5t + 7t + 6 > 30 0.5t 2 − 7t + 24 < 0 t 2 − 14t + 48 < 0 (t − 6)(t − 8) < 0 +

− 6

22(ii)

= =

+ 8

22

20(iv) Axis of symmetry = average of t − intercepts

23

6+8 2

=7 Maximum point = y|t=7 = −0.5(7)2 + 7(7) + 6 = 30.5 ✓ 20(v) y = −0.5(t − 7)2 + 30.5 ✓ (a) 𝑦

20(v) (b)

𝑡

𝑦

𝑦 = −0.5𝑥 2 + 7𝑥 + 30.5 (7,30.5)

6

𝑥

𝑂

No. The graphs have different domain. 1st: t ≥ 0 2nd : t ∈ ℝ 21

22(i)

β=

23

2a

=−

2a

+

)

(−b)2 −(b2 −4ac) 4a2 4ac 4a2 c a

[shown] ✓

Yes? Regardless b2 − 4ac is less than 0 or not, LHS is proven to equal RHS A: B: 2 y y = −0.3x + 3 = −0.2x 2 + 1.8 𝑦 𝑦 3

C: y = −0.08x 2 +2.4 𝑦 2.4

1.8

𝑥

3

𝑥

𝑂√30

𝑥

Comparing x-intercepts, C >A >B √30 > √10 > 3 ⇒ C will send water the farthest ✓

y = 10 −0.5x 2 + 2x + k = 10 2 −0.5x + 2x + k − 10 = 0 x 2 − 4x − 2k + 20 =0 (−2k i.e. a = 1, b = −4, c = + 20) For water jet to send water to height of 10 m, b2 − 4ac ≥0 (−4)2 − 4(1)(−2k + 20) ≥ 0 16 − 4(−2k + 20) ≥0 4 − (−2k + 20) ≥0 2 − (−k + 10) ≥0 2 + k − 10 ≥0 k ≥8

2a −b−√b2 −4ac

−b+√b2 −4ac

2a

23(iii) Compare modulus of coefficient of x 2 : A >B >C |−0.3| > |−0.2| > |−0.08| ⇒ A will produce the narrowest path ✓

−b+√b2 −4ac

α+β =

−b−√b2 −4ac

23(ii) Comparing y-intercepts, A> C >B 3 > 2.4 > 1.8 ⇒ A will send water the highest ✓

Incorrect. ✓ He forgets to include m ≠ 0 m = 0 ⇒ Linear equation α=

)(

𝑂 23(i)

✓

2a

𝑂 √10

(7,30.5) 6 𝑂

−b+√b2 −4ac

=( =

∴6< t<8✓

=

αβ

−b−√b2 −4ac 2a

2b 2a b

=− ✓ a


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28


Ex 1.4 1(f)

Ex 1.4 1(a)

x2 − 4 ≤0 (x + 2)(x − 2) ≤ 0

(x − 1)(x + 2) < 0

+

− −2

+ − + −2 1

+ 2

−2 ≤ x ≤ 2

−2 < x < 1 ✓ 2

−2 1

−2 1(b)

✓

2

(x + 3)(x − 4) ≤ 0

✓

2x 2 − 4x − 3 > x 2x 2 − 5x − 3 > 0 (2x + 1)(x − 3) > 0

+ − + −3 4

+

−

−

−3 ≤ x ≤ 4

1

+ 3

2

1

x < − or x > 3 ✓ 2

4

−3 1(c)

✓

3(a)

(2x + 3)(x − 2) > 0 +

−

−

3

x(x − 2) <3 x 2 − 2x − 3 < 0 (x − 3)(x + 1) < 0 + − + −1 3

+ 2

2

−1 < x < 3 ✓

3

x < − or x > 2 2

− 1(d)

3(b)

3

2

2

✓

+ − + −2 6

x(x − 5) ≥ 0 +

x2 > 4x + 12 2 x − 4x − 12 > 0 (x − 6)(x + 2) > 0

− 0

x < −2 or x > 6 ✓

+ 5

3(c)

x ≤ 0 or x ≥ 5

4x(x + 1) ≤3 2 4x + 4x − 3 ≤0 (2x + 3)(2x − 1) ≤ 0 +

0 1(e)

5

x 2 − 4x x(x − 4) +

−

✓

0

3(d)

+

4

0≤x≤4

0

4

+ 1

2

2

3

1

2

2

− ≤x≤ ✓

≤0 ≤0 −

− 3

(1 − x)2 ≥ 17 − 2x 2 x − 2x + 1 ≥ 17 − 2x x 2 − 16 ≥0 (x + 4)(x − 4) ≥ 0 + − + −4 4

✓


x ≤ −4 or x ≥ 4 ✓

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29

A math 360 sol (unofficial) 3(e)

(x + 2)2 x 2 + 4x + 4 2x 2 − 36 x 2 − 18 (x + √18)(x − √18) +

−

Ex 1.4 < x(4 − x) + 40 < 4x − x 2 + 40 <0 <0 <0

6(b)

Discriminant For two real roots: b2 − 4ac ≥0 2 (2p) − 4(9)(1) ≥ 0 4p2 − 36 ≥0 p2 − 9 ≥0 (p + 3)(p − 3) ≥ 0

+

−√18 √18 −√18 < x < √18 ✓ 4

S = 600 + 520T − 20T 2 S > 3800 2 600 + 520T − 20T > 3800 20T 2 − 520T + 3200 < 0 T 2 − 26T + 160 <0 (T − 16)(T − 10) <0 +

− 10

+

6(c)

+

−

< 150 < 150 <0 <0

+

+

4

18

5

6(a)

− −1

5 18

Quadratic equation px 2 − 2x + 2p + 1 = 0 i.e. a = p, b = −2, c = 2p + 1 Discriminant For no real roots: b2 − 4ac (−2)2 − 4(p)(2p + 1) 4 − 4p(2p + 1) 1 − 2p2 − p 2p2 + p − 1 (2p − 1)(p + 1)

C = 5x 2 − 38x + 222

+

+ 3

p ≤ −3 or p ≥ 3 ✓

16

C 5x 2 − 38x + 222 5x 2 − 38x + 72 (5x − 18)(x − 4)

− −3

10 < T < 16 ✓ 5

Quadratic equation 9x 2 + 2px + 1 = 0 i.e. a = 9, b = 2p, c = 1

<0 <0 <0 <0 <0 >0

+ 1 2 1

p < −1 or p > ✓ 2

Quadratic equation x 2 − px + p = 0 i.e. a = 1, b = −p, c = p Discriminant For two distinct real roots: b2 − 4ac >0 (−p)2 − 4(1)p > 0 p2 − 4p >0 p(p − 4) >0 +

− 0

+ 4

p < 0 or p > 4 ✓


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30


Ex 1.4

Quadratic equation 2x 2 − 2px + p2 + p = 0 i.e. a = 2, b = −2p, c = p2 + p Discriminant For real roots: b2 − 4ac (−2p)2 −4(2)(p2 + p) 4p2 −8(p2 + p) p2 −2(p2 + p) p2 −2p2 − 2p −p2 − 2p p2 + 2p p(p + 2)

8

Quadratic equation 3kx 2 + (k − 5)x = 5x 2 + 2 2 (3k − 5)x + (k − 5)x − 2 = 0 i.e. a = (3k − 5), b = (k − 5), c = −2 Discriminant For no real roots: b2 − 4ac (k − 5)2 −4(3k − 5)(−2) k 2 − 10k + 25 +8(3k − 5) k 2 − 10k + 25 +(24k − 40) k 2 + 14k − 15 (k + 15)(k − 1)

≥0 ≥0 ≥0 ≥0 ≥0 ≥0 ≤0 ≤0

<0 <0 <0 <0 <0 <0

+ − + −15 1

+ − + −2 0

−15 < x < 1 ✓

−2 ≤ p ≤ 0 ✓ 9 7(a)

1st inequality x 2 + ax
Discriminant For real roots: b2 − 4ac [−(2p + 4)]2 −4(p)(10 − p) 4p2 + 16p + 16 −4p(10 − p) p2 + 4p + 4 −p(10 − p) p2 + 4p + 4 −10p + p2 2p2 − 6p + 4 p2 − 3p + 2 (p − 1)(p − 2)

(x + 2)(x − 4) < 0 x 2 − 2x − 8 <0 Compare x: a = −2 ✓ Compare x 0 : b = 8 ✓ 7(b)

Quadratic equation px 2 − p + 10 = 2(p + 2)x 2 px − p + 10 = (2p + 4)x 2 px − (2p + 4)x + 10 − p = 0 i.e. a = p, b = −(2p + 4), c = 10 − p

1st inequality 2x 2 + a > bx 2 2x − bx + a > 0

+

− 1

≥0 ≥0 ≥0 ≥0 ≥0 ≥0 ≥0 ≥0

+ 2

p ≤ 1 or p ≥ 2 2nd inequality x < −2 or x > 3 is the solution: k[x − (−2)](x − 3) > 0 k(x + 2)(x − 3) >0 2 Compare x : k = 2

The opposite is true ⇒ p cannot lie between 1 and 2 ✓

2(x + 2)(x − 3) >0 2 2(x − x − 6) >0 2 2x − 2x − 12 >0 Compare x: a = −12 ✓ Compare x 0 : b = 2 ✓


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31


Ex 1.4

Quadratic equation (p + 2)x 2 − 12x + 2(p − 1) = 0 i.e. a = p + 2, b = −12, c = 2(p − 1) Discriminant For real and distinct roots: b2 − 4ac (−12)2 −4(p + 2)[2(p − 1)] 144 −8(p + 2)(p − 1) 18 −(p2 + p − 2) −p2 − p + 20 p2 + p − 20 (p + 5)(p − 4)

12

−(1)

2xy + 6 = 0 −(2) sub (1) into (2): 2x(2x + c) + 6 = 0 4x 2 + 2cx + 6 =0 i.e. A = 4, B = 2c, C = 6 Discriminant For line not to intersect curve: B 2 − 4AC <0 2 (2c) − 4(4)(6) <0 2 4c − 4(4)6 <0 2 c − 24 <0 (c + √24)(c − √24) < 0 +

−

x 2 − xy + y 2 = 1

−(2)

Discriminant For two distinct points: B 2 − 4AC (−3k)2 −4(3)(k 2 − 1) 9k 2 −12(k 2 − 1) 3k 2 −4(k 2 − 1) 3k 2 −4k 2 + 4 k2 − 4 (k + 2)(k − 2)

−5 < x < 4 ✓ Line & curve y = 2x + c

−(1)

sub (1) into (2): x 2 − x(2x − k) +(2x − k)2 x 2 − 2x 2 + kx +(4x 2 − 4kx + k 2 ) −x 2 + kx +(4x 2 − 4kx + k 2 ) 3x 2 − 3kx + k 2 3x 2 − 3kx + (k 2 − 1) i.e. A = 3, B = −3k, C = k 2 − 1

>0 >0 >0 >0 >0 <0 <0

+ − + −5 4

11

Line & curve 2x − y = k y = 2x − k

=1 =1 =1 =1 =0

>0 >0 >0 >0 >0 <0 <0

+ − + −2 2 −2 < k < 2 ✓ 13(i)

d ≤ 80 2 0.15v + v ≤ 80 2 3v + 20v − 1600 ≤ 0 [shown] ✓

13(ii) (3v + 80)(v − 20) ≤ 0 +

+

−

−√24 √24 −

−√24 < c < √24 ✓

− 80

+ 20

3

80 ≤ v ≤ 20 3

Max speed = 20ms −1 ✓


sleightofmath.com

32


Ex 1.4

14

Compare x: 𝑟

ℎ=6

Total surface area 2πr 2 + 2πr(6) 2πr 2 + 12πr − 224π r 2 + 6r − 112 (r + 14)(r − 8)

Compare x 0 : 16 = 4k k =4✓

> 224π > 224π >0 >0 >0

+ − + −14 8

≥ 3(x − 5) ≥ 3x − 15 ≥ −8 ≥ −4 ✓

Quadratic equation (k − 6)x 2 − 8x + k = 0 i.e. a = (k − 6), b = −8, c = k Discriminant For two distinct points: b2 − 4ac >0 2 (−8) − 4(k − 6)(k) > 0 64 − 4(k 2 − 6k) >0 16 − k 2 + 6k >0 2 k − 6k − 16 <0 (k − 8)(k + 2) <0

✓

15 x(2x + 1) >6 2 (i)(b) 2x + x − 6 >0 (2x − 3)(x + 2) > 0

+ − −2

−(2)

= −12 ✓ 17

−4

−(1)

sub (2) into (1): p = −(4 + 2(4))

r < −14 (rej ∵ r > 0) or r > 8 ✓ 15 5x − 7 (i)(a) 5x − 7 2x x

p = −(4 + 2k)

+ − + −2 8

+

−2 < k < 8

3 2

For minimum point, (k − 6) > 0 k >6

3

x < −2 or x > ✓ 2

−2

3 2

✓

Combine inequalities, −2 < k < 8 and k > 6 ⇒ 6
15 (ii) 3

−4 −2 −4 ≤ x < −2 16

2

or

3

18(a) x 2 + 2x < 0 x(x + 2) < 0

and x 2 − x >2 2 x −x−2 >0 (x − 2)(x + 1) > 0 + − + −2 0 + − + −1 2 −2 < x < 0 x < −1 or x > 2

x> ✓ 2

1st inequality 2x 2 + px + 16 < 0 2nd inequality 2 < x < k is solution ⇒ A(x − 2)(x − k) Compare x 2 : A = 2 2(x − 2)(x − k) 2(x 2 − 2x − kx + 2k) 2[x 2 − (2 + k)x + 2k] 2x 2 − 4(4 + 2k)x + 4k


−2 −1

<0

0

2

∴ −2 < x < −1 ✓ <0 <0 <0 <0

sleightofmath.com

33


Ex 1.4

18(b) x 2 ≥4 2 x −4 ≥0 (x + 2)(x − 2) ≥ 0

and x 2 − 6 > 5x 2 x − 5x − 6 > 0 (x − 6)(x + 1) > 0

+ − + −2 2

2

+

− 0

+ − + −258.36 258.03 −258.36 ≤ x ≤ 258.03 ✓ 19(iii) 1 + 2n + 3 n(n − 1) ≤ 100 000 2

⇒ −258.36 ≤ n ≤ 258.03 Largest integer satisfying the above inequality ⇒ 258 ✓

+ 4

0≤x≤4 20 4

∴3< x≤ 4✓ 18(d) x ≤ x 2 < 9 x ≤ x2 x2 − x ≥ 0 x(x − 1) ≥ 0 +

− 0

x 2 − 2x − 3 > 0 (x + 1)(x − 3) > 0 𝐱 + 𝟏 > 𝟎 or x − 3 > 0 ⇒ x > −1 x > 3 3rd step is incorrect ✓

and x 2 <9 2 x −9 <0 (x + 3)(x − 3) < 0

+

We cannot indiscriminately distribute the same inequality sign to both factors. ✓

+ − + −3 3

1

x ≤ 0 or x ≥ 1

−3

≤ 100 000

⇒ 3x 2 + x − 199 998 ≤ 0

18(c) 0 < x(x − 3) ≤ x 0 < x(x − 3) and x(x − 3) ≤ x x 2 − 4x ≤ 0 + − + x(x − 4) ≤ 0 3 0

3

2(3)

x = −258.36 or x = 258.03 ✓

∴ x ≤ −2 or x > 6 ✓

0

−1±√12 −4(3)(−199 998)

19(ii) 1 + 2x + 3 (x − 1)x

6

x < 0 or x > 3

2

x=

x < −1 or x > 6

2

3

1 + 2x + (x − 1)x = 100 000 2 + 4x + 3(x − 1)x = 200 000 2 + 4x + 3x 2 − 3x = 200 000 3x 2 + x − 199 998 = 0

+ − + −1 6

x ≤ −2 or x ≥ 2

−2 −1

19(i)

Correct Solution x 2 − 2x − 3 > 0 (x + 1)(x − 3) > 0

−3 < x < 3

0

1

+ − + −1 3

3

∴ −3 < x ≤ 0 or 1 ≤ x < 3 ✓


x < −1 or x > 3 ✓

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34


Rev Ex 1 A3(i)

Rev Ex 1 A1

x + 2y = 5 2y = −x + 5 y

1

5

2

2

=− x+

−(1) Sum of roots

2x + y = 2xy sub (1) into (2):

−(2)

2

x+

1

5

1

5

2

2

2

2

2

5 5

2

2

x − x+

2x − 7x + 5 (2x − 5)(x − 1) x=

5 2

or 1 5

5

2 2 5

2

y|x=5 = − ( ) +

a

c a

=− =−

1 2 5 2

5

−2 (− )

2

2

1

=5 ✓ 4

=0 =0

✓

2

1 2

= (− )

=0

2

=

b

a2 + β2 = (α + β)2 −2αβ

= −x 2 + 5x

2 7

=α+β =−


2x + (− x + ) = 2x (− x + ) 3

1st equation 2x 2 + x − 5 = 0 i.e. a = 2, b = 1, c = −5 Roots: α & β, where α < β

A3(ii) (α − β)2 = α2 − 2αβ + β2 = (α2 + β2 ) −2αβ

x =1✓ 1

5

2

2

y|x=1 = − (1) +

=− ✓

=2✓

4

=5 =

1

5

−2 (− )

4 41

2

4 1

= 10 ✓ 4

A2

Perimeter 2x + 2y = 36 x+y = 18 y = 18 − x

𝑦 𝑥 −(1)

A3(iii) α − β = √41 or α − β = − √41 ✓ 2

2

(rej ∵ α < β) A3(iv) 2nd equation

Square of diagonal length 2

1

Recall α + β = − , 2

αβ = −

5 2

(√x 2 + y 2 ) = 164 x2 + y2

= 164

−(2)

sub (1) into (2): x 2 +(18 − x)2 −164 =0 2 2 x +(x − 36x + 324) −164 = 0 (2x 2 − 36x + 324) − 164 =0 2 2x −36x + 160 =0 2 x − 18x + 80 =0 (x − 8)(x − 10) =0 x=8 or x = 10 y|x=8 = 18 − (8) y|x=10 = 18 − (10) = 10 =8 Dimensions are 8 m by 10 m ✓

Roots: 2α & − 2β Sum of roots = 2α + (−2β) = 2(α − β) = 2 (−

√41 ) 2

= −√41

Product of roots = (2α)(−2β) = −4αβ 5

= −4 (− ) = 10 2

x 2 − (SOR)x + (POR) = 0 x 2 − (−√41)x +10 = 0 x2 + √41x +10

=0✓

35


Rev Ex 1

A4(a) Quadratic equation x2 + 3 = 2x + p x 2 − 2x + 3 − p = 0 i.e. a = 1, b = −2, c = 3 − p Discriminant For real roots: b2 − 4ac (−2)2 − 4(1)(3 − p) 4 − 4(3 − p) 1+p−3 p

A5(a) x 2 − 5x + 3 > 5 − 4x x2 − x − 2 >0 (x − 2)(x + 1) > 0 + − + −1 2 x < −1 or x > 2 ✓

≥0 ≥0 ≥0 ≥0 ≥2✓

A4(b) Quadratic equation 2x 2 + 2√3x + p = p(x 2 + 2) 2x 2 + 2√3x + p = px 2 + 2p (2 − p)x 2 + 2√3x − p = 0 i.e. a = 2 − p, b = 2√3, c = −p

−1 2 A5(b) Quadratic inequality 3x 2 − 6x + c > 4 for all real values of x 2 3x − 6x + c − 4 > 0 i.e. A = 3, B = −6, C = c − 4 2 conditions (i) A > 0 3>0

a ≠0 (2 − p) ≠ 0 p ≠2 Discriminant For distinct real roots: b2 − 4ac 2

(2√3) −4(2 − p)(−p) 12 −4(p2 − 2p) 12 −4p2 + 8p p2 − 2p − 3 (p − 3)(p + 1)

(ii) B 2 − 4AC (−6)2 − 4(3)(c − 4) 36 − 12(c − 4) 3 − (c − 4) 3−c+4 7−c −c c

>0 >0 >0 >0 <0 <0

+ − + −1 3 −1 < p < 3 Combine inequalities −1 < p < 3 and p ≠ 2 ⇒ −1 < p < 2 or 2 < p < 3 ✓

A6(i)

<0 <0 <0 <0 <0 <0 < −7 >7✓

Quadratic equation x 2 + px + 8 =p 2 x + px + 8 − p = 0 i.e. a = 1, b = p, c = 8 − p Discriminant b2 − 4ac = (p)2 − 4(1)(8 − p) = p2 − 32 + 4p = p2 + 4p − 32 = (p + 8)(p − 4) For equal roots: b2 − 4ac =0 (p + 8)(p − 4) =0 p = −8 or p = 4 ✓

36


Rev Ex 1

A6(ii) For distinct real roots: b2 − 4ac >0 ⇒ (p + 8)(p − 4) > 0

A7(b) Line & curve y = px − 3 −(1) 2 y = 4x + 5 −(2) sub (1) into (2): px − 3 = 4x 2 + 5 2 4x − px + 8 = 0 i.e. a = 4, b = −p, c = 8

+ − + −8 4 p < −8 or p > 4 ✓ A6(iii) For a positive & a negative root: POR < 0 αβ <0 c <0

Discriminant For line to intersect curve: b2 − 4ac ≥0 2 (−p) − 4(4)(8) ≥0 2 p − 128 ≥0 (p + √128)(p − √128) ≥ 0

a

8−p <0 −p < −8 p >8✓

+ 2

A7(a) 2x − 4x + 5 = 2(x 2 − 2x) +5 2 2 = 2[(x − 1) − 1 ] +5 = 2(x − 1)2 +3 ⇒ turning pt (1,3) ⇒ y − intercept = 5 ⇒ ∪ shape 𝑦 𝑦 = 2(𝑥 − 1)2 + 3 5 (1,3) 𝑂

p ≤ −√128 or p ≥ √128 ⇒ largest negative integer is − 12 ✓ A8(i)

Discriminant For line to be tangent to the curve: B 2 − 4AC =0 2 (−6) − 4(2)(5 − c) = 0 36 − 8(5 − c) =0 36 − 40 + 8c =0 8c − 4 =0 2

=α+β =−

Product of roots = αβ ✓

1

1st equation x 2 − 4x + 6 = 0 i.e. a = 1, b = −4, c = 6 Roots: α & β Sum of roots

𝑥

= ✓

+

−√128 √128

Line & curve y = 2x 2 − 4x + 5 −(1) y = 2x + c −(2) sub (1) into (2): 2x 2 − 4x + 5 = 2x + c 2 2x − 6x + 5 − c = 0 i.e. A = 2, B = −6, C = 5 − c

c

−

=

c a

b a

=− =

6 1

−4 1

=4 =6

α2 − αβ + β2 = (α2 + β2 ) −αβ = [(α + β)2 − 2αβ] − αβ = (α + β)2 − 3αβ [shown] ✓ A8(ii) α3 + β3 = (α + β) (α2 − αβ + β2 ) = (α + β) [(α + β)2 − 3αβ] [42 = (4) − 3(6)] = −8 ✓ A8(iii) 2nd equation Roots: α3 & β3 Sum of roots = α3 + β3 = −8 Product of roots = α3 β3 = (αβ)3 = (6)3 = 216 x 2 − (SOR)x +POR = 0 x 2 − (−8)x +(216) = 0 x2 + 8x + 216 =0✓

37

A math 360 sol (unofficial) B1

Rev Ex 1 B2(b) 3 − 2x − x 2 ≥ 0 x 2 + 2x − 3 ≤ 0 (x + 3)(x − 1) ≤ 0

x + 2y = 10 2y = −x + 10 1

y

=− x+5 2

2

2y − 7y + x = 1 sub (1) into (2):

+ − + −3 1

−(2)

−3 ≤ x ≤ 1 ✓

2

1

−(1)

1

2 (− x + 5) − 7 (− x + 5) + x 2

=1

2

1

7

4

2

2 ( x 2 − 5x + 25) + ( x − 35) + x 1

9

2

2

( x 2 − 10x + 50) + ( x − 35) 1 2 x 2 1 2 x 2 2

− −

11 2 11 2

=1

x + 15

=1

x + 14

=0

x − 11x + 28 (x − 4)(x − 7) x=4

=0 =0

y|x=4 = − (4) + 5 =3

=

⇒ A(4,3) ✓

3 2 3

b a −2(p+1)

=−

3α

= 2p + 2 2p+2

−(1)

3

a p2 +p

2 ( ) (p + 1) 3 8 (p2 9 2

−(2) = p2 + p

)

2

+ 2p + 1)

d+c cd

=

√12 1

(1)

A

= √12

=1

= 2√3 [shown] ✓

B3(ii) 2nd equation Roots:

1 c

&

1 d

Sum of roots

1

1

c

d

= +

= 2√3

1

1

1

c

d

cd


1

= (1) = 1

⇒ 4α

=

α

=

=p +p

= α(3α) =

2

2

2

8p + 16p + 8 = 9p + 9p 2 p − 7p − 8 =0 (p + 1)(p − 8) = 0 p = −1 or p=8 α|p=−1 = 0 a|p=8 = 6 (rej ∵ roots are non − zero)

−2k 6

k 3 k 12

−(1)

Product of roots

2

=p +p

=0✓

B4(a) 6x 2 − 2kx + k = 0 Roots: α & 3α Sum of roots = α + 3α = −

1 2

2p+2 2

1

d

+ =

x2 − 2√3x + 1

2α2 =p +p sub (1) into (2): 3 2 2

(−√12)

x 2 − (SOR)x + (POR) = 0

1

Product of roots c = α(2α) =

2(

C

Product of roots = cd =

c

2

⇒ 3α

=

B

A

Sum of roots = c + d = − = −

1

⇒ B (7, ) ✓

= α + 2α = −

⇒ 2α

x 2 − √12x + 1 = 0 i.e. A = 1, B = −√12, C = 1 Roots: c & d

2

B2(a) x 2 − 2(p + 1)x + p2 + p = 0 Roots: α & 2α Sum of roots

=

1st equation

1

y|x=7 = − (7) + 5

2

α

B3(i)

or x = 7

1

2

=1

⇒ 3α

=

k 6 k 6

−(2)

sub (1) into (2): k 2

3( ) 12

k2 48 2

= =

k 6 k 6

k = 8k k 2 − 8k = 0 k(k − 8) = 0 k = 0 (rej ∵ k is positive) or k = 8

38


Rev Ex 1

B4(b) x 2 = 3x + 5 2 x − 3x − 5 = 0 i.e. a = 1, b = −3, c = 5 Roots: α & β Sum of roots

B5(a) Quadratic inequality y = −x 2 + 2(k − 3)x − 25 < 0 for all x i.e. a = −1, b = 2(k − 3), c = −25

=α+β

Product of roots = αβ 1 α2

+

1

=

β2

= =

=− =

b a

c a

α2 +β2 α2 β2 (α+β)2 −2αβ (αβ)2 32 −2(−5) (−5)2

=

19 25

✓

−(1)

=− =−

−3 1 5 1

=3

2 conditions (i) a < 0 −1 < 0

= −5 (ii) b2 − 4ac <0 [2(k − 3)]2 −4(−1)(−25) < 0 4(k 2 − 6k + 9) −4(25) <0 2 (k − 6k + 9) −25 <0 2 k − 6k − 16 <0 (k − 8)(k + 1) <0 + − + −1 8

Show 𝛂𝟒 = 𝟓𝟕𝛂 + 𝟕𝟎 αβ = −5 β =−

5

−(2)

α

sub (2) into (1): 1 α2 1 α2

+ +

1 5 2 (− ) α α2

25 4

25 + α α4 α4 α4 α4

= =

19 25 19 25

= 19α2 = 19α2 − 25 = 19(3α + 5) − 25 = 57α + 95 − 25 = 57α + 70 ✓

−1 < k < 8 ✓ B5(b) y = (x − 3)(x + 1) (i) = x 2 − 2x − 3 = (x − 1)2 − 12 − 3 = (x − 1)2 − 4 ⇒ turning pt (1,4) ⇒ ∪ shape ⇒ y − intercept = −3 𝑦 𝑦 = (𝑥 − 1)2 − 4 −1

Method 2 ∵ α is a root ⇒ α2 = 3α + 5

−(1)

(1) × α2 : α4 = 3α3 + 5α2

−(2)

𝑂 −3

3

𝑥

(1, −4) ✓

sub (1) into (2): α4 = 3α(3α + 5) + 5(3α + 5) α4 = 9a2 + 15α + 15α + 25 α4 = 9α2 + 30α + 25 −(3) sub (1) into (3) α4 = 9(3α + 5) + 30α + 25 α4 = 27α + 45 + 30α + 25 α4 = 57α + 70 ✓

39

A math 360 sol (unofficial) B5(b) Quadratic equation (x − 3)(x + 1) = p (ii) x 2 − 2x − 3 =p 2 x − 2x − 3 − p = 0 i.e. a = 1, b = −2, c = −3 − p Discriminant For equal real roots: b2 − 4ac =0 (−2)2 −4(1)(−3 − p) = 0 4 +12 + 4p = 0 4p = −16 p = −4 ✓ Quadratic equation with 𝐩 = −𝟒 x 2 − 2x − 3 − (−4) = 0 x 2 − 2x + 1 =0 (x − 1)2 =0 x =1✓ B6(a) Quadratic equation (x + 1)2 = h(x + 2) 2 x + 2x + 1 = hx + 2h 2 (2 x + − h)x + 1 − 2h = 0 i.e. a = 1, b = (2 − h), c = 1 − 2h Discriminant For real root x: b2 − 4ac ≥0 2 (2 − h) − 4(1)(1 − 2h) ≥ 0 (h2 − 4h + 4) − 4(1 − 2h) ≥ 0 (h2 − 4h + 4) − 4 + 8h ≥0 2 h + 4h ≥0 h(h + 4) ≥0 + − + −4 0 h ≤ −4 or h ≥ 0 The opposite is true ∴ h cannot lie between − 4 and 0 if x is real ✓

Rev Ex 1 B6(b) Line & curve y = 3x + k −(1) y2 = 1 − x2

−(2)

sub (1) into (2): (3x + k)2 = 1 − x2 9x 2 + 6kx + k 2 = 1 − x2 10x 2 + 6kx + k 2 − 1 = 0 i.e. a = 10, b = 6, c = k 2 − 1 Discriminant For 2 distinct points: b2 − 4ac (6k)2 − 4(10)(k 2 − 1) 36k 2 − 40(k 2 − 1) 9k 2 − 10(k 2 − 1) 9k 2 − 10k 2 + 10 −k 2 + 10 k 2 − 10 (k + √10)(k − √10) +

−

>0 >0 >0 >0 >0 >0 <0 <0

+

−√10 √10 −√10 < k < √10 ✓ Tangent ⇒ k = ±√10 ✓ B7(a) (1 + x)(6 − x) (x + 1)(x − 6) x 2 − 5x − 6 x 2 − 5x − 14 (x − 7)(x + 2)

≤ −8 ≥8 ≥8 ≥0 ≥0

+ − + −2 7 x ≤ −2 or x ≥ 7 ✓

−2

7

✓

40


Rev Ex 1

B7(b) 2x(x + 2) < (x + 1)(x + 3) 2 2x + 4x < x 2 + 4x + 3 2 x −3 <0 (x + √3)(x − √3) < 0

+

−

B8(ii) Discriminant b2 − 4ac = (4)2 − 4(1)(−28) = 128 ≥0 ⇒ real roots ✓

+

−√3 √3

B8(iii) product of roots = αβ = −28 < 0 ⇒ one root is positive and the other is negative. ⇒ time can only be positive and the negative root is rejected hence the rate is 40 mg/s only at one particular time ✓

−√3 < x < √3 ✓ −√3 √3 B8(i)

✓

Quadratic equation At r = 40: (t − 10)2 − 2(t − 4)2 (t 2 − 20t + 100) − 2(t 2 − 8t + 16) (t 2 − 20t + 100) − 2t 2 + 16t − 32 −t 2 − 4t + 68 t 2 + 4t − 28 ✓

= 40 = 40 = 40 = 40 = 0 [shown]

B8(iv) r = −t 2 − 4t +68 2 = −[t + 4t] +68 2 2 = −[(t + 2) − 2 ] +68 = −(t + 2)2 +72 ⇒ turning pt (−2,72) ⇒ ∩ shape ⇒ y − intercept = 68 𝑟 (−2,72) 68 𝑂

𝑡 𝑟 = −(𝑡 + 2)2 + 72

✓

41


Ex 2.1 4

Ex 2.1

(7 + 3√2) 1(a)

√18 = √9 × 3 = 3√2 ✓

2

−(5 − 2√5)

= (7)2 + 2(7)(3√2) +(3√2)

2

2 2

−[(52 ) −2(5)(2√5) +(2√5) ] 1(b)

1(c)

1(d)

2(a) 2(b)

√300 = √100 × 3 = 10√3 ✓

= 49 +42√2 +9(2) −[25 −20√5 +4(5)]

√80 = √16 × 5 = 4√5 ✓

−(45 −20√5)

= 67 +42√2

−45 +20√5

= 22 + 42√2 − 20√5 ✓

√48 = √16 × 3 = 4√3 ✓

5(a)

6 √3

4√2

−√8

= =

2√3 + 5√3 − 3√3 = 4√3 ✓

6

×

√3 6√3

+√50

5(b)

√2 2√3

=

√2 2√3

=

√6 2(3)

=

√6 6

2

(2 + √5)(2 − √5)= (2)2 − (√5) =4 −5 = −1 ✓

5(c)

12 2√5−4

2

(1 − 2√7)

=

(1)2

−2(1)(2√7) +(2√7)

=1 −4√7 = 29 − 4√7 ✓ 3(d)

2

(5 + √18)

√3

✓

=

= =

2

√3

×

(1 + √3)(2 − √3) = 1(2 − √3) +√3(2 − √3) = 2 − √3 +2√3 − 3 = −1 + √3 ✓

3(c)

√3

3

= 3(b)

√3

= 2√3 ✓

= 4√2 −√4 × 2 +√25 × 2 = 4√2 −2√2 +5√2 = 7√2 ✓ 3(a)

= 67 +42√2

12 2√5−4

×

2√5+4 2√5+4

24√5+48 2

(2√5) −(4)2 24√5+48 20−16 24√5+48 4

= 6√5 + 12 ✓

+4(7) 6(a)

3√2−4 4+3√2

2

=

= (5)2 +2(5)√18 +(√18)

=

= 25 +10√18 +18 = 43 +10√9 × 2 = 43 +10(3√2)

= =

= 43 +30√2 ✓

3√2−4 4+3√2

×

4−3√2 4−3√2

3√2(4−3√2)

−4(4−3√2)

2 (4)2 −(3√2)

12√2 −9(2)

−16+12√2

16−9(2) −34+24√2 −2

= 17 − 12√2 ✓


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42


√3+2√2 √3−2√2

= = = = =

6(c)

8 (√6+2)

2

√3+2√2 √3−2√2 (√3)

2

×

Ex 2.1

√3+2√2

+2√3(2√2) 2

(√3) −(2√2) 3

6(d)

√3+2√2

+(2√2)

√50 − √48

2

2

+4√6

√18 − √12

=

+4(2)

3−4(2)

=

11+4√6

√18 − √12 √50 − √48

×

5

= = = = =

=

(√6)

✓

= =

+(2)2

+2√6(2)

8 6

− √12(√50 + √48) 2

(√50) − (√48)

8 2

√50 + √48

√18(√50 + √48)

−5 −11−4√6

√50 + √48

+ 4√6

=

+4

8 10 + 4√6 8 10 + 4√6

×

80 − 32√6 (10)2 − (4√6) 80 − 32√6 100 − 16(6)

80 − 32√6 = 4

30 + √25 × 33

− 10√6 − 24 2

6 + √25 × 33

− 10√6 2

6 + (22.5 )(31.5 ) 2

− 10√6

=

6 + (22 √2)(3√3) 2

− 10√6

6 + 12√6

− 10√6

=

2

− √600 − √576 50 − 48

= 10 − 4√6 10 − 4√6

√900 + √864

2

=

2 6 + 2√6 2

= 3 + √6 ✓ 7(a)

= 20 − 8√6 ✓

x√3 − x = 2 x(√3 − 1) = 2 x

= = = = =

2 √3−1 2

×

√3+1

√3−1 √3+1 2√3+2 2

(√3) −(1)2 2√3+2 3−1 2√3+2 2

= √3 + 1 ✓


7(b)

√2x − 3 = 5 2x − 3 = 25 𝑥 = 14 ✓

7(c)

2√1 − x = √3 4(1 − x) = 3

sleightofmath.com

3

1−x

=

x

= ✓

4 1 4

43


8(a)

√8 − x = √x − 2 8−x =x−2 2x = 10 x =5✓

Ex 2.1 9(ii)

= √1800 = √100 × 18 = 10√18 = 10√9 × 2

a + b√3 = 2 − √3

= 10(3√2) = 30√2 ✓

Equate rational terms: a=2✓ Equate irrational terms: b = −1 ✓ 8(b)

8(c)

a + b√7 = 4√7 − 12 − √7 = −12 + 3√7 Equate rational terms: a = −12 ✓ Equate irrational terms: b=3✓

Change in velocity = 30√2 − 20√2 = 10√2 ✓ 10(i) √80 − 3 3 + 2√5 Total Length = 2(square perimeter) +4(height)

a + b√2 = 3(1 − √2) + 4√2 = (3 − 3√2) + 4√2 = 3 + √2 Equate rational terms: a=3✓ Equate irrational terms: b=1✓

9(i)

v|s=90 = √20(90)

v = √20s

= 2(4)(3 + 2√5)

+4(√80 − 3)

= 8(3 + 2√5)

+4√16 × 5 − 12

= 24 + 16√5

+4(4√5) − 12

= 24 + 16√5

+16√5 − 12

= 12 + 32√5 ✓

10(ii) Volume = (square base)

(height)

2

= (3 + 2√5)

v|s=40 = √20(40)

(√80 − 3) 2

= √800

= [(3)2 + 2(3)(2√5) + (2√5) ] (√16 × 5 − 3)

= √100 × 8

= (9

= 10√8

+ 12√5

+ 20)

(4√5 − 3)

= 10√4 × 2

= (12√5 + 29)

(4√5 − 3)

= 10(2√2)

= 12√5(4√5 − 3)

+29(4√5 − 3)

= 48(5) − 36√5

+116√5 − 87

= 20√2 ✓

= 153 + 80√5 ✓ 11(a) 4√12

2

− √3 3 2

= 4√4 × 3 − √3 3 = 4(2√3) = 8√3

2

− √3 3 2

− √3 3

− − −

18 √3 18 √3

×

√3 √3

18√3 3

−6√3

4

= √3 ✓ 3 © Daniel & Samuel A-math tuition 📞9133 9982

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44


4 √2

= = =

4 √2

×

√2 √2

4√2

−

√128 3

+

−

√64×2 3

+

−

2 4√2

−

2

Ex 2.1

8√2

+

3 8√2

+

3 8

= 2√2

− √2 3

= 2√2

− √2 3

+

8

1

12(b)

2

(2√7−3)

√8 2

=

√4×2 2

=

2√2 1

=

√2 1 √2

×

√2

=

√2

1

+ √2 2

=

1

= − √2 ✓ 6 11(c)

2

(

√6

= =

2 √6

×

√6

(

√6

2√6

(

6 1

= √6 3

(

1

= √6 3

(

1

4 √27 4 √9×3 4

4 3√3

×

( √3 9

1

37

= = =

37 37 37

= = = =

√81×3 2

−

9√3

−

2 9

+ √3 2 +

−

9√3

−

2 9

)

5 √4×3 5 2√3

= )

=

)

5 2√3 5√3 2(3)

×

√3 √3

1 4(7)

−12√7

−

+9

1

−

37−12√7 1 37−12√7

×

37+12√7

−

37+12√7

37+12√7

−

2 (37)2 −(12√7)

37+12√7

−

1369−144(7) 37+12√7

−

361 24√7 361

1 2

(2√7) +2(2√7)(3)+(3)2 1 4(7)

37+12√7 1 37+12√7

4√3−2

4√3+2 4√3+2

4√3+2 2

(4√3) −(2)2 4√3+2

37−12√7 (37)2 −(12√7) 37−12√7

37−12√7 361

38

= 4√4 × 3 +3√9 × 2

−

= 4(2√3) +2(3√2)

−

= 8√3

+9√2

−

= 8√3

+9√2

−(5√2 + 2√3)

= 8√3

+9√2

−5√2 − 2√3

)

4√3+2

− −

−

44

2

1369−144(7)

−

13(a) (3 − √3)2

−

16(3)−4

37−12√7

5√2−2√3 38 5√2−2√3

×

5√2+2√3 5√2+2√3

38(5√2+2√3) 2

(5√2) −(2√3)

2

38(5√2+2√3) 25(2)−4(3) 38(5√2+2√3) 38

= 4√2 + 6√3 ✓

− ×

37−12√7

×

= 4√12 +3√18

)

+9

1

−

− √3) 6

√2 ✓

1

+12√7

✓

5

+ √3 2

2

12(c) √6(4√2 + 3√3)

(3√2)

27

1

=

+

5 √12

(2√7)

−

−2(2√7)(3) +(3)2

√3)

4√3−2

=

−

√9 × 2

27

9

9

√243 2

1 2

1 (2√7+3)

√18

27

37

√3

3(3)

= √6 3

=

√3

4√3

(

+

+

3√3

4

= √6 3

12(a)

=

−

2

−

2 4√3+2 2 4√3+2

×

8√3−4 2

(4√3) −(2)2

5

−

= 12 − 6√3

−

= 12 − 6√3

−10 − 5√3

2−√3

×

2+√3

= 9 − 6√3 + 3 4√3−2 4√3−2

5 2−√3

2+√3

10+5√3 4−3

= 2 − 11√3 ✓

8√3−4 16(3)−4 8√3−4 44

6−4√3 44 3 22

−

1 11

√3 ✓


sleightofmath.com

45


s2 +1 s+2

= = = = = = =

(1−√5)

2

Ex 2.1 15(b) √6 − 5x − x = −2x = −x √6 − 5x 6 − 5x = x2 x 2 + 5x − 6 = 0 (x + 6)(x − 1) = 0 x = −6 or x = 1 (rej) ✓

+1

(1−√5)+2 (1−2√5+5) +1 3−√5 7−2√5 3−√5

×

3+√5 3+√5

7(3+√5)

−2√5(3+√5)

(3)2 −(√5) 21+7√5

2

15(c) √x + 5 + x =1 =1−x √x + 5 x+5 = x 2 − 2x + 1 x 2 − 3x − 4 = 0 (x − 4)(x + 1) = 0 x = −1 or x = 4 (rej) ✓

−6√5+2(5) 9−5

11+√5 4 11 4

14(i)

1

+ √5 ✓ 4 4+

6

16

√3

(4 +

= (5 − √4 × 3) (4 + = (5 − 2√3) = (5 − 2√3)

(4 +

6 √3 6 √3

×

6√3 3

𝑥

) √3 √3

= =

)

)

=

(4 + 2√3)

= 8 + 2√3 ✓ 14(ii) By Pythagoras Theorem, Square of the length of diagonal 2

+ (4 +

√10 √40 − √5

×

√40 + √5 √40 + √5

√400 + √50 2

=

20 + √25 × 2 40 − 5

=

20 + 5√2 35

=

4 + √2 7

=

𝑎 + √𝑏 [given] 7

−8√3 − 4(3)

= (5 − √12)

√10 √40−√5

2

(√40) − (√5)

= 5(4 + 2√3) −2√3(4 + 2√3) = 20 + 10√3

= x√5 + √10

𝑥(√40 − √5) = √10

5 − √12 Area = (5 − √12)

x√40

6 √3

2

= (5)2 − 2(5)√12 + (√12) + (16 +

)

2

48

+

√3 48

36 3

= 25 − 10√12 + 12

+ (28 +

= 37 − 10√4 × 3

+ (28 +

= 37 − 10(2√3)

+ (28 +

= 37 − 20√3

+(28 + 16√3)

√3

)

)

48 √3

×

48√3 3

√3 √3

)

Equate rational terms: a=4✓ Equate irrational terms: b=2✓

)

= 65 − 4√3 ✓ 15(a) √2 + x − x =0 =x √2 + x 2+x = x2 x2 − x − 2 =0 (x − 2)(x + 1) = 0 x = −1 (rej) or x = 2 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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46


Ex 2.1

17(a) 3 + a√5 = b(3 − √5) −2(3 + √5) = 3b − b√5

18

−6 − 2√5

a + b√7

= 3b − 6 −(b + 2)√5 Equate rational terms: 3 = 3b − 6 3b = 9 b =3✓ −(1) Equate irrational terms: a = −(b + 2) −(2) sub (1) into (2): a = −((3) + 2)

= = =

17(b) a − b√7 = (2 + √7)(2 − √7) +√7 = (2)2 − (√7)

2

=4−7 = −3 + √7 Equate rational terms: a = −3 ✓ Equate irrational terms: −b = 1 𝑏 = −1 ✓

=

+√7 +√7

2

2

324 2

(4)2 − 2(4)√7 + (√7) 324 16

− 8√7 324

+7

23 − 8√7 324 23 − 8√7

×

23 + 8√7 23 + 8√7

324(23 + 8√7) 2

(23)2 − (8√7)

=

324(23 + 8√7) 529 − 64(7)

=

324(23 + 8√7) 81

Equate rational terms: a = 92 ✓ Equate irrational terms: b = 32 ✓

+b√8 2

13

)

= 92 + 32√7

= (3)2 − 2(3)(2√2) + (2√2)

+b√4 × 2

=9

+b(2√2)

= 17 − 12√2 = 17 +(2b − 12)√2 Equate rational terms: a = 17 ✓ Equate irrational terms: 1 = 2b − 12

18 4−√7

= 4(23 + 8√7)

17(c) a + √2 = (3 − 2√2)2 − 12√2

18 4−√7

=( =

= −5 ✓

b=

√a + b√7 =

+ 4(2)

+2b√2 19(a) (a + 3√2)(3 − 4√2)

= −18 + b√2

a(3 − √2) +3√2(3 − 4√2) = −18 + b√2 3a − a√2 +9√2 − 12(2) 3a − 24 +(9 − 4a)√2

✓

= −18 + b√2 = −18 + b√2

Equate rational terms: 3a − 24 = −18 3a =6 a =2 ✓ −(1) Equate irrational terms: 9 − 4a = b −(2) sub (1) into (2): 9 − 4(2) = b b =1✓


sleightofmath.com

47

A math 360 sol (unofficial) 19(b) (3 − 2√3)(4 + a√3) 3(4 + a√3)

Ex 2.1 = b − 23√3

19(d)

−2√3(4 + a√3) = b − 23√3

12 + 3a√3 −8√3 − 2a(3) 12 − 6a +(3a − 8)√3 Equate rational terms: 12 − 6a = b −(1)

2+4√7

= 4 − √7

a+b√7

2 + 4√7 = (4 − √7)(a + b√7)

= b − 23√3 = b − 23√3

= 4(a + b√7) = 4a + 4b√7 = 4a − 7b Equate rational terms: 2 = 4a − 7b 4a = 7b + 2

Equate irrational terms: 3a − 8 = −23 3a = −15 a = −5 ✓ −(2) sub (2) into (1): 12 − 6(−5) = b b = 42 ✓

a

=

7b+2

Equate irrational terms: 4 = 4b − a −(2) sub (1) into (2): 4 = 4b − (

7b+2 4

)

= 6 − 4√3

a(5 − 3√3)

+b√3(5 − 3√3)

= 6 − 4√3

16 = 16b − 7b − 2 9b = 18 b = 2✓

5a − 3a√3 +5b√3 − 3b(3) 5a − 9b +(5b − 3a)√3 Equate rational terms: 5a − 9b = 6

= 6 − 4√3 = 6 − 4√3

Put b = 2 into (1):

=

9b+6 5

a|b=2 =

20(i)

9b+6 5

)

= −4

2

2

(2 − √3) = (2)2 − 2(2)√3 + (√3) +3

20(ii) 28 − 16√3= 4(7 − 4√3) 2

25b − 27b − 18 = −20 −2b = −2 b =1✓

= 4(2 − √3)

The two square roots are ±2(2 − √3) i.e. 4 − 2√3 or −4 + 2√3 ✓

Put b = 1 into (1): a|b=1 =

4

=4 − 4√3 = 7 − 4√3 ✓

sub (1) into (2): 5b − 3 (

7(2)+2

=4✓

−(1)

Equate irrational terms: 5b − 3a = −4 −(2)

−a√7 − b(7) +(4b − a)√7

−(1)

4

19(c) (a + b√3)(5 − 3√3)

a

−√7(a + b√7)

9(1)+6 5

=3✓


sleightofmath.com

48


Ex 2.1

20(iii) √34 − 24√2 = c + d√2

21(ii) 2

1

=

R

34 − 24√2 = (c + d√2)

1 2

= (c)2 +2(c)(d√2) +(d√2) = c 2 +2cd√2 = c 2 + 2d2 Equate rational terms: 34 = c 2 + 2d2 −(1) Equate irrational terms: −24 = 2cd c

12

=−

10

+d2 (2) +2cd√2

6√2+7√3

= (−

34

=

34d2

144 d2

=

2

60√2−70√3 36(2)−49(3) 60√2−70√3 −75 4

) + 2d2

+ 2d2

= 144 + 2(d2 )2

14

+ 1

+

√3+3√2 1 √3+3√2

×

√3−3√2 √3−3√2

√3−3√2 2

(√3) −(3√2)

2

+

=

√3−3√2 −15

+

− √2 + √3 5 15

=−

√3 − √2

=

R2

=

1

1

15

√3 + 5 √2

1

=

R2

= √3 + √2 ✓

c|d=−3 = −

−3

= −4 =4

=

V = IR = 5√6 [

1 10

(6√2 + 7√3)]

1

= √6(6√2 + 7√3) 2 = 3√12 = 3√4 × 3

7

+ √18 2

1 R2 1 R2

√3−√2

×

√3+√2

√3+√2 3−2

= √a + 1 − √a =

∴ −4 + 3√2 or 4 − 3√2 ✓

1 R2

√3+√2

R2

3 12

1 R2

1

=0

12

1 R2

√3−√2

(d2 )2 − 17d2 + 72

c|d=3 = −

1 R2

1

=

1

+

1 R2

R2

R2

√a+√a+1

+

√3−3√2 3−9(2)

=0

22(i)

+

=

2(d2 )2 − 34d2 + 144

(d2 − 9)(d2 − 8) =0 2 2 d =9 d =8 d = ±3 d = ±2√2 (rej ∵ non-surds)

21(i)

=

6√2−7√3

2

12 2 d

6√2−7√3

60√2−70√3

sub (2) into (1): 34

×

(6√2) −(7√3)

−(2)

d

=

1 (6√2+7√3) 10

1 R1

1 √a + √a + 1 1 √a + √a + 1

=

√a − √a + 1 a − (a + 1)

=

√a − √a + 1 −1

×

√a − √a + 1 √a − √a + 1

= √a + 1 − √a [shown] ✓

7

+ √9 × 2 2

22(ii)

1

+

1

7

√1+√2

2

= (√2 − √1)

= 3(2√3)

+ (3√2)

= 6√3

+

21 2

√2 ✓

√2+√3

+

1 √3+√4

+ ⋯+

1 √8+√9

Method of Difference [ ] Telescopic Sum Concept

+(√3 − √2) +(√4 − √3) +⋯ +(√9 − √8) = √9 − √1 =3−1 =2✓


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49


Ex 2.1

√x 2 − x − 2 + 3√x 2 + x − 6 = 0 √x 2 − x − 2

= −3√x 2 + x − 6

x2 − x − 2

= 9(x 2 + x − 6)

x2 − x − 2

= 9x 2 + 9x − 54

8x 2 + 10x − 52

=0

4x 2 + 5x − 26

=0

(4x + 13)(x − 2) = 0 x=−

13 4

(rej) or x = 2 ✓


sleightofmath.com

50


Ex 2.2 2(b)

Ex 2.2

when y = 16: 3

2x 2 = 16 1(a) 1(b)

(23 × 32 )0 = 1 ✓ 4

4

(2√3)

3

x2 = 8 2

4

x

= (8)3

4

x x

= (23 )3 = 4✓

= 2 (√3) 1 2

= 24 (3 )

2

= 16(3)2 = 144 ✓ 1(c)

1

1

1

93 ×33 6

3(a)

16x = (24 )x = (2x )4 = u4 ✓

3(b)

22x+3 = (22x ) (23 ) = (2x )2 (23 ) = 8u2 ✓

3(c)

4x−1 = (22 )x−1 = 22x−2 = 22x × 2−2

1

(32 )3 ×33

=

6 1

(27)3

=

6 1

(33 )3

=

6 3

=

6 1

= ✓ 2

1(d)

1

1

254

= (2x )2 ×

1

× 53 × 56

= (52 )

1 4

1 3

1 6

1 3

1 6

=5

4

×5 ×5

3(d)

1 1 1 52+3+6

= = 51 =5✓ 1(e)

1

1

1

33 ×30 ×93 2 273

=

1

×30

33

1

1

42−x = (22 )2−x = 21−2x = (21 ) (2)−2x (2x )−2 =2 (u)−2 =2

×(32 )3

=

2 (33 )3 1

=

2

33

×1

4

1

= u2 ✓

×5 ×5

1 2

1

×33

4(a)

32

2 u2

✓

× 2x ÷ √2y

8

y

= 23 × 2x ÷ 22

1 2 33+3−2

= = 3−1

y

= 23+x−2 ✓

1

= ✓ 3

2(a)

y=

4(b)

2x

× 42x+1 x

3 2x 2 3

y|x=9 = 2(9)2 3

2

× 2−1 (24x )

=2

×

= 2x

× 24x+2

× 2−1+4x

= 2x

+4x+2

−1+4x

= 29x+1 ✓

= 2(32 )2 = 2(33 ) = 54 ✓


(22 )2x+1

1

× (16x )

sleightofmath.com

51


1

×√

4 x−1

Ex 2.2 6(b)

8y

6

2( √a)+( √a)

32

1

=

1 22x−2

=

9

1

25

=

3 5 y− 2 2

=

1

2x+1

7(a)

=

(23 )

x+2

1

3

15 2

−17 (2 )

7(b)

17 √2

×

4

1

1

1

2

1

1

2

2

1

1

2

+a3 b 3 − a3 b 3 + b

3

3

3

= √a3 =a✓

17√2 2

√2 ✓ 7(c)

3

3

3

3

3

3

3

3

3

3

3

3

− √8

= √33

3

− √23

=3

−2

3

3

3

3

− √2( √9 + √6 + √4) 3

= √27 + √18 + √12

3

3

− √18 − √12 − √8

3

= √27

3

=1✓

+ (4)(5)(52x ) 5x+2 + 4(52x+1 )

3

( √3 − √2)( √9 + √6 + √4) = √3( √9 + √6 + √4)

7(d)

1

1

1

1

1

(a2 + √2a4 b 4 + b 2 ) 1

1

1

1

1

(a2 − √2a4 b 4 + b 2 ) 1

1

1

1

1

1

= [(a2 + b 2 ) + √2a4 b 4 ] [(a2 + b 2 ) − √2a4 b 4 ] 1

1

2

1

1

= (a2 + b 2 )

1

1

2

− (√2a4 b 4 )

= (a + 2a2 b 2 + b)

1

1

−2a2 b 2

=a+b✓

= 5−x ✓ (√a)

2

3

5

5x+2 52x+2

1

= √(a3 + b 2 ) − b 2

√2

x+2

=

2

( √√a3 + b 2 + b) ( √√a3 + b 2 − b)

√2

× 52−2x + 20(52x )

5x+2 5(52x+1 )

1

3

53x 52x+1

=

1

= √(√a3 + b 2 + b)(√a3 + b 2 − b)

52x+1

52x+1

2

17 √2

× (52 )1−x + 3(52x ) + 17(52x )

52x+1

1

=a+b✓

1 − 2

53x × 251−x 52x+1 + 3(25x ) + 17(52x ) 53x

1

= a − a3 b 3 + a3 b 3

23x+2

= 16√2 −

2

2

2

− 17(23x+1 )

= 16√2 −

1

= a3 (a3 − a3 b 3 + b 3 ) +b 3 (a3 − a3 b 3 + b 3 )

23x+6

= 16√2 −

=

1

(a3 + b 3 ) (a3 − a3 b 3 + b 3 )

−17(2)(23x ) 3 2x+1 (22 )

=2

6(a)

2

= 3a9 ✓

1 4 2

=

3a3 a9

=

=

2

1

(√8)

=

+ a3 a9

3 5 22−2x+2y−2

8x+2 −34(23x )

=

1

2a3 1

3

5(b)

1

(a2 )9

(23 )y

= 22y−2x−2 ✓ 5(a)

+(a6 )

× √23y−5

= 22−2x × 2 =

2(a3 )

2

√a2

= (22 )x−1 × √

1 2

1 3

12

4

× ( √a ) 4

1

12

÷ a4

= (a2 ) × (a4 )

÷ a4

= a2 =a✓

÷ a4

× a3


sleightofmath.com

52


7

(√3) +(√3) 33.5

5

Ex 2.2

+(√3)

+32.5

3

12

+42(√3) = 3k

+31.5

+42√3

= 3k

33 √3 +32 √3 +3√3

+42√3

= 3k

√3(33 +32

+42)

= 3k

+3

2n−1

P = (2 + √5)

2n 2n

=

k

√3(81)

=3

=

1 2

(2 + √5)

2 + √5 (4 + 4√5 + 5) 2 + √5

(3 ) (34 ) = 3k 342

= 3k

∴k

= ✓

=

9

4

n

× √2 ×

33

1

3

4

11 6

17 6

5

5

1

6

2

5 2 5

2 )2

(23

÷

= 2x 3y pz

×3×p

−2 )2

x y z

=2 3 p

14(i) (a)

(2−1 ⋅ x −1 )r

u = 2x v = yx 2x+1 +y 2 (y x−2 ) = 11 2x ⋅ 2 +y x = 11 2u +v = 11 ✓

= 2x 3y pz 1

−(1)

,y = 3 ,z = 8✓

(2n )21

2

−3(2n )

+4(2n+2 )

n)

n )22

−3(2

+4(2

= 2n [21

−3

+4(22 )]

= 2n (2

−3

+16)

14(i) (b)

(22 )

1 x+1 2

2x ⋅ 22 4u

10(ii) 2n+1 − 3(2n ) + 4(2n+2 ) = 15(2n ) ∵ 15 is a factor, it is divisible by 15 ✓ 4n +5(4n+1 ) = 4n +5(4n )(4) = 4n +20(4n ) = 4n +20(4n ) = 19(4)n

1

42x+1

2x+2

= 15(2n ) ✓

11

x

3

÷ (22 × 32 × p−3 ) = 2x 3y pz

7

2n+1 =

=

(23 2 )8−r

= 224−4r x16−3r [shown] ✓

3 2

2 2 32 p8 2

1 r

(8x 2 )8−r ( )

= 224−3r ⋅ x16−2r (2−r ⋅ x −r )

3

9

[shown] ✓

2x

✓

÷ (24p−2 )

(210 × 35 × p5 )

11

13

=2 3 p

6

×3 ×p

= 2x 3y pz x y z

5

2

=2 3 p

13 2

∴ x = 1 ,y = 2 ,z = 6 (144p2 )

×

= (√5 − 2)(9 + √80)

x y z

p5

× 22 × 32 × p 2

2 ×3 ×p

∴x=

(9 + 4√5) (2 − √5) 4−5

= (√5 − 2)(9 + √16√5)

= 2x 3y pz

× √54p5

2 3 × 3 3 × p4

10(i)

2 − √5

n

2

11

2 − √5

n

2

(36p6 )3

(24

×

= (√5 − 2)(9 + 4√5)

2

(22 × 32 × p6 )3

9(b)

n

n

1

9(a)

−1

= (2 + √5) (2 + √5)

+5−1 y x = 4 1

+ yx 5 1 + yx 5 1 + yx 5 1

+ v 5

=4 =4 =4 =4✓

−(2)

14(ii) 2(1) − (2): 1

2v − v = 22 − 4 5

9 5

v

−22n+1 −(22n )(2) −(22 )n (2) −4n (2)

v

= 18 = 10 ✓

−(3)

sub (3) into (1): 2u + 10 = 11 u

1

= ✓ 2

∵ 19 is a factor, it is divisible by 19 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

53

A math 360 sol (unofficial) 15(ii) ax = b y = (a)w

14(iii) (4y)x = 4x y x = (2x )2 y x = (u)2 v 1 2

Ex 2.2

b

ax = b y

(10)

=( ) 2

y

a = bx

5

−(1)

= ✓ 2

15(i)

a w

by = ( )

sub (1) into (2):

y

a

= bx

−(1)

y

−(2) by

sub (1) into (2):

b

y

= [b

]

yz ( +z) x

by = b yz ⇒y = +z

y

= z(

xy y+x

z

x

w

y

= w ( − 1)

y

= w(

y−x

y

x y+x

y

= (b x−1 )

y xy

x

= z ( + 1)

b

x

z

y

=( )

wy

= [(b ) b] y +1 x

bx

b y = b x −w wy ⇒y = −w

z

y x

w

y

b b y = (ab)z

by

−(2)

b

ax = b y

w

x y−x x

)

=w =

xy y−x

[shown] ✓

)

=z =

xy y+x

[shown] ✓


sleightofmath.com

54


Ex 2.3 1(a)

Ex 2.3

32x = 27 32x = 33 ⇒ 2x = 3 3 x = ✓ 2

1(b)

4x = 32 22x = 25 ⇒ 2x = 5

1(c)

2

3x

3 x 2

3(i) 1 −3

= −2 ✓ 1 x+2

( )

=3 3 = 31 ⇒ −2x − 4 = 1 −2x−4

22x 2x−1

2x+1

= 10 = 10 = 101 =1 1 =

2(c)

3(2x−1 ) = 2x + 4 x )(2−1 ) 3(2 = 2x + 4 3(2x ) ( )

= 2x + 4

3

(2x )

= 2x + 4

(2x )

=4

2

1

sub u = 3x : u

= (23 )2−x = 26−3x

= ✓

5(a)

4

2

4x −6 − 16x+1 = 0 2 4x −6 = 16x+1 2 4x −6 = (42 )x+1 2 4x −6 = 42x+2 ⇒ x2 − 6 = 2x + 2 x 2 − 2x − 8 = 0 (x − 4)(x + 2) = 0 x = 4 or x = −2 ✓


3

1

= 4 −3 ( ) u

u2 = 4u −3 2 u − 4u + 3 =0 (u − 3)(u − 1) = 0 u=3 or u = 1 x sub u = 3 : sub u = 3x : x 3 =3 3x = 1 3x = 31 3x = 30 ⇒ x =1✓ ⇒x =0✓

= 82−x

=5

x

3x = 4 −3 ( x )

2

5

x

=0

3x = 4 −3(3−x )

⇒ x + 1 = 6 − 3x 4x

x2 + x − 2

4

2

2x−1

=2−x

33x

2x = 8 2x = 23 ⇒ x =3✓

5

4x

⇒ x2

32x+2

=

3(ii)

=− ✓

x

2(b)

= 32−x

27x

2 =8 sub u = 2x : u=8✓

(3−2 )x+2

4x (52x ) 22x (52x ) 102x ⇒ 2x x

2

2

=3

9

2(a)

3x

2 1

2

1(d)

2

1

8

2 =2 3 ⇒ x = −3 x

3x

=

x = −2 or x = 1 ✓

= ✓

(√2) =

2

(x + 2)(x − 1) = 0

5

x

9x+1

3x

sleightofmath.com

52x −6(5x ) +5 = 0 (5x )2 −6(5x ) +5 = 0 sub u = 5x : u2 − 6u + 5 =0 (u − 1)(u − 5) = 0 u=1 or u=5 x sub u = 5 : sub u = 5x : x 5 =1 5x = 5 5x = 51 5x = 50 ⇒ x =1✓ ⇒ x =0✓

55


6

7(a)

22x −10(2x ) +16 (2x )2 −10(2x ) +16 sub a = 2x : a2 −10a +16 (a − 2)(a − 8) a=2 or sub a = 2x : 2x = 2 ⇒ x =1✓

2x+1

Ex 2.3 =0 =0

7(b)

3x 9y

=0 =0 a=8 sub a = 2x : 2x = 8 2x = 23 ⇒ x =3✓

3x 32y

3x−2y

x

=

35x (32y ) = 5x+2y

= 33 = 33

= 2y + 3

−(1)

2nd eqn 1 42x (26y ) =

2 = 3(2 + 2 22x ⋅ 2 = 3(2x ) + 2 (2x )2 ⋅ 2 = 3(2x ) + 2 x sub u = 2 : 2u2 = 3u + 2 2u2 − 3u − 2 = 0 (2u + 1)(u − 2) = 0 1 u=2 u=− or 2 sub u = 2x : sub u = 2x : 2x = 2 1 2x = − 2 2x = 21 x (rej ∵ 2 > 0) ⇒ x = 1[shown] ✓ 5x (252y ) =1 x ((52 )2y ) 5 = 50 x (54y ) 5 = 50 5x+4y = 50 ⇒ x + 4y =0 x = −4y

= 27

⇒ x − 2y = 3

x)

35x (9y )

1st eqn

4

24x+6y = 2−2 ⇒ 4x + 6y = −2 sub (1) into (2): 4(2y + 3) + 6y 8y + 12 + 6y 14y y

−(2)

= −2 = −2 = −14 = −1 ✓

Put y = −1 into (1): x|y=−1 = 2(−1) + 3 =1✓ 8(i)

1

n

√2 × 4m = (2 × 4m )n 1

= (2 × 22m )n 1

= (22m+1 )n

−(1)

=2

1

2m+1 n

[shown] ✓

9 1 32 −2

3 =3 ⇒ 5x + 2y = −2

−(2)

sub (1) into (2): 5(−4y) + 2y = −2 −18y = −2 1 y = ✓ 9

Put y =

1 9

into (1): 1

x|y=1 = −4 ( ) 9

9

4

=− ✓ 9


sleightofmath.com

56

A math 360 sol (unofficial) 8(ii)

Ex 2.3

1st eqn n √2 × 4m = 8 2m+1 n

9(c)

3

2 =2 2m+1 ⇒ =3 n

n

2m+1

=

−(1)

3

27m

= 10(3x )

(32 )x+1

+1

= 10(3x )

32x+2

+1

= 10(3x )

(32x )(32 ) +1

= 10(3x )

9(3x )2

= 10(3x )

+1

= 81

9u2

= 34

9u2 − 10u + 1 = 0

3 = 34 ⇒ 3m − 2n − 2 = 4 3m − 2n =6

(9u − 1)(u − 1) = 0

9n+1 33m 32n+2 3m−2n−2

u= −(2)

3m − 2 ( 3m − (

2m+1

3 4m+2

)

)

3

9m − 4m − 2 5m m

=6

u=1

x

sub u = 3x : 3x = 1 3x = 30 ⇒ x =0✓

=6 9(d)

= 18 = 20 = 4✓

2(4)+1 3

x

sub u = 4 : 1 4x = 2

22x = 2−1 ⇒ 2x = −1 1 x =− ✓

9(e)

2(3x )

= 34 − 3x

3(3x )

= 34

3x

= 33

⇒x

=3

(√9)

2x

− 3x+2

32x − 3x+2

= 3x − 9

(3𝑥 )2 − (3𝑥 )(32 )

= 3𝑥 − 9

(3𝑥 )2 − 9(3𝑥 )

= 3𝑥 − 9

u2 − 10u + 9

=0

(u − 9)(u − 1) = 0 𝑢= 9 or x sub u = 3 : 3x = 9 3x = 32 ⇒ x =2✓

2

10(a) √8x (4y )

5(7x )

=5

7x

=1

2 2 22y

7x

= 70

2 2 +2y

⇒x

=0✓

3x

3x

⇒

3x 2

sleightofmath.com

𝑢= 1 sub u = 3x : 3x = 1 3x = 30 ⇒ x =0✓

= 32√2 1

= 25 (22 ) 1

= 252

+ 2y = 5

3x + 4y © Daniel & Samuel A-math tuition 📞9133 9982

= 3x − 9

sub u = 3x : x

= 2(7x ) + 3

7(7x ) −2 = 2(7x ) + 3

= 34 − 3x

(3𝑥 )2 − 10(3𝑥 ) + 9 = 0

sub u = 4 : 4x = 2 22x = 21 ⇒ 2x = 1 1 x = ✓

2

6(3x−1 )

6(3x )(3−1 ) = 34 − 3x

2(16x ) = 5(4x ) − 2 2(4x ) = 5(4x ) − 2 x )2 2(4 = 5(4x ) − 2 sub u = 4x : 2u2 = 5u − 2 2 2u − 5u + 2 = 0 (2u − 1)(u − 2) = 0 1 u=2 u= or

7x+1 −2

or

9

9

=3✓

2

1

= 10u

3x = 3−2 ⇒ x = −2 ✓

Put m = 4 into (1): n|m=4 =

+1

sub u = 3 : 1 3x =

sub (1) into (2):

9(b)

+1

sub u = 3x :

2nd eqn

9(a)

9x+1

1 2

= 11 [shown] ✓

−(1) 57


3y

Ex 2.3

1

12

= (3x )

2

9x 3y

3

−1 (3x )

=3

2

32x y−2x2

3 = 3−1+x 2 ⇒ y − 2x = −1 + x y = 2x 2 + x − 1 [shown] ✓ 10(c) sub (2) into (1): 3x + 4(2x 2 + x − 1) 3x + (8x 2 + 4x − 4) 8x 2 + 7x − 4 8x 2 + 7x − 15 (8x + 15)(x − 1) x=−

15 8

✓

y|x=−15 = 4 8

11

r2 4 r2 4 r2 4

32

2

(

3r ⋅ x r

( x −2 )

3 ⋅x

9x2

)

2

9 2 6−r

r

( )

2 6−r

r 2 (3r ) ( ) 9

2 6−r

r 2 (3r ) ( ) 9

9

=

6−r

k

u2

x3

27 2

9 4(3x ) 9

x 3r−12

= kx −3

=k

2 6−(3)

−(1)

13

(27) ( )

=k

9

3

9

−1

3

3

=k

4u

x 2 − 8x −2 = 7 8

x2 −

−(2)

(3)2 (3(3) ) ( )

9 2 3

=

−1

u = 12u − 27 2 u − 12u + 27 = 0 (u − 3)(u − 9) = 0 u=3 u=9 sub u = 3x : sub u = 3x : 3x = 3 or 3x = 9 x 1 3 =3 3x = 32 ⇒x=1✓ ⇒x=2✓ y|x=1 = 2(1) − 3 y|x=2 = 2(2) − 3 = −1 ✓ =1✓

= kx −3

x 2r−12 = kx −3

Compare power of x: 3r − 12 = −3 3r =9 r =3✓ sub (2) into (1):

k

=

27

Compare coefficient:

9

3x

sub u = 3x : 6−r

−(1)

= 4 ( ) −1

27 (3x )2

y|x=1 = 2 ✓

(3x)r

r

32x

x=1 ✓

✓

= 16x = 24x = 24x = 4x = 4x − 6 = 2x − 3

2nd eqn 3y = 4(3x−2 ) − 1 −(2) sub (1) into (2): 32x−3 = 4(3x−2 ) −1 2x −3 3 ⋅3 = 4(3x ⋅ 3−2 ) −1

= 11 = 11 = 11 =0 =0

or 5

−(2)

1st eqn 64(4y ) 26 (22y ) 26+2y ⇒ 6 + 2y 2y y

=7

3

x2

3

sub u = x 2 : 8 u− =7 u

u2 − 7u − 8 = 0 (u − 8)(u + 1) = 0 u=8 or u=1

2

= ✓ 3

3

sub u = x 2 : 3

x2 = 8 3

x 2 = 23 x = 22 x = 4✓ 14

3

sub u = x 2 : 3

x2 = 1 x =1✓

No. Counter example when a = 1 & b = 1, (1)x (1)y = (1)2 (1)5 LHS = RHS when x, y ∈ ℝ


sleightofmath.com

58


Ex 2.3

Method 1 (Substitution) x − 3√x − 4 =0 sub u = √x: u2 − 3u − 4 =0 (u − 4)(u + 1) = 0 u=4 or u=1 sub u = √x: sub u = √x: √x = 4 √x = 1 x = 16 ✓ x = 1 (rej) ✓ Method 2 (Square both sides) x − 3√x − 4 =0 x−4 = 3√x ⇒ x − 4 > 0, x > 0 2 (x − 4) = 9x x 2 − 8x + 16 = 9x x 2 − 17x + 16 = 0 (x − 1)(x − 16) = 0 x = 1 (rej) or x = 16 ✓


sleightofmath.com

59


Ex 2.4 2(c)

Ex 2.4 1(a)

1 x

y = 4x ∵ base 4 > 1, graph slopes up

= 2( ) 3

1(b)

𝑦 = 4𝑥 𝑥 ✓

1 x y=( ) 3 1 ∵ 0 < base < 1, graph slopes down

2(d)

3

3 𝑂

1 𝑥

1(c)

𝑦=( ) 3

𝑥

✓

𝑦

1(d)

3(i)

y = ex ∵ base e > 1, graph slopes up 𝑦 = 𝑒𝑥

1 𝑂

y = 3(5.1)x ∵ base 5.1 > 1, graph slopes up 𝑦

𝑦 1 𝑂

𝑦 = 3(5.1)𝑥 𝑥 ✓

y = 50 000(1.04)x y|x=0 = 50 000(1.04)0 = $50 000 ✓

3(ii)

y|x=5 = 50 000(1.04 )5 = $60 833 ✓

4(i)

y = 40(1.4)x

𝑥 ✓

1

∵ 0 < base < 1, graph slopes down 3 𝑦 𝑦 = 2(3−𝑥 ) 2 𝑥 𝑂 ✓

𝑦 1 𝑂

y = 2(3−x ) = 2(3−1 )x

y = πx ∵ base π > 1, graph slopes up

y|x=0 = 40(1.4)0 = 40 ✓

𝑦 𝑦 = 𝜋𝑥

1 𝑂 2(a)

y|x=6 = 40(1.4)6 = 301 ✓

4(iii)

y|x=24 = 40(1.4)24 = 128 568 ✓

5(i)

T = 90(0.98)x

𝑥 ✓

y = 6ex ∵ base e > 1, graph slopes up 𝑦 𝑦 = 6𝑒 𝑥

6 𝑂 2(b)

4(ii)

T|x=0 = 90(0.98)0 = 90° ✓

𝑥 ✓

5(ii)

T|x=10 = 90(0.98)10 = 73.5° ✓

5(iii)

T|x=60 = 90(0.98)60 = 26.8° ✓

−x

y = 2e = 2(e−1 )x 1 x

= 2( ) e

1

∵ 0 < base < 1, graph slopes down e 𝑦 𝑦 = 2𝑒 −𝑥 2 𝑥 𝑂 ✓


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60


Ex 2.4

y = 3(5x ) ∵ base 5 > 1, graph slopes up

9(ii)

y = 3(0.2)x ∵ 0 < base 0.2 < 1, graph slopes down 𝑦 𝑦 = 3(5𝑥 ) 3

O

𝑦 = 3(0.2) 𝑥

y|x=10 = 45 000(1.02)10 = 54 855 ✓

y = 3(2x ) ∵ base 2 > 1, graph slopes up 𝑦 𝑦 = 3(2𝑥 ) 6 3 𝑥

O

𝑦 =6−𝑥 ✓

7(iii)

One intersection ⇒ One solution ✓

8(i)

P = 600(2 + e−0.2t )

y = 45 000(1.02)x ∵ base 1.02 > 1, graph slopes up 𝑦 𝑦 = 45 000(1.02)𝑥

𝑥

✓ f(x) = 3(5x ) f(−x) = 3(5−x ) = 3(5−1 )x = 3(0.2)x y = f(x) and y = f(−x) are reflection of each other in the y-axis ✓ 7(i) 7(ii)

45 000

P|t=12 = 600(2 + e−0.2(12) ) = 1254 ✓

8(iii)

t → ∞, e−0.2t → 0, P ⇒ 600(2 + 0) = 1200 ✓

9(i)

y = 45 000(1.02)

𝑥

𝑂 10(i)

✓

C = 4.86e−0.047t C|t=0 = 4.86e0 = 4.86μg/ml ✓

10(ii) C|t=10 = 4.86e−0.047(10) = 3.04μg/ml ✓ 10(iii) C|t=24 = 4.86e−0.047(24) = 1.57μg/ml ✓ 10(iv) t ⇒ ∞, e−(0.047t) ⇒ 0, C⇒ 0✓ 10(v) C = 4.86e−0.047t = 4.86(e−0.047 )t = 4.86 (

P|t=0 = 600(2 + e0 ) = 1800 ✓ 8(ii)

end of 2021: x = 2021 − 2011 = 10

1 e0.047

∵ 0 < base

)

t

1

e0.047

< 1, graph slopes down

𝐶 4.86 𝑂

𝐶 = 4.86𝑒 −0.047𝑡 𝑡 ✓

x

end of 2015: x = 2015 − 2011 =4 y|x=4 = 45 000(1.02)4 = 48 709 ✓


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61


Ex 2.4 1 x

12(iv) m

y = 2−x = (2−1 )x = ( ) 1

405 ( ) = 80

∵ 0 < base < 1, graph slopes down 2

2 t

3 2

𝑦 = 8(2𝑥 )

8

13(i)

𝑦 = 2−𝑥

1

𝑥

𝑂

3

✓

16

( )

=

( ) 3 ⇒t

=( ) 3 =4

3 2 t

y = 8(2x ) ∵ base 2 > 1, graph slopes up 𝑦 (− , 2√2)

= 80 2 t

2

81 2 4

V = Ae−kt Initial value of the car is $60 000: V|t=0 = 60 000 −k(0) Ae = 60 000 A = 60 000

−x

y=2 −(1) y = 8(2x ) −(2) sub (1) into (2): 2−x = 8(2x ) 2−2x = 8 2−2x = 23 ⇒ −2x = 3 3 x =− −(3)

Value of the car is $39 366 after 4 years: V|t=4 = 39366 Ae−k(4) = 39366 −4k 60 000e = 39366 ∵ A = 60 000 39 366 −4k e = (e

2

60 000 9 4

−k 4

=( )

)

−k

⇒e

=

sub (3) into (1): 3 −(− )

V = 60 000 ( ) 10

= 60 000(0.9)t [shown] ✓

2

13(ii) V = 60 000(0.9)t ∵ 0 < base 0.9 < 1, graph slopes down

2 t

m = 405 ( ) 3

2 0

𝑉

m|t=0 = 405 ( ) 3

= 405g ✓ 12(ii)

10

9 t

y = 2 2 = 21.5 = 2√2 3 ⇒ (− , 2√2) 12(i)

10 9

60 000 𝑂

2 3

m|t=3 = 405 ( ) 3

𝑉 = 60 000(0.9)𝑡 𝑡 ✓

= 120g ✓ 12(iii) Amount decayed in 3 years = m|t=0 −m|t=3 = 405 −120 = 285g ✓


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62


Rev Ex 2 A2(b)

Rev Ex 2 A1(i)

1

(√5 − 2)x = √5 + 2 x

=

√5+2

=

√5+2

=

×

√5−2

√5+2 √5+2

(√5) +2(√5)(2) +(2)2 (√5) 5

2

−(2)2

+4√5

+4

5−4

= 9 + 4√5 ✓ A1(ii) x

+

= 9 + 4√5

+

1 x 1 9+4√5 1

= 9 + 4√5

+

= 9 + 4√5

+

= 9 + 4√5

+

= 9 + 4√5

+9 − 4√5

9+4√5

×

9−4√5 9−4√5

9−4√5 (9)2 −(4√5)

2

= 4√2

9−4√5

−√50

−

−√25 × 2

−

3

+

√2 3 √2

×

3

−5√2

− √2 2

= 4√2

−5√2

− √2 2

= 4√2

−5√2

− √2 2

= 4√2

−5√2

− √2 2

23 12

√96 3

−

√2 2

×(

√16×6 3

−

√2 2

×(

√2 2

× ( √6 − √6 3 3

√2 2

× ( √6 − √6 3 3

√2 2

× ( √6 − √6 3 3

√2 2

× ( √6 − √6 3 3

√2 2

× (10√6)

√2 √2

4√6 3

2 √6

√6

×

√6

2√6

+

6 1

− √6 3

4 4

+

1

+

1

4

1

4

1

+

+

3√216 2

)

= k√3

3√23 ×33 2 3√23 √33 2

)

= k√3

)

= k√3

3(21.5 )(31.5 ) 2

)

3(2√2)(3√3) 2 18√6

)

= k√3 = k√3

)

= k√3

+9√6)

= k√3

+9√6)

= k√3

2

= k√3

5√12

= k√3

5(√4 × 3)

= k√3

5(2√3)

= k√3

10√3

= k√3

⇒ k = 10 ✓

= 4√2

=−

×(

×

+

√6

81−16(5)

= 18 ✓ A2(a) 4√2

−

1

2

=

√12√8 3

√2

√5−2

2

×(

√2

3 3 3

√2 √2

+ + + + +

A3(a) √x 2 − 7 =3 2 x −7 =9 2 x − 16 =0 (x + 4)(x − 4) = 0 x = −4 or x = 4 ✓

7 √72 7 √36×2 7 6√2 7 6√2 7 6(2) 7 12

×

√2 √2

√2

√2

A3(b) 2x + √3 − 4x √3 − 4x 3 − 4x 4x 2 + 4x − 3 (2x + 3)(2x − 1)

√2 ✓

3

1

2

2

x = − or x = A3(c)

x √1−8x

=0 = −2x = 4x 2 =0 =0 (rej) ✓

=

1 3

3x = √1 − 8x 2 9x = 1 − 8x 2 9x + 8x − 1 = 0 (9x − 1)(x + 1) = 0 x=


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1 9

or x = −1 (rej) ✓

63

A math 360 sol (unofficial) A3(d)

4x

A5(b) 4x−1 + 16x = 66 (4x )(4−1 ) + (42 )x = 66

= √2(2x−1 )

8 22x

1

= 22 (2x−1 )

23

1 4

1 x− 2

2x−3

2

= 66

4

2

4u2 + u − 264 =0 (4u + 33)(u − 8) = 0

5 2

u=−

1st eqn 8 × 4y 23 × 22y 23+2y ⇒ 3 + 2y 2x x

= 66

sub u = 4 : u + u2

1

= ✓

x

(4x ) + (4x )2 x

=2

⇒ 2x − 3 = x −

A4

Rev Ex 2

2x−1

=2 = 22x−1 = 22x−1 = 2x − 1 = 4 + 2y =y+2

33

or

4

sub u = 4x : 4x = −

u=8 sub u = 4x :

33

4x = 8

4 x

4x = 23 22x = 23 ⇒ 2x = 3

(rej ∵ 4 > 0) −(1)

3

= ✓

x 2nd eqn 3y √3x

A6(i)

= 81 x

x

= 34 =4

2

−(2)

sub (1) into (2): y+2

y+

y

=4

2

2

2

y

A6(ii) m|t=2 = 300e−0.85(2) = 54.8mg ✓ A6(iii) y = 300e−0.85t = 300(e−1 )0.85t

y + + 1= 4 3

1 0.85t

=3

y

m = 300e−0.85t m|t=0 = 300e0 = 300mg ✓

(3y ) (32 ) = 34 3y+2 x ⇒y+

2

= 300 ( ) e

=2

1

∵ 0 < base < 1, graph slopes down e

Put y = 2 into (1): x|y=2 = (2) + 2

𝑚

=4✓ x)

A5(a) 2(4

x)

2(4

x+2

+4 +

(4x )(42 )

2(4x ) + 16(4x ) x)

18(4

=

= 9(4 = 9(

2

sub u = 4x : 18u

=

u

=

1 4 0.5 1 √4

B1(a) √x − 8 × √x √(x − 8)x

9

4 1

4x ⇒x

= 4−1 = −1 ✓

4


=3 =3

√x 2 − 8x =3 x 2 − 8x =9 2 x − 8x − 9 = 0 (x − 9)(x + 1) = 0 x = 9 or x = −1 (rej) ✓

sub u = 4x : =

✓

A6(iv) p = 300 − 300e−0.85t = 300(1 − e−0.85t ) ✓

2 1

4x

𝑡

)

= 9( )

9

300 𝑂

−0.5 )

𝑚 = 300𝑒 −0.85𝑡

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64


Rev Ex 2

B1(b) √18 − √x − 1 = 4 18 − √x − 1 = 16 =2 √x − 1 x−1 =4 x =5✓ B1(c)

√11+√4 x 2

=

B3(i)

1 2 1 2

x

33 (32 )

=

(BC)h

√11−√4

= =

3

x 2

3x 3

√9 3x

=3

⇒3+

2

x

x B2

=x− =

2

=

ℎ B C (8√3 − 2√2)

92 8√3−2√2

92 8√3 − 2√2

×

8√3 + 2√2 8√3 + 2√2

92(8√3 + 2√2) 2

2

=

92(8√3 + 2√2) 64(3) − 4(2)

=

92(8√3 + 2√2) 184

=

8√3 + 2√2 2

1

x−

x

= 46

(8√3) − (2√2)

(32 )3 3+

A

(8√3 − 2√2)h = 46

h=

x

x = −√7 or x = √7 ✓ =

= 46

(8√3 − 2√2)h = 92

x = 11 − 4 2 x −7 =0 (x + √7)(x − √7) = 0

B1(d) 27(√3)x

Area

2 3

2 3

11 3 22 3

= 4√3 + √2 ✓

✓

(a − 6√5)(2 + b√5)

B3(ii) By Pythagoras’ Theorem,

= −82

1

AC = √( BC)

a(2 + b√5) −6√5(2 + b√5) = −82

2

+ h2

2

2a + ab√5 −12√5 − 6b(5) = −82 2a − 30b +(ab − 12)√5 Equate rational terms: 2a − 30b = −82 a − 15b = −41 a = 15b − 41

1

= √[ (8√3 − 2√2)]

= −82

2

2

2

= √(4√3 + √2)

2

+ (4√3 − √2) + (4√3 − √2)

2

−(1)

2

2

[(4√3) + 2(4√3)(√2) + (√2) ]

=√

2

2

+ [(4√3) − 2(4√3)(√2) + (√2) ] Equate irrational terms: ab − 12 = 0 −(2) sub (1) into (2): (15b − 41)b − 1 = 0 15b2 − 41b − 12 = 0 (15b + 4)(b − 3) = 0 b=−

4

or

15

a|b=− 4 = 15 (− 15

4

) − 41

15

= −45 ✓


= √16(3) + 8√6 + 2

+ (16(3) − 8√6 + 2)

= √100 = 10 b=3✓ a|b=3 = 15(3) − 41 =4✓

sleightofmath.com

Perimeter = AB +AC +BC = 10 +10 +8√3 − 2√2 = 20 +8√3 −2√2 ✓

65


At (3,4):

Rev Ex 2 = a(3)n − 23 = a(3)n = a(3)n =a = 33−n

4 27 33 33−n a

At (9,220): 220 243 243 a(32n )

−(1)

= a(9)n − 23 = a(9)n = a(32n ) = 243 −(2)

At (−1, k): k = a(−1)n − 23

B5(b) 9x + 10(3x ) (32 )x + 10(3x ) (3x )2 + 10(3x ) sub u = 3x : u2 + 10u u2 + u − 12 (u + 4)(u − 3) u = −4 or sub u = 3x : 3x = −4 (rej ∵ 3x > 0)

−(3)

= 3x+2 + 12 = (3x )(32 ) + 12 = 9(3x ) + 12 = 9u + 12 =0 =0 u=3 sub u = 3x : 3x = 3 3x = 31 ⇒x=1✓ t

sub (1) into (2): (33−n )32n = 243 33+n = 35 ⇒3+n =5 n =2✓

B6(i)

Put n = 2 into (1): a|n=2 = 33−2 = 3 ✓

B6(ii) θ| = 20 + 100(0.8)86 t=8 ≈ 94.3 ✓

θ = 20 + 100(0.8)6 θ|t=0 =X 0

20 + 100(0.8)6 20 + 100(1) X

Put a = 3, n = 2 into (3): k = 3(−1)2 − 23 = −20 ✓

B6(iii) θ

= 84

20 + 100(0.8) x

x−1 )

B5(a) 5 + 5 = 30(5 5x + 5 = 30(5x )(5−1 )

100(0.8)

1

5x + 5 = 30(5x ) ( ) 5

5x + 5 = 6(5x ) sub u = 5x : u + 5= 6u 5 = 5u u =1 sub u = 5x : 5x = 50 ⇒ x =0✓


=X =X = 120°C ✓

t 6

= 84

t 6

= 64

(0.8)

t 6

= 0.64

(0.8)

t 6

= (0.8)2

⇒

t

=2

6

= 12s ✓

t B6(iv)

sleightofmath.com

t

t → ∞, 100(0.8)6 → 0 θ → 20°C ✓

66


Ex 3.1 3(b)

Ex 3.1 1(a)

Yes ✓ Are exponents  Non-negative?  Integers?

= 15x 4 +5x 3 −10x 2 +6x 3 +2x 2 −4x 2 −12x −4x +8 = 15x 4 + 11x 3 − 20x 2 − 8x + 8 ✓

True True 4(a)

∵ both conditions are met, 2 x − 1 is a polynomial 3

1(b)

(2x 2 − x + 1)(3x − 2) Coefficient of x 2 = (2)(−2) + (−1)(3) = −7 ✓

Yes ✓ Are exponents  Non-negative? True  Integers? True

4(b)

∵ both conditions are met, 4x 2 − 2x is a polynomial . 1(c)

(5x 2 + 2x − 4)(3x 2 + x − 2)

(x 2 + 3x + 2)(8x 2 − 5x − 4) Coefficient of x 2 = (1)(−4) + (3)(−5) +(2)(8) = −4 −15 +16 = −3 ✓

No ✓ Are exponents  Non-negative? True  Integers? False

4(c)

(2x 2 − 2x + 5)(−x 2 − 3x + 1)

1

√x has non-integer power of 2.

Coefficient of x 2 = (2)(1) + (−2)(−3) +(5)(−1) =2 +6 −5 =3✓

∵ both conditions are not met, 4x 3 + √x + 3 is not a polynomial. 1(d)

No ✓

4(d)

Are exponents  Non-negative? False  Integers? True 3 x2

Coefficient of x 2 = (−1)(6) + (−2)(3) = −6 − 6 = −12 ✓

has a negative power of −2.

∵ both conditions are not met, 3 1 + 2 is not a polynomial.

5(a)

x

2(i)

Q(x) − P(x) = (2x 2 − 3x + 2) −(x 2 + x + 1) = x 2 − 4x + 1 ✓

2(ii)

P(x) + 2Q(x) = (x 2 + x + 1) +2(2x 2 − 3x + 2) = (x 2 + x + 1) +(4x 2 − 6x + 4) = 5x 2 − 5x + 5 ✓

3(a)

(4x 3 − x 2 + 7x − 2)(2x 3 + 3x 2 + 6)

a(x − 2) + b = 5 − 3x sub x = 2: a(2 − 2) + b = 5 − 3(2) b = −1 ✓ sub x = 0: a(0 − 2) + b = 5 − 3(0) −2a + b =5 −2a + (−1) = 5 a = −3 ✓

∵ b = −1

(7x − 3)(2x 2 + 4x − 1) = 14x 3 +28x 2 −7x −6x 2 −12x +3 3 = 14x + 22x 2 − 19x + 3 ✓


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67


Ex 3.1

a(x − 1) + b(x + 3) = 3x + 1

6(i) (3x 2 + 2)

sub x = 1: a(1 − 1) + b(1 + 3)= 3(1) + 1 4b =4 b =1✓

(5x + 10) 1

A(x) = (5x + 10)(3x 2 + 2) 2 1

= [5x(3x 2 + 2) +10(3x 2 + 2)] 2 1

sub x = −3: a(−3 − 1) + b(−3 + 3) = 3(−3) + 1 −4a = −8 a =2✓ 5(c)

= (15x 3 + 10x +30x 2 + 20) 2 1

= (15x 3 + 30x 2 + 10x + 20) =

2 15 3 x 2

+ 15x 2 + 5x + 10

a(x − 2) + b(x − 4) = x + 2 Yes. Exponents are non-negative integers ✓ sub x = 2: a(2 − 2) + b(2 − 4)= 2 + 2 −2b =4 b = −2 ✓

6(ii)

= √(3x 2 + 2)2

sub x = 4: a(4 − 2) + b(4 − 4)= 4 + 2 2a =6 a =3✓ 5(d)

5(e)

= √3x 4 + 37x 2 + 100x + 104 P(x) = (5x + 1) + (3x 2 + 2) +√3x 4 + 37x 2 + 100x + 104

a =4✓ b =0✓ c = −1 ✓ 3 = −c + d 3 = −(−1) + d d=2✓

= 3x 2 + 5x + 3 +√3x 4 + 37x 2 + 100x + 104 No because √3x 4 + 37x 2 + 100x + 104 will have non-integer exponent ✓

x 3 − 6x 2 + 14x − 8 = (x − 2)3 + ax

7(i) (a)

sub x = 2: (2)3 − 6(2)2 + 14(2) − 8 = (2 − 2)3 + 2a 4 = 2a a =2✓ 5(f)

7(i) (b)

= a(2 − 2)2 + b(2 + 1)3 +


2P(x) − [Q(x)]2 = 2(x + 1) − (x 2 − x + 1)2 = 2(x + 1) − (x 2 − x + 1)(x 2 − x + 1) Coefficient of x 2 = −[(1)(1) +(−1)(−1) +(1)(1)] = −3 ✓

= 27b + 8 = 27b = −1 ✓

sub x = −1: −2(−1)2 − 7(−1) + 3 1)3 + (−1)3 8 9 a

M(x) = P(x)Q(x) = (x + 1)(x 2 − x + 1) Coefficient of x 2 = (1)(−1) +(1)(1) =0✓

−2x 2 − 7x + 3 = a(x − 2)2 + b(x + 1)3 + x 3 sub x = 2: −2(2)2 − 7(2) + 3 (2)3 −19 −27 b

+ (5x + 10)2

= √(3x 4 + 12x 2 + 4) + (25x 2 + 100x + 100)

ax 3 − x + 3 = 4x 3 + bx 2 + c(x − 1) + d Compare x 3 : Compare x 2 : Compare x: Compare x 0 :

By Pythagoras Theorem Hypotenuse

= a(−1 − 2)2 + b(−1 +

7(i) (c)

= 9a − 1 = 9a =1✓

Q(x)[3x 2 + P(x)] = (x 2 − x + 1)(3x 2 + x + 1) Coefficient of x 2 = (1)(1) +(−1)(1) +(1)(3) =1 −1 +3 =3✓

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68


Ex 3.1 9(c)

Deg (M(x)) = 3 ✓ Deg [2P(x) − [Q(x)]2 ] = 4 ✓ Deg [Q(x)[3x 2 + P(x)]] = 4 ✓

sub x = 2: 3(2)2 − 5(2) + 4 = a(2 − 2)2 + b(2 − 2) + c 6 =c c =6✓

8(i)

V(x) = (2x + 3)(3x − 1) (2x + 5) = (6x 2 + 7x − 3) (2x + 5)

sub x = 0: 3(0)2 − 5(0) + 4 4 4

= 12x 3 +14x 2 −6x +30x 2 +35x −15 3 = 12x + 44x 2 + 29x − 15 ✓ 8(ii)

9(a)

A(x) = 2[(2x + 3)(3x − 1) +(2x + 3)(2x + 5) +(3x − 1)(2x + 5)] = 2[(6x 2 + 7x − 3) +(4x 2 + 16x + 15) +(6x 2 + 13x − 5)] = 2(16x 2 + 36x + 7) = 32x 2 + 72x + 14 ✓

4 2b b 9(d)

x 3 − 6x 2 − x + c = (x − 3)(ax 2 − 3x + b) 3

Compare x : a = 1 ✓ sub x = 3: (3)3 − 6(3)2 − 3 + c = (3 − 3)(a(3)2 − 3(3) + b) −30 + c =0 c = 30 ✓ sub x = 0: (0)3 − 6(0)2 − (0) + c b) c 30 b 9(b)

= (0 − 3)(a(0)2 − 3(0) + = −3b = −3b ∵ c = 30 = −10 ✓

x 3 + cx 2 + x + 6 = (x + 1)(x − 2)(ax + b) Compare x 3 : a = 1 ✓ sub x = 0: (0)3 + c(0)2 + (0) + 6 = (0 + 1)(0 − 2)(a(0) + b) 6 = −2b b = −3 ✓ sub x = 2: (2)3 + c(2)2 + 2 + 6 4c + 16 c+4 c


3x 2 − 5x + 4 = a(x − 2)2 + b(x − 2) + c Compare x 2 : a = 3 ✓

= a(0 − 2)2 + b(0 − 2) + c = 4a − 2b + c = 4(3) − 2b + 6 ∵ a = 3, c = 6 = 18 − 2b = 14 =7✓

x 3 + 3x 2 − 2x + 16 − c(x + 2) = ax 2 (x − 1) + b(x − 2)2 (x − 1) sub x = 1: (1)3 + 3(1)2 − 2(1) + 16 − c(1 + 2) = a(1)2 (1 − 1) + b(1 − 2)2 (1 − 1) 18 − 3c =0 −3c = −18 c =6✓ sub x = 0: (0)3 + 3(0)2 − 2(0) + 16 − c(0 + 2) = a(0)2 (0 − 1) + b(0 − 2)2 (0 − 1) 16 − 2c = −4b 16 − 2(6) = −4b ∵ c = 6 4 = −4b b = −1 ✓ sub x = 2: (2)3 + 3(2)2 − 2(2) + 16 − c(2 + 2) = a(2)2 (2 − 1) + b(2 − 2)2 (2 − 1) 32 − 4c = 4a 32 − 4(6) = 4a ∵ c = 6 8 = 2a a =2✓

= (2 + 1)(2 − 2)(a(2) + b) =0 =0 = −4 ✓

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69


𝐱 𝟑 yz 2

Ex 3.1

+(𝐱 − z)2 y

2

+(2𝐱 − √y)

+y 2 z 4

= 𝐱 𝟑 yz 2 +(x 2 − 2𝐱z + z 2 )y +4𝐱 𝟐 − 4𝐱√y + y +y 2 z 4 = 𝐱 𝟑 yz 2 +𝐱 𝟐 y − 2𝐱zy + z 2 y +4𝐱 𝟐 − 4𝐱√y + y +y 2 z 4 = yz 2 𝐱 𝟑 +y𝐱 𝟐 − 2yz𝐱 + yz 2 +4𝐱 𝟐 − 4√y𝐱 + y +y 2 z 4 = yz 2 𝐱 𝟑 +y𝐱 𝟐 −2yz𝐱 +yz 2 +4𝐱 𝟐 −4√y𝐱 +y +y 2 z 4 = yz 2 𝐱 𝟑 +(y + 4)𝐱 𝟐 −(2yz + 4√y)𝐱 +yz 2 + y + y 2 z 4 Yes, all exponents of x are non-negative integers ✓ Degree is 3 ✓ 10(ii) x 3 𝐲z 2

+(x − z)2 𝐲 +(2x − √𝐲)

2

+𝐲 𝟐 z 4

= x 3 z 2 𝐲 +(x − z)2 𝐲 +4x 2 − 4x√𝐲 + 𝐲 +z 2 𝐲 𝟐 = x3z2 𝐲 +(x − z)2 𝐲 +𝐲

+4x 2 −4x√𝐲 +z 2 𝐲 𝟐

= [x 3 z 2 + (x − z)2 + 1]𝐲 +4x 2 −4x√𝐲 +z 2 𝐲 𝟐 = z 2 𝐲 𝟐 +[x 3 z 2 + (x − z)2 + 1]𝐲

−4x√𝐲 +4x 2

No, exponents of y include fractions. √y has non − integer exponent ✓ 10(iii) x 3 y𝐳 𝟐

+(x − 𝐳)2 y

2

+(2x − √y)

+y 2 𝐳 𝟒 2

+y 2 𝐳 𝟒

2

+y 2 𝐳 𝟒

2

+y 2 𝐳 𝟒

= x 3 y𝐳 𝟐 +(x 2 − 2x𝐳 + 𝐳 𝟐 )y +(2x − √y) = x 3 y𝐳 𝟐 +x 2 y − 2xy𝐳 + 𝐳 𝟐 y +(2x − √y) = x 3 y𝐳 𝟐 +x 2 y − 2xy𝐳 + y𝐳 𝟐 +(2x − √y) = x 2 y𝐳 𝟐 +x 2 y +y𝐳

𝟐

−2xy𝐳 2

+(2x − √y)

+y 2 𝐳 𝟒 2

= [x 2 y + y]𝐳 𝟐 +x 2 y + (2x − √y) −2xy𝐳 = y 2 𝐳 𝟒 +[x 2 y + y]𝐳 𝟐

−2xy𝐳

+y 2 𝐳 𝟒

+x 2 y + (2x − √y)

2

Yes, all exponents of z are non-negative integers ✓ Degree is 4 ✓


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70


Ex 3.1

A(t) = −t 3 + at 2 + p B(t) = bt(t − 2)2 + 2t 2 + ct + q

12(ii) (x − a)(x − b)(x − c) (b) = (x 2 − (a + b)x + ab)(x − c)

Yes. Exponents of t are non-negative integers ✓

= x 3 −(a + b)x 2 +abx −cx 2 +c(a + b)x −abc 3 2 (a = x − + b + c)x + [ab + c(a + b)]x − abc = x 3 − (a + b + c)x 2 + (ab + ac + bc)x − abc ✓

11(ii) A(0) = 300 p = 300 ✓ B(0) = 0 q=0✓ 11(iii) A(t) would decrease until it becomes empty at zero. A(t) cannot be negative ✓ 11(iv) 300 − A(t) 300 − (−t 3 + at 2 + p) t 3 − at 2 − p + 300 Sub p = 300, q = 0: t 3 − at 2 − 300 + 300 t 3 − at 2

12(ii) (x − a)(x − b) … (x − z) = (x − a)(x − b) … (x − x) … (x − z) = (x − a)(x − b) … (0) … (x − z) =0✓ 13

= B(t) = bt(t − 2)2 + 2t 2 + ct + q = bt(t − 2)2 + 2t 2 + ct + q = bt(t − 2)2 + 2t 2 + ct + 0 = bt(t − 2)2 + 2t 2 + ct

A = 3, B = −2, C = 5 and D = −2 3(x − 1) − 2(x − 1)(x + 1) + 5x(x 2 − 1) − 2 𝑦 𝑦 = 3(𝑥 − 1) − 2(𝑥 − 1)(𝑥 + 1) +5𝑥(𝑥 2 − 1) − 2

𝑦 = 3𝑥 3 − 2𝑥 2 + 𝑥 − 4

Compare t 3 : b = 1 ✓ t 3 − at 2 = t(t − 2)2

The graphs are not aligned therefore James’ answers are wrong.

+2t 2 + ct

t 3 − at 2 = t(t 2 − 4t + 4) +2t 2 + ct

3x 3 − 2x 2 + x − 4 = A(x − 1) + B(x − 1)(x + 1) + Cx(x 2 − 1) + D

t 3 − at 2 = t 3 − 4t 2 + 4t +2t 2 + ct t 3 − at 2 = t 3 − 2t 2 + (4 + c)t −at 2

Compare x 3 : C = 3 ✓

= −2t 2 + (4 + c)t

Compare t 2 : −a = −2 a =2✓

sub x = 1:

D = −2 ✓

sub x = −1:

−10 −10 2A A

Compare t: 4 + c = 0 c = −4 ✓ 12(i) (a)

𝑥

𝑂

(x − a)(x − b) = x 2 − (a + b)x + ab ✓

sub x = 0:

= −2A + D = −2A + (−2) =8 =4✓

∵ D = −2

−4 = −A − B + D B =4−A+D = 4 − (4) + (−2) ∵ A = 4, D = −2 = −2 ✓


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71


Ex 3.2 2(b)

Ex 3.2 1(a)

1(b)

3x +4 2 3x −2 9x +6x +2 −(9x 2 −6x) 12x +2 −(12x −8) 10 Quotient = 3x + 4 ✓ Remainder = 10 ✓ +3x 2x 2 3 2x +1 4x +8x 2 −(4x 3 +2x 2 ) 6x 2 −(6x 2

1(c)

+2 +7x −5

+8x 5x 4 x −2x −3 5x −2x 3 −(5x 4 −10x 3 8x 3 −(8x 3 2

4x 2

+37 +6x 2 −15x 2 ) +21x 2 −16x 2 37x 2 −(37x 2

−4x

−4

−x 2 −3x 2 ) +8x 2x 2 2 +2x) −(2x 6x −4 −(6x +6) −10

+

2x 2 +2x −1 8x 4 +0x 3 −(8x 4 +8x 3 −8x 3 −(−8x 3

11 2

−x 2 −4x 2 ) +3x 2 −8x 2 11x 2 −(11x 2

+0x

+5

+0x +4x) −4x +5 +11x − 11) 2

−15x + 21 2

+4x

−3

8x 4 − x 2 + 5 = (2x 2 + 2x − 1) (4x 2 − 4x +

+4x −24x) +28x −3 −74x −111) 102x +108

+1 2x 2 3 2 x −2 2x −4x +x −2 −(2x 3 −4x 2 ) +x −2 −(x −2) 0 2x 3 − 4x 2 + x − 2 = (x − 2)(2x 2 + 1) ✓


+6 +8x

2(c)

Quotient = 5x 2 + 8x + 37 ✓ Remainder = 102x + 108 ✓ 2(a)

+2x −x 2

3x 4 − x 2 + 8x − 4 = (x + 1)(3x 3 − 3x 2 + 2x + 6) − 10 ✓

+7x +3x) 4x −5 −(4x +2) −7 2 Quotient = 2x + 3x + 2 ✓ Remainder = −7 ✓ 2

3x 3 −3x 2 x +1 3x 4 +0x 3 4 −(3x +3x 3 ) −3x 3 −(−3x 3

11

21

2

2

) − 15x +

✓

3(i) Deg(Dividend) = Deg(Divisor) +Deg(Q(x)) 4 3 2 Deg (x + 2x − 2x ) = Deg(x + 2) +Deg(Q(x)) −2x + 4 (4) = (1) +Deg(Q(x)) Deg (Q(x))

=3✓

Deg(R(x)) < Deg(Divisor) < Deg(x + 2) <1 =0✓ −2x x3 4 3 x +2 x +2x −2x 2 −(x 4 +2x 3 ) −2x 2 −(−2x 2

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+2 −2x +4 −2x −4x) 2x +4 −(2x +4) 0

72


Ex 3.2

3(ii) Deg(Dividend) = Deg(Divisor) +Deg(Q(x)) 4 3 2 Deg (x + 2x − 2x ) = Deg(x + 2) +Deg(Q(x)) −2x + 4 (4) = (2) +Deg(Q(x)) Deg (Q(x))

=2

Deg(R(x)) < Deg(Divisor) < Deg(x 2 + x + 3) <2✓ +x x2 2 +x +3 x 4 x +2x 3 −(x 4 +x 3 x3 −(x 3

−6 −2x 2 +3x 2 ) −5x 2 +x 2 −6x 2 −(−6x 2

3(iv) Deg(Dividend)

(4)

= (3)

Deg (Q(x))

=1

+Deg(Q(x))

Deg(R(x)) < Deg(Divisor) < Deg(x 3 − x 2 + x − 1) <3✓ −2x +4 x −2x +3x) −5x +4 −6x −18 ) x +22

3

−x

4(a)

−2x 2 +x 2 −3x 2 −3x 2

−2x +4 −x) −x +3x +4) −4x +4

= (2x 2 − 1)(3x + 2) = 6x 3 + 4x 2 − 3x − 2

Dividend

=x−2 +16x 6x 3 x −2 6x +4x 2 −(6x 3 −12x 2 ) 16x 2 −(16x 2

+29 −3x

−2

−3x −32x) 29x −2 −(29x −58) 56

(2x 2 − 1)(3x + 2) = (x − 2)(6x 2 + 16x + 29) + 56 ✓

−2x +4 −2x −2x) +0x +4 +0x +1) 3

x +3 4 +x −1 x +2x 3 −(x 4 −x 3 3x 3 −(3x 3

2

Deg(R(x)) < Deg(Divisor) < Deg(x 2 − 1) <2✓ −1 −2x 2 −x 2 ) −x 2 +0x −x 2 −(−x 2

2

Divisor

=2

+2x x2 2 +0x −1 x 4 x +2x 3 −(x 4 +0x 3 2x 3 −(2x 3

+Deg(Q(x))

4 3 2 3 2 Deg (x + 2x − 2x ) = Deg (x − x ) +Deg(Q(x)) −2x + 4 +x − 1

3(iii) Deg(Dividend) = Deg(Divisor) +Deg(Q(x)) 4 3 2 Deg (x + 2x − 2x ) = Deg(x 2 − 1) +Deg(Q(x)) −2x + 4 (4) = (2) +Deg(Q(x)) Deg (Q(x))

= Deg(Divisor)

4(b)

Dividend Divisor

x

2

= 3x 3 − x 2 + 2x + 3 = (x + 1)(x − 1) = x 2 − 1

3x −1 3 +0x −1 3x −x 2 −(3x 3 +0x −x 2 −(−x 2

+2x −3x) +5x +0x 5x

+3 +3 +1 ) +2

3x 3 − x 2 + 2x + 2 = (x + 1)(x − 1)(3x − 1) + 5x + 2 ✓


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73

A math 360 sol (unofficial) 4(c) Dividend Divisor

4

3

Ex 3.2 5(c) Dividend = 6x 4 − 3x 3 − 28x 2 − x − 10 Divisor = 3 − 2x = −2x + 3

2

= 10x − 7x − 4x + 8x + 7 = (x + 1)(2x − 3) = 2x 2 − x − 3 −x +5 5x 2 2 −x −3 10x 4 3 2x −7x −4x 2 +8x +7 −(10x 4 −5x 3 −15x 2 ) 3 −2x +11x 2 +8x +3x) −(−2x 3 +x 2 2 +5x +7 10x 2 −5x −15) −(10x 10x +22 4

3

−3x 3 −3x 2 + 19 x

+

2

2

−

−11x 4x 3 2x +1 8x −18x 2 −(8x 3 +4x 2 ) −22x 2 −(−22x 2

+6 +x

2 59 2

x

−(−

59 2

−10

−10 x + −

Dividend = 8x 3 − 18x 2 + x + 6 Divisor = 2x + 1 2

4

−2x +3 6x 4 −x −3x 3 −28x 2 4 3 −(6x −9x ) 6x 3 −28x 2 3 −(6x −9x 2 ) −x −19x 2 2 −(−19x + 57 x)

10x − 7x − 4x + 8x + 7 = (2x 2 − x − 3)(5x 2 − x + 5) + 10x + 22 ✓ 5(a)

59

No, there is a remainder − 6

+6

+x −11x) 12x +6 −(12x +6) 0

217 4

177

)

4 217 4

✓

2x 5 − x 3 + x 2 − 4 = (x 2 − x − 2)Q(x) + ax + b 2x 5 − x 3 + x 2 − 4 = (x − 2)(x + 1)Q(x) + ax + b sub x = 2: 64 − 8 + 4 − 4 = 2a + b 56 = 2a + b b = 56 − 2a

−(1)

2

Yes, it is 4x − 11x + 6 ✓ sub x = −1: −2 + 1 + 1 − 4 = −a + b −4 = −a + b −(2) (1) (2): sub into −4 = −a + (56 − 2a) 3a = 60 a = 20 ✓

5(b) Dividend = 6x 4 − 25x 3 + 5x 2 + 60x − 36 Divisor = (x − 2)(3x − 2) = 3x 2 − 8x + 4 −3x −9 2x 2 2 −8x +4 6x 4 +60x −36 3x −25x 3 +5x 2 4 3 2 −(6x −16x +8x ) 3 +60x −9x −3x 2 3 2 −12x) −(−9x +24x 2 +72x −36 −27x −(−27x 2 +72x −36) 0

Put a = 20 into (1): b|a=20 = 56 − 2(20) = 16 ✓ 7(i)

Deg(Q(x)) = 1 ✓

Yes, it is 2x 2 − 3x − 9 ✓


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74


Ex 3.2

x 4 − 2x 3 + ax 2 − 2 = x(x − 1)(x + 1)Q(x) + 4x 2 + bx + c sub x = 0:

c = −2 ✓

sub x = 1:

1−2+a−2 =4+b+c a−3 =4+b−2 a =b+5

sub x = −1:

9 (i)(a)

−(2)

Remainder = 4x 2 + bx + c = 4x 2 + (−2)x − 2 = 4x 2 − 2x − 2 ✓

8(ii)

32 − 24 + 1 = 2a − 3 9 = 2a − 3 a =6✓

4x 3 − 6x 2 + 1 = (x − 2)Q(x) + 6x − 3 3 2 4x − 6x − 6x + 4 = (x − 2)Q(x) Q(x)

=

+16x +0x 2

+26 −6x

−5

9 Remainder when P(x) + Q(x) is divided by x − 2 (i)(b) = 3 + 47 = 50 ✓

4x 3 − 6x 2 + 1 = (x − 2)Q(x) + ax − 3 sub x = 2:

2x 3 +8x 2 x −2 2x 4 +4x 3 −(2x 4 −4x 3 ) 8x 3 −(8x 3

−9

+0x 2 −16x 2 ) −6x 16x 2 2 −32x) −(16x 26x −5 −(26x −52) 47 Remainder when Q(x) is divided by x − 2 = 47 ✓

Put b = −2 into (1): a|b=−2 = (−2) + 5 =3✓

8(i)

+6 +4x

+4x −2x) 6x −9 −(6x −12) 3 Remainder when P(x) is divided by x − 2 =3✓

−(1)

1+2+a−2 =4−b+c a+1 =4−b−2 a =1−b sub (1) into (2): b + 5= 1 − b 2b = −4 b = −2 ✓

+x 3x 2 3 x −2 3x −5x 2 3 −(3x −6x 2 ) x2 −(x 2

Remainder of 2P(x) − Q(x) is divided by x − 2 = 2(3) − 47 = −41 ✓

4x3 −6x2 −6x+4

+2x 4x 2 3 x −2 4x −6x 2 −(4x 3 −8x 2 ) 2x 2 −(2x 2

x−2

−2 −6x

+4

−6x −4x) −2x +4 (−2x +4) 0

Q(x) = 4x 2 + 2x − 2 ✓


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75


Ex 3.2

9(ii)(a) 2

10(ii) For x = 7: Dispenser capacity

3x −2 3 −x −2 3x −5x 2 −(3x 3 −3x 2 −2x 2 −(−2x 2

+4x −9 −6x) +10x −9 +2x +4) 8x −13 2 Remainder when P(x) is divided by x − x − 2 = 8x − 13 ✓ x

+6x 2x 2 2 −x −2 2x 4 x +4x 3 4 −(2x −2x 3 6x 3 −(6x 3

+10 +0x 2 −4x 2 ) +4x 2 −6x 2 10x 2 −(10x 2

Number of Bottles

−5 11

−6x −12x) +6x −5 −10x −20) 16x +15 2 Remainder when Q(x) is divided by x − x − 2 = 16x + 15 ✓

2x+y

12

Remainder when 3P(x) + Q(x) is divided by x 2 − x − 2 = 3(8x − 13) +(16x + 15) = 24x − 39 +16x + 15 = 40x − 24 ✓ Dispenser capacity = (3x 4 + x 3 + 6x + 2) Number of bottles = (x 2 + 1) +x −3 3x 2 2 +0x +1 3x 4 +6x +2 x +x 3 +0x 2 4 3 2 −(3x +0x +3x ) +6x x3 −3x 2 3 2 +x) −(x +0x 2 +5x +2 −3x 2 +0x −3) −(−3x 5x +5

= 5(7) + 5 = 40 cm3 ✓

3x −4y 2 2x +y 6x −5xy −4y 2 2 +3xy) −(6x −8yx −4y 2 −(−8yx −4y 2 ) 0 6x2 −5xy−4y2

9(ii)(b) Remainder when Q(x) − P(x) is divided by x 2 − x − 2 = (16x + 15) − (8x − 13) = 8x + 28

10(i)

= (7)2 + 1 = 50 ✓

Capacity of each bottle = 3(7)2 + 7 − 3 = 151 cm3 ✓ Amount of drink left

−6x

= 3(7)4 + (7)3 + 6(7) + 2 = 7590 cm3 ✓

= 3x − 4y ✓

−5yx 3x 2 3 x +2y 3x +yx 2 −(3x 3 +6yx 2 ) −5yx 2 −(−5yx 2

+7y 2 −3y 2 x

−y 3

−3y 2 x −10y 2 x) 7y 2 x −y 3 −(7y 2 x +14y 3 ) −15y 3 Quotient = 3x 2 − 5yx + 7y 2 ✓ Remainder = −15y 3 ✓

Capacity of each bottle = 3x 2 + x − 3 ✓ Amount of drink left = 5x + 5 ✓


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76


Ex 3.2

3x 3 − ax 2 − bx + 72 = (x 2 + 5x + c)(dx + e) Compare x 3 : d = 3 3x 3 − ax 2 − bx + 72 = (x 2 + 5x + c)(3x + e) = 3x 3 +15x 2 +3cx +ex 2 +5ex +ce 3 2 = 3x + (15 + e)x +(3c + 5e)x +ce Compare x 2 : −a = 15 + e

−(1)

Compare x:

−(2)

−b = 3c + 5e

Compare x 0 : 72 = ce ⇒ let c = 1, e = 72 ✓ Put e = 72 into (1): −a = 15 + 72 a = −87 ✓ Put c = 1, e = 72 into (2): −b = 3(1) + 5(72) b = −363 ✓ 3x +72 3 x +5x +1 3x +87x 2 −(3x 3 +15x 2 72x 2 −(72x 2 2


+363x +72 +3x) +360x +72 +360x +72) 0

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77


Ex 3.3 1(a)

Ex 3.3

Let f(x) = x 4 − 4x 3 + 3x 2 + x + 3 By Remainder Theorem, Remainder = g(2)

By Remainder Theorem, Remainder = f(−1) =1+4+3−1+3 = 10 ✓ 1(b)

g(x) = 2x 3 − bx 2 + 2ax − 4 = 2x 3 −(3a − 5)x 2 +2ax −4

Let f(x) = [x(x − 1)(1 − 2x)2 + x 2 − 3]

= 2(2)3 −(3a − 5)(2)2 +2a(2)

−4

= 16 −(3a − 5)(4) +4a

−4

= 16 −12a + 20 +4a

−4

= 32 −8a ✓ By Remainder Theorem, Remainder = f(2) = 2(1)(9) + 4 − 3 = 19 ✓ 1(c)

5

Let f(x) = [3(x + 4)2 − (1 − x)3 ] By Remainder Theorem, Remainder = f(0) = 3(4)2 − 1 = 47 ✓

2

3

Let f(x) = 8x 3 + ax 2 + bx − 9 f(x) has remainder of −95 when divided by x + 2 By Remainder Theorem, f(−2) = −95 −64 + 4a − 2b − 9 = −95 4a − 2b = −22 4a = 2b − 22 a

=

Let f(x) = x + 3x − kx + 4

3

−(1)

=3

2

By Remainder Theorem, f(2) =k 8 + 12 − 2k + 4 = k 3k = 24 k =8✓

9

3

4 3

2

27 + a + b − 9 9

Let f(x) = x 4 + x 3 + 2ax 2 − 14a4

=3

a+ b

= −15

9a + 6b 3a + 2b

= −60 = −20

4

2

−(2)

sub (1) into (2): 3(

By Remainder Theorem, f(−2a) = 32 4 3 3 4 16a − 8a + 8a − 14a = 32 2a4 = 32 4 a = 16 a = 2 or a = −2 ✓

b−11 2

) + 2b

(3b − 33) + 4b 7b b

= −20 = −40 = −7 = −1✓

put b = −1 into (1): a|b=−1 =

4(i)

2

f(x) has remainder of 3 when divided by 2x − 3 By Remainder Theorem,

2

f( )

3

b−11

f(x) = ax 2 + bx − 6

(−1)−11 2

= −6 ✓

By Remainder Theorem, f(−3) =9 9a − 3b − 6 = 9 3b = 9a − 15 b = 3a − 5 ✓


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78


Ex 3.3

f(x) = x 3 + ax 2 + bx − 3 f(x) has remainder of 27 when divided by x − 2, By Remainder Theorem, f(2) = 27 8 + 4a + 2b − 3 = 27 4a = 22 − 2b a

=

11−b 2

7(i)

−(1)

f(x) has remainder of 3 when divided by x + 1, By Remainder Theorem, f(−1) =3 −1 + a − b − 3 = 3 a−b =7 −(2)

f(x) has a remainder of p + 5 when divided by x−2 By Remainder Theorem, f(2) =p+5 8 + 4a + 7 =p+5 4a = p − 10 −(2)

sub (1) into (2):

sub (1) into (2): 8p − 24 = p − 10 7p = 14 p =2✓

11−b 2

−b

=7

11 − b − 2b = 14 3b = −3 b = −1 ✓

Put p = 2 into (1): a|p=2 = 2(2) − 6

Put b = −1 into (1): a|b=−1 =

= −2 ✓

11−(−1) 2

=6✓ 6(ii)

f(x) = x 3 + ax 2 + 7 f(x) has remainder of 2p when divided by x + 1 By Remainder Theorem, f(−1) = 2p 3 2 (−1) + a(−1) + 7 = 2p −1 + a + 7 = 2p a = 2p − 6 −(1)

7(ii)

f(x) = x 3 + 6x 2 − 1x − 3

f(x) = x 3 − 2x 2 + 7 By Remainder Theorem, Remainder = f(−p) = f(−2) = (−2)3 − 2(−2)2 + 7 = −8 − 8 + 7 = −9 ✓

By Remainder Theorem, Remainder = f(1) = (1)3 + 6(1)2 − 1(1) − 3 = 3✓ 8

Let f(x) = ax 3 + bx 2 + 2x + c By Remainder Theorem, f(1) = 2f(−1) a + b + 2 + c = 2(−a + b − 2 + c) a + b + 2 + c = −2a + 2b − 4 + 2c c = 3a − b + 6 [shown] ✓


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79


Ex 3.3

Let f(x) = 2x 2 + 6x + 3

11(iii) P(x) ÷ (3x 2 − x − 2) ⇒ Remainder = ax + b ⇒ P(x) = (3x 2 − x − 2)Q(x) +ax + b = (x − 1)(3x + 2)Q(x) +ax + b

By Remainder Theorem, f(−p) = f(2q) 2(−p)2 + 6(−p) + 3 = 2(2q)2 + 6(2q) + 3 2p2 − 6p + 3

10

p2 − 3p

= 4q2 + 6q

3p + 6q

= p2 − 4q2

3(p + 2q)

= (p + 2q)(p − 2q)

p − 2q

=3✓

−1 + a

= 20 = 20 = 20 = 20 =0 =0

12

3

3

=1

f(x) ÷ (x 2 + x − 2) ⇒ Remainder = ax + b ⇒ f(x) = (x 2 + x − 2)Q(x) +ax + b = (x − 1)(x + 2)Q(x) +ax + b

when f(x) is divided by x + 2, remainder is −2 By Remainder Theorem, f(−2) = −2 −2a + b = −2 −(2) sub (1) into (2): −2(4 − b) + b = −2 −8 + 2b + b = −2 3b =6 b =2 a|b=2 = 4 − (2) =2

11(ii) P (− 2) = − 2 a + b −

5

= a

When f(x) is divided by x − 1, remainder is 4 By Remainder Theorem, f(1) = 4 a + b= 4 a =4−b −(1)

P(1) =a+b 9−3−5−2 =a+b −1 = a + b [shown] ✓ −(1)

3

3

Remainder = ax +b = (1)x +(−2) =x−2✓

P(x) = 9x 3 − 3x 2 − 5x − 2 = (x − 1)(3x + 2)Q(x) + ax + b

8

8

Put a = 1 into (1): −1 = 1 + b b = −2

f(x) = x 2 − 2x + p g(x) = px 2 + x + 3 By Remainder Theorem, f(−1)g(−1) [(−1)2 − 2(−1) + p][p(−1)2 + (−1) + 3] (p + 3)(p + 2) p2 + 5p + 6 p2 + 5p − 14 (p + 7)(p − 2) p = −7 or p = 2 ✓

11(i)

(1) − (2):

= 8q2 + 12q + 3

3 2

= − a + b [shown] ✓ −(2) 3

Remainder = 2x + 2 ✓ 13

f(x) = 4x 3 (x − 2)5 − x 2 + x√x − 7 f(x) is not a polynomial because not all exponents are non-negative integers 3

x√x has exponent of ✓ 2


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80


Ex 3.4 3(ii)

Ex 3.4 1(a)

3

2

P(x) = 2x − 4x + x − 2 P(2) = 16 − 16 + 2 − 2 =0

+5x 2x 2 3 x −2 2x +x 2 −(2x 3 −4x 2 ) 5x 2 −(5x 2

By Factor Theorem, Yes ✓ 1(b)

3

+2 −8x

−4

−8x −10x) 2x −4 (2x −4) 0

g(x) = 2x 3 + x 2 − 8x − 4 = (x − 2)(2x 2 + 5x + 2) = (x − 2)(2x + 1)(x + 2) ✓

2

P(x) = 16x + 8x − 6x − 9 1

P (− ) = −2 + 2 + 3 − 9 2

= −6 ≠0 By Factor Theorem, No ✓

4(a)

x 3 − 64 = x 3 − 43 = (x − 4)[x 2 + x(4) + 42 ] = (x − 4)(x 2 + 4x + 16) ✓

4(b)

27a3 + 125x 3 = (3a)3 + (5x)3

1(c)

P(x) = x 4 − 2x(x 2 − 8) − 12x 2 + 11 P(−3) = 81 − 2(−3)(9 − 8) − 12(9) + 11 = −10 ≠0

= (3a + 5x)[(3a)2 − 3a(5x) + (5x)2 ] = (3a + 5x)(9a2 − 15ax + 25x 2 ) ✓ 4(c)

8 + 27x 3 = 23 + (3x)3 = (2 + 3x)[22 − 2(3x) + (3x)2 ]

By Factor Theorem, No ✓ 2

2

= (2 + 3x)(4 − 6x + 9x 2 ) ✓

P(x) = 3x − 4ax − 4a

x + 2 is a factor of P(x) By Factor Theorem, P(−2) 3(−2)2 − 4a(−2) − 4a2 12 + 8a − 4a2 a2 − 2a − 3 (a − 3)(a + 1) a = 3 or a = −1 ✓ 3(i)

4(d)

2

432x 3 − 2y 3 = 2(216x 3 − y 3 ) = 2[(6x)3 − y 3 ] = 2(6x − y)[(6x)2 + 6x(y) + y 2 ]

=0 =0 =0 =0 =0

= 2(6x − y)(36x 2 + 6xy + y 2 ) ✓

g(x) = 2x 3 + x 2 + px − 4 x − 2 is a factor of g(x) By Factor Theorem, g(2) 2(2)3 + (2)2 + p(2) − 4 16 + 4 + 2p − 4 2p p


=0 =0 =0 = −16 = −8 ✓

sleightofmath.com

81


Ex 3.4

f(x) = ax 3 + bx 2 − 5x + 2a Quadratic factor: x 2 − 3x − 4 = (x − 4)(x + 1) x − 4 is a factor of f(x) By Factor Theorem, f(4) =0 64a + 16b − 20 + 2a = 0 66a + 16b = 20 33a + 8b = 10 x + 1 is a factor of f(x) By Factor Theorem, f(−1) =0 −a + b + 5 + 2a= 0 a+b+5 =0 a = −5 − b

6(ii)

6(iii)

p = −2 ✓

7

f(x) = 2x 2 + 3px − 2q g(x) = x 2 + q

−(2)

Put b = −7 into (2): a|b=−7 = −5 − (−7) =2✓

2x −1 3 −3x −4 2x −7x 2 −(2x 3 −6x 2 −x 2 −(x 2

−(1)

x − a is a factor of g(x) By Factor Theorem, g(a) =0 a2 + q = 0 a2 = −q

−(2)

a = 0 (rej ∵ a ≠ 0) or a = − −5x +4 −8x) +3x +4 +3x +4) 0

3p 4

−(3)

Method 1 sub (3) into (1): 2 (−

∴ The third factor is (2x − 1) ✓ 6(i)

x − a is a factor of f(x) By Factor Theorem, f(a) =0 2 2a + 3pa − 2q = 0

sub (2) into (1) 2a2 + 3pa + 2a2 = 0 4a2 + 3pa = 0 a(4a + 3p) =0

f(x) = 2x 3 − 7x 2 − 5x + 4

x2

=0 =0 =0 =0 =0 =0

−(1)

sub (2) into (1): 33(−5 − b) + 8b = 10 −165 − 33b + 8b = 10 −25b = 175 b = −7 ✓

5(ii)

By Factor Theorem, f(−2) (2p + 1)4 − 2p + 2p2 8p + 4 − 2p + 2p2 2p2 + 6p + 4 p2 + 3p + 2 (p + 1)(p + 2) p = −1 or p = −2 ✓

2( 8

−

By Factor Theorem, f(1) =0 2 (2p + 1) + p + 2p = 0 2p2 + 3p + 1 =0 (2p + 1)(p + 1) =0

4

9p2

9p2

f(x) = (2p + 1)x 2 + px + 2p2

3p 2

16

−

9p2 8 2

) + 3p (−

)− 9p2 4

9p2 4

− 2q

− 2q

− 2q

3p 4

) − 2q = 0 =0 =0 =0

9p + 16q

= 0 [shown]✓

Method 2 sub (3) into (2):

1

p = − or p = −1 ✓

(−

2

3p 2

9p2 16

4

)

= −q = −q

2

9p + 16q = 0 [shown] ✓


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82


Ex 3.4 10(a) 2x 4 + 54x = 2x(x 3 + 27) = 2x(x 3 + 33 ) = 2x(x + 3)[x 2 − x(3) + 32 ] = 2x(x + 3)(x 2 − 3x + 9) ✓

Part I f(x) = 3x 3 − 5ax 2 + ka2 x + 4a3 , a ≠ 0 By Remainder Theorem, f(−2a) 3(−8a3 ) − 5a(4a2 ) + ka2 (−2a) + 4a3 −24a3 − 20a3 − 2ka3 + 4a3 −40a3 − 2ka3 2ka3 + 8a3 ka3 + 4a3 a3 (k + 4) a = 0 (rej ∵ a ≠ 0) or k = −4 ✓

= −32a3 = −32a3 = −32a3 = −32a3 =0 =0 =0

10(b) (1 + x)3 + 64 = (1 + x)3 + 43 = [(1 + x) + 4][(1 + x)2 (1 + 2x + x 2 = (x + 5) = (x + 5)(x 2 − 2x + 13) ✓

− (1 + x)(4) + (4)2 ] − 4 − 4x + 16)

⇒ f(x) = 3x 3 − 5ax 2 − 4a2 x + 4a3 Part II Quadratic factor: x 2 − ax − 2a2 = (x − 2a)(x + a)

f(2a) = 3(2a)3 − 5a(2a)2 − 4a2 (2a) + 4a3 = 24a3 − 20a3 − 8a3 + 4a3 =0 By Factor Theorem, (x − 2a) is a factor of f(x) f(−a) = 3(−a)3 − 5a(−a)2 − 4a2 (−a) + 4a3 = −3a3 − 5a3 + 4a3 + 4a3 =0 By Factor Theorem, (x + a) is a factor of f(x) ∴ x 2 − ax − 2a2 is a factor of f(x) ✓ 9

f(a) = a3 − ba2 − 4b2 a + 4b3 f(2b) = (2b)3 − b(2b)2 − 4b2 (2b) + 4b3 = 8b3 − 4b3 − 8b3 + 4b3 =0 By Factor Theorem, (a − 2b) is a factor of f(a) ✓ when f(a) is divided by a + b, By Remainder Theorem, Remainder = f(−b) = (−b)3 − b(−b)2 − 4b2 (−b) + 4b3 = −b3 − b3 + 4b3 + 4b3 = 6b3 ✓


sleightofmath.com

83


Ex 3.4

10(c) (x + y)3 − (x − y)3 = [(x + y) − (x − y)] [(x + y)2 + (x + y)(x − y) + (x − y)2 ] 2 2 = 2y [x + 2xy + y +x 2 − y +(x 2 − 2xy + y 2 )] = 2y [2x 2 + 2xy + (x 2 − 2xy + y 2 )] = 2y(3x 2 + y 2 ) ✓ 10(d) 1000x 3 − y 6 = (10x)3 − (y 2 )3 = (10x − y 2 )[(10x)2 + 10x(y 2 ) + (y 2 )2 ] = (10x − y 2 )(100x 2 + 10xy 2 + y 4 ) ✓ 11(i)

x 6 − 64 = (𝑥 3 )2 − 82 = (x 3 + 8)(x 3 − 8) = (x 3 + 23 )(x 3 − 23 ) = (x + 2)[x 2 − x(2) + 22 ] (x − 2)[x 2 + x(2) + 22 ] = (x + 2)(x 2 − 2x + 4) (x − 2)(x 2 + 2x + 4) ✓

11(ii) x 6 − 64

= (x 2 + 4)2 − 4x 2

(x + 2)(x 2 − 2x + 4)(x − 2)(x 2 + 2x + 4)

= (x 2 + 4)2 − (2x)2

(x 2 + 2x + 4)(x 2 − 2x + 4)(x + 2)(x − 2)

= [(x 2 + 4) + 2x][(x 2 + 4) − 2x]

(x 2 + 2x + 4)(x 2 − 2x + 4)(x + 2)(x − 2)

= (x 2 + 2x + 4)(x 2 − 2x + 4)

(x 2 + 2x + 4)(x 2 − 2x + 4)(x 2 − 4 − 1)

=0

(x 2 + 2x + 4)(x 2 − 2x + 4)(x 2 − 5)

=0

(x 2 + 2x + 4)(x 2 − 2x + 4)(x − √5)(x + √5) = 0 x = −√5 or x = √5 ✓ 12(i)

2

1

x 3 − 2x 3 + 3 = 0 has roots α & β 1

sub y = x 3 : 1

1

y 2 − 2y + 3 = 0 has roots α3 & β3 1

Sum of roots

1

= α3 + β3 = − 1 3

Product of roots = α β

1 3

=

1

1

c a

b a

=−

(−2)

(3)

= (1)

(1)

=2✓ =3✓

12(ii) α + β 1

3

3

1

= (α3 ) + (β3 ) 1

1

1

2

1

2

= (α3 + β3 ) [(α3 ) − (α3 ) (β3 ) + (β3 ) ] ∵ sum of cubes factorization 1

1

2

1

1

2

= (α3 + β3 ) (α3 − α3 β3 + β3 ) 2

2

1

= 2 (α3 − 3 + β3 ) 1

1

2

1

1

1

∵ α3 + β3 = 2, α3 β3 = 3 1

1

= 2 [(α3 + β3 ) − 2α3 β3 − 3]

∵ a2 + b2 = (a + b)2 − 2ab

= 2[(2)2 − 2(3) − 3] = 2(4 − 6 − 3) = −10 ✓

∵ α3 + β3 = 2, α3 β3 = 3


1

1

1

1

sleightofmath.com

84


Ex 3.4

(2n + 1)3 − (2n − 1)3 = [(2n + 1) − (2n − 1)][(2n + 1)2 + (2n + 1)(2n − 1) + (2n − 1)2 ] 2 (4n + 4n + 1 + 4n2 − 1 =2 + 4n2 − 4n + 1) = 2(12n2 + 1) ✓ difference between the cube of two consecutive positive odd numbers 4 =

(2n + 1)3 − (2n − 1)3 4

=

2(12n2 + 1) 4

=

12n2 + 1 2

12n2 has to be an even number 12n2 + 1 has to be an odd number ⇒ (12n2 + 1) cannot be divisible by 2 ∴ difference between the cube of two consecutive positive odd numbers can never be divisible by 4.✓ 14

Difference in cubes = (large cube) − (small cube) = a3 − b3 red cuboid = (a)(a − b)(a) = (a − b)a2

𝑎 𝑎−𝑏

𝑎 green cuboid = a(b)(a − b) = (a − b)ab blue cuboid = (a − b)(b)(b) = (a − b)b2

𝑎

𝑎−𝑏

𝑎−𝑏 𝑏

𝑏 𝑏

sum of cuboids = (a − b)a2 + (a − b)ab + (a − b)b2 = (a − b)(a2 + ab + b2 ) ✓ Resultant Volume = sum of cuboids a3 − b3 = (a − b)(a2 + ab + b2 ) [shown] ✓


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85


Ex 3.5 3(b)

Ex 3.5 1(a)

1(b)

2(i)

x 3 − 2x 2 − 4x + 8 ) … (x − 2)( ) … (x − 2)(x 2 2 = (x − 2)(x −4) = (x − 2)(x + 2)(x − 2) = (x + 2)(x − 2)2 ✓ 3

(x − 3)(4x − 3)(x + 2) = 0 x = 3 or x =

2

3x − 10x + 9x − 2 … (x − 1)( ⬚) 2 ) … (x − 1)(3x 2 (x … − 1)(3x +2) = (x − 1)(3x 2 −7x +2) = (x − 2)(3x − 1)(x − 2) ✓

f(x) x 3 − 3x 2 − 4x x 3 − 3x 2 − 4x + 12 (x − 2)( ⬚) 2 (x − 2)(x ⬚) (x − 2)(x 2 −6) 2 (x − 2)(x −x −6)

x 3 − 4x 2 + x + 6 (x − 2)( (x − 2)(x 2 (x − 2)(x 2 (x − 2)(x 2 − 2x

or x = −2 ✓

3(d)

x(x + 3)(x − 1) x(x 2 + 2x − 3) x 3 + 2x 2 − 3x x 3 + 2x 2 − 4x − 8 (x − 2)( ⬚) 2 (x − 2)(x +4) (x − 2)(x 2 +4x +4) (x − 2)(x + 2)2 x = 2 or x = −2 ✓

4

(x − 2)(x − 3)(x + 2) = 0 x = 2 or x = 3 or x = −2 ✓ 3(a)

4

x3 + 4 = x(x + 4) 3 x +4 = x 2 + 4x 3 2 x − x − 4x + 4 =0 (x − 1)( ⬚) 2 (x − 1)(x ⬚) (x − 1)(x 2 − 4) = 0 (x − 1)(x + 2)(x − 2) =0 x = 1 or x = −2 or x = 2 ✓

= −12 = −12 =0

=0

3

3(c)

f(x) = x 3 − 3x 2 − 4x f(x) =0 2 x(x − 3x − 4) = 0 x(x − 4)(x + 1) = 0 x = 0 or x = 4 or x = −1 ✓

2 (ii)

4x 3 + 18 = 7x 2 + 21x 4x 3 − 7x 2 − 21x + 18 = 0 (x − 3)( ⬚) 2 (x − 3)(4x ⬚) 2 (x − 3)(4x − 6) (x − 3)(4x 2 + 5x − 6) = 0

=0 ⬚) ⬚) − 3) − 3) = 0

=x+8 =x+8 =x+8 =0

=0 =0

P(x) = k[x − (−4)] [x − (−1)] [x − (3)] = k(x + 4)(x + 1)(x − 3) = 5(x + 4)(x + 1)(x − 3) ∵ coefficient of x 3 is 5 = 5(x 2 + 5x + 4)(x − 3) = 5(x 3 +5x 2 +4x −3x 2 −15x −12) = 5(x 3 +2x 2 −11x −12) = 5x 3 +10x 2 −55x −60 ✓

(x − 2)(x − 3)(x + 1) = 0 x = 2 or x = 3 or x = −1 ✓


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86

A math 360 sol (unofficial) 5(a)

x 3 − 9x 2 + 25x − 21 =0 (x − 3)( ⬚) (x − 3)(x 2 ⬚) 2 (x − 3)(x + 7) 2 (x − 3)(x −6x + 7) = 0 x = 3 or x = =

5(b)

2(1) 6±√8

=

2

6±2√2 2

=

2

= 3 ± √2 ✓

8

1 3 x 6 3

=

1±√13 6

✓

=

4

7±√17 4

9(i)

1

1

3

2

8

4

2

1

+ x

= 35

3

= 210 =0

−4±√(4)2 −4(1)(30) 2(1) −4±√16−120 2

=

−4±√−104 2

(not real) ✓

f(x) = 2x 3 + 9x 2 + 7x + 3 f(k) =9 3 2 2k + 9k + 7k + 3 = 9 2k 3 + 9k 2 + 7k − 6 = 0 [shown] ✓

✓

Let f(x) = 2x 3 − x 2 + 3x + 2 1

−

= 35 1 2 x 2 2

=

2(2)

=

3

x = 7 or x =

−(−7)±√(−7)2 −4(2)(4) 7±√49−32

1

2

(x − 7)( ⬚) (x − 7)(x 2 ⬚) (x − 7)(x 2 + 30) (x − 7)(x 2 +4x + 30) = 0

2x 3 + 6x − 6 = (13x − 6)(x − 1) 3 2x + 6x − 6 = 13x 2 − 19x + 6 3 2 2x − 13x + 25x − 12 = 0 (x − 3)( ⬚) 2 (x − 3)(2x ⬚) (x − 3)(2x 2 + 4) (x − 3)(2x 2 −7x + 4) (x − 3)(2x 2 − 7x + 4) = 0 x = 3 or x =

1

6

x − 3x + 2x x 3 − 3x 2 + 2x − 210

2(3) 6

1

N(x) = x 3 − x 2 + x N(x)

−(−1)±√(−1)2 −4(3)(−1) 1±√1+12

y = P(x) cuts the x-axis at −2,1 & 2: P(x) = k[x − (−2)][x − (1)] [x − (2)] = k(x + 2)(x − 1)(x − 2) P(x) passes through (0,12): 12 = P(0) = k[(0) + 2][(0) − 1][(0) − 2] = k(2)(−1)(−2) = 4k ⇒k=3 ∴ P(x) = 3(x + 2)(x − 1)(x − 2) ✓

6±√36−28

3x 3 + 5x 2 = 3x + 2 3 2 3x + 5x − 3x − 2 = 0 (x + 2)( ⬚) (x + 2)(3x 2 ⬚) (x + 2)(3x 2 − 1) (x + 2)(3x 2 −x − 1) = 0

=

6(i)

7

−(−6)±√(−6)2 −4(1)(7)

x = −2 or x =

5(c)

Ex 3.5

9(ii)

2k 3 + 9k 2 + 7k − 6 = 0

f (− ) = 2 (− ) − − + 2 (k + 2)( (k + 2)(2k 2 (k + 2)(2k 2 (k + 2)(2k 2

=0 By Factor Thm, (2x + 1) is factor ✓ 6(ii)

2x 3 + 4 2x 3 + 4 2x 3 − x 2 + 3x + 2

= (x − 1)(x − 2) = x 2 − 3x + 2 =0

(k + 2)(2k − 1)(k + 3) = 0 k = −2 or k =

(2x + 1)( (2x + 1)(x 2 (2x + 1)(x 2 (2x + 1)(x 2 −x 1

x = − or x =

⬚) ⬚) + 2) + 2) = 0

1 2

or k = −3 ✓

−(−1)±√(−1)2 −4(1)(2)

2

=

⬚) ⬚) − 3) +5k − 3) = 0

2(1) 1±√1−8 2


=

1±√−7 2

(not real) ✓ sleightofmath.com

87


Ex 3.5

y = 2x 3 −(1) (2 y = − x)(5x + 6) −(2) sub (1) into (2): 2x 3 = (2 − x)(5x + 6) 3 2x + (x − 2)(5x + 6) = 0 2x 3 + 5x 2 − 4x − 12 =0 (x + 2)( ⬚) (x + 2)(2x 2 − 6) 2 (x + 2)(2x +x − 6) = 0 (x + 2)(2x − 3)(x + 2) x = −2

or

y|x=−2 = 2(−2)3

11(i)

+5x x2 2 −x −2 x 4 x +4x 3 −(x 4 −x 3 5x 3 −(5x 3

=0 x=

3 2 3 3

y|x=3 = 2 ( ) =

=0 =0

+6 −x 2 −2x 2 ) +x 2 −5x 2 6x 2 −(6x 2

⇒( ,

4

−16x −12 −16x −10x) −6x −12 −6x −12) 0

27 4

(x − 2)(x + 1)(x 2 + 5x + 6) = 0 (x − 2)(x + 1)(x + 2)(x + 3) = 0 x = 2 or x = −1 or x = −2 or x = −3 ✓

3 27 2

=0 =0

2

2

= −16 ⇒ (−2, −16) ✓

11(ii) P(x) x 4 + 4x 3 − x 2 − 16x − 12 ∵ a = 4, b = −16 (x − 2)(x + 1)Q(x) (x 2 − x − 2)Q(x)

)✓

P(x) = x 4 + ax 3 − x 2 + bx − 12 P(x) is divisible by (x − 2) By Factor Theorem, P(2) =0 16 + 8a − 4 + 2b − 12 = 0 8a + 2b =0 b = −4a P(x) is divisible by (x + 1) By Factor Theorem, P(−1) =0 1 − a − 1 − b − 12 = 0 a + b + 12 =0 a + (−4a) + 12 =0 ∵ b = −4a a = 4✓

12(i)

f(x) = 2x 3 + ax 2 − 7a2 x − 6a3 f(−a) = 2(−a)3 + a(−a)2 − 7a2 (−a) − 6a3 = −2a3 + a3 + 7a3 − 6a3 =0 By Factor Theorem, (x + a) is a factor of f(x) ✓

12(ii) f(x) =0 3 2 2 3 2x + ax − 7a x − 6a =0 (x + a)( ⬚) (x + a)(2x 2 ⬚) (x + a)(2x 2 − 6a2 ) (x + a)(2x 2 −ax − 6a2 ) = 0 (x + a)(2x + 3a)(x − 2a) = 0 3

x = −a or x = − a or x = 2a ✓ 2

b|a=4 = −4(4) = −16 ✓ 13(i)

x 2 (9 − 2x) 9x 2 − 2x 3 2x 3 − 9x 2 + 2 (2x − 1)( (2x − 1)(x 2 (2x − 1)(x 2 (2x − 1)(x 2 −4x 1

x = or x =


sleightofmath.com

⬚) ⬚) − 2) − 2) = 0

−(−4)±√(−4)2 −4(1)(−2)

2

=

=2 =2 =0

2(1) 4±√16+8 2

=

4±√24 2

=

4±2√6 2

= 2 ± √6 ✓

88


Ex 3.5

13(ii) x 2 (3 + √2x)(3 − √2x) = 2 x 2 (9 − 2x) =2 1

x = or x = 2 + √6 or x = 2 − √6 2

( 14

rej ∵ √2x ) ✓ would not exist

15(iii) P(x) = 0 has only 1 real root ⇒ 0 real roots in quadratic factor b2 − 4ac <0 2 a − 4(2)(a)< 0 a2 − 8a <0 a(a − 8) <0

Intersects x = −1 or x = 3 only ⇒ P(x) = k(x + 1)(x − 3)2

+ 0

+ 8

0
P(x) passes through (1,16): 16 = P(1) = k(1 + 1)(1 − 3)2 = k(2)(4) 2 =k

16

∴ P(x) = 2(x + 1)(x − 3)2 P(−3) = −144 ✓ ⇒ P(x) = k(x + 1)2 (x − 3)

5 − 32x 2 5 − 32x 2 12x 3 + 32x 2 + 3x − 5 (2x + 1)( ⬚) (2x + 1)(6x 2 ⬚) (2x + 1)(6x 2 − 5) 2 (2x + 1)(6x +13x − 5)

= 3x(4x 2 + 1) = 12x 3 + 3x =0

=0

(2x + 1)(2x + 5)(3x − 1) = 0

y = P(x) passes through (1,16): 16 = P(1) = k(1 + 1)2 (1 − 3) = k(4)(−2) −2 = k ∴ P(x) = −2(x + 1)2 (x − 3) P(−3) = 48 ✓ 15(i)

−

1

5 1

2

2 3

x = − ,− , ✓ 17(i)

29.7

√𝑥2 + 𝑦2 21 𝑦

P(x) = 4x 3 + 2ax 2 − 2x 2 + ax − a

𝑥 Compare area,

1

1 3

1 2

1 2

1

2

2

2

2

2

1

A(x) = xy

P ( ) = 4 ( ) + 2a ( ) − 2 ( ) + a ( ) − a =

1 2

1

+ a 2

−

1 2

1

+ a−a 2

Compare width,

=0

√x 2 + y 2 + y = 21

By Factor Theorem, (2x − 1) is a factor of P(x) ✓ 15(ii) P(x) … (2x − 1)( … (2x − 1)(2x 2 … (2x − 1)(2x 2 = (2x − 1)(2x 2

−(1)

2

) ) +a) +ax +a) ✓

√x 2 + y 2 x2 + y2 42y

= 21 − y = 441 − 42y + y 2 = 441 − x 2

y

=

441−x2 42

−(2)

sub (2) into (1): 1

A(x) = x ( =

441−x2

2 42 x(441−x2 ) 84

)

✓

17(ii) 21 is the width of rectangle and this results in no leftover area for A(x)✓


sleightofmath.com

89

A math 360 sol (unofficial) 17(iii)

x(441−x2 )

= 15

84

Ex 3.5 3 7

2)

x(441 − x = 1296 x 2 − 441x + 1296= 0 x

= =

−(−441)±√(−441)2 −4(1)(1296) 2(1) 441±√189297 2

x = 2.96 or x = 438 (rej ∵ x < 21) ✓ 𝑦 (12.4,42.4) 𝑂

𝑥 𝑦 = 𝐴(𝑥)

✓

17(iv) A(x) = 42.4 cm2 is max area ✓


sleightofmath.com

90


Ex 3.6 2(a)

Ex 3.6 1(a)

3x+1 (x−2)(x+5)

A

≡

+

x−2

B

x+2 (x+1)2

≡

Cover-up rule:

x = −5:

B

+ (x+1)2

Cover-up rule:

x+5

x = −1: x = 2:

A x+1

7 =A ( )(7) −14 =B (−7)( )

1 ( )2

⇒A=1

Substitution:

⇒B=2

x = 0:

2

=B

⇒B=1

A

B

1

1

= +

1

A =1 3x+1

1

∴ (x−2)(x+5) = 1(b)

x−2

+

2

✓

x+5

x+2

∴ (x+1)2 =

x−2 A B ≡ + (x + 2)(x + 1) x + 2 x + 1

2(b)

Cover-up rule: −4 =A ( )(−1) −3 =B (1)( )

x = −2: x = −1: ∴ 1(c)

x−2 (x+2)(x+1)

x+5 x2 +x

=

=

x+5

4 x+2

≡

x(x+1)

A x

−

3 x+1

x = −1: ∴ 1(d)

x+5 x2 +x

x+10 x2 −4

=

5

4

x

x+1

∴

x+10 x2 −4

2 x+2

+

C

+ (x−2)2

=A

⇒A=4

x = 2:

32 (4)( )2

=C

⇒C=8

x = 0:

x+1

44 (2)(4)

⇒ B = −4

2(c)

14+7x−3x2 x2 (x+2)

≡

A

B

2

−2

= +

∴ (x+2)(x−2)2 =

⇒A=5

A x

4 x+2

+

−

B x2

3

C

⇒ B = −3

4 8

x−2

+

+

+ (x−2)2 ✓

C x+2

Cover-up rule: A x+2

+

8 =A ( )(−4) 12 =B (4)( )

=−

B x−2

Substitution:

✓ ≡

+

64 ( )(16)

B

x = 0:

14 ( )2 (2)

x = −2:

−12 (4)( )

3 x−2


=7

⇒B=7

x−2

Cover-up rule:

x = 2:

A x+2

x2 −8x+44

x+10 (x+2)(x−2)

x = −2:

≡

x = −2:

✓

5 =A ( )(1) 4 =B (−1)( )

= −

x2 −8x+44 (x+2)(x−2)2

⇒ B = −3

Cover-up rule: x = 0:

1

+ (x+1)2 ✓

Cover-up rule:

⇒A=4

B

+

1 x+1

⇒ A = −2

Substitution:

⇒B=3

x = 1:

✓

∴

sleightofmath.com

18

⇒ C = −3

A

B

C

1

1

3

3

✓

= + +

3

14+7x−3x2 x2 (x+2)

=C

=

7 x2

−

x+2

⇒A=0

91


4x2 −9x+7 x(x−1)2

A

≡

B

+

x

Ex 3.6 3(c)

C

x−1

+ (x−1)2

Cover-up rule:

7x2 −9x+29 (x−3)(x2 +4)

A

≡

+

x−3

Bx+C x2 +4

Cover-up rule:

x = 0:

7 ( )(1)

x = 1:

2 (1)( )2

=A =C

65 ( )(13)

⇒A=7

x = 3:

⇒C=2

Substitution:

=A

29 (−3)(4)

x = 0:

=

⇒A=5

A −3

+

C 4

⇒ C = −3

Substitution: 5 (2)(1)2

x = 2:

∴ 3(a)

4x2 −9x+7 x(x−1)2

7

3

x

x−1

= −

8x2 −11x+5 (2x−3)(x2 +1)

A

≡

A

B

C

2

1

1

= + +

2x−3

⇒ B = −3

∴ (x−3)(x2

Bx+C

3(d)

x2 +1

Cover-up rule: 13 2 13 )( ) 4

3

x= : 2

(

x = 1:

∴ 3(b)

=A

=

2

=

(−1)(2)

8x2 −11x+5 (2x−3)(x2 +1)

x(x2 +3)

A

≡

x

⇒

A

+

−3 A

+

−1

C

=

+

2 2x−3

⇒ C = −1

1 B+C

2x−3

⇒B=2

✓

x2 +4

Bx+C x2 +5

⇒A=2

2

⇒B=3

+

3x−1 x2 +1

∴ (x−1)(x2

✓ 4(a)

x2 +3

=A

4(b) Bx+C

x

x2 +3 2

= +

x(x2 +3) 2

Compare coefficients: x 2 : 1 = 5 + B ⇒ B = −4 x: C=2 5

−4x+2

x

x2 +3

= +

=

2

+

C 5

A

2B+C

1

9

= +

x−1

+

4−x x2 +5

⇒C=4 ⇒ B = −1

✓

Deg(numerator) = 1 Deg(denominator) = 2

Deg(numerator) = 1 Deg(denominator) = 2 ∵ Deg(numerator) < Deg(denominator), the fraction is proper ✓

4(c)

Deg(numerator) = 2 Deg(denominator) = 2 ∵ Deg(numerator) = Deg(denominator), the fraction is improper ✓

✓ 4x2 +3 x2 −2


+5)

A −1

∵ Deg(numerator) < Deg(denominator), the fraction is proper ✓

⇒A=5 5

=

20 (1)(9)

x = 2:

Bx+C

x2 +2x+15

x(x2 +3)

5

A

(−1)(5)

x2 +5x+6

15 ( )(3)

x2 +2x+15

+

+

x−1

6

x = 0:

x + 2x + 15 = 5(x + 3) +(Bx + C)x 2 = 5x + 15 +Bx 2 + Cx = (5 + B)x 2 + Cx + 15

∴

A

≡

12 = ( )(6)

x = 1:

⇒A=2


5 x−3

B+C

Substitution:

5 (−3)(1)

x2 +2x+15

x2 +5x+6 (x−1)(x2 +5)

=

+

−2

Cover-up rule:

Substitution: x = 0:

+4)

A

=

(−2)(5)

7x2 −9x+29

2

+ (x−1)2 ✓

+

27

x = 1:

sleightofmath.com

=

4(x2 −2)+11 x2 −2

=4+

11 x2 −2

✓

92


Ex 3.6

Deg(numerator) = 4 Deg(denominator) = 3

5(b)

x+3 (x−2)(x2 −4)

x2 (2x−3)

x+3

1 2 3

2

2x −3x x

x

+

∴ 5(a)

x2 (2x−3)

x+3 (x−2)(x−4)

≡

+0x 2

1

5

2

4

= x+ + A

+

x−2

−2x −2

+0x 2

x = 2: x = 4:

7 (2)( )

x+3

∴ (x−2)(x−4) = =

1 ( )(16)

=A

⇒A=

x = 2:

5 (4)( )2

=C

⇒C=

x = 0: −2x −2

15 2 x −2x−2 4 x2 (2x−3)

3 (−2)(−4)

= − +

⇒B

=−

✓

x−4

=

⇒A=−

=B

x−2 7

+

2(x−4)

⇒B=

5

5(c)

2

≡

1 16(x+2) A

Cover-up rule:

2

x = 2:

5 ( )(8)

+

x+2

x−2

7

7 2

1 16 5 4

+

A

B

C

2

2 1

4

−

16

1 16

x−2

−

5 4 + (x−2) 2

1 16(x−2)

+

5 4(x−2)2

✓

Bx+C x2 +4

=A

⇒A=

5 8

Substitution:

x−4

−

x+3 (x−2)(x2 +4)

5 2(x−2)

✓

x = 0:

3 (−2)(4)

x = 1:

4 (−1)(5)

x+3

∴ (x−2)(x2

+4)

= =


1 16

x+3

B

5 2

C

+ (x−2)2

x = −2:

∴ (x+2)(x−2)2 =

=A

−

B x−2

Substitution:

15 2 x ) 4 15 2 x 4

−

Cover-up rule: 5 ( )(−2)

+

4

+x 3

3 −(x 4 − 2 x 3 ) 5 3 x 2 5 −( x 3 2

(x+1)(x3 −2)

x+2

Cover-up rule:

2x3 −3x2 5

4

A

≡

x4 +x3 −2x−2

=

(x−2)(x+2)(x−2)

= (x+2)(x−2)2

∵ Deg(numerator) > Deg(denominator), the fraction is improper (x+1)(x3 −2)

x+3

=

sleightofmath.com

=

A

+

−2

= 5

8(x−2) 5 8(x−2)

+ −

C

⇒C=

4

A −1

+

B+C 5

1 4

⇒B=−

5 8

5 1 8 4 x2 +4

− x−

5x+2 8(x2 +4)

✓

93


2x3 +15x2 +39x+33

=

(x+2)(x+3)

Ex 3.6

2x3 +15x2 +39x+33

2x +5 3 x +5x +6 2x +15x 2 −(2x 3 +10x 2 5x 2 −(5x 2 2

2x3 +15x2 +39x+33

6(c)

x2 +5x+6

+39x +12x) +27x +25x 2x

6x3 −5x2 −19x+28 (2x+3)(x−1)2

=

+33

=

+33 +30) +3

2x

3

−x

2x+3

= 2x + 5 + (x+2)(x+3)

(x+2)(x+3)

≡ 2x + 5 +

A x+2

B

+

2

6x3 −5x2 −19x+28 (2x+3)(x−1)2

x+3

6(b)

−1 ( )(1)

x = −3:

−3 (−1)( )

=A

2x3 +15x2 +39x+33 (x+2)(x+3)

4x3 −4x2 −16x+7 (x+1)(x−2)

x2

=

(x+1)(x−2)

6x3 −5x2 −19x+28 2x3 −x2 −4x+3

−2x2 −7x+19

= 3 + (2x+3)(x−1)2 A 2x+3

+

B x−1

C

+ (x−1)2

⇒ A = −1 Cover-up rule:

=B

⇒B=3

= 2x + 5 −

1 x+2

+

3

25

2

25 4

x=− : 3 x+3

x = 1:

✓

4x3 −4x2 −16x+7

=A

10 (5)( )2

⇒A=4

=C ⇒C=2

Substitution:

x2 −x−2

x = 0:

4x −x −2 4x 3 −4x 2 −16x +7 −(4x 3 −4x 2 −8x) −8x +7

4x3 −4x2 −16x+7

6x3 −5x2 −19x+28 2x3 −4x2 +2x+3x2 −6x+3

≡3+

x = −2:

−2x+1)

3 −4x +3 6x 3 −5x 2 −19x +28 −(6x 3 −3x 2 −12x +9) −2x 2 −7x +19

Cover-up rule:

∴

6x3 −5x2 −19x+28

= (2x+3)(x2

∴

19 (3)(1)

6x3 −5x2 −19x+28 = (2x+3)(x−1)2

=

A 3

3+

+

B −1

4 2x+3

+

−

C 1 3

x−1

⇒ B = −3 2

+ (x−1)2 ✓

7−8x

= 4x + (x+1)(x−2) ≡ 4x +

A

+

x+1

B x−2

Cover-up rule: x = −1:

15 ( )(−3)

x = 2:

−9 3( )

∴

4x3 −4x2 −16x+7 (x+1)(x−2)

=A

=B

= 4x −


⇒ A = −5 ⇒ B = −3 5 x+1

−

3 x−2

✓

sleightofmath.com

94


Ex 3.6

6(d)

8

4x5 +2x4 +3x3 −x2 −x+1 x(x2 +1)

=

4x5 +2x4 +3x3 −x2 −x+1 x3 +x

+2x 4x 2 3 2 +x 5 +2x 4 x +0x 4x 5 +0x 4 −(4x 2x 4 −(2x 4

4x5 +2x4 +3x3 −x2 −x+1 x(x2 +1)

−1 +3x 3 +4x 3 ) −x 3 +0x 3 −x 3 −(−x 3

−x 2

−x +1

−x 2 +2x 2 ) −3x 2 −x +0x 2 −x) −3x 2 +0x +1

= 4x 2 + 2x − 1 +

4x3 +12x2 −x−4

=x+3+ ≡x+3+

Bx+C

x

x2 +1

1

x= : =A

2

⇒A=1

1

x=− : 2

⇒

−3x2 +1

=

x(x2 +1)

1

+

x

Bx+C

9(i)

x2 +1

−3x 2 + 1 = (x 2 + 1) +(Bx + C)x = x2 + 1

+Bx 2 + Cx

7

x(x2 +1) 6x2 −21x+25 x(2x−5)

=

=

1

4x

x

x2 +1

= 4x 2 + 2x − 1 + −

D 2x+1

−1 =C ( )(2) −1 =D (−2)( )

1

⇒C=− ✓ 2

1

⇒D= ✓ 2

x

a 3x2 +8x+1 x2 +2x+1

2

3 +2x +1 3x 2 −(3x 2

+8x +6x 2x

+1 +3) −2

6x2 −21x+25 2x2 −5x

2x−2

3 −21x +25 2x −5x 6x 2 2 −15x) −(6x −6x +25 ⇒ A=3✓ x(2x−5)

+

V

✓

h = 3 + (x+1)2 ✓ ≡3+

2

6x2 −21x+25

C 2x−1

V = 3x 2 + 8x + 1 a = (x 2 + 2x + 1)

h=

Compare coefficients: x 2 : −3 = 1 + B ⇒ B = −4 x: C = 0 4x5 +2x4 +3x3 −x2 −x+1

−1 (2x − 1)(2x + 1)

V = ah

= (1 + B)x 2 + Cx + 1

∴

−1 4x2 −1

Cover-up rule:


=x+3+

4x2 −1

x(x2 +1) A

−x −4 −x) +0x −4 +0x −3) −1

⇒ A = 1, B = 3 ✓

−3x2 +1

≡ 4x 2 + 2x − 1 + + 1 ( )(1)

x +3 3 4x +0x −1 4x +12x 2 3 −(4x +0x 2 12x 2 −(12x 2 2

=3+

9(ii)

A

B

x+1

+ (x+1)2


−4 ( )2

=B

⇒ B = −4

−6x+25 x(2x−5) B C

≡3+ + x


2x−5

−2 1

A

B

1

1

= +

⇒A=2

Cover-up rule: 5

25 ( )(−5) 10

2

( )( )

x = 0: x= :

5 2

=B

=C


∴ h= 3 +

⇒ B = −5✓ ⇒ C=4✓

10(i)

sleightofmath.com

2 x+1

4

− (x+1)2 ✓

x 3 − 3x 2 + 4 = (x + 1)(x 2 − 4x + 4) = (x + 1)(x − 2)2 ✓

95


Ex 3.6

10(ii) x 2 +3x x −3x +0x +4 x 5 +0x 4 −(x 5 −3x 4 3x 4 −(3x 4 3

10(ii)

x2 +8

11(ii) 2

x5 −36x2 +53x+18

+9 +0x 3 +0x 3 +0x 3 −9x 3 9x 3 −(9x 3

−36x 2 +4x 2 ) −40x 2 +0x −40x 2 −27x 2 −13x 2

= (x 2 + 3x + 9) +

x3 −3x2 +4

= (x 2 + 3x + 9) + (x 2

≡

+ 3x + 9) +

≡− +53x +12x) +41x +18 +0x +36) +41x −18

x = 2:

12 (3)( )2

=C

−18

A

=

1(4)

1

+

x3 −3x2 +4 −13x2 +41x−18 (x+1)(x−2)2 A B x+1

+

x−2

2 (4)( )

=B

⇒B=

x2 +8

∴

=−

+

100x+250 x2 +5x

=

x+1

≡

A x−4

+

∴

5 x−2

4

+ (x−2)2 ✓

12(iii)

B

x = −2:

6 (−6)( )

11(i)

2x+16

+

2

−

2

(x−4)(x+2) x+2 2(x+8) 2 + (x−4)(x+2) x+2 2 1 2

2(

x−4

4 x−4 4 x−4

4 x

−

−

)+

x+2

2 x+2

+

⇒A=2

=B

x−4

x+2 2

x+2

x2 +2

+ −

1 2

x+2 1

+

2(x+2)

1 2 1 2

x−2

+

1 2(x−2)

✓

x

x+5

+

=

50 x

=A =B

+

50 x+5

⇒ A = 50 ⇒ B = 50 ✓

50 x 50

≡ Time for to − journey ✓ ≡ Time for return − journey ✓

x+2

=A

x+8

∴ (x−4)(x+2) =

x2 +5x

x+5

Cover-up rule: 12 ( )(6)

100x+250

x2 +2 1

−

2

12(ii) 50km ✓

13(i) x = 4:

x(x+5) B

−250 (−5)( )

x 5 − 36x 2 + 53x + 18 x 3 − 3x 2 + 4 8 = (x 2 + 3x + 9) − − x+8 (x−4)(x+2)

A

x = −5: ⇒ B = −5

1

1

100x+250

250 ( )(5)

∴

11

=−

(x2 −4)(x2 +2)

x = 0:

4

x−2

x = 2:

⇒C=4

C

B

+

⇒A=−

Cover-up rule:

+

x+2

=A

⇒ A = −8

B

A

+

x2 +2

−13x2 +41x−18

12(i)

−2

2

+ (x+2)(x−2)

x2 +2 1

2 ( )(−4)


x2 +2

x = −2:

≡ =A

1

−

Cover-up rule:

Cover-up rule: −72 ( )9

x2 −4 1

=−

+53x +18

C (x−2)2

x = −1:

2

=

(x2 −4)(x2 +2)

x 3 − 8 = (x)3 − (2)3 = (x − 2)[x + x(2) + 22 ] = (x − 2)(x 2 + 2x + 4) ✓

⇒ B = −1 1

x+2

✓

=1 =1 =1 =1 =1 =x−4 =8✓


sleightofmath.com

96


x2 +4 x3 −8

Ex 3.6

x2 +4

= (x−2)(x2 A

≡

+

x−2

14(iii)

1

+

2(3) 1

x2 +2x+4

2 1

3 1

3 1

4 1

4

5

= − + −

Cover-up rule: 8 ( )(12)

x = 2:

=A

⇒A=

+ −

2

+

x = 0:

4 (−2)(4)

=

x = 1:

5 (−1)(7)

=

∴

2 3

=

x3 −8

x−2 2

= 14(i)

+

1 (x+1)(x+2)

≡

A

+

x+1

+

A −1

+

C

⇒C=−

4

B+C

⇒B=

7

+

2 3

1

x−2 3(x2 +2x+4)

✓

1 ( )(1)

x = −2:

1 (−1)( )

1

∴ (x+1)(x+2) =

15(i)

14(ii)

1 2(3)

1

1

2

3

1

1

+⋯+

4(5)

101(102)

−

1 101 1 102

1

2 102 25 81

✓

5x2 −3x+8 (x−1)(x2 +4) 2

=

2

+

x−1

3x+1 x2 +4

2

5x − 3x + 8 = 2(x + 4) + (3x + 1)(x − 1)

x+2

Compare constants: 8=8−1 8 = 7 [inconsistent] ✓ ⇒A=1 15(ii)

=B

x+1

+

B

=A

1

101

−

1


1 100 1

= −

3

=

1 2 x− 3 3 x2 +2x+4

+

3(x−2)

A −2

3(4) 1

+⋯

3

Substitution:

x2 +4

1

+2x+4) Bx+C

−

⇒ B = −1 1

x+2

5x2 −3x+8 (x−1)(x2 +4)

x−1

+

Bx+C x2 +4

Cover-up rule:

✓

x = 1:

= − ✓

10 ( )(5)

=A

⇒A=2


8 (−1)(4)

x = 2:

22 (1)8

5x2 −3x+8

∴ (x−1)(x2


A

≡

sleightofmath.com

+4)

−1

+

A

2B+C

1

8

= +

=

A

=

2 x−1

+

3x x2 +4

C 4

⇒C=0 ⇒B=3

✓

97


Rev Ex 3 A3(i)

Rev Ex 3 A1(i)

g(x) = x 3 + ax 2 + x + 5 By Remainder Theorem, g(2) = 31 3 2 2 + a(2) + 2 + 5 = 31 8 + 4a + 2 + 5 = 31 4a = 16 a =4✓

Difference in volumes is 189 cm3 : (x + 3)3 − x 3 = 189 9(x 2 + 3x + 3) = 189 x 2 + 3x + 3 = 21 2 x + 3x − 18 = 0 (x + 6)(x − 3) = 0 x = −6 or x = 3 ✓ (rej x ≥ 0) x+3=6✓

=c+5 =c+5 =c+5 =6✓

g(2) 8 + 4a + 2 + 5 15 + 4(4) 31 b A2(i)

A4(a)

= 2(1)(2 − b) + 2c + 5 = 2(2 − b) + 2c + 5 = 2(2 − b) + 2(6) + 5 = 2(2 − b) + 17 = −5 ✓

1−x2

=

≡

−7x+3 x2 −1 −7x+3 (x+1)(x−1) A B x+1

x−1

x = −1:

10 ( )(−2)

x = 1:

−4 (2)( )

∴

7x−3 1−x2

−5

=

x+1 2

= =0 =0 =0 =0 =3✓

+

Cover-up rule:

f(−1) = (−1)3 + (k − 2) −(k − 7) − 4 = −1 +k − 2 −k + 7 − 4 =0 By Factor Theorem, (x + 1) is a factor of f(x) ✓

A4(b)

1−x

10x−5 (x+1)(x2 +4)

⇒ A = −5

=B

⇒ B = −2

−2

+

x−1 5

−

x+1

A

≡

=A

x+1

+

✓ Bx+C x2 +4

Cover-up rule: −15 = ( )(5)

x = −1:

f(x) = x 3 + (3 − 2)x 2 + (3 − 7)x − 4 = x 3 + x 2 − 4x − 4 … (x + 1)( ⬚) 2 … (x + 1)(x ⬚) 2 = (x + 1)(x − 4) (x = + 1)(x + 2)(x − 2)

A

⇒ A = −3


−5 (1)(4)

= +

x = 1:

5 (2)(5)

= +

10x−5

∴ (x+1)(x2

∴ the third factor (x − 2) ✓


7x−3

=

f(x) = x 3 + (k − 2)x 2 + (k − 7)x − 4

A2(ii) By Factor Theorem, f(−2) −8 + (k − 2)4 + (k − 7)(−2) − 4 4k − 8 − 2k + 14 − 12 2k − 6 k

+ x2] + x2 )

A3(ii) x ≡ shorter side x + 3 ≡ longer side

A1(ii) g(x) = x(x − 1)(x − b) + cx + 5 g(1) 1+a+1+5 1+4+1+5 c

(x + 3)3 − x 3 = [(x + 3) − x][(x + 3)2 + (x + 3)x 2 (x + 6x + 9 + x 2 + 3x =3 (3x 2 + 9x + 9) =3 = 9(x 2 + 3x + 3) ✓

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+4)

=

−3 x+1

A

C

1

4

A

B+C

2

5

+

⇒C=7

3x+7 x2 +4

⇒B=3 ✓

98

A math 360 sol (unofficial) A4(c)

x3 (x−4)2

=

Rev Ex 3

x3

x +8 3 x −8x +16 x +0x 2 3 −(x −8x 2 8x 2 −(8x 2 2

x3 (x−4)2

A6 f(x) = 4x 3 − 16x 2 − 9x + 40 (a)(i) By Remainder Theorem, f(4) = 4(4)3 − 16(4)2 − 9(4) + 40 =4✓

x2 −8x+16

+0x +16x) −16x −64x 48x

+0 +0 +128) −128

A6 f(x) = x (a)(ii) x = 4 is a root ✓ 4x 3 − 16x 2 − 9x + 40 4x 3 − 16x 2 − 10x + 40 2x 3 − 8x 2 − 5x + 20 (x − 4)( ⬚) (x − 4)(2x 2 ⬚) (x − 4)(2x 2 − 5) 2 (x − 4)(2x − 5)

48x−128 (x−4)2 A B + (x−4)2 x−4

=x+8+ ≡x+8+


48(4)−128 ( )2

=𝐵

⇒ B = 64


−128 16

=

A −4

+

B 16

5

5

2

2

=x =0 =0

=0

(x − 4) (x + √ ) (x − √ ) = 0 ⇒ A = 48

5

x = 4 or x = ±√ ✓ x3

∴ (x−4)2 = x + 8 + A5(i)

48 x−4

2

64

+ (x−4)2 ✓

A6(b) P(x) = x 3 + 2x 2 y − 5xy 2 − 6y 3

x 3 + 3x 2 − 6x − 8 =0 (x − 2)( ⬚) 2 (x − 2)(x ⬚) (x − 2)(x 2 + 4) 2 (x − 2)(x +5x + 4) = 0 (x − 2)(x + 4)(x + 1) =0 x = 2 or x = −4 or x = −1 ✓

P(2y) = (2y)3 + 2(2y)2 y − 5(2y)y 2 − 6y 3 = 8y 3 + 8y 3 − 10y 3 − 6y 3 =0 By Factor Theorem, (x − 2y) is a factor of P(x) P(x) … (x − 4)( ⬚) 2 … (x − 4)(x ⬚) … (x − 4)(x 2 + 3y 2 ) … (x − 4)(x 2 +4xy + 3y 2 ) = (x − 2y)(x + 3y)(x + y)

A5(ii) f(x) = 22x 3 + 15x 2 − 4x − 6 (a) g(x) = 2 + 14x − 12x 2 − 5x 3 By Remainder Theorem, f(a) = g(a) 22a3 + 15a2 − 4a − 6 = 2 + 14a − 12a2 − 5a3 27a3 + 27a2 − 18a − 8 = 0 [shown] ✓

The other 2 factors are (x + 3y) & (x + y) ✓

A5(ii) (3a)3 + 3(3a)2 − 6(3a) − 8 = 0 (b) sub y = 3a: y 3 + 3y 2 − 6y − 8 = 0 y=2 y = −4 y = −1 sub y = 3a: sub y = 3a: sub y = 3a: 3a = 2 3a = −4 3a = −1 a=

2 3

a=−


4 3

1

a=− ✓ 3

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99

A math 360 sol (unofficial) A7

Rev Ex 3

f(x) ÷ (x 2 − x − 2) ⇒ Remainder = ax + b ⇒ f(x) = (x 2 − x − 2)Q(x) + ax + b = (x + 1)(x − 2)Q(x) + ax + b f(x) has a remainder of −5 when divided by x + 1 By Remainder Theorem, f(−1) = −5 −a + b = −5 b =a−5 −(1) f(x) has a remainder of 7 when divided by x − 2 By Remainder Theorem, f(2) =7 2a + b = 7 −(2) sub (1) into (2) 2a + (a − 5) = 7 3a = 12 a =4 b|a=4 = (4) − 5 = −1

B2(a) y = x(16x − 19) −(1) 3 y = 4x − 6 −(2) sub (1) into (2): x(16x − 19) = 4x 3 − 6 16x 2 − 19x = 4x 3 − 6 4x 3 − 16x 2 + 19x − 6 = 0 (x − 2)( ⬚) 2 (x − 2)(4x ⬚) 2 (x − 2)(4x + 3) (x − 2)(4x 2 − 8x + 3) = 0 (x − 2)(2x − 1)(2x − 3) = 0 x = 2 or x =

Sub back x = a2 & y = b2 : g(a) = (a2 − 4b2 )(a2 + 3b2 ) = (a + 2b)(a − 2b)(a2 + 3b2 )✓ B3(i)

B1(b) f(x) = 3(x + 2)5 + (x + k)2


=0 =0 =0 =0

Sub x = a2 & y = b2 : g(a) = x 2 − xy − 12y 2 = (x − 4y)(x + 3y)

1+1−4 =A−5 A =3

By Remainder Theorem, f(−1) =7 5 2 (k 3(1) + − 1) = 7 k 2 − 2k + 1 =4 2 k − 2k − 3 =0 (k − 3)(k + 1) =0 k = 3 or k = −1 ✓

2

With k = −12, g(a) = a4 − a2 b2 − 12b4

−4 = (−1)(−1) + B B = −5

Remainder = Ax + B = 3x − 5 ✓

3

or x = ✓

g(2b) 16b4 − (2b)2 b2 + kb4 16b4 − 4b4 + kb4 b4 (12 + k) b = 0 or k = −12 ✓

B1(a) x 3 + x − 4 = (x 2 + x − 1)(x − 1) + Ax + B

x = 1:

2

B2(b) g(a) = a4 − a2 b2 + kb4

Remainder = ax + b = 4x − 1 ✓

x = 0:

1

B3(ii)

81x 3 + 24y 3 = 3[(3x)3 + (2y)3 ] = 3(3x + 2y)[(3x)2 − (3x)(2y) + (2y)2 ] = 3(3x + 2y)(9x 2 − 6xy + 4y 2 ) ✓ 81x3 +24y3 9x2 −6xy+4y2

=

3(3x+2y)(9x2 −6xy+4y2 ) 9x2 −6xy+4y2

= 3(3x + 2y) ✓

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100

A math 360 sol (unofficial) B4(i) (a)

2x2 −7x+14 (x+2)(x−4)

Rev Ex 3

2x2 −7x+14

=

B5(a) Let f(x) = x 2 + 5px + p2 + 5

x2 −2x−8 2(x2 −2x−8)−3x+30

=

=2

f(x) is divisible by x + 2 By factor Theorem, f(−2) =0 (−2)2 + 5p(−2) + p2 + 5 = 0 4 − 10p + p2 + 5 =0 2 p − 10p + 9 =0 (p − 9)(p − 1) =0 p = 1 or p = 9 f(x) is not divisible by x + 3

(x+2)(x−4) 30−3x + (x+2)(x−4) A B

≡2+

+

x+2

x−4


36 ( )(−6)

x = 4:

18 (6)( )

∴ B4(i) (b)

2x2 −7x+14 (x+2)(x−4)

42 (2x−5)(x+1)2

⇒ A = −6

=B

=2− ≡

=A

A

⇒B=3

6 x+2

+

3 x−4

✓

B

2x−5

By Factor Theorem, f(−3) 9 + 5p(−3) + p2 + 5 p2 − 15p + 14 (p − 14)(p − 1) p ≠ 14 or p ≠ 1 ∴p= 9✓

C

+ (x+1) + (x+1)2

Cover-up rule: 5

42

2

( )( )

x= :

7 2 2

x = −1:

=A

42 (−7)( )2

24

⇒A=

=C

7

B5 (b)

⇒ C = −6

Substitution: 42 A B C = + + (−5)(1)2 −5 1 1 12

x = 0:

B

=−

42

∴ (2x−5)(x+1)2 = B4(ii)

2x4 −7x2 +14 (x2 +2)(x2 −4)

7

24 7(2x−5)

=2− =2− ≡2−

6 x2 +2 6 x2 +2 6 x2 +2

− +

12

− (x+1)2 ✓

3 x2 −4 3

A x+2

+

B x−2


3 ( )(−4)

x = 2:

3 (4)( )

∴

2x4 −7x2 +14 (x2 +2)(x2 −4)

=A

⇒A=−

=B

=2− =2−

⇒B= 6 x2 +2 6 x2 +2

+ +

−

x 2 (x + 3) = 10x + 24 3 2 x + 3x = 10x + 24 x 3 + 3x 2 − 10x − 24 = 0 (x − 3)( ⬚) 2 (x − 3)(x ⬚) 2 (x − 3)(x + 8) (x − 3)(x 2 +6x + 8) = 0 (x − 3)(x + 2)(x + 4) = 0 x = 3 or x = −2 or x = −4 ✓

6

7(x+1)

+ (x+2)(x−2) +

3 4

x+2 3

+

4(x−2)

3 4

3

u = −4 sub u = 3x : 3x =

4

3 4

x−2

−

B5(b) 9x (3x + 3) = 10(3x ) + 24 (32 )x (3x + 3) = 10(3x ) + 24 (i) (3x )2 (3x + 3) = 10(3x ) + 24 sub u = 3x : u2 (u + 3) = 10u + 24 u=3 u = −2 x sub u = 3 : sub u = 3x : 3x = 3 3x = −2(NA) −4 (NA) ✓ ⇒x=1

3 4(x+2)

✓

B5 25x 2 (5x + 3) = 50x + 24 (b)(ii) (5x)2 (5x + 3) = (10)5x + 24 sub y = 5x y 2 (y + 3) = 10y + 24 y=3 or y = −2 or y = −4 5x = 3 5x = −2 5x = −4 x=


≠0 ≠0 ≠0 ≠0

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3 5

x=−

2 5

4

x=− ✓ 5

101

A math 360 sol (unofficial) B6(i)

f(x) = a[x − (−1)][x − (3)][x − (k)] = a(x + 1)(x − 3)(x − k) = 2(x + 1)(x − 3)(x − k) ∵ highest power term is 2x 3

Rev Ex 3 B7(i)

f(x) − g(x) (x 4 − x 3 − 7x 2 + x + 6) −(x 4 − 2x 3 − 10x 2 + 5x + 18) x 3 + 3x 2 − 4x − 12 (x − 2)( ⬚) (x − 2)(x 2 ⬚) (x − 2)(x 2 + 6) 2 (x − 2)(x +5x + 6) (x − 2)(x + 2)(x + 3) x = 2 or x = −2 or x = −3 ✓

By Remainder Theorem, f(4) = 20 2(5)(1)(4 − k) = 20 4−k =2 k =2✓ B6(ii) f(x) = 2(x + 1)(x − 3)(x − 2) f(−2) = 2(−1)(−5)(−4) = −40 ✓

f(x) = x 4 − x 3 − 7x 2 + x + 6 g(x) = x 4 − 2x 3 − 10x 2 + 5x + 18 =0 =0 =0

=0 =0

B7(ii) f(x) − g(x)= 0 f(x) = g(x) f(2) f(−2) f(−3)

= g(2) = −12 = g(−2) = 0 = g(−3) = 48

By Factor Theorem: (x + 2) is a factor of f(x) & g(x) ∴ α = −2 ✓


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102

A math 360 sol (unofficial) Ex 4.1 1(a)

|5 − 9| = |−4| =4 ✓

1(b)

|3 − π − 1| = |2 − π| = −(2 − π) =π−2✓

1(c)

2|−5| − |8| = 2(5) −8 = −2 ✓

2(i)

Given: y = |x 2 − 8x + 2|

Ex 4.1 4(ii)

y = 18: 2|3 − x| + 4 = 18 |3 − x| =7 3 − x= 7 or 3 − x = −7 x = −4 ✓ x = 10 ✓

5(a)

|x − 4| = 6 x − 4 = 6 or x = 10 ✓

5(b)

|2x + 3| = 5 2x + 3 = 5 2x + 3 = −5 2x =2 2x = −8 x =1✓ x = −4 ✓

5(c)

|x − 2| + 3 = 5 |x − 2| =2 x − 2= 2 x − 2= 2 x =4✓ x =4✓

6(a)

y = |f(x)|

y|x=1 = |(1)2 − 8(1) + 2| = |−5| =5✓ 2(ii)

y|x=−1 = |(−1)2 − 8(−1) + 2| = |11| = 11 ✓

2(iii)

y|x=2 = |(2)2 − 8(2) + 2| = |−10| = 10 ✓

3(a)

x − 4 = −6 x = −2 ✓

𝑦 (1,4)

|2 − √2| +| − 1|

(3,5)

(−2,3)

= 2 − √2 +(1) = 3 − √2 ✓ 3(b)

|2 − √5|

𝑥

= −(2 − √5) +3 − √5 = √5 − 2 =1✓ 3(c)

2|√6 − √3|

✓

+|3 − √5| +3 − √5

6(b)

y = |f(x)| 𝑦

3

+|√3 − √6|

(1,4)

−1

3

𝑥

✓

= 2(√6 − √3) +[−(√3 − √6)] = 2(√6 − √3) +(√6 − √3)

4(i)

6(c)

y = |f(x)|

= 3(√6 − √3)

𝑦

= 3√6 − 3√3 ✓

3 1

Given: y = 2|3 − x| + 4 O

✓

y|x=8 = 2|3 − (8)| +4 = 2|−5| +4 [ ] = 2 −(−5) +4 = 2(5) +4 = 14 ✓


𝑥

sleightofmath.com

103


Ex 4.1

y = |f(x)|

10

𝑦 1

7(a)

c <0 −c > 0 6 − c> 6 >0 ⇒ |6 − c| = 6 − c

𝑥

O

✓

y = |x − 3| for 0 ≤ x ≤ 4 y (−1,4) 3

∴ f(c) = (6 − c) − c 2 = 6 − c − c2

(4,1)

O

𝑥

(3,0)

f(x) |6 − x| − x 2

✓ 7(b)

y = |x + 2| for − 3 ≤ x ≤ 3 (3,5)

2 𝑥

(−2,0) O

✓ 8

y = |3x − 5| 𝑦

5 𝑂

5

𝑥

3

✓ x= 9

=0 =0

For x < 0, 6 − x − x2 =0 2 x +x−6 =0 (x + 3)(x − 2) = 0 x = −3 or x = 2 (rej ∵ x < 0) ✓

y

(−3,1)

f(x) = |6 − x| − x 2 f(c) = |6 − c| − c 2

5 3

f(x) = x + 2 − |3 − 2x| f(a) = a + 2 − |3 − 2a| a >2 −2a < −4 3 − 2a < −1 <0 ⇒ |3 − 2a| = −(3 − 2a)

11(a) 2|x + 1| + 3 = 9 2|x + 1| =6 |x + 1| =3 x + 1 = 3 or x + 1 = 3 x =2✓ x =2✓

11(b) 11 − 2|x + 3| = 1 2|x + 3| = 10 |x + 3| =5 x + 3 = 5 or x + 3= 5 x =2✓ x =2✓ 12(a) |2x − 3| = x 2x − 3 = x or x =3✓ 12(b) |5x − 8| = 3x 5x − 8 = 3x or 2x =8 x =4✓

2x − 3 = x x =3✓

5x − 8 = 3x 2x =8 x =4✓

∴ f(a) = a + 2 −[−(3 − 2a)] = a + 2 +(3 − 2a) = 5 − a [shown] ✓


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104


Ex 4.1 14(c) |x(x − 2)| = 2x − 3 x(x − 2) = 2x − 3 or x 2 − 2x = 2x − 3 x 2 − 4x + 3 =0 (x − 1)(x − 3) = 0 x = 1 (rej) or x = 3 ✓

12(c) 4|x − 6| − 5x = 0 |x − 6| x − 6= x

= 5x 4

5x 4

or

x − 6= x

= −6

4

13

4

= −6

4

x = −24 (rej ∵ x ≥ 0)

5x

x = −24 (rej ∵ x ≥ 0)

14(d) Method 1 2|x| = x 2 − 4x

1

|x − y| = |y − 5| 2

|x| y >5 y−5 >0 ⇒ |y − 5| = y − 5

x

1 2

1 2 1

5

2

2

x − y= y − 3 2

y

y

=x+

2

3

3

2 1

5

2

2

x − y= y − 3

= x+ ✓

=

x2 −4x 2

y

=x+

5 2

2

5

3

3

= x+ ✓

y

x

=

x2 −4x 2

2x = x 2 − 4x x 2 − 6x = 0 x(x − 6) = 0 x = 0 or x = 6 ✓

x = −√6 or x = √6 ✓ 15

y = |2x − 6| 𝑦

14(b) Method 1 |x|2 + x 2 =4 2 2 x +x =4 2 2x =4 2 x −2 =0 (x + √2)(x − √2) = 0

6 3

x = −√2 or x = √2 ✓ Method 2 |x|2 + x 2 =4 |x 2 | + x 2 =4 2 |x | = 4 − x2 x2 = 4 − x 2 or x 2 2x 2 = 4 2x 2 2 x =2 x2 x = ±√2 x


or

Method 2 2|x| = x 2 − 4x 2 4|x| = (x 2 − 4x)2 4x 2 = (x 2 − 4x)2 4x 2 = x 4 − 8x 3 + 16x 2 x 4 − 8x 4 + 12x 2 =0 x 2 (x 2 − 8x 2 + 12) = 0 x 2 (x − 2)(x − 6) = 0 x = 0 ✓ or x = 2 or x = 6 ✓ (rej)

14(a) |x 2 − 4| = 2 x2 − 4 = 2 or x 2 − 4 =2 2 2 x −6 =0 x −6 =0 (x + √6)(x − √6) = 0 (x + √6)(x − √6) = 0 x = −√6 or x = √6 ✓

2

1

2 5

x2 −4x

x − y = (y − 5)

5

2

=

2x = x 2 − 4x x 2 − 6x = 0 x(x − 6) = 0 x = 0 or x = 6 ✓

∴ |x − y| = (y − 5) x − y = (y − 5) or

x(x − 2) = 2x − 3 2 x − 2x = 2x − 3 2 x − 4x + 3 =0 (x − 1)(x − 3) = 0 x = 1 (rej) or x = 3 ✓

𝑥

✓ 15(i)

=4−x =4 =2 = ±√2

2

y = 2|x − 3| = |2x − 6| ✓

15(ii) y = |6 − 2x| = |(2x − 6)(−1)| = |2x − 6||−1| = |2x − 6| ✓

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105


Ex 4.1

16(a) y = |1 − x 2 | = |(1 + x)(1 − x)| 𝑦 (−2,3) (0,1) 𝑥 (−1,0) 𝑂 (1,0)

17(ii) y = |2x − 4| 𝑦

4

16(b) y = |x 2 − 4| = |(x + 2)(x − 2)| 𝑦 (0,4) 𝑂 (−2,0)

✓ 18(i)

✓

16(c) y = |(x − 1)(x − 3)| 𝑦

(5,8)

(2,1) 𝑂 𝑥 (−1,0) (3,0)

y = 2|x + c| − 5 At (−1,3): 3 = 2|(−1) + c| − 5 8 = 2|c − 1| |c − 1| = 4 c − 1 = 4 or c − 1 = −4 c =5✓ c = −3 ✓

𝑥

(2,0)

18(ii) y = |2x + 5|

17(i)

𝑥 (4,0)

y = |2x − 3| 𝑦

𝑦

✓

5

16(d) y = |x 2 − 4x| = |x(x − 4)| 𝑦 (2,4) 𝑂

𝑥

2

✓

−

5

𝑂

3 𝑂

𝑥

3 2

2

✓ 19(i) ✓

y = |2x + c| At (1,2), 2 = |2(1) + c| 2 = |2 + c| |c + 2| = 2 c + 2 = 2 or c + 2 = −2 c =0 c = −4 ∴ c = −4 ✓

𝑥 ✓

y = |3 − kx| (5,12): 12 = |3 − k(5)| |3 − 5k| = 12 3 − 5k = 12 or 3 − 5k = 12 5k = −9 5k = −9

At (0,4), 4 = |2(0) + c| |c| = 4 c = ±4

k 19 (ii)(a)

9

=− ✓ 5

𝑦 (−1,6) 3

O 1

9

=− ✓

k

5

𝑦 = |3 − 3𝑥| (5,12) 𝑥 ✓


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106

A math 360 sol (unofficial) |3 − kx| = 3 19 (ii)(b) |3 − 3x| = 3 3|1 − x| = 3 |1 − x| = 1 1 − x= 1 or x =0✓ 20(i)

Ex 4.1 22 k = 0 or k > 4 ✓ (ii)(a) 22 k=4✓ (ii)(b)

1 − x= 1 x =0✓

22 0
𝑦 𝑦 = |2𝑥 − 5| 𝑦 =𝑥+2

23(a) No. It can be 0 ✓

5 −2

2

O

23(b) |x| ≥ 0 ✓

𝑥

5 2

23(c) ℝ ✓ ✓ 24

20(ii) 2 intersections ⇒ 2 solutions ✓ 21(i)

y = |x 2 + 2x − 3| = |(x + 3)(x − 1)| 𝑦 𝑦 = |𝑥 2 + 2𝑥 − 3| (−1,4) 3 𝑦=2 𝑥 −3 𝑂 1

Method 2 x 2 − |x| − 2 = 0 |x| = x2 − 2 x = x 2 − 2 or x = x2 − 2 x2 − x − 2 =0 x2 − x − 2 =0 (x − 2)(x + 1) = 0 (x − 2)(x + 1) = 0 x = 2 or x = −1 x = 2 or x = −1 (rej) ✓ (rej) ✓

✓

2

21(ii) |x − 1|

= |x+3|

|x − 1||x + 3| = 2 |x 2 + 2x − 3| = 2

25 √x 2 = x (i)(a) Not true √x 2 = |x| ✓

Draw y = 2 4 intersections ⇒ 4 solutions ✓ 22(i)

y = |x 2 − 6x + 5| for − 1 ≤ x ≤ 7 = |(x − 1)(x − 5)| 𝑦 (−1,12) (3,4) 5 𝑂 1 5 |x 2

Method 1 x 2 − |x| − 2 =0 |x|2 − |x| − 2 =0 (|x| − 2)(|x| + 1) = 0 |x| = 2 |x| = 2 or or x = ±2 ✓ x = ±2 ✓

25 √x 2 = ±x (i)(b) By definition, square root results in only positive number ✓ 25(ii) |x| ✓

(7,12) 𝑥 ✓

− 6x + 5| = k


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107


Ex 4.2 2(a)

Ex 4.2 1(a)

5

y = 3x ∵ power of 5 is positive & odd (except for 1), graph is similar to y = x 3

1

y = x4 (0 < power

1 4

< 1)

𝑦

𝑦 𝑥

O 𝑥

O

2(b) ✓

1(b)

2

y = x3 2 (0 < power < 1) 3

𝑦

y = −3x 4 ∵ power of 4 is positive & even, graph is similar to y = −x 2 𝑦

𝑥

O 2(c)

𝑥

O

✓ 3

y = x2 3 (power > 1) 2

✓ 1(c)

✓

𝑦

2

y = 3 = 2x −3 x ∵ power of − 3 is negative & odd, 1 graph is similar to y =

𝑥

O

x

✓

𝑦 2(d)

O

𝑥

3

y = 3x 2 3 (power > 1) 2

𝑦 ✓

1(d)

2

y = − 4 = −2x −4 x ∵ power of − 4 is negative & even, 1 graph similar to y = 2 x ∵ coefficient is negative, reflect in x − axis 𝑦

O

𝑥

O 2(e)

✓ 3

y = −3x 2 3 (power > 1) 2 ∵ coefficient − 3 is negative, reflect in x − axis 𝑦

𝑥 ✓

O

𝑥

✓

108


Ex 4.2

1

4(c)

y = x −3 1 (power − < 0)

x = πy 2 y=√

3

𝑦

= 𝑥

O 3(b)

=

x π

√x √π 1 √π

1

x2 1

(0 < power < 1)

✓

2

𝑦

2

y = x −3 2 (power − < 0) 3

𝑦

5(i)

𝑥

O 3(c)

✓

y= 3 (power − < 0)

=

𝑦

= 5(ii)

𝑥

1

2π √10

x2 ✓

=

5(iii)

✓

4 3

5(iv)

√4√10

1

1

x2

= √10 2

x

= (10)

x

= 2.5 ✓

1 4

y = 2π√ =

✓

√10

x2 = π

1

𝑥

√10 2π

(40)2

y = π:

√10

𝑥

𝑦

1

2π

= 2π√4 = 4π ✓

y = πx 3

O

√x

y|x=40 =

2π

4(b)

√10

y = 4πx 2

O

√10

2π

✓

𝑦

x 10

√x

= 2π

2

4(a)

✓

y = 2π√

3 x −2

O

𝑥

O

2π √10

x 10 1

x2 1

(0 < power < 1) 2

𝑦

O

𝑥 ✓

109


Ex 4.2 9(b)

1

𝑦

𝑦 = 2𝑥 6 𝑥

O

1

𝑦 = −2𝑥 6

𝑦

✓ Reflection of each other in the x-axis ✓ 7

2

y=

1

3

y = − √8x = −2x 3 1 (0 < power < 1) 3 ∵ coefficient − 2 is negative, reflect in x-axis

1

= 2x −2

𝑥

O

√x 1 (power − < 0)

✓

2

𝑦

𝑦2 = 2𝑥 − 3

9(c)

2

𝑦1 = √𝑥 𝑥

O

y=

4

4

=

5

√32x2

2 2x5

2

2

= 2x −5

(power − < 0) 5

𝑦

−3 ✓ 1 sol ✓ 8

y= 1 (0 < power < 1) 3

10(i)

𝑦 1

4

𝑦1 = 4𝑥 3 𝑥 𝑦2 = 4 − 𝑥 2

O

✓ 1 3 1 3

2

4x + x = 4

✓

y = 2x + 1 −(1) 1 y= −(2) x (1) (2): sub into 1 2x + 1 = x 2x 2 + x =1 2x 2 + x − 1 =0 (2x − 1)(x + 1) = 0 1 x= or x = −1 ✓ 2

4x =4−x y1 = y2 ∴ 1 sol ✓ 9(a)

𝑥

O

1 4x 3

2

1

= 2( ) + 1

y|x=1

=2✓ 10(ii)

1

= −1 ✓

𝑦

y = √6x = √6x 2 1 (0 < power < 1)

𝑦 = 2𝑥 + 1 1

2

1

𝑦 (−1, −1)

O

y|x=−1 = 2(−1) + 1

2

2

( , 2) 2

𝑦=

1 𝑥

O

𝑥 ✓

𝑥 ✓

11(i)

1

1

5x 2

= 2x 6 1

1

5x 2 − 2x 6 1 6

=0

1 3

x (5x − 2) = 0 1

x6 = 0 x =0✓

1

or

x6 = 0 x =0✓

110


Ex 4.2

11(ii) y|x=0 = 0 ⇒ (0,0) y|x= 2√2 √5

8 125

×

⇒(

12(iii) Sub P = 18:

= 5(

√5

1 23 2

1 2

8 125

) = 5 ( 3) 5

21.5 51.5

=5

2√2 5√5

,

2

=

L

2(18) L

36

=√

L

36 L 36

= 2 I = 36I −2 ∵ power of − 2 is negative & even, 1 it is similar to y = 2

)

5

x

1

𝑦 = 5𝑥12 𝑦 = 2𝑥 6 8 2√10 ( , ) 125 5 𝑥 (0,0)

𝐿

✓

Note: Verify which graph is higher at x-coordinates 8 bigger & smaller than 125

12(i)

I

=

2√10

=

𝑦

𝑂

=5

5 √5 8 2√10

125

=√

I

𝐼

O 13(i)

✓

𝑦 1

𝑦2 = 4𝑥 5 I=√

2P L

O L = 8: 2P

P

√P

4 1

√4

I = √(8) = √ =

1

𝑦1 =

3

𝑥

𝑥

✓

1

= P2 2

13(ii) 1 sol ✓

(0 < power < 1) 2

𝐼

𝑝

O

✓

12(ii) Sub P = 18: I=√

2(18) L

=√

36 L

1

= 6L−2

1

(power − < 0) 2

𝐼

O

𝐿 ✓

111

A math 360 sol (unofficial) 14(i) y = −

4 √x

Ex 4.2 15(iii) 2x = 4x −12

for 0.5 ≤ x ≤ 4

3

x2 = 2

x 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 y −5.66 −4.00 −3.27 −2.83 −2.53 −2.31 −2.14 −2.00

2

x

= 23 2

y|

5

= 2(2)3 = 23

2 x=23 2 3

5

⇒ (2 , 23 ) ✓ 16(i)

2x − 3√x − 2 =0 (2√x + 1)(√x − 2) = 0 1

√x = − 2 (rej ∵ √x ≥ 0) ✓ 16(ii) y = 2x − 2 y = 3√x

or

√x = 2 x =4✓

−(1) −(2)

sub (1) into (2): 2x − 2 = 3√x 2x − 3√x − 2 = 0 ⇒x=4 y|x=4 = 2(4) − 2 =6 ⇒ (4,6) 𝑦

𝑦 = 2𝑥 − 2 𝑦 = 3√𝑥

(4,6)

𝑥

O −2

✓ 1 3x 2 1

y = 3√x = (0 < power < 1) 2

17(i) 14(ii) (1 − x)√x = −4 1−x

=

At (8,1),

At (8,1),

1 = a √8

−4 √x

b

y=

3

1 = a(2)

1 sol ✓ 15(i)

3

y = a √x

1

a= ✓

1=

2x

b 2(8)

b = 16 ✓

2

𝑦 𝑦1 = 2𝑥 1

𝑦2 = 4𝑥 −2 𝑥

O

✓

1 4x −2

y= 1 (power − < 0) 2

15(ii) 1 solution ✓ 112


Ex 4.2 17(iii) (−8, −1) ✓

𝑦 𝑦=

8 𝑥

(8,1) 𝑦 = 1 3√𝑥 2

𝑂

𝑥

✓ Symmetrical about O ✓

113


Rev Ex 4 A2(d) | 3√x − 1| = 3

Rev Ex 4

3

3

or √x − 1 = 3 3 =4 √x x = 43 x = 64 ✓

A1(a) f(x) = |3x − 1| f(−2) = |3(−2) − 1| = |−7| =7✓

√x − 1 = −3 = −2 √x x = (−2)3 x = −8 ✓ 3

A3(a) y = |−3x − 2| for − 2 ≤ x ≤ 1 f(2) = |3(2) − 1| = |5| =5✓

𝑦 (1,5) (−2,4)

f(x) =3 |3x − 1| = 3 3x − 1 = 3 or 3x =4 x

−

3x − 1 = −3 3x = −2

4

= ✓

x

3

2 𝑥

𝑂

2 3

2

✓

=− ✓ 3

A3(b) y = |9 − x 2 | A1(b) g(x) =

|x 2

− 2x|

𝑦 (0,9)

g(1) = |(1)2 − 2(1)| = |1 − 2| = |−1| =1✓

𝑂

(−3,0)

(3,0)

✓ A3(c) y = |4x 2 − 1| for − 1 < x < 1

g(3) = |(3)2 − 2(3)| = |9 − 6| = |3| =3✓

𝑦

(−1,3)

𝑂

1

(− , 0)

x=


2 3

1

𝑥

( , 0) 2

✓

A3(d) y = |(2x + 1)(x − 2)|

2x − 3 = −x 3x =3 x =1✓

𝑦 3

1

4

8

( ,3 )

A2(b) |x + 1| = 2x − 3 x + 1 = 2x − 3 or x + 1 = −(2x − 3) x + 1 = −2x + 3 x =4✓ 3x = 2

A2(c) |x 2 + 6| = 5x x 2 + 6 = 5x or x 2 − 5x + 6 = 0 (x − 2)(x − 3) = 0 x = 2 or x = 3 ✓

(1,3)

(0,1) 2

A2(a) |2x − 3| = x 2x − 3 = x or x =3✓

𝑥

1

(− , 0)

𝑂 (2,0) 𝑥

2

A4

(rej) ✓

x2 + 6 = −5x x 2 + 5x + 6 = 0 (x + 2)(x + 3) = 0 x = −2 or x = −3 (rej) (rej)

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✓

y = |x − 2| − 5 At x − axis, y =0 |x − 2| − 5 = 0 |x − 2| =5 x − 2 = 5 or x =7 ⇒ P(7,0) ✓

x − 2 = −5 x = −3 ⇒ Q(−3,0) ✓

114

A math 360 sol (unofficial) A5(i)

Rev Ex 4

1

A6(ii) A6(iii)

y = 2x 3 1

(0 < 𝑝𝑜𝑤𝑒𝑟 < 1) 3

1

y=

16√x

1

1

=

x −2

16 1

𝑦 16 𝑦2 = |𝑥 − 16| 1 (25,9) 𝑦1 = 2𝑥 4 (25,4.5) 𝑥 16

O

(𝑝𝑜𝑤𝑒𝑟 − < 0)

✓

2

𝑦

𝑦=

1 16√𝑥

𝑦 = 2𝑥 𝑥

O

✓

A5(ii) y = 2x 13

−(1)

1

y=

16√x 1 −1

=

16

x

−(2)

2

sub (1) into (2): 2x

1 3

1

=

16 1

1 1

x 3+2 = x

5 6

A6(iv) 2x 14 = |x − 16| ⇒ 2 solutions ✓

1 3

x

B1(a) |2x + 4| = x 2 + 1 2x + 4 = x 2 + 1 or x 2 − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1 or x = 3 ✓

2x + 4 = −(x 2 + 1) x 2 + 2x + 5 = 0 x= =

B1(b) |x 2 − 1| = x 2 x2 − 1 = x2 or −1 = 0 (NA)

1 − 2

= = 2−6

1

y|x= 1 = 2(2−6 )

2x 4

or x =

1 √2

✓

1

=8

1 4

1 3

x =4 x = 44 x = 256 ✓

64

= 2(2−2 )

⇒(

√2

2

2x 4 − 3 = −5

1

1

1

1

1

2x 4 − 3 = 5 or

sub (3) into (1):

=

(NA)

B1(c) |2x 14 − 3| = 5 −(3)

64

=

x=−

6 5

1

=

2

x2

32

(2−5 )

x

2(1) −2±√−16

x 2 − 1 = −x 2 2x 2 =1

= 2−5

x x

−(2)±√(2)2 −4(1)(5)

1

2x 4 x

= −2

1 4

= −1

(NA ∵

1 x4

≥ 0)

2

1

, )✓

64 2

1 A6 y = 2x 4 (i)(a)

y = 4: 1

2x 4 = 4 1

x4 = 2 x = 24 x = 16 ✓

B1(d) |√x − 2| = x − 4 = x − 4 or √x − 2 = −(x − 4) √x − 2 x − √x − 2 =0 x + √x − 6 =0 2

2

(√x) − √x − 2 = 0 (√x − 2)(√x + 1) = 0

(√x) + √x − 6 = 0 (√x + 3)(√x − 2) = 0

√x = 2 x =4✓

√x = −3 (rej)

√x = −1 (rej)

√x = 2 x=4

1 A6 y|x=25 = 2(25)4 (i)(b) ≈ 4.5 ✓


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115


Rev Ex 4 B3(b) 2x + 3y = 19 3y = 19 − 2x

𝑦 𝑦 = |2𝑥 − 5|

y

5 4 5 2

𝑦 =4−𝑥 𝑥 4

=

|x − ( |x −

B2(ii) y = |2x − 5| −(1) y=4−x −(2)

19−2x 3

19

5

| x−

3 19

3 5

3

3

2

|

B3(a) 2x + y = 3 y = 3 − 2x

y|x=1 = 4 − (1) =3 ⇒ (1,3) ✓

=3

=3

x

=

x

2x − 5 = −(4 − x) 2x − 5 = −4 + x x =1✓

=

y|x=28 =

B4(ii)

or

28 3 28 5

5 3 5 3

✓ 28 5

19−2( )

x−

13 5

19

= −3

3

x

=

10 3

=2✓

x y|x=2 =

19−2(2) 3

=5✓

3

5

B4(i)

−(2)

+ x| = 3

3 5

x−

−(1)

)| = 3

3 3 19

= y|x=2 = 4 − (3) =1 ⇒ (3,1) ✓

3

|x − y| = 3 sub (1) into (2): ✓

sub (1) into (2): |2x − 5| = 4 − x 2x − 5 = 4 − x or 3x =9 x =3✓

19−2x

✓

For unhealthy temperature, |x − 36.9| ≥ 0.9 ✓ 𝑦

−(1)

𝑦 = |𝑥 − 36.9|

y = |2x − 1| −(2) sub (1) into (2): |2x − 1| = 3 − 2x 2x − 1 = 3 − 2x or 2x − 1 = −(3 − 2x) 4x =4 2x − 1 = −3 + 2x 2 = 0 (NA) x =1✓ y|x=1 = 3 − 2(1) =1✓

36.9

𝑦 = 0.9 𝑥

36.0 36.9 37.8

|x − 36.9| = 0.9 ⇒ x − 36.9 = −0.9 and x = 36

x − 36.9 x

= 0.9 = 37.8

For healthy temperature, 36 < x < 37.8 ✓ B5(i)


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|x 2 − 10x + 8| = 8 x 2 − 10x + 8 = 8 or x 2 − 10x =0 x(x − 10) =0 x = 0 or x = 10 ✓

x 2 − 10x + 8 = −8 x 2 − 10x + 8 = −8 x 2 − 10x + 16 = 0 (x − 2)(x − 8) = 0 x = 2 or x = 8 ✓

116

A math 360 sol (unofficial) B5(ii)

𝑦

Rev Ex 4 B6(b) y = 2√x + 1

𝑦 = |𝑥 2 − 10𝑥 + 8|

y=

(5,17) 8 𝑥 𝑂 5 − √17 5 + √17

=

2(1)

2

10±2√17

=

−(2)

√x

2√x + 1

✓

−(−10)±√(−10)2 −4(1)(8)

10±√68

−(1)

sub (1) into (2):

|x 2 − 10x + 8| = 0 x 2 − 10x + 8 = 0 x=

6

2

=

=

6 √x

2x + √x =6 2x + √x − 6 =0 (2√x − 3)(√x + 2) = 0

10±√100−32 2

3

√x = 2 or √x = −2 (rej ∵ √x ≥ 0)

= 5 ± √17

x=

B5(iii) Solutions for x in (i) are the x-coordinates of the intersection points between the graph in (ii) and the line y = 8

9 4 9

sub x = into (1): 4

9

3

4

2

y|x=9 = 2√ + 1 = 2 ( ) + 1 = 4

B6(a)

4

9

⇒ ( , 4) ✓

𝑦 𝑦2 =

4

1 2

6

= 6𝑥 1 √𝑥 𝑦1 = 2√𝑥 = 2𝑥 2 (3,2√3) 𝑥

O

✓

y = 2√x =

1 2x 2 1

(0 < power < 1) 2

y=

6

1

= 6x −2

√x 1 (power − < 0) 2

y1 = 2√x y2 =

−(1)

6

−(2)

√x

sub (1) into (2): 2√x =

6 √x

x=3

−(3)

sub (3) into (2): y2 |x=3 =

6 √3

=

6 √3

×

√3 √3

=

6√3 3

= 2√3

⇒ (3,2√3) ✓


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117


Ex 5.1

Ex 5.1 1(a)

2! × 3! = (2 × 1)(3 × 2 × 1) = (2)(6) = 12 ✓

1(b)

6! 3!

=

6⋅5⋅4⋅3! 3!

=6⋅5⋅4 = 120 ✓ 1(c)

5! 5 ( ) = (5−1)!(1)! 1 5! = 4!

=5✓ 1(d)

9! 9 ( ) = (9−2)!(2)! 2 9! =

=

7!2! 9×8 2

= 36 ✓ 2(a)

2(b)

n 0 1 2 3 4 5

Binomial Coefficients 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

(1 − 2x)4 = [1 + (−2x)]4 = 1 +(4)(−2x)1 = 1 +(4)(−2x) = 1 −8x

+(6)(−2x)2 +(6)(4x 2 ) +24x 2

+(4)(−2x)3 +(4)(−8x 3 ) −32x 3

+(1)(2x)4 +(1)(16x 4 ) +16x 4 ✓

(1 + 3x)5 = 1 +(5)(3x)1 = 1 +(5)(3x) = 1 +15x

+(10)(3x)2 +(10)(9x 2 ) +90x 2

+(10)(3x)3 +(10)(27x 3 ) +270x 3

+(5)(3x)4 +(5)(81x 4 ) +405x 4


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+(1)(3x)5 +(1)(243x 5 ) +243x 5 ✓

118


n 0 1 2 3 4 5 1 6 1 7 1 7 8 1 8 9 1 9 36 10 1 10 45 (1 + 2x)9 = 1 +(9)(2x) = 1 +18x

3(b)

4(a)

Ex 5.1 Binomial Coefficients 1 1 1 1 2 1 1 3 3 1 4 6 4 5 10 10 6 15 20 15 21 35 35 28 56 70 56 84 126 126 120 210 252 210

+(36)(4x 2 ) +144x 2

1 5

1 6

21

1 7

28 84

1 8

36 120

1 9

45

1 10

1

+(84)(8x 3 ) + ⋯ +672x 3 + ⋯ ✓

(1 − x 2 )10 = [1 + (−x 2 )]10 = 1 +(10)(−x 2 ) +(45)(x 4 ) = 1 −10x 2 +45x 4

+(120)(−x 6 ) + ⋯ −120x 6 + ⋯ ✓

(1 + x)10 10 (x)1 ) 1 +(10)(x) +10x +(

=1 =1 =1

10 + ( ) (x)2 2 +(45)(x 2 ) +45x 2

10 (x)3 ) +⋯ 3 +(120)(x 3 ) + ⋯ +120x 3 + ⋯ ✓ +(

4(b)

(1 − 3x)8 = [1 + (−3x)]8 8 8 8 =1 + ( ) (−3x)1 + ( ) (−3x)2 + ( ) (−3x)3 + ⋯ 3 1 2 =1 +(8)(−3x) +(28)(9x 2 ) +(56)(−27x 3 ) + ⋯ =1 −24x +252x 2 −1512x 3 + ⋯ ✓

4(c)

(1 − 2x 2 )7 = [1 + (−2x 2 )]7 7 7 =1 + ( ) (−2x 2 )1 + ( ) (−2x 2 )2 1 2 =1 +(7)(−2x 2 ) +(21)(4x 4 ) =1 −14x 2 +84x 4

4(d)

1

1

7 + ( ) (−2x 2 )3 + ⋯ 3 +(35)(−8x 6 ) + ⋯ −280x 6 + ⋯ ✓

16

(1 − x 3 ) 2

1

16

= [1 + (− x 3 )] 2

1 2 1 1 16 16 = 1 + ( ) (− x 3 ) + ( ) (− x 3 ) 2 2 1 2 1 3 1 6 = 1 +(16) (− x ) +(120) ( x ) 2

= 1 −8x 3


4

+30x 6

3 1 16 ) (− x 3 ) + ⋯ 2 3 1 +(560) (− x 9 ) + ⋯

+(

8

−70x 9 + ⋯ ✓

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119


n n! n(n−1)(n−2)! ( )= (n−2)!2! = (n−2)!(2×1) 2 (

= (

n(n−1)

=

2

✓

(n+1)! (n+1)! n+1 ) = [(n+1)−2]!(3)! = (n−2)!(3)! 3

=

5(ii)

Ex 5.1

(n+1)(n)(n−1)(n−2)! (n−2)!(3.2.1) (n+1)(n)(n−1)

✓

6

n = 4( ) 2

n+1 ) 3

(n+1)(n)(n−1)

= 4[

6

(n + 1)(n)(n − 1)

n(n−1) 2

]

= 12n(n − 1)

(n)(n − 1)[(n + 1) − 12] = 0 (n)(n − 1)(n − 11) =0 n=0 or n = 1 or n = 11 ✓ (rej ∵ n ≥ 2) (rej ∵ n ≥ 2) 6(i)

5

(1 + √x) − (1 − √x)

5

5

5

= (1 + √x) − [1 + (−√x)]

0 1 2 3 4 5 5 5 5 5 5 5 = [( ) (√x) + ( ) (√x) + ( ) (√x) + ( ) (√x) + ( ) (√x) + ( ) (√x) ] 0 3 5 1 2 4 0 1 2 3 4 5 5 5 5 5 5 5 −[( ) (−√x) + ( ) (−√x) + ( ) (−√x) + ( ) (−√x) + ( ) (−√x) + ( ) (−√x) ] 0 3 5 1 2 4

= [1 −[1 = (1 −(1

+(5)(√x) +(5)(−√x)

+(10)(x)

+(10)(x√x)

+(5)(x 2 )

+(1)(x 2 √x)]

+(10)(x)

+(10)(−x√x)

+(5)(x 2 )

+(1)(−x 2 √x)]

+5√x −5√x

+10x +10x

+10x√x −10x√x

+5x 2 +5x 2

+x 2 √x) −x 2 √x)

=10√x +20x√x 6(ii)

+2x 2 √x [shown] ✓

5

5

(1 + √2) − (1 − √2) = 10√2 = 10√2 = 58√2 ✓

+20(2)√2 +40√2


+2(2)2 √2 +8√2

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1

(2 − x) (1 + x)

Ex 5.1

8

2

8 1 0 8 1 3 8 1 1 8 1 2 = (2 − x)[( ) ( x) + ( ) ( x) + ( ) ( x) + ( ) ( x) + ⋯ ] 0 2 3 2 1 2 2 2 1 1 2 1 = (2 − x)[1 +(8) ( x) +(28) ( x ) +(56) ( x 3 ) + ⋯ ] 2

= (2 − x)(1

4

+4x

+7x

2

8

3

+7x + ⋯ )

= 2 +8x +14x 2 +14x 3 −x −4x 2 −7x 3 + ⋯ = 2 +7x +10x 2 +7x 3 + ⋯ ✓ 7(ii)

Substitution 1.9 × (1.05)8 = (2 − 0.1) (1 + 0.05)8 1

= [2 − (0.1)] [1 + (0.1)]

8

2

= 2 +7(0.1) +10(0.1)2 +7(0.1)3 + ⋯ = 2 +0.7 +0.1 +0.007 + ⋯ ≈ 2.807 ✓ 8(i)

(1 + x)10 000 10 000 (x)0 10 000 (x)1 ) ) =( +( +⋯ 0 1 =1 +10 000x + ⋯ ✓

8(ii)

Substitution (1.1)10 000 = [1 + (0.1)]10 000 = 1 + 10 000(0.1) + ⋯ = 1 + 1000 + ⋯ = 1001 + ⋯ > 1000 ✓ 10 000 ∴ 1.1 is larger


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121


Ex 5.1

(1 + x)20 20 20 20 = ( ) (x)0 + ( ) (x)1 + ( ) (x)2 + ⋯ 0 1 2 2 = 1 + 20x + 190x + ⋯

10(ii) (1 + 4x)6 (1 − 2x)14 = (1 + 24x + 240x 2 + ⋯ )(1 − 28x + 364x 2 + ⋯ )

Estimated amount (at r = 4, t = 5) = 1 −28x +24x

) ]|

400

= 20 000 [(1 + = 20 000 [1 + (

4t

r

= [20 000 (1 +

4 400 1

100

)

t=5,r=4 (4×5)

]

)]

≈ 20 000 [1 + 20 (

20

1

= 1 − 4x − 68x 2 + ⋯ ✓

) + 190 (

100

+364x 2 −673x 2 +240x 2 + ⋯

1 100

2

) ]

≈ $24 380 ✓ 9(ii)

E = 24 380 A = 20 000 [1 + (

1

)]

20

100

= 24 403.8008 A−E A

× 100%

(24 403. 8008) − (24 380) × 100% (24403. 8008) ≈ 0.10% ✓ =

9(iii)

Yes ✓

10(i) (a)

(1 + 4x)6 6 6 = ( ) (4x)0 + ( ) (4x)1 0 1

6 + ( ) (4x)2 + ⋯ 2

=1

+6(4x)

+15(16x 2 ) + ⋯

=1

+24x

+240x 2 + ⋯ ✓

10(i) (b)

(1 − 2x)14 = [1 + (−2x)]14 14 14 14 = ( ) (−2x)0 + ( ) (−2x)1 + ( ) (−2x)2 0 1 2 +⋯ =1

+14(−2x)

+91(4x 2 ) + ⋯

=1

−28x

+364x 2 + ⋯ ✓


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122


Ex 5.1

(1 − x)6 = [1 + (−x)]6 6 6 6 = ( ) (−x)0 + ( ) (−x)1 + ( ) (−x)2 0 1 2 =1 +6(−x) +15x 2 2 3 = 1 − 6x + 15x − 20x + ⋯ ✓

6 + ( ) (−x)3 + ⋯ 3 +20(−x 3 ) + ⋯

11(ii) Method 1 (factorize & multiply) (1 + x − 2x 2 )6 = [(1 − x)(1 + 2x)]6 = (1 − x)6 (1 + 2x)6 6 6 6 6 = (1 − 6x + 15x 2 − 20x 3 + ⋯ )[( ) (2x)0 + ( ) (2x)1 + ( ) (2x)2 + ( ) (2x)3 + ⋯ ] 0 3 1 2 = (1 − 6x + 15x 2 − 20x 3 + ⋯ )[1 +(6)(2x) +(15)(4x 2 ) + (20)(8x 3 ) + ⋯ ] = (1 − 6x + 15x 2 − 20x 3 + ⋯ )(1 + 12x + 60x 2 + 160x 3 + ⋯ )

= 1 +12x +60x 2 −6x −72x 2 +15x 2

+160x 3 −360x 3 +180x 3 −20x 3 = 1 + 6x + 3x 2 − 40x 3 + ⋯ Method 2 (substitution) (1 + x − 2x 2 )6 = [1 − (−x + 2x 2 )]6 = 1 − 6(−x + 2x 2 )

+15(−x + 2x 2 )2

−20(−x + 2x 2 )3

= 1 − 6(−x + 2x 2 )

+15[−x(1 − 2x)]2

−20[−x(1 − 2x)]3

= 1 − 6(−x + 2x 2 )

+15(−x)2 (1 − 2x)2

−20(−x)3 (1 − 2x)3

= 1 − 6(−x + 2x 2 )

+15x 2 (1 − 2x)2

+20x 3 (1 − 2x)3

= 1 − 6(−x + 2x 2 )

+15x 2 (1 − 4x + ⋯ )

+20x 3 (1 + ⋯ )

= 1 +6x −12x 2 +15x 2

−60x 3 +20x 3 + ⋯

= 1 + 6x + 3x 2 − 40x 3 + ⋯ ✓


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123


Ex 5.1

(1 − x)(1 + ax)6 = 1 + bx 2 + ⋯

14(i) (1 − x)4

(1 − x)(1 + ax)6 6 6 6 = (1 − x)[( ) (ax)0 + ( ) (ax)1 + ( ) (ax)2 + ⋯ 0 1 2 = (1 − x)[1 +(6)(ax) +(15)(a2 x 2 ) + ⋯ ] = (1 − x)(1 +6ax +15a2 x 2 + ⋯ ) +15a2 x 2 −6ax 2

= 1 +6ax −x

6

Compare x 2 : 15a2 − 6a

+(4)(−x 3 ) +(1)(x 4 )

= 1 − 4x

+6x 2

−4x 3

+x 4 ✓

=b

1 2

1

1

6

6

6

15 ( ) − 6 ( ) = b ∵ a =

13(i)

+(6)(x 2 )

S = y 4 − 4y 3 + 6y 2 − 4y + 1 = 1 − 4y + 6y 2 − 4y + 1 = (1 − y)4

1

= ✓

b

= 1 + (4)(−x)

Sub y = 1 − x 3 :

6a − 1 = 0 a

4 4 4 4 = 1 + ( ) (−x)1 + ( ) (−x)2 + ( ) (−x)3 + ( ) (−x)4 3 1 2 4

14(ii) S = (1 − x 3 )4 − 4(1 − x 3 )3 + 6(1 − x 3 )2 − 4(1 − x 3 ) + 1

= 1 +(6a − 1)x +(15a2 − 6a)x 2 + ⋯ Compare x:

= [1 + (−x)]4

=−

7 12

✓

Sub back y = 1 − x 3 : S = [1 − (1 − x 3 )]4 = (x 3 )4 = x12

n n (1 − 2x)n = 1 + ( ) (−2x)1 + ( ) (−2x)2 + ⋯ 1 2 n = ( ) (4x 2 ) + ⋯ 2 n =4 ( ) x2 + ⋯ 2 n ∴ Coefficient of x 2 = 4 ( ) [shown] ✓ 2

13(ii) Coefficient of x 2 = 24 n 4( ) = 24 2 n ( ) =6 2 n! (n−2)!2!

=6

n(n−1)(n−2)! (n−2)!2!

=6

n(n−1) 2

=6

n(n − 1)

= 12

n2 − n

= 12

n2 − n − 12

=0

(n − 4)(n + 3) = 0 n = 4 or n = −3 (rej ∵ n > 0) ✓


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124


Ex 5.1

(1 + x)5 (1 − 4x)4 = (1 + x)5 [1 + (−4x)]4 4 4 4 5 5 5 = [( ) (x)0 + ( ) (x)1 + ( ) (x)2 + ⋯ ] [( ) (−4x)0 + ( ) (−4x)1 + ( ) (−4x)2 + ⋯ ] 0 1 2 0 1 2 2) 2) [ ] [ (5)(x) (10)(x (4)(−4x) (6)(16x = 1 + + +⋯ 1 + + +⋯] 2 2 (1 − 16x + 96x + ⋯ ) = (1 + 5x + 10x + ⋯ )

= 1 −16x +96x 2 +5x −80x 2 +100x 2 + ⋯ ≈ 1 − 11x + 26x 2 [shown] ✓ 15(ii) Factor & multiply polynomials (a) (1 + x)5 (1 − 4x)5 = [(1 + x)5 (1 − 4x)4 ] (1 − 4x) = [1 − 11x + 26x 2 + ⋯ ] (1 − 4x) = 1 −11x +26x 2 −4x +44x 2 + ⋯ = 1 − 15x + 70x 2 + ⋯ ✓ 15(ii) Substitute (1 − x)5 (1 + 4x)4 = [1 + (−x)]5 [1 − 4(−x)]4 (b) ≈ 1 −11(−x) +26(−x)2 ≈ 1 +11x

+26x 2 ✓

15(ii) Expand & substitute (1 + x 2 )5 (1 − 2x)4 (1 + 2x)4 = (1 + x 2 )5 (1 − 4x 2 )4 (c) = [1 + (x 2 )]5 [1 − 4(x 2 )]4 ≈ 1 −11(x 2 ) +26(x 2 )2 ≈ 1 −11x 2


+26x 4 ✓

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125


Ex 5.1

16 28x 6 − 30x 4 − 1 = 0 sub x = 1 + h: 28(1 + h)6 − 30(1 + h)4 − 1 =0 6 (h)0 6 (h)1 6 (h)2 28 [( ) +( ) +( ) +⋯] 0 1 2 =0 4 4 4 −30 [( ) (h)0 + ( ) (h)1 + ( ) (h)2 + ⋯ ] − 1 0 1 2 28(1 + 6h + 15h2 + ⋯ ) −30(1 + 4h + 6h2 + ⋯ ) − 1

18(ii) Selective expansion (1 + y)7 sub y = x(1 + x): [1 + x(1 + x)]7 = 1 +7x(1 + x) +21[x(1 + x)]2 +35[x(1 + x)]3 +35[x(1 + x)]4 +21[x(1 + x)]5 +7[x(1 + x)]6 +[x(1 + x)]7 (1 + x)4 = 35x 4 +21x 5 (1 + x)5 +7x 6 (1 + x)6 (1 + x)7 + ⋯ +x 7

=0

28 + 168h + 420h2 =0 −30 − 120h − 180h2 − 1 + ⋯ 240h2 + 48h − 3 80h2 + 12h − 1 (20h − 1)(4h + 1) h=

1

1

20

or h = − (rej ∵ root greater than 1) 4

⇒ root x = 1 + 17(i)

4 = 35x 4 [… ( ) (x)3 … ] 3 5 [… (5) (x)2 +21x …] 2 6 +7x 6 [… ( ) (x)1 … ] 1 7 7 [… ( ) (x)0 ] + ⋯ +x 0

≈0 ≈0 ≈0

1 20

= 1.05 ✓

= [35(4) + 21(10) + 7(6) + 1]x 7 + ⋯ = 393x 7 + ⋯

(1 + 2x)2n 2n 2n 2n = ( ) (2x)0 + ( ) (2x)1 + ⋯ + ( ) (2x)r 0 1 r 2n (2x)2n +⋯+ ( ) ✓ 2n

17(ii) 22n = [1 + (1)]2n =(

=(

2n (1)0 2n 2n ) + ( ) (1)1 + ( ) (1)2 0 1 2 2n + ⋯ + ( ) (1)2n 2n 2n ) 0

⇒ coefficent of x 7 = 393 ✓ 18(iii) 1st Observation (1 + y)7 = 1 + 7y + 21y 2 + 35y 3 + 35y 4 + 21y 5 + 7y 6 + y7 ✓

2n 2n 2n + ( ) + ( ) + ⋯ + ( ) [proven] 1 2 2n

✓ 18(i) (1 + y)7 7 7 7 7 = ( ) (y)0 + ( ) (y)1 + ( ) (y)2 + ( ) (y)3 0 3 1 2 7 (y)4 7 (y)5 7 (y)6 7 +( ) +( ) +( ) + ( ) (y)7 5 6 4 7 2 3 4 5 = 1 + 7y + 21y + 35y + 35y + 21y + 7y 6 + y 7 ✓


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All the coefficients of the terms from the 2nd term onwards are divisible by 7 except the last term. 2nd Observation Last Term of y = [x(1 + x)]7 = x 7 (1 + x)7 7 7 7 7 ( ) (x)0 + ( ) (x)1 + ( ) (x)2 + ( ) (x)3 0 3 1 2 = x7 [ ] 7 7 7 7 (x)4 +( ) + ( ) (x)5 + ( ) (x)6 + ( ) x 7 5 6 7 4 = x 7 [(1)x + (7)x 2 + (21)x 2 + (35)x 3 + (35)x 4 + (21)x 5 + (7)x 6 + (1)x 7 ]

126


Ex 5.2

Ex 5.2 1(a)

(2 − x)3 = [2 + (−x)]3 = 23

3 3 3 + ( ) (2)3−1 (−x)1 + ( ) (2)3−2 (−x)2 + ( ) (2)3−3 (−x)3 3 1 2

=8

+(3)(4)(−x)

+(3)(2)(x 2 )

+(−x)3

= 8 − 12x + 6x 2 − x 3 ✓ 1(b)

(x + 2y)4 = x4

4 4 4 4 + ( ) (x)4−1 (2y)1 + ( ) (x)4−2 (2y)2 + ( ) (x)4−3 (2y)3 + ( ) (x)4−4 (2y)4 3 1 2 4

= x4

+(4)(x 3 )(2y)

+(6)(x 2 )(4y 2 )

+(4)(x)(8y 3 )

+(16y 4 )

= x 4 + 8x 3 y + 24x 2 y 2 + 32xy 3 + 16y 4 ✓ 1(c)

(2 + x 2 )5 = 25

5 5 5 5 5 + ( ) (2)5−1 (x 2 )1 + ( ) (2)5−2 (x 2 )2 + ( ) (2)5−3 (x 2 )3 + ( ) (2)5−4 (x 2 )4 + ( ) (2)5−5 (x 2 )5 3 5 1 2 4

= 25

+(5)(16)(x 2 )

+(10)(8)(x 4 )

+(10)(4)(x 6 )

+(5)(2)(x 8 )

+x10

= 32 + 80x 2 + 80x 4 + 40x 6 + 10x 8 + x10 ✓ 2(a)

x 5

1

2

2

(4 − ) = [4 + (− x)]

5

1 1 5 + ( ) (4)5−1 (− x) 2 1

= 45

1

= (1024) +(5)(256) (− x) 2

2 1 5 + ( ) (4)5−2 (− x) 2 2

3 1 5 + ( ) (4)5−3 (− x) + ⋯ 2 3

1

1

+(10)(64) ( x 2 )

+(10)(16) (− x 3 )

4

8

+⋯

= 1024 − 640x + 160x 2 − 20x 3 + ⋯ ✓ 2(b)

1

12

( + x2 ) 2

1 12 2

= = 2(c)

(

1 2x

12 1 12−1 (x 2 )1 12 1 12−2 (x 2 )2 +( )( ) +( )( ) 2 1 2 2

=( ) 1

+(12) (

4096 1 4096

8

− 2x 2 ) = [

+

1 2x

3 512

1 8 1 1


x4 +

55 128

+(66) (

1 1024

) (x 4 )

−

1 8x5

+

12 1 12−3 (x 2 )3 )( ) +⋯ 3 2

+(220) (

1

) (x 6 )

512

+⋯

x6 + ⋯ ✓

8

+(8) (

256x8 256x8

) (x 2 )

8 1 8−1 (−2x 2 )1 8 1 8−2 (−2x 2 )2 +( )( ) +( )( ) 1 2x 2 2x

2x

=

33 512

+ (−2x 2 )]

=( ) =

x2 +

1 2048

+(

1 128x7

7 4x2

) (−2x 2 )

+(28) (

1 64x6

) (4x 4 )

8 1 8−3 (−2x 2 )3 +( )( ) +⋯ 3 2x +(56) (

1 32x5

) (−8x 6 ) + ⋯

− 14x + ⋯ ✓

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127


Ex 5.2

For (x + 2y)5

5

5 Tr+1 = ( ) (x)5−r (2y)r ✓ r 3(b)

3(c)

Tr+1 4(a)

1 13 2y

)

= [3x 2 + (−

1 2y

13

4

For coefficient of x 3 ,

1 r

13 = ( ) (3x 2 )13−r (− ) ✓ 2y r

r=6

6(a)

10 (2)10−6 (x)6 ) 6 = (210)(16)(x 6 ) = 2260x 6 ✓

(rej)

18

1

For ( − x 2 )

For term independent of x, 3r − 18 = 0 r =6

r=3

Term independent of x 18 = ( ) (−1)6 (x)0 6 = (18 564)(1)(x)0 = 18 564 ✓

9 4th term = ( ) (3x)9−3 (−2)3 3 = (84)(729x 6 )(−8) = 489 888x 6 ✓ For (y − 2x)10 = [y + (−2x)]10

6(b)

10 (y)10−r (−2x)r ) r

For middle term,

r=

10 2

= [x −1 + (−x 2 )]18

18 (x −1 )18−r (−x 2 )r ) r 18 (−1)r x 2r = ( ) x r−18 r 18 = ( ) (−1)r x 3r−18 r

For (3x − 2)9 = [3x + (−2)]9

Tr+1 = (

2

Tr+1 = (

9 Tr+1 = ( ) (3x)9−r (−2)r r

4(c)

3

2r = 6 ⇒ r = 3 1 3 15 10 Coefficient of x 6 = [( ) (− ) ] = (− ) ✓ 4 8 3

7th term = (

For 4th term,

⇒r=

For coefficient of x 6 ,

x

4(b)

2r = 3

Coefficient of x 3 = 0 ✓

10 = ( ) (2)10−r (x)r r

For 7th term,

10

1

= [1 + (− x 2 )]

)]

For (2 + x)10 Tr+1

4

10

)

r 1 10 ) (− x 2 ) 4 r 1 r 10 = ( ) (− ) (x 2 )r 4 r 1 r 10 = ( ) (− ) x 2r 4 r

10 (2x)10−r (−3)r ) ✓ r

For (3x 2 −

x2

Tr+1 = (

For (2x − 3)10 = [2x + (−3)]10 Tr+1 = (

For (1 −

For (x +

1 2x2

12

)

1

= [x + (2 x−2 )]

12 12−r )x r 12 = ( ) x12−r r 12 1 r = ( )( ) r 2

Tr+1 = ( =5

10 Middle term = ( ) (y)10−5 (−2x)5 5 = (252)(y 5 )(−32x 5 ) = −8064x 5 y 5 ✓

For middle term,

1

( x −2 )

12

r

2

1 r

( ) x −2r 2

x12−3r r=

12 2

=6

12 1 6 (x)12−3(6) )( ) 6 2 231 = 6 ✓

Middle term = (

16x


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128


Ex 5.2

(px − 3)n = [px + (−3)]n n = ( ) (px)n−0 (−3)0 0 = (px)n

n + ( ) (px)n−1 (−3)1 + ⋯ 1 +(n)(px)n−1 (−3) + ⋯

= pn x n

−3npn−1 x n−1 + ⋯

= 512x 9 −qx 8 + ⋯ [given] Compare 1st term: Compare 2nd term: −3npn−1 x n−1 = −qx 8 pn x n = 512x 9 −3(9)29−1 x 9−1 = −qx 8 ⇒ n =9✓ ∵ n = 9, p = 2 8 )x 8 −27(2 = −qx 8 ⇒ pn = 512 ⇒q = 6912 ✓ ∵ n = 9, p9 = 512 p =2✓ 8(i)

x 7 (2 − ) 2

7 1 = [2 + (− x)] 2 1 1 7 = (2)7 + ( ) (2)7−1 (− x) 2 1 1 = 128 +(7)(64) (− x) 2 2

2 1 7 + ( ) (2)7−2 (− x) 2 2 1 2 +(21)(32) ( x ) 4

3 1 7 + ( ) (2)7−3 (− x) + ⋯ 2 3 1 3 +(35)(16) (− x ) + ⋯ 8

3

= 128 − 224x + 168x − 70x + ⋯ ✓ 8(ii)

(1.995)7 = (2 − 0.005)7 = (2 −

0.01 7 2

)

= 128 − 224(0.01) + 168(0.01)2 − 70(0.01)3 + ⋯ ≈ 125.7767 ✓ 9(i)

(1 − 2x)9 = [(1) + (−2x)]9 9 9 =1 + ( ) (−2x)1 + ( ) (−2x)2 + ⋯ 1 2 =1 +(9)(−2x) +(36)(4x 2 ) + ⋯ = 1 − 18x + 144x 2 + ⋯ ✓ 5 5 + ( ) (2)5−1 (x)1 + ( ) (2)5−2 (x)2 1 2 = 32 +(5)(16)(x) +(10)(8)(x 2 ) = 32 + 80x + 80x 2 + ⋯ ✓

(2 + x)5 = 25

9(ii)

(1 − 2x)9 (2 + x)5 = (1 − 18x + 144x 2 + ⋯ )(32 + 80x + 80x 2 + ⋯ )

= 32 +80x +80x 2 −576x −1440x 2 +4608x 2 + ⋯ = 32 −496x +3248x 2 + ⋯ ✓


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129


Ex 5.2

(2x − 1)6 = [(2x) + (−1)]6 6 6 6 = (2x)6 + ( ) (2x)6−1 (−1)1 + ( ) (2x)6−2 (−1)2 + ( ) (2x)6−3 (−1)3 3 1 2 = (64x 6 ) +(6)(32x 5 )(−1) +(15)(16x 4 )(1) +(20)(8x 3 )(−1) = 64x 6 −192x 5 +240x 4 −160x 3 + ⋯ ✓

10(ii) (2x − 1)6 (x 2 − 2x + 3) = (64x 6 − 192x 5 + 240x 4 − 160x 3 + ⋯ )(x 2 − 2x + 3) = (−192x 5 )(3) +(240x 4 )(−2x) +(−160x 3 )(x 2 ) + ⋯ = −576x 5 −480x 5 = −1216x 5 + ⋯

−160x 5 + ⋯

∴ Coefficient of x 5 = −1216 ✓ 11(i)

1

( − 2x)

5

1

= [ + (−2x)]

2

5

2

1 5

5 1 5−1 (−2x)1 +( )( ) 1 2 1 +(5) ( ) (−2x)

=( ) = =

2 1

32 1 32

5 1 5−2 (−2x)2 +( )( ) 2 2 1 +(10) ( ) (4x 2 )

16

5

2

8

5 1 5−3 (−2x)3 +( )( ) +⋯ 3 2 1 +(10) ( ) (−8x 3 ) + ⋯ 4

3

− x + 5x − 20x + ⋯ ✓ 8

11(ii) Expansion 1

(1 + ax + 3x 2 ) ( − 2x) 2

= (1 + ax + 3x 2 ) ( = 5x 2 5a 2 x 8 3 + x2 32 5a

= (5 − 163 32

5

− x + 5x 2 − 20x 3 + ⋯ ) 8

−20x 3 +5ax 3

−

=(

1 32

5

+

8

−

15 3 x +⋯ 8 3 ) x 2 + (5a 32

−

5a 8

) x2

− 20 −

+ (5a −

175 8

15 8

) x3 + ⋯

) x3 + ⋯

Coefficients Coefficient of x 2 = (

163

32 5a

−

5a 8

)

2

=

[given] 13 2

=−

8

a

=−

Coefficent of x = 5a −

32 9 4

3

8

= 5 (− ) − 4 265 8

45

175 9

=−

13

175 8

∵a=−

9 4

✓


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Ex 5.2

For (x + my)8 8 Tr+1 = ( ) x 8−r (my)r ✓ r

13(iii) For constant term,

Constant Term = (

12(ii) if m = 2: 8 (a) Tr+1 = ( ) x 8−r (2y)r r 8 8−r r r = ( )x 2 y r 8 = ( ) 2r x 8−r y r r

14(i)

For (x 3 −

2 10 x2

)

x

k r 9 Tr+1 = ( ) x 9−r ( ) x r 9 9−r (kx −1 )r = ( )x r 9 9−r r −r = ( )x k x r 9 r 9−2r = ( )k x r For coefficient of x 3 , For coefficient of x 3 , 9 − 2r = 3 9 − 2r = 3 −2r = −6 −2r = −6 r =3 r =3

r=5 8 5 = ( )2 5 = 1792 ✓

= −1512x 5 y 3

Coefficient of x 3 Coefficient of x 3 9 9 = ( ) k3 = ( ) k3 3 3 3 = 84k = 84k 3 Equate coefficients Coefficient of x 3 = Coefficient of x [given] 84k 3 = 126k 4 2k 3 = 3k 4 3 4 2k − 3k =0 3 k (2 − 3k) =0

= −1512x 5 y 3 = −1512x 5 y 3 = −1512x 5 y 3 = −1512 = −3 ✓ = [x 3 + (−2x −2 )]10

10 (x 3 )10−r (−2x −2 )r ) r 10 = ( ) x 30−3r (−2)r x −2r r 10 (−2)r =( ) x 30−5r r

2

k = 0 (rej ∵ k is positive)or k = ✓

Tr+1 = (

13(i)

Specific coefficients k 9

For term (−1512x 5 y 3 ), r=3

13

= 13 440

For (x + ) ,

8 8−r (my)r 12(ii) Recall: T r+1 = ( ) x r (b)

T4 8 ( ) x 8−3 (my)3 3 8 ( ) x 5 m3 y 3 3 56m3 x 5 y 3 56m m

10 (−2)6 30−5(6) ) x 6

✓

For coefficient of x 3 y 5 , Coefficient of x 3 y 5

30 − 5r = 0 −5r = −30 r =6

3

For term in x10 ,

Term in x10

30 − 5r = 10 −5r = −20 r =4 10 (−2)4 10 =( ) x = 3360x10 ✓ 4

13(ii) For coefficient of 1 , 5

30 − 5r = −5

x

−5r r Coefficient of

1 x5

=(

= −35 =7

10 (−2)7 ) = −15 360 ✓ 7


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Ex 5.2

14(ii) Put k = 2, 3

(1 − 6x

2)

2 3

9

(x + ) x

= (1 − 6x 2 )[(term in x 3 ) + (term in x 5 ) + ⋯ ] [to be continued] Specific terms 2

9

9

2

For (x + 3 ) = [x + ( x −1 )] , x

3

r 2 9 Tr+1 = ( ) x 9−r ( x −1 ) 3 r 9 9−r 2 r −r ( ) x = ( )x 3 r r 9 2 = ( ) ( ) x 9−2r r 3

For term in x 3 , 9 − 2r = 3 2r =6 r =3 2 3

Term in x 3 = 84 ( ) x 3 = 3

224 3 x 9

For term in x 5 , 9 − 2r = 5 2r =4 r =2 2 2 9 Term in x 5 = ( ) ( ) x 5 = 16x 5 2 3 Expansion (1 − 6x

2)

2 3

9

(x + ) x

= (1 − 6x 2 ) (

224 3 x 9

+ 16x 5 + ⋯ )

= (1)(16x 5 ) + (−6x 2 ) ( =

400 5 − x 3

224 3 x ) 9

+⋯

+⋯

∴ Coefficient of x 5 = −


400 3

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Ex 5.2

x n

(2 − ) 2

1

= [2 + (− x)]

n

2

0 1 n n 1 1 = ( ) (2)n−0 (− x) + ( ) (2)n−1 (− x) 2 2 0 1 1

2 n 1 + ( ) (2)n−2 (− x) + ⋯ 2 2 n(n−1)

1

= 2n

+n(2n−1 ) (− x)

= 2n

−n(2n−1 ) ( ) x

+

= 2n

−n(2n−1 )(2−1 )x

+n(n − 1)(2−3 )2n−2 x 2 + ⋯

= 2n

−n(2n−2 )x

+n(n − 1)2n−5 x 2 + ⋯ ✓

2

1 2

+

2

(2n−2 ) x 2 + ⋯

n(n−1) n−2 2 2 x 8

4

+⋯

15(ii) (1 + 2x) (2 − x)n 2

= (1 + 2x)[2n −n(2n−2 )x +n(n − 1)2n−5 x 2 + ⋯ ] = 2n

−n(2n−2 )x +(2n+1 )x

+n(n − 1)2n−5 x 2 −(n)2n−1 x 2 + ⋯

= 2n +[2n+1 − n(2n−2 )]x = a + bx 2 + ⋯ [given]

+[n(n − 1)2n−5 − (n)2n−1 ]x 2 + ⋯

Compare x: 2n+1 −n(2n−2 ) =0 n−2 3 n−2 (2 )2 −n(2 ) = 0 2n−2 (23 − n) =0 n =8✓ 15(iii) Compare x 0 : a = 2n = 28 ∵ n = 8 = 256 ✓ Compare x 2 : b = n(n − 1)2n−5 − (n)2n−1 = 8(8 − 1)28−5 − (8)28−1 ∵ n = 8 = −576 ✓


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Ex 5.2

Expansion (1 + x)(a − bx)12 = (1 + x)[(term in x 7 ) + (term in x 8 ) + ⋯ ] [to be continued]

17(i)

Specific terms For (a − bx)12 = [a + (−bx)]12 , 12 12−r (−bx)r )a r 12 = ( ) a12−r (−b)r x r r

For coefficient of x 2 , r = 2 n n Coefficient of x 2 = ( ) 32 = 9 ( ) 2 2 Equate coefficients (Coefficient of x 3 ) = 6(Coefficient of x 2 ) n n 27 ( ) = 6 ⋅ 9( ) 3 2 n n ( ) = 2 ( ) [shown] ✓ 3 2

Tr+1 = (

For term in x 7 ,

Find specific coefficients For (1 + 3x)n n Tr+1 = ( ) (3x)r r n r r = ( )3 x r For coefficient of x 3 , r = 3 n n Coefficient of x 3 = ( ) 33 = 27 ( ) 3 3

r=7

12 Term in x 7 = ( ) a5 (−b)7 x 7 7 = −792a5 b7 x 7

17(ii)

For term in x 8 , r = 8 12 4 (−b)8 8 )a x 8 = 495a4 b8 x 8

Term in x 8 = (

n! (n−3)!3! n(n−1)(n−2)(n−3)! (n−3)!3! n(n−1)(n−2) 3! (n)(n−1)(n−2) 6

Expansion (1 + x)(a − bx)12 = (1 + x)[(−792a5 b7 x 7 ) + (495a4 b8 x 8 ) + ⋯ ] = 1(495a4 b8 x 8 ) + x(−792a5 b7 x 7 ) + ⋯ = 495a4 b8 x 8 − 792a5 b7 x 8 + ⋯

n!

= 2 ⋅ (n−2)!2! =2⋅ =2⋅ =2⋅

n(n−1)(n−2)! (n−2)!2! n(n−1) 2! (n)(n−1) 2

(n)(n − 1)(n − 2) = 6(n)(n − 1) (n)(n − 1)[(n − 2) − 6] = 0 (n)(n − 1)(n − 8) =0 n=0 or n = 1 or n = 8 ✓ (rej ∵ n ≥ 2) (rej ∵ n ≥ 2)

= (495a4 b8 − 792a5 b7 )x 8 + ⋯ Coefficient of x 8 = 0 [given] 4 8 5 7 495a b − 792a b = 0 5a4 b8 − 8a5 b7 =0 5a4 b8 = 8a5 b7 a b

5

= ✓


8

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Ex 5.2

(2 + p)5 5 5 5 + ( ) (2)5−1 (p)1 + ( ) (2)5−2 (p)2 + ( ) (2)5−3 (p)3 + ⋯ 3 1 2 2 = (32) +(5)(16)(p) +(10)(8)(p ) +(10)(4)(p3 ) +⋯ 2 3 = 32 +80p +80p +40p +⋯✓ = 25

18(ii) (2 + x − 2x 2 )5 = [2 + (x − 2x 2 )]5 = 32 +80(x − 2x 2 ) +80(x − 2x 2 )2 = 32 +80(x − 2x 2 ) +80(x 2 − 4x 3 + ⋯ ) = 32 +80x −160x 2 2 +80x −320x 3 +40x 3 + ⋯ = 32 +80x 19

−80x 2

+40(x − 2x 2 )3 + ⋯ +40(x 3 + ⋯ )

−280x 3 + ⋯ ✓

(a + bx + cx 2 )4 = 81 + 216x + 108x 2 + dx 3 + ⋯ [given] (a + bx + cx 2 )4 = [a + (bx + cx 2 )]4 4 4 = a4 + ( ) (a)4−1 (bx + cx 2 )1 + ( ) (a)4−2 (bx + cx 2 )2 1 2

4 + ( ) (a)4−3 (bx + cx 2 )3 + ⋯ 3

= a4 + 4a3 (bx + cx 2 )

+6a2 (b2 x 2 + 2bcx 3 + ⋯ )

+4a(b3 x 3 + ⋯ ) + ⋯

= a4 + 4a3 bx +4a3 cx 2

+6a2 b2 x 2 +12a2 bcx 3

+4ab3 x 3 + ⋯

= a4 + 4a3 bx + (4a3 c + 6a2 b2 )x 2 + (12a2 bc + 4ab3 )x 3 + ⋯ = 81 + 216x + 108x 2 + dx 3 + ⋯ [given] Compare x 0 : a4 = 81 a = 3 ✓ or a = −3 Compare x:

(rej ∵ a > 0)

4a3 b = 216 a3 b = 54 ∵ a = 3, (3)3 b = 54 b =2✓

Compare x 2 : 4a3 c + 6a2 b2 = 108 3 2 2 2a c + 3a b = 54 ∵ a = 3, b = 2, 2(3)3 c + 3(3)2 (2)2 = 54 54c + 108 = 54 c = −1 ✓ Compare x 3 : 12a2 bc + 4ab3 =d ∵ a = 3, b = 2, c = −1, 12(3)2 (2)(−1) + 4(3)(2)3 = d d = −120 ✓


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135


For (2x 2 −

1 n √x

Ex 5.2 1

) = [2x 2 + (−x −2 )]

n

21(i)

1 r n Tr+1 = ( ) (2x 2 )n−r (−x −2 ) r 1 n n−r 2n−2r (−1)r x −2r = ( )2 x r 5 n n−r = ( ) 2 (−1)r x 2n−2r r

For (a + b)n , n TR+1 = ( ) an−R bR R For 2nd term, R = 1 n 2nd term: p = ( ) an−1 b1 [given] 1 = nan−1 b For 3rd term: R = 2 n 3rd term: q = ( ) an−2 b2 [given] 2

For term independent of x: 5

n!

2n − r = 0

= (n−2)!2! an−2 b2

2

5 2

r

= 2n 5

n

4

n(n−1) n−2 2 a b 2

n!

= (n−3)!3! an−3 b3

For coefficient of x 7 √x: 5

10 − r = 7.5 2

r

=

For 4th term: R = 3 n 4th term: r = ( ) an−3 b3 [given] 3

5 5−r (−1)r 2(5)−52r 20(ii) T x r+1 = ( ) 2 r 5 5 = ( ) 25−r (−1)r x10−2r r

2

n(n−1)(n−2)! n−2 2 a b (n−2)!2!

= r

∵ r is non − negative integer, smallest n = 5 ✓

5

=

r

=

5 2

=

n(n−1)(n−2)(n−3)! n−3 3 a b (n−3)!3!

=

n(n−1)(n−2) n−3 3 a b 6

pr q2

=1

nan−1 b

Coefficient of x 7 √x 5 = ( ) (2)5−1 (−1)1 1 = (5)(16)(−1) = −80 ✓

=

n(n − 1)(n − 2) n−3 3 [ a b ] 6 n(n − 1) n−2 2 [ a b ] 2

2

1 2 n (n − 1)(n − 2)a2n−4 b4 6 = 1 2 n (n − 1)2 a2n−4 b 4 4 =

2(n − 2) [shown] 3(n − 1)

✓ 21(ii) when p = 8, q = 24, r = 36: (8)(36) (24)2 1 2

= =

2(n−2) 3(n−1) 2(n−2) 3(n−1)

3(n − 1) = 4(n − 2) 3n − 3 = 4n − 8 n =5✓


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Ex 5.2

22 3

3

Given: a = √2 + √5 + √2 − √5 Prove: a3 = 4 − 3a LHS = a3 3

3

= [( √2 + √5) + ( √2 − √5)] 3

= ( √2 + √5)

3

3

3−1 3 1 3−2 3 2 3 3 3 3 ( √2 − √5) + ( ) ( √2 + √5) ( √2 − √5) + ( ) ( √2 + √5) 1 2 3

2

3

3

+3 ( √2 + √5) ( √2 − √5)

= 2 + √5 3

3

3

3

+3 ( √2 + √5) ( √2 − √5)

3−3 3 3 3 3 ( √2 − √5) + ( ) ( √2 + √5) 3

2

+2 − √5

3

= 4 +3 ( √2 + √5) ( √2 − √5) [( √2 − √5) + ( √2 + √5)] 3

= 4 +3( √4 − 5)

(a)

= 4 +3(−1)

(a)

= 4 −3

(a)

3

3

∵ a = √2 + √5 + √2 − √5

= 4 −3a = RHS [proven] ✓ a3 = 4 − 3a 3 a + 3a − 4 =0 (a − 1)( + + ⬚) (a − 1)(a2 + + ⬚) 2 (a − 1)(a + + 4) 2 (a − 1)(a + a + 4) = 0 a = 1 or a =

−(1)±√(1)2 −4(1)(4) 2(1)


=

−1±√−15 2

(rej ∵ discriminant < 0) ✓

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Rev Ex 5 A2(b)

Rev Ex 5 A1(i) (a)

(1 + 3x)6 6 6 6 = 1 + ( ) (3x)1 + ( ) (3x)2 + ( ) (3x)3 + ⋯ 3 1 2 = 1 + (6)(3x) +(15)(9x 2 ) +(20)(27x 3 ) + ⋯ = 1 + 18x + 135x 2 + 540x 3 + ⋯ ✓

A1(i) (b) (1 − 4x)5

For (x −

2 12 x2

)

= [x + (−2x −2 )]12

12 12−r (−2x −2 )r )x r 12 = ( ) x12−r (−2)r x −2r r 12 = ( ) (−2)r x12−3r r

Tr+1 = (

For coefficient of x 3 , 12 − 3r = 3 −3r = −9 r =3

= [1 + (−4x)]5 5 5 5 = 1 + ( ) (−4x)1 + ( ) (−4x)2 + ( ) (−4x)3 + ⋯ 3 1 2

Coefficient of x 3 12 = ( ) (−2)3 3 = −1760 ✓

= 1 + (5)(−4x) +(10)(16x 2 ) +(10)(−64x 3 ) + ⋯ = 1 − 20x + 160x 2 − 640x 3 + ⋯ ✓ A1(ii) (1 + 3x)6 (1 − 4x)5 = (1 + 18x + 135x 2 + ⋯ )(1 − 20x + 160x 2 + ⋯ ) = (1)(160x 2 ) +(18x)(−20x)

+(135x 2 )(1) + ⋯

= 160x 2

+135x 2 + ⋯

−360x 2

= −65x 2 + ⋯ Coefficient of x 2 = −65 ✓ A2(a)

x 8

For (1 + ) , 2

8 x r Tr+1 = ( ) ( ) r 2 8 1 r = ( ) ( x) r 2 8 1 r = ( ) ( ) xr r 2 For coefficient of x 3 , r=3 Coefficient of x 3 8 1 3 = ( )( ) 3 2 =7✓


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Rev Ex 5

2 8

(x 2 − ) x

2 8 = [x 2 + (− )] x = (x 2 )8

2 1 2 2 8 8 + ( ) (x 2 )8−1 (− ) + ( ) (x 2 )8−2 (− ) x x 1 2

= x16

+(8)(x14 ) (− )

2 x

2 3 8 + ( ) (x 2 )8−3 (− ) x 3

4

8

+(28)(x12 ) ( 2 )

+(56)(x10 ) (− 3 )

x

x

= x16 − 16x13 + 112x10 − 448x 7 + ⋯ ✓ A3(ii)

2 8

(x 3 + 1)2 (x 2 − ) x = (x 6 + 2x 3 + 1)(x16 − 16x13 + 112x10 − 448x 7 + ⋯ ) = (x 6 )(−448x 7 ) +(2x 3 )(112x10 ) = (−448x13 ) +(224x13 ) 13 = −240x + ⋯

+(1)(−16x13 ) + ⋯ +(−16x13 ) + ⋯

Coefficient of x13 = −240 ✓ A4(i)

x n

1

2

2

(1 − ) = [1 + (− x)]

n

0 1 n n 1 1 = ( ) (− x) + ( ) (− x) 2 2 0 1 1

= (1)

+(n) (− x)

=1

− nx

2

1

2

+( +

2

A4(ii) (1 − x)n = 1 + ax + 7x 2 + ⋯

2 n 1 + ( ) (− x) + ⋯ 2 2 n(n−1)

1

2

4

) ( x2 ) + ⋯

n(n−1) 2 x 8

+⋯✓

[given]

Compare x 2 : n(n−1)

=7

8 2

n −n = 56 2 n − n − 56 =0 (n − 8)(n + 7) = 0 n = 8 or n = −7 (rej ∵ n > 2) ✓ A4(iii) Compare x: 1

− n 2 1

=a

− (8) = a 2

a

= −4 ✓


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Rev Ex 5

(1 + p)4 4 4 4 4 = ( ) (p)0 + ( ) (p)1 + ( ) (p)2 + ( ) (p)3 0 3 1 2 =1

+6p2

+4p

+4p3

B2

4 + ( ) (p)4 4 4 +p ✓

B2(i)

8

1

1 1

= [a2 x −2 + (− x 2 )]

8

a

r

8

For middle term,

r= =4 2

B2(ii) For coefficient of 1, x

= 1 + 4x + 10x 2 + 16x 3 + ⋯ ✓

Coefficient of

A5(iii) (1.11)4 = [1 + (0.1) + (0.1)2 ]4 ≈ 1 + 4(0.1) +10(0.1)2 + 16(0.1)3 ≈ 1 + 0.4 +0.1 +0.016 ≈ 1.516 ✓ B1(i)

√x ) a

8 Middle term = ( ) a16−12 (−1)4 x 4−4 4 = 70a4 ✓

= 1 +4x +4x 2 +6(x 2 + 2x 3 + ⋯ ) +4(x 3 + ⋯ ) + ⋯ = 1 +4x +4x 2 +6x 2 +12x 3 +4x 3 + ⋯

√x

−

8−r

= [(1) + (x + x 2 )]4 +4(x + x 2 )3 + ⋯

a2

1 1 1 8 (− x 2 ) Tr+1 = ( ) (a2 x −2 ) a r 1 1 8 = ( ) a16−2r x 2r−4 (−1)r a−r x 2r r 8 = ( ) a16−3r (−1)r x r−4 r

A5(ii) (1 + x + x 2 )4 = 1 +4(x + x 2 ) +6(x + x 2 )2

For (

1 x

r=3

8 = ( ) a16−3(3) (−1)(3)−4 3 = (56)a7 (−1) = −56a7 ✓

(2 − x)7 = [2 + (−x)]7 7 7 = ( ) (2)7−0 (−x)0 + ( ) (2)7−1 (−x)1 0 1 7 + ( ) (2)7−2 (−x)2 + ⋯ 2 = 128

+(7)(64)(−x) +(21)(32)(x 2 ) + ⋯ = 128 − 448x + 672x 2 + ⋯ ✓ B1(ii) 1.997 = (2 − 0.01)7 = 128 −448(0.01) +672(0.01)2 + ⋯ = 128 −4.48 +0.0672 + ⋯ ≈ 123.587 ✓ B1(iii) (k − x)(2 − x)7 = (k − x)(128 − 448x + 672x 2 + ⋯ ) = k(672x 2 ) +(−x)(−448x) + ⋯ = 672kx 2 +448x 2 + ⋯ = (672k + 448)x 2 + ⋯ Coefficient of x 2 = 616 [given] 672k + 448 = 616 672k = 168 k

1

= ✓ 4


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Rev Ex 5

B3(i) 5

(2 + √3) 0 5 = ( ) (2)5−0 (√3) 0

1 2 3 4 5 5 5 5 5 5 + ( ) (2)5−1 (√3) + ( ) (2)5−2 (√3) + ( ) (2)5−3 (√3) + ( ) (2)5−4 (√3) + ( ) (2)5−5 (√3) 3 5 1 2 4

= (32)

+(5)(16)(√3)

+(10)(8)(3)

+(10)(4)(3√3) +(5)(2)(9)

+(1)(1)(9√3)

= (32 + 240 + 90) + (80 + 120 + 9)√3 = 362 + 209√3 ✓ B3(ii) (2 − √3)5 = 362 − 209√3 ✓ 5

Show: (2 − √3) =

1 (2+√3)

5 (2+√3)

LHS = (2 − √3) B4(i)

(2+√3)

5

5 5

=

(4−3)5 (2+√3)

5

=

1 (2+√3)

5

= RHS [shown] ✓

(a − x)(1 + 2x)n = 3 + 47x + bx 2 + ⋯ sub x = 0: n

(a − 0)(1 + 2(0)) = 3 + 47(0) + b(0)2 a

=3✓

B4(ii) (3 − x)(1 + 2x)n = 3 + 47x + bx 2 + ⋯ [given] (3 − x)(1 + 2x)n

= (3 − x) [1

n n + ( ) (2x)1 + ( ) (2x)2 + ⋯ ] 1 2 n(n−1) (4x 2 ) + ⋯ ] + (n)(2x) +

= (3 − x)(1

+ 2nx

= (3 − x) [1

= 3 +6nx −x

2

+ 2n(n − 1)x 2 + ⋯ )

+6n(n − 1)x 2 −2nx 2

= 3 +(6n − 1)x +(6n2 − 6n − 2n)x 2 + ⋯ = 3 +(6n − 1)x +(6n2 − 8n)x 2 + ⋯ = 3 +47x + bx 2 + ⋯ [given] Compare x: (6n − 1) = 47 6n = 48 n =8✓ B4(iii) Compare x 2 : (6n2 − 8n) = b 6(8)2 − 8(8) = b ∵ n = 8 b = 320 ✓


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Rev Ex 5

Expansion (1 + ax)6 (2 + bx)5 6 6 = [1 + ( ) (ax)1 + ( ) (ax)2 + ⋯ ] 1 2 5 (2)5−1 (bx)1 5 5 [2 + ( ) + ( ) (2)5−2 (bx)2 + ⋯ ] 1 2 = (1 + 6ax + 15a2 x 2 + ⋯ )(32 + 80bx + 80b2 x 2 + ⋯ )

+80b2 x 2 +480abx 2 +480a2 x 2 + ⋯ +(80b2 + 480ab + 480a2 )x 2 + ⋯

= 80bx +192ax = (80b + 192a)x

Equate coefficients Coeff. of x = −112 [given] 80b + 192a = −112 b

= =

−112−192a 80 −7−12a

−(1)

5

Coeff. of x 2 = 80 [given] 2 2 80b + 480ab + 480a = 80 b2 + 6ab + 6a2 =1 −(2) sub (1) into (2): (

−7−12a 2

)

5 144a2 +168a+49 25

+6a ( +

−7−12a

5 −42a−72a2

) + 6a2 = 1

5

+6a2 = 1

(144a2 + 168a + 49) −210a − 360a2 + 150a2

= 25

(144a2 + 168a + 49) −210a − 210a2

= 25

−66a2 −42a +24 11a2 +7a +4 (11a + 4)(a + 1)

=0 =0 =0

a=−

4 11

(rej ∵ a ∈ ℤ)

or

a = −1 ✓ b|a=1 =

−7−12(−1) 5

=1✓


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142


Ex 6.1 4(ii)

Ex 6.1 1(a)

Point:

D(1,4) or E(−3,1) (4)−(1)

Gradient: mDE = (1)−(−3) =

A(4,5) B(6,9) MAB = (

4+6 5+9

,

2

2

y − (4)=

= (5,7) ✓ 1(b)

−5+11 3+(−7)

,

2

2

y

2(b)

4

4

✓

C = MAB 9+b (−6)+b2 (3, −4) = ( 1 , )

2

3 = 2 a1 = 0 ⇒ A(0,3) ✓

4 13

A(9, −6) C(3, −4) B(b1 , b2 )

M = MAB a +6 a +7 (3,5) = ( 1 , 2 ) 2 (a1 )+(6)

3

4 3

5(i) A(a1 , a2 )

M(3,5) B(6,7)

[x − (1)]

4 3

= x+

)

= (3, −2) ✓ 2(a)

3

y−4 = x−

A(−5,3) B(11, −7) MAB = (

4

y − y1 = mDE (x − x1 )

DE:

)

3

9+b1

(a2 )+(7)

and

5 = a2 = 3

2

(−6)+b2

3 = and −4 = 2 2 b1 = −3 b2 = −2 ∴ B(−3, −2) ✓

2

5(ii)

M(−2,6) B(−4, −8) A(a1 , a2 )

2

Radius = |BC| = √[(−3) − 3]2 + [(−2) − (−4)]2

M = MAB a +(−4) a2 +(−8) (−2,6) = ( 1 ) , 2 (a1 )+(−4)

2

−2 = 2 a1 = 0 ∴ A(0,20) ✓ 3(a)

and

= √40

(a2 )+(−8)

6 = a2 = 20

= √4 × 10 = 2√10 ✓

2

6(i)

A(2a, −a) B(4a, 5a) MAB = (

= √36 + 4

|AC| = √(5 − 6)2 + [3 − (−4)]2

2a+4a −a+5a

,

2

)

2

= √1 + 49 = √50

= (3a, 2a) ✓ 3(b)

|BC| = √[5 − (−2)]2 + (3 − 4)2

A(2t, 5) B(4,1 − 2t) MAB = (

= √49 + 1 = √50

(2t)+4 5+(1−2t)

,

2

2

)

∵ |AC| = |BC|, △ ABC is isosceles ✓

= (t + 1,3 − t) ✓ 4(i)

A(−1,6) B(3,2)

C(−5, −4)

6(ii)

MAB = (

6+(−2) (−4)+4 2

,

2

)

= (2,0) ✓

D = MAB =(

A(6, −4) B(−2,4) C (5,3)

(−1)+3 6+2

,

2

2

)

= (1,4) ✓ E = MAC =(

(−1)+(−5) 6+(−4) 2

,

2

)

= (−3,1) ✓


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143


Ex 6.1

area of △ ABC

8

=

1 |AB||MAB C| 2

=

1 [6 − (−2)]2 √(5 − 2)2 + (3 − 0)2 √ 2 +[(−4) − 4]2 1

= √64 + 64 2

y = x 2 + 2x − 3

√9 + 9

1 √128√18 2 1 = √2304 2 1 = (48) 2

x= =

= 24 unit 2 ✓

(

2 p2 +q2 2 2

2

=

−4±4√2

2

= −2 ± √2

2

y|x=−2+√2 = 1 − 2(−2 + √2)

) = (5,1)

=5

−4±√16×2

−4±√32

⇒ A(−2 − √2, −3 + 2√2)

= (5,1) 2

=

2(1)

= 5 + 2√2

p2 +q2 p+q

,

−(4)±√(4)2 −4(1)(−4)

y|x=−2−√2 = 1 − 2(−2 − √2)

A(p2 , p) B(q2 , q) MAB

−(2)

sub (1) into (2): 1 − 2x = x 2 + 2x − 3 2 x + 4x − 4 = 0

=

7

Points A & B At A & B, 2x + y = 1 intersects y = x 2 + 2x − 3. 2x + y = 1 y = 1 − 2x −(1)

and

2

p + q = 10 −(1) sub (2) into (1): p2 + (2 − p)2 = 10 2 2) (4 p + − 4p + p = 10 2p2 − 4p − 6 =0 2 p − 2p − 3 =0 (p + 1)(p − 3) =0

p+q 2

= 5 − 2√2 =1

⇒ B(−2 + √2, −3 − 2√2)

q = 2 − p −(2)

p = −1 or q|p=−1 = 3 A = (p2 , p) = ((−1)2 , −1) = (1, −1)

p=3 q|p=3 = −1 A = (p2 , p) = ((3)2 , 3) = (9,3) ✓

B = (q2 , q) = ((3)2 , 3) = (9,3)

B = (q2 , q) = ((−1)2 , −1) = (1, −1) ✓

Midpoint of AB MAB = (

(−2−√2)+(−2+√2) (5+2√2)+(5−2√2)

,

2

2

)

= (−2,5) ✓ 9

B(−2,5) C(c1 , c2 )

A(3,7) MAC = (

(3)+(c1 ) (7)+(c2 ) 2

,

2

)

MAC lies on x-axis (y = 0): 7+c2 2

c2

=0 = −7

MBC = (

(−2)+(c1 ) (5)+(c2 ) ) , 2 2

MBC lies on y − axis (x = 0): −2+c1 2

c1

=0 =2

∴ C(2, −7) ✓


sleightofmath.com

144


Ex 6.1

A(6,8) B(8, −4) C(−2, −2) Point M M = MAB = (

6+8 8+(−4) 2

,

2

12(i)

O(0,0) P(4, r) Q(q1 , q2 ) R(3,4) For rhombus OPQR: |OP| = |OR|

) = (7,2) ✓

√(4 − 0)2 + (r − 0)2 = √(3 − 0)2 + (4 − 0)2

Point N N = MBC = ( Line MN Points:

8+(−2) (−4)+(−2)

,

2

2

M(7,2) or N(3, −3) (2)−(−3)

Gradient: mMN =

=

(7)−(3)

= √9 + 16 √16 + r 2 r2 =9 r = 3 ✓ or r = −3 (NA)

) = (3, −3) ✓

12(ii) O(0,0) P(4,3) Q(q1 , q2 ) R(3,4)

5 4

MPR = (

y − y1 = mMN (x − x1 )

MN:

5

35

4

4

+2

12(iii) O(0,0) P(4,3) Q(q1 , q2 ) R(3,4) MOQ

Point P 5

27

4

4

At P, MN (y = x − y 5

x−

4 5

4

=

x

=

⇒ P(

27 5

) cuts x-axis (y = 0).

27 4 27

2

q2

7

= 2 =7

For parallelogram PQRS, MPR = MQS

27 5

(

M = MAC = (

1+6 3+2

2 1+6

, 0) N(3, −3)

2

A(3, −2) B(b, 3) C(6,2) 3+6 (−2)+2 2

,

2

,

=

) =(

2 3+s1 2

3+s1 5+s2

and

s1 = 4 ∴ S(4,0) ✓

2

,

2 3+2 2

s2

) 5+s

2 = 2 =0

D(7, d) 9

) = ( , 0) ✓ 2

=M

b+7 3+d

2

and

2 2 0+q2

13(a) P(1,3) Q(3,5) R(6,2) S(s1 , s2 )

By similar triangles (using x-coordinates) MP: PN = (2 − 0) : [0 − (−3)] = 2: 3 ✓

2 b+7

2

7 7

)=( , )

5

M(7,2) P (

(

=

2 7

q1 = 7 ∴ Q(7,7) ✓

, 0) ✓

11(ii) MBD

,

2

10(ii) Ratio 𝐌𝐏: 𝐏𝐍

11(i)

= MPR

0+q1 0+q2

2 0+q1

=0

x

4

(

=0 27

)

2

2 2

5 27 x− 4 4

=

,

7 7

4

= x−

2

=( , )✓

5

y − (2)= ( ) [x − (7)] y

4+3 4+3

,

=

2 9 2

9

) = ( , 0) 2

and

b =2✓

3+d 2

=0

d = −3 ✓


sleightofmath.com

145


Ex 6.1

13(b) P(1,3) Q(3,5) R(6,2) S(4,0) T(t1 , t 2 )

15(i)

T(t1 , t 2 ) lies on y = 2x: t 2 = 2t1 ⇒ T(t1 , 2t1 ) PT

M = MPR = ( N = MQS = ( 15(ii) M

= √(3 − t1

)2

+ [5 − (2t1

)]2

(

5(t1 )2 − 14t1 + 10 12t1 t1

2

= 5(t1 )2 − 26t1 + 34 = 24 =2

r

r−4 7

6+0 (−2)+s

,

2

2

s−2

) = (3,

2

, )✓

2

2

)✓

2

) =(

,

=

2 0+d1

(−1)+d1 6+d2

=

,

2

2 6+1 2

) =(

2 0+3 2

(−1)+0 6+0

,

=

2

) 7

and

2

= 10 ✓

s−2

=

2

s =9✓

𝑦

=

2

0+b a+0

,

2

b a

) =( , )

2

2 2

b 2

a 2

2

2

=√

b2 4

+

a2 4

0+d2 2 b 2

a 2

2

2

b 2

a 2

2

2

|AM| = √(0 − ) + (a − )

=√

|BM| = √(b − ) + (0 − )

=√

b2 4

b2 4

+

+

a2 4

a2 4

0+3 0+1

and

) =(

2 3+d1

𝑥

|OM| = √(0 − ) + (0 − )

)

2

,

2 6+d2

) =

2

d2 d1 =4 ∴ D(4, −5) ✓ Case 3 For parallelogram ACBD, MAB = MCD 2 (−1)+0

s−2

𝑂 B(b, 0)

d1 =2 ∴ D(2,7) ✓ Case 2 For parallelogram ABDC, MAD = MBC

2

=3

M = MAB = (

0+d1 0+d2

and

2

,

2

16

d2 = 7

2 (−1)+d1

, ) = (3,

D(d1 , d2 )

C(3,1)

(−1)+3 6+1

2 (−1)+3

=N

PQRS is a parallelogram ✓

Case 1 For parallelogram ABCD, MAC = MBD

2

)=(

2

A(0, a)

A(−1,6) B(0,0)

(

,

2

S(0, s)

15(iii) ∵ MPR = MQS ,

⇒ T(2,4) ✓

(

R(r, 5)

(−4)+r 2+5

r−4 7

2 r−4

(t1 )2 − 2t1 + 1 (t1 )2 − 6t1 + 9 = + 4(t1 )2 − 12t1 + 9 +4(t1 )2 − 20t1 + 25

(

Q(6, −2)

= QT

(1 − t1 )2 √ +[3 − (2t1 )]2

14

P(−4,2)

and

d1 = −4 ∴ D(−4,5) ✓

,

17(i)

O(0,0)

P(2a, 0) Q(2b, 2c)

A = MOP = (

(0)+(2a) (0)+(0)

,

2

2

R(2d, 2e)

)

= (a, 0) [shown] ✓

2 6+0 2

2

= −5

3+d1 1+d2 2

∵ |OM| = |AM| = |BM|, M is equidistant from the three vertices ✓

0+1

)

=

B = MPQ

1+d2 2

=(

(2a)+(2b) (0)+(2c)

,

2

2

)

= (a + b, c) [shown] ✓

d2 = 5

C = MQR = (

(2b)+(2d) (2c)+(2e)

,

2

2

)

= (b + d, c + e) [shown] ✓ D = MRO = (

(2d)+(0) (2e)+(0) 2

,

2

)

= (d, e) [shown] ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

146


Ex 6.1

17(ii) A(a, 0) B(a + b, c) C(b + d, c + e) D(d, e) MAC = ( MBD = (

a+(b+d) 0+(c+e)

,

2 2 (a+b)+d c+e

,

2

2

)= (

)= (

a+b+d c+e

,

2 2 a+b+d c+e 2

,

2

19 O(0,0) B(2,4) (a)(ii) Let A be (x, y)

) OA

)

√ ∵ MAC = MBD , ABCD is a parallelogram ✓ 18

𝑦 C(0, n)

B(m, n)

O

A(m, 0)

𝑥

MAC = (

0+m 0+n

,

2 2 m+0 0+n 2

,

2

=√

x2 + y2 4x x

= x 2 + y 2 − 4x − 8y + 20 = −8y + 20 = −2y + 5 −(1)

m n 2 2 m n

√

) =( , ) 2 2

∵ MOB = MAC , diagonals of rectangle OABC bisect each other ✓

= OM 1)2

+ (y −

(x 2 − 2x + 1) +(y 2 − 4y + 4)

2)2

= √(0 − 1)2 + (0 − 2)2 = √12 + 22

(x 2 − 2x) + (y 2 − 4y) + 5 (x 2 − 2x) + (y 2 − 4y)

=5 =0

−(2)

sub (1) into (2): (−2y + 5)2 − 2(−2y + 5) +y 2 − 4y = 0 (4y 2 − 20y + 25) + (4y − 10) +y 2 − 4y = 0 (4y 2 − 16y + 15) +y 2 − 4y = 0 5y 2 − 20y + 15 = 0 y 2 − 4y + 3 =0 (y − 3)(y − 1) =0 y=3 or y=1 x = −2(3) + 5 x = −2(1) + 5 = −1 =3 A(−1,3) or A(3,1) C(3,1) C(−1,3)

19 As OABC is a square, 0+2 0+4 (a)(i) M ) = (1,2) ✓ , AC = MOB = (


(x 2 − 4x + 4) +y 2 − 8y + 16

√x 2 + y 2

AM

) =( , )

2

= AB

(x − 0)2 = √(x − 2)2 + (y − 4)2 +(y − 0)2

√(x − MOB = (

M(1,2)

2

sleightofmath.com

147

A math 360 sol (unofficial) 19(a) (ii)

Ex 6.1 19 (a) (iii)

𝑦 𝐵(2,4) 𝐴

𝐶

𝑥

𝑂

19(b) M(1,2) ⇒ (m + 1, m + 2) ✓ A(−1,3) ⇒ (m − 1, m + 3) ✓ C(3,1) ⇒ (m + 3, m + 1) ✓

⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐎𝐁 OB⊥ ≡ ⊥ bisector of OB Point: MOB (1,2) Gradient: mOB⊥ =

−1

=

mOB

−1

=−

4−0 2−0

20

1 2

y − y1 = mOB⊥ (x − x1 )

OB⊥ :

y − (2)= (− ) (x − 1)

(

2

1

2 a+b

2 1

2 5

2

2

2

=− x+

2

5 2

1

5

2

2

⇒ A (a1 , − a1 + ) |AM|

= |OM|

√(a1 − 1)2 + [(− 1 a1 + 5) − 2] 2

2

1

5

2

2

(a1 − 1)2 + [(− a1 + ) − 2] 1

1 2

2

2

2

(a1 − 1)2 + (− a1 + )

1

1

1

4

2 5

4

5

5 a − a1 + 4 1 2 4 5 5 15 (a1 )2 − a1 − 4 2 4 5(a1 )2 − 10a1 − 15 (a1 )2 − 2a1 − 3

(a1 + 1)(a1 − 3) a1 = −1 or 1

5

2

2

a2 = − (−1) + =3 ⇒ A(−1,3) ⇒ C(3,1)


2

) = (2,6)

=2

2

=√

(0 − 1)2 +(0 − 2)2

=5

and

2a+b+3 2

=6

2a + b + 3 = 12 2a + b =9 b = 9 − 2a −(2) sub (1) into (2): b = 9 − 2(4 − b) b = 9 − 8 + 2b b = 1 + 2b −b = 1 b = −1 a|b=−1 = 4 − (−1) =5 ⇒ A(5,10) ⇒ B(−1,2)

=5

(a1 )2 − 2a1 + 1 + (a1 )2 − a1 +

,

a + b= 4 a =4−b −(1)

Point A A(a1 , a2 ) lies on OB⊥ : 1

= (2,6)

a+b 2a+b+3

1

y−2 =− x+

a2 = − a1 +

A lies on y = 2x ⇒ A(a, 2a) B lies on y = x + 3 ⇒ B(b, b + 3) MAB

1

y

M(1,2) ⇒ (4,5) ✓ A(−1,3) ⇒ (2,6) ✓ C(3,1) ⇒ (6,4) ✓

=5 =5 =0 =0 =0 =0 a1 = 3 1

5

2

2

a2 = − (3) + =1 ⇒ A(3,1) ✓ ⇒ C(−1,3) ✓

sleightofmath.com

148


Ex 6.2 3

Ex 6.2

P(−1,2) Q(5,0)

R(7,4)

S(1,6)

𝑆 1(a)

A(−3,5) B(2,7)

𝑄

(5)−(7)

mAB = (−3)−(2) tan θ =

1(b)

(2)−(0)

mPQ = (−1)−(5) = −

2

(6)−(4)

5

mSR = (1)−(7) = −

A(4, −3) B(4,6)

mPS = (−1)−(1) = 2 (0)−(4)

(4)−(4)

∵ mPQ = mRS and mPS = mQR , P, Q, R and S are vertices of a parallelogram ✓

A(5, −4) B(7, −4)

A(1,0)

4

(1)−(−5)

B(4,2)

(0)−(2) 6

C(0, −3) D(3, −1)

−2

−3 c

2 3

(−3)−(−1) (0)−(3)

=

B(2,7)

(1)−(c)

= (0)−(−1) =

1−c 1

=1−c =4✓

2 3

5(a)

Point: A(−1,3) Gradient: ∵ line ∥ y = 4x − 1, m=4 Line: y − y1 = m (x − x1 ) y − (3)= (4)[x − (−1)] y − 3 = 4x + 4 y = 4x + 7 ✓

5(b)

Point:

∵ mAB = mCD , AB ∥ CD ✓ A(1,5)

B(2, −5) C(−1, c)

A,B,C lie on same straight line, mAB = mAC

5−7

(0)−(2)

2(b)

A(0,1)

(−4)−(−4)

mAB = (1)−(4) = mCD =

3

mQR = (5)−(7) = 2

(−3)−(6)

tan θ = 0 θ = 0° ✓ 2(a)

3

(2)−(6)

tan θ → ∞ θ = 90° ✓

mAB =

1

1

θ ≈ 21.8° ✓

mAB =

1(c)

𝑅

𝑃

C(0,4)

D(1,3)

(5)−(7)

mAB = (1)−(2) = 2 (4)−(3)

A(0,1)

mCD = (0)−(1) = −1 Gradient: ∵ line ∥ 2x + y = 3 y = −2x + 3 m = −2

∵ mAB ≠ mCD , AB is not ∥ CD ✓

Line:


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y − y1 = m (x − x1 ) y − (1)= (−2)[x − (0)] y − 1 = −2x y = −2x + 1 ✓

149


Ex 6.2

Point:

A(−2,0)

7

Gradient:

∵ line ∥ x + 3y = 12 3y = 12 − x

Line:

b

1

y m=−

=− x+4

2nd line dx + ey + f = 0 ey = −dx − f

3

= m(x − x1 )

y

1

y − (0)= − [x − (−2)]

5(d)

1

2

3

3

=− x− ✓

a

− =− b

A(−2,0)

Gradient:

∵ line ∥ BC, m = mBC =

Line:

6(a)

Point:

a

8 3−6

=

−6 −3

e

d e

[shown] ✓

P(k 2 , 3k) Q(k, k − 2) R(k, k + 2)S(1,1) For PQ ∥ RS: mPQ

y − y1 = m (x − x1 ) y − (1)= (2)[x − (3)] y − 1 = 2x − 6 y = 2x − 5 ✓

= mRS

3k−(k−2)

=

k2 −k 2k+2

=

k2 −k 2(k+1)

(1,2)

=

k(k−1) k+1 2

( − 1)

k+1 2−k

y − y1 = m(x − x1 ) y − (2)= (4)[x − (1)] y − 2 = 4x − 4 y = 4x − 2 ✓

(

k

(k+2)−1 k−1 k+1 k−1 k+1 k−1

=0

k−1 k k−1

)

=0

(k + 1)(2 − k) = 0 k = −1 or k = 2 ✓ 9

6(b)

e

=2

Gradient: ∵ line ∥ y = 4x − 3, m=4 Line:

d

=

b

(−2)−4

f

e

Lines are parallel: m1 = m2

A(3,1) B(3, −2) C(6,4) Point:

d

=− x−

3

y

b

3

1

y − y1

1st line ax + by + c = 0 by = −ax − cy a c y =− x−

P(a + b, a) Q(a − b, 2a)

R(b, c)

y − intercept: c = 3 Gradient: ∵ line ∥ y = 3x + 4, m=3

P, Q, R are collinear points, mPQ = mQR

Line:

(a+b)−(a−b)

(a)−(2a)

y = mx + c = 3x + 3 ✓

−a 2b 2ab−a2 2b


sleightofmath.com

(2a)−(c)

= (a−b)−(b) =

2a−c a−2b

= 2a − c 2ab−a2

c

= 2a −

c

= 2a − a +

c

=a+

a2 2b

2b a2 2b

✓

150


Ex 6.2

Gradient of AP: 3−(−5) (−2)−p

mAP =

=

11(ii) Line Point:

8

Gradient: ∵ Line ∥ 3x + 2y − 6 = 0 2y = −3x − 6

Line & its gradient: 4x + 3y − 5 = 0 3y = −4x + 5 y

A(−1, −1)

−2−p

4

5

3

3

=− x+

3

y ⇒m=−

=− x−3 2

3 2

4

⇒ mline = −

3

Line:

y − y1

= m (x − x1 ) 3

y − (−1) = (− ) [x − (−1)]

∵ AP ∥ line mAP = mline 8 −2−p

24 p

=−

2

= 8 + 4p =4✓

10(ii) Line AP Point:

12(i) A(−2,3) or P(4, −5) 8

mAP =

AP:

y − y1 = mAP (x − x1 )

−2−(4)

=−

4

Gradient:

4 3

y−3 =−

4 3

(x + 2)

4

8

3

3

4

1

3

3

y−3 =− x− y

=− x+ ✓

Point A At A, x + y + 2 = 0 intersects 3x − 2y + 1 = 0: x+y+2 =0 y = −x − 2 −(1) 3x − 2y + 1 = 0

= − (x + 1)

y+1

=− x−

y

=− x− ✓

2 3

3

2 3

2 5

2

2

A(3, −1) or P

Gradient:

∵ AP ∥ y = 2x + 3, mAP = 2

AP:

y − y1 = mAP (x − x1 ) y − (−1) = 2 (x − 3) y+1 = 2x − 6 y = 2x − 7 ✓

12(ii) Point P At P, AP (y = 2x − 7) intersects y = 3x − 11. 2x − 7 = 3x − 11 −x = −4 x =4 y|x=4 = 2(4) − 7 =1 ⇒ P(4,1) ✓

−(2) 13(i)

sub (1) into (2): 3x − 2(−x − 2) + 1 = 0 3x + 2x + 4 + 1 =0 5x = −1 x = −1 −(3) y|x=−1 = −(−1) − 2 = −1 ⇒ A(−1, −1) ✓


Line AP Point:

3

y − (3)= (− ) [x − (−2)]

11(i)

y+1

4 3

3

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Point B B = (6 − 3,2) = (3,2) ✓

151

A math 360 sol (unofficial) 13(ii) Line BC Point: Gradient:

Ex 6.2

B(3,2) or C ∵ BC ∥ 2y + x = 0 2y = −x

BC:

C or D(0,5)

Gradient:

∵ CD ∥

=− x 2

1

y

2

1

mCD =

y − y1 = mBC

1

3

2

2

1

7

2

2

y−2 =− x+

y − 5 = (x − 0) 2 1

= x+5✓

y

2

=− x+ ✓ Point C 1

1

7

2

2

2

At C, CD (y = x + 5) meets BC (y = − x + )

A(6,2)

1 2

∵ AD ∥ BC,

1

7

2 3

2

x+5=− x+

x

mAD = mBC = − AD:

2

1

2

Gradient:

2

y − y1 = mCD (x − x1 )

CD:

1

13(iii) Line AD Point:

1

= x

(x − x1 )

y − (2)= − (x − 3)

y

2y − x = 0 2y =x

1

y mBC = −

13(iv) Line CD Point:

=−

1

2

2 1

3

2 17

2

y|x=−3 = (− ) + 5

y − y1 = mAD (x − x1 )

2

=

1

4 3 17

y − (2)= (− ) (x − 6) 2

⇒ C (− , 2

1

y−2 =− x+3

4

)✓

2

y

1

=− x+5✓ 2

Point D At D, AD cuts y − axis (x = 0): y|x=0 = 5 ⇒ D(0,5) ✓


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152


Ex 6.3 4(ii)

Ex 6.3 1

A(3,7) B(6,1) C(20,8) (7)−(1)

6

mAB = (3)−(6) =

−3

(1)−(8)

mBC = (6)−(20) =

=

−14

1 2

−5

3

) (− )

= −1 = −1

2t − t 2 = −15 t 2 − 2t − 15 = 0 (t − 5)(t + 3) = 0 t = 5 or t = −3 ✓

2

∴ AB ⊥ BC ✓ A(2, −1) B(5,4) C(15, −2) 5 mAB = mBC =

(−1)−(4)

−5

=

(2)−(5) (4)−(−2)

−3 6

=

(5)−(15)

5

=

3

=−

−10

5

3

3

5

3

mBC =

a−2 (−3)−1 2−10

= =

1

6

(− ) (k)= −1 2

a−2 −4

1

−8

2

=

k

AB ⊥ BC: (mAB )(mBC ) = −1 (

6

1

)( )

a−2 6

2

6

= −1 = −1

2a−4

6 2a a

mBC =

2−0 1−9 0−t 9−6

= =

2 −8 −t 3

=− =−

t

4

3

1st line & its gradient y = ax + b m1 = a

(1, −2) lies on y = ax + b −2 = a + b −(1)

1 4 t

lines are ⊥: m1 ⋅ m2 = −1 a ⋅ (3) = −1

3

∡ABC = 90°: (mAB )(mBC ) = −1 1

=2✓

2nd line & its gradient y − 3x = 4 y = 3x + 4 m2 = 3

= 4 − 2a = −2 = −1 ✓

A(1,2) B(9,0) C(6, t) mAB =

2

Lines are ⊥: m1 m2 = −1

A(a, 3) B(2, −3) C(10,1) 3−(−3)

1

=− x+1

2nd line y − kx + 4 = 0 y = kx − 4

∴ AB ⊥ BC ∡ABC = 90° ✓

mAB =

1st Line x + 2y − 2 = 0 2y = −x + 2 y

5

(mAB ) (mBC ) = ( ) (− ) = −1

4(i)

t

−5 2t−t2 15

(mAB )( mBC ) = (−2) ( ) = −1

3

2−t

(

1

2

2−t

AC ⊥ BC: (mAC )(mBC ) = −1

= −2 −7

(2)−(t)

mAC = (1)−(6) =

a

1

= − ✓ −(2) 3

(− ) (− ) = −1 t 12

t

sub (2) into (1):

= −1

1

−2 = (− ) + b

= −12 ✓

3

5

b =− ✓ 3


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153


Ex 6.3

(4,5)

Point:

10

Gradient: ∵ Line ⊥ x + 2y − 4 = 0 2y = −x + 4 1

y ⇒m=

1 (− ) 2

=− x+2 2

y 2 = 6x − 32

=2

Point A At A, y = x − 1 intersects y = x 2 − x x−1 = x2 − x x 2 − 2x + 1 = 0 (x − 1)2 =0 x =1

⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB

y|x=1 = (1) − 1 =0 ⇒ A(1,0) Line Point: A(1,0) Gradient: ∵ line ⊥ y = x − 1

9

11(i)

A(3,3) B(7,3) AB⊥ ≡ ⊥ bisector of AB AB⊥ :

x=

3+7

Gradient:

mAB⊥ =

AB⊥ :

y − y1 = mAB⊥ (x − x1 ) y − (−3) = 1(x − 11) y+3 = x − 11 y = x − 14 ✓

2

,

2 −1 mAB

2

=

=1

A(3,6) or F

Gradient:

∵ F is foot of ⊥ from A to BC,

AF:

−1 mBC

=

−1 (−1)−(7) ( (2)−(6) )

=−

1 2

y − y1 = mAF (x − x1 ) 1

y − 6 = − (𝑥 − 3) 2 1

3

2 1

2 15

2

2

y−6 =− x+ y


−1 (2)−(−8)

( (6)−(16) )

Line AF Point:

mAF =

=5✓

) = (11, −3)

MAB = (

1

y − y1 = m (x − x1 ) y − 0 = −1(x − 1) y = −x + 1 ✓

6+16 2+(−8)

Point:

m = − (1) = −1 Line:

−(2)

sub (1) into (2): (8 − x)2 = 6x − 32 x 2 − 16x + 64 = 6x − 32 x 2 − 22x + 96 = 0 (x − 6)(x − 16) = 0 x=6 or x = 16 y = 8 − 6 y = 8 − 16 =2 = −8 ⇒ A(6,2) ⇒ B(16, −8)

y − y1 = m(x − x1 ) y − 5 = 2 (x − 4) y = 2x − 3 ✓

Line:

8

−1

Points A & B At A & B, x + y = 8 meets y 2 = 6x − 32: x + y= 8 y = 8 − x −(1)

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=− x+

✓

154

A math 360 sol (unofficial) 11(ii) Line BC Point:

Ex 6.3

B(2, −1) or C(6,7) (−1)−7

Gradient:

mBC =

BC:

y − y1 = mBC (x − x1 ) y − (−1) = 2 [x − (2)] y+1 = 2x − 4 y = 2x − 5

2−6

=2

12(ii) Point P At P, AP (y = −4x + 13) cuts x − axis (y = 0). y =0 −4x + 13 = 0 x ⇒ P(

1

15

2

2

).

1

15

2

2

2x − 5 = − x + 5 2

x

x

=

13 4

P(

2

=5

, 0) ✓

13 4

, 0) A(3,1)

PA: AQ = (

11(iii) Perpendicular distance AF A(3,6) F(5,5) 13(i)

= √4 + 1 = √5 units ✓

Line AB Point:

AB: A(3,1) or P

y

y − y1 y−1 y−1 y

4

− 3) : (3 − 0)

1

:3

4

A(4,13) or B(9,3) (13)−(3)

Gradient: mAB =

Gradient: ∵ AP ⊥ x − 4y = 8 −4y = −x + 8

AP:

13

= 1: 12 ✓

|AF| = √(3 − 5)2 + (6 − 5)2

mAP =

Q(0,13)

By similar triangles (using x-coordinates) [diagram?]

=

Line AP Point:

4

12(iii) Ratio 𝐏𝐀: 𝐀𝐐

25

y|x=5 = 2(5) − 5 =5 ⇒ F(5,5) ✓

12(i)

13

Point Q At Q, AP(y = −4x + 13) cuts y-axis (x = 0). y|x=0 = 13 ⇒ Q(0,13) ✓

Point F At F, BC (y = 2x − 5) intersects AF (y = − x +

=

−1 1 4

( )

1

= x−2 4

= −4

13(ii) Line Point:

(4)−(9)

=

= −2

= mAB (x − x1 ) = −2(x − 4) = −2x + 8 = −2x + 21 ✓

y − y1 y − 13 y − 13 y

C(10,8)

Gradient: ∵ Line ⊥ y − 4x = 5 y = 4x + 5

= mAP (x − x1 ) = −4(x − 3) = −4x + 12 = −4x + 13 ✓

m=− Line:

1 4

y − y1 = m (x − x1 ) 1

y − 8 = − (x − 10) 4 1

5

4 1

2 21

4

2

y−8 =− x+ y


10 −5

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=− x+

✓

155


Ex 6.3

13(iii) Point P 1

21

4

2

At P, y = − x + 1

21

4

2

− x+ 7 4

meets AB (y = −2x + 21):

= −2x + 21

x

=

x

14(iii) Ratio 𝐀𝐌: 𝐌𝐃 A(3,1) M(4,3) D(−2,6) AM: MD = √(3 − 4)2 + (1 − 3)2

21

: √[4 − (−2)]2 + (3 − 6)2

2

=6

= √5 : √45

y|x=6 = −2(6) + 21 =9 ⇒ P(6,9)

= √5 : √9 × 5 = √5 : 3√5 =1

14(i)

Line DM Point: Gradient:

D(−2,6) ∵ DM ⊥ AB (y = 2x − 5) −1

mDM =

=−

mAB

15(i)

2

AB⊥ ≡ ⊥ bisector of AB

1

y − 6 = − [x − (−2)] 2 1

y−6 =− x−1 2 1

=− x+5✓

y

⊥ bisector of AB A(5,4) B(3, −2)

1

y − y1 = mDM (x − x1 )

DM:

:3✓

2

5+3 4+(−2)

) = (4,1)

Point:

MAB = (

Gradient:

mAB⊥ =

AB⊥ :

y − y1 = mAB⊥ (x − x1 )

2

,

−1 mAB

2

=

−1 (4)−(−2)

( (5)−(3) )

=

−1 6 2

( )

=−

1 3

1

y − 1 = − (x − 4)

14(ii) Point M 1

3 1

4

3 1

3 7

3

3

At M, DM (y = − x + 5) intersects

y−1 =− x+

AB (y = 2x − 5).

y

2

=− x+ ✓

1

− x + 5 = 2x − 5 2 5

− x

15(ii) Point P

= −10

2

x

1

7

3

3

At P, AB⊥ (y = − x + ) intersects y = x + 5.

=4

1

7

3 4

3

− x+ =x+5 1

8

y|x=4 = − (4) − 5

− x

=

=3 ⇒ M(4,3) ✓

x

= −2

2

3

y|x=−2 = (−2) + 5 =3 ⇒ P(−2,3) ✓

Point B Let B be (b1 , b2 ) M

= MAB

(4,3) = ( 4 =

3

3+b1 1+b2 2

3+b1 2

b1 = 5 ∴ B(5,5) ✓

,

2

and

) 3 =

1+b2 2

b2 = 5


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156


Ex 6.3

⊥ bisector of AB A(5,2) B(3,6) AB⊥ ≡ ⊥ bisector of AB MAB = (

Point:

−1

18(i)

2

=

,

) = (4,4)

2 −1

m=

AB⊥ :

y − y1 = m (x − x1 )

(2)−(6)

((5)−(3))

=

−1 −4 ) 2

(

=

2

1

Point C At C, BC(y = 5x + 6) cuts y − axis (x = 0). y|x=0 = 6 ⇒ C(0,6)

2 1

y−4 = x−2 2 1

= x+2✓ 2

16(ii) Point P At P, AB⊥ cuts x-axis (y = 0) 1 2

Point Q At P, AB⊥ cuts y-axis (x = 0)

18(ii) Line AC Point:

∵ BD is ⊥ bisector of AC, mAC =

2

x = −4 ⇒ P(−4,0) ✓

A or C(0,6)

Gradient:

1

y|x=0 = (0) + 2

x+2=0

y − y1 = mBC (x − x1 ) y − (1)= (5) [x − (−1)] y − 1 = 5x + 5 y = 5x + 6

BC:

1

y − 4 = (x − 4)

y

B(−1,1) or C ∵ BC ∥ y = 5x mBC = 5

5+3 2+6

Gradient:

mAB

Line BC Point: Gradient:

=2 ⇒ Q(0,2) ✓

AC:

−1 mBD

=

−1 (1)−(7)

((−1)−(8))

=−

3 2

y − y1 = mAC (x − x1 ) 3

y − 6 = − (x − 0)

17(a) ⊥ bisector of PQ P(3,5) Q(5,9) PQ ⊥ ≡ ⊥ bisector of PQ

2 3

y−6 =− x

3+5 5+9

MPQ = (

Gradient:

m=

PQ ⊥ :

y − y1 = m (x − x1 )

−1 mPQ

=

,

2 −1

(5)−(9) ((3)−(5))

=−

1

18(iii) Point A 1

2

At A, AC (y = − x + 6) cuts x − axis (y = 0): 3

y

=0 3

1

y − 7 = − (x − 4)

− x + 6= 0

y−7 =− x+2

− x

2 3

2 1

y

2

) = (4,7)

Point:

2

2 3

=− x+6✓

y

2

2 1

= −6

x =4 ⇒ A(4,0) ✓

=− x+9✓ 2

17(b) Point At point where y = 6x intersects

Midpoint of AC A(4,0) C(0,6)

1

PQ ⊥ (y = − x + 9). 2

M = MAC = (

1

6x = − x + 9

4+0 0+6 2

,

2

) = (2,3) ✓

2

13 2

x

x= 9 =

18 13 18

y|x=18 = 6 ( ) 13

=

13 108

⇒ Point (

13 18 108 13

,

13

)✓


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157


Ex 6.3

18(iv) Area of quadrilateral 𝐀𝐁𝐂𝐃 A(4,0) B(−1,1) C(0,6) D(8,7)

20(i)

|AC| = √(4 − 0)2 + (0 − 6)2 = √52 = √4 × 13 = 2√13

Line AB Point:

A(1, −1) or B(5,3)

Gradient:

mAB =

AB:

y − y1 = mAB (x − x1 ) y − (−1) = (1) [x − (1)] y+1 =x−1 y =x−2✓

|BD| = √[(−1) − 8]2 + (1 − 7)2 = √117 = √9 × 13 = 3√13

M = MAB = ( 20(ii) Line PQ Point:

Line CD A(2,3) B(6,7) C(7, t) MAB = (

Point:

2+6 3+7 2

,

2

) = (4,5)

−1

CD:

y − y1 y−5 y−5 y

mAB

1

1

(2)−(6)

−4

= − (3)−(7) = − −4

= mCD (x − x1 ) = (−1)[x − (4)] = −x + 4 = −x + 9 ✓

(1)+(5) (−1)+(3) 2

−4 −4

= −1

,

2

) = (3,1) ✓

Gradient:

mPQ =

PQ:

y − y1 y−1 y−1 y

−1 mAB

=

−1 1

= −1

= mPQ (x − x1 ) = (−1)(x − 3) = −x + 3 = −x + 4 ✓ Value of q Q(7, q) lies on PQ(y = −x + 4), (q) = −(7) + 4 q = −3 ✓

20(iii) Area of quadrilateral APBQ A(1, −1) P(0.5,3.5) B(5,3) Q(7, −3)

19(ii) Value of t C(7, t) lies on y = −x + 9: (t) = −(7) + 9 t =2✓

|AB| = √(1 − 5)2 + [(−1) − 3]2

19(iii) Point D C(7,2) D(d1 , d2 )

area of APBQ = |AB||PQ|

= √32 = √16 × 2 = 4√2 169

|PQ| = √(0.5 − 7)2 + [3.5 − (−3)]2 = √

4 =

7+d1 2+d2 2

7+d1 2

,

2

2

=

13 √2

1 2 1

13

2

√2

= (4√2) ( )

AB is ⊥ bisector of CD, MAB = MCD (4,5) = (

=1

M(3,1) or P(p, 3.5) or Q(7, q)

Value of p P(p, 3.5) lies on PQ(y = −x + 4), (3.5) = −(p) + 4 p = 0.5 ✓

Gradient: ∵ CD is ⊥ bisector of AB, mCD =

=

(1)−(5)

Midpoint of AB

Area of ABCD 1 = |AC||BD| 2 1 = (2√13)(3√13) 2 = 3(13) = 39 unit 2 ✓ 19(i)

(−1)−(3)

= 26 ✓

)

and

d1 = 1

5 =

2+d2 2

d2 = 8

∴ D(1,8) ✓


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158


Ex 6.3

⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 A(−4,3) B(8, −3) AB⊥ ≡ perpendicular bisector of AB

21(iii) C(p, q) lies on AB⊥ : q = 2p − 4 ⇒ C(p, 2p − 4)

C A(−4,3)

MAB B(8, −3) C

Point:

MAB = (

(−4)+(8) (3)+(−3)

Gradient: mAB⊥ = − =− AB⊥ :

,

2 1 mAB 1 6 −12

=−

2

) = (2,0)

MAB = (

−4+8 3+(−3)

,

2

1 (3)−(−3)

|CMAB | = √(p − 2)2 + [(2p − 4) − 0]2

((−4)−(8))

= √(p2 − 4p + 4) + (4p2 − 16p + 16)

=2

= √5p2 − 20p + 20

y − y1 = mAB⊥ (x − x1 ) y − 0 = 2 (x − 2) y = 2x − 4 ✓

|AB| = √[(−4) − 8]2 + [3 − (−3)]2 = √180 = √36 × 5 = 6√5

21(ii) Show If (10,16) lies on AB⊥ (y = 2x − 4), (16) = 2(10) − 4 16 = 16 [consistent] ∴ (10,16) lies on AB⊥

△ ABC area 1 2 1 2

=6

|CMAB ||AB|

3√5p2 − 20p + 20√5

=6

√5p2 − 20p + 20√5

=2

√25p2 − 100p + 100

=2

25p2 − 100p + 100

=4

25p2 − 100p + 96

=0

(5p − 8)(5p − 12)

=0

8

or

5

2p − 4 = − 8

4

5

5

4 5

⇒ C( ,− )

sleightofmath.com

=6

√5p2 − 20p + 20(6√5) = 6

p=


2

) = (2,0)

p=

12 5

2p − 4 = ⇒ C(

4 5

12 4 5

, )✓ 5

159


Ex 6.3

21(iii) Shoelace formula A C B C(p, q) lies on AB⊥ (y = 2x − 4) q = 2p − 4 ⇒ C(p, 2p − 4)

22

P(0,4)

−4 | | 2 3

2

−4 8 || 3 −3

p 2p − 4 p 2p − 4

Q

−4 || 3

Line PQ Pt: P(0,4) or Q

=6

Grad: mPQ =

−4 || 3

= 12

|30p − 60|

p

=

PQ:

= 12

p

5 12

⇒ C(

5

4

=−

5

12 4

8

4

5

5

5

5 4

1 2

(− )

=2

y − y1 = mPQ (x − x1 ) y − (4)= (2) [x − (0)] y = 2x + 4

Line PR Pt: P(0,4) or R

5

, ) or C ( , − ) ✓ 5

−1

y|x=−4 = −4 ⇒ Q(−4, −4) ✓

2p − 4 = 2 ( ) − 4

5

=

8

8

2p − 4 = 2 ( ) − 4 =

=

−1 ml3

Point Q At Q, PQ meets y = x 2x + 4 = x x = −4

=2 or 5p − 10 = −2 5p =8

12

R

l1 : x = 0

|12 + (16p − 32) + 3p = 12 −24 − (−3p) − (−8p + 16)| |5p − 10| 5p − 10 = 2 5p = 12

𝑥

𝑂 =6

8 −3

l2 : y = x

1

l3 : y = − x

△ ABC area 1

𝑦

Grad: mPR = −

1 ml2

1

= − (1) = −1

PR: y − y1 = mPR (x − x1 ) y − (4)= (−1)[x − (0)] y = −x + 4 Point R 1

At R, PR meets y = − x 2

1

−x + 4 =− x 2

1

− x

= −4

x

=8

2

1

y|x=8 = − (8) 2

= −4 ⇒ R(8, −4) ✓


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160


Ex 6.3

∵ AO = OC, (radius) △ AOC is isosceles △ ∵ BO = OC, (radius) △ BOC is isosceles △ ✓

A

23(ii) α = ∡AOC = ∡OCA β = ∡OBC = ∡OCB sum of ∡s in △

= 180°

23(iii) A(1,0) B(5,2) C(a, b)

C

A

O = MAB = ( B

O

C 𝛼 𝛽 𝛼 O

= 180°

2α + 2β

= 180°

2(α + β)

= 180°

α+β

= 90°

∡ACB

= 90°

∴AC ⊥ BC ✓


,

2

) = (3,1)

= radius

|OC|

= (diameter AB)

|OC|

= |AB|

1 2 1

3)2

+ (b −

1)2

2 1

= √(1 − 5)2 + (0 − 2)2 2 1

√(a − 3)2 + (b − 1)2 = √20 2

B

∡AOC + ∡OCA = 180° +∡OBC + ∡OCB α +α +β + β

2

|OC|

√(a − 𝛽

1+5 0+2

1

(a − 3)2 + (b − 1)2

= (20)

(a − 3)2 + (b − 1)2

= 5 [shown] ✓

4

Note: Alternate approach is to use right angle triangle in semicircle. By Pythagoras’ Theorem, |AC|2 + |BC|2 = |AB|2 . 23(iv) At (2,3), [(2) − 3]2 +[(3) − 1]2 = 5 (−1)2 +22 =5 1 +4 =5 5 = 5 [consistent] ∴ (2,3) lies on circle ✓

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161


Ex 6.4 3(i)

Ex 6.4 1(a)

A(2,3) B(5,6) C(−1,4)

△ ABC area

△ ABC area

=| |

1 2 = | 2 3

5 6

1 2

−1 4

2 | 3

2 1

= (12) 2

A(5,2) B(1,6) C(−2, −3) △ ABC area 1 −2 6 −3

A, B and C are collinear ✓

4(a)

O(0,0) A(4,1) B(6,4) C(−2,4)

5 | 2

Area of OABC =

1 2

2

2 4

−2 4

0 | 0

2

=

3 6 5 2

4(b) 2 || 4

1 = |10 + 6 + 24 2 1 = |−6| 2

1 (42) 2

= 21 unit 2 ✓

△ ABC area 1

4 6 1 4

= [0 + 16 + 24 + 0 −0 − 6 − (−8) − 0]

A(2,4) B(3,5) C(6,2)

=| |

1 0 | 2 0 1

1 (48) 2

= 24 unit 2 ✓ 2(a)

− 4 − 0 − (−10)|

3(ii)

= [30 + (−3) + (−4) −2 − (−12) − (−15)] =

−5 || −2

= 0 unit 2 ✓

= 6 unit 2 ✓

5 2

−2 1 0 2

1 |0 + (−4) + (−2) 2 1 = |0| 2

= [12 + 20 + (−3) −15 − (−6) − 8]

=|

−5 −2

=

1

1(b)

A(−5, −2) B(−2,0) C(1,2)

P(1,4) Q(−4,2) R(1, −2) S(4,0) Area of PQRS 1 1 −4 1 4 1 | | 2 4 2 −2 0 4 1 = [2 + 8 + 0 + 16 − (−16) − 2 − (−8) − 0] 2 1 = (48) 2 =

− 12 − 30 − 4|

= 3 unit 2 ✓

= 24 unit 2 ✓ 2(b)

A(−4, −2) B(−2,4) C(6,0) △ ABC area 1

=| | 2

−4 −2

−2 6 4 0

−4 || −2

1 |−16 + 0 + (−12) 2 1 = |−56| 2 =

− 4 − 24 − 0|

= 28 unit 2 ✓


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162


Ex 6.4

A(2, −3) B(3, −1) C(2,0) D(−1,1) E(−2, −1)

6(iii)

Area of △ ABC 1 2 3 2 2 | = | 2 −3 −1 0 −3 1 = [−2 + 0 + (−6) −(−9) − (−2) − 0] =

2 3

1 2 1 2

2 9

7(i)

unit 2 ✓

2

√5(5) sin ∡BAC

A(3,5)

1

| | 2

||

= 5 unit 2 ✓

9

2

2

= 11 unit 2 ✓ 𝑦

3 5

3 5

7(ii) 𝑥

𝑂

A

C(9, −1)

1 2

4 −1

B(−5,9) C(k, k + 2)

−5 9

= 18

k k+2

−5 9

k k+2

3 || 5

= 18

3 || 5

= 36

5 9 1 −1

= 36

F

B

CF ≡ perpendicular distance of C from AB Area of △ ABC = 18

Area of △ ABC =| |

✓

C

B(5,1)

A(4, −1)

2 √5

|−2k + 6| =6 −2k + 6 = 6 or −2k + 6 = −6 −2k =0 −2k = −12 k =0✓ k =6✓

Area of pentagon 3

=

|−12k + 36|

= + +5

1

4 || −1

2

|AB||CF|

|CF|

1

=

= [4 + (−5) + (−9) −(−5) − 9 − (−4)] 2 1

=

= |−10| 2

= 5 unit 2 ✓ 6(ii)

=5

|27 + (−5k − 10) + 5k = 36 −(−25) − 9k − (3k + 6)|

2

6(i)

(AB)(AC) sin ∡BAC = 5

Area of △ ABC

Area of △ ADE 1 2 −1 −2 2 | = | 2 −3 1 −1 −3 1 = [2 + 1 + 6 −3 − (−2) − (−2)]

5(ii)

=5

sin ∡BAC

Area of △ ACD 1 2 2 −1 2 | = | 2 −3 0 1 −3 1 = [0 + 2 + 3 −(−6) − 0 − 2] =

Area of △ ABC

unit 2 ✓

2

AC = 9 − 4 =5

8(i)

AB = √(4 − 5)2 + [(−1) − 1]2

= 18

36 |AB| 36 √80

= =

36 √[3−(−5)]2 +(5−9)2 36 √5×16

=

36 4√5

=

9 √5

9

= √5 ✓ 5

A(2, t) B(3 + t, 2) C(3,4) Area of △ ABC 1 2 3+t 3 2 | = | 2 t 2 4 t 1 = (4 + 12 + 4t + 3t − 3t − t 2 − 6 − 8)

= √1 + 4 = √5 units [shown] ✓

2 1

= (−t 2 + 4t + 2) unit 2 ✓ 2


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163


Ex 6.4

Area of △ ABC = 2 1

(−t 2 + 4t + 2) =

2

1

11(i)

2

5

Area of △ ABC 1 3 −4 6 3 | = | 2 4 2 −1 4 1 = [6 + 4 + 24 − (−16) − 12 − (−3)] 2 1 = (41) 2 = 20.5 unit 2 ✓

2

−t 2 + 4t + 2 =5 2 t − 4t + 3 =0 (t − 1)(t − 3) =0 t = 1 or t = 3 ✓ 9(i)

A(1,3) B(5,1) C(3, r) Area of △ ABC =4 1 1 5 3 1 | | =4 2 3 1 r 3 1 (1 + 5r + 9 − 15 − 3 − r) = 4 2 1

(4r − 8)

2

A(3,4) B(−4,2) C(6, −1) D(p, 3)

=4

2r − 4 2r r

=4 =8 =4✓

11(ii) Area of ABCD 1 3 −4 6 p 3 | = | 2 4 2 −1 3 4 1 = [6 + 4 + 18 + 4p − (−16) − 12 − (−p) − 9] 2 1 = (23 + 5p) unit 2 ✓ 2

11(iii) (Area of ABCD) = 3(Area of △ ABC) 1 2

9(ii)

Area of △ ACB =4 1 1 3 5 1 | | =4 2 3 r 1 3 1 (r + 3 + 15 − 9 − 5r − 1) = 4 2 1

(−4r + 8)

2

=4

−2r + 4 −2r r 10(i)

12(i)

2

=9 = −3x + 9

y

= x−

3

9

2

2

3 2

AD ⊥ line ⇒ mAD =

−1 mline

=

−1

= MBD

2+1 1+4

2 2+1

= 123 = 100 = 20 ✓

3x − 2y −2y mline =

A(2,1) B(b1 , b2 ) C(1,4) D(0,2)

(

= 3(20.5)

23 + 5p 5p p

=4 =0 =0✓

MAC

(23 + 5p)

,

=

) =(

2 b1 +0 2

b1 +0 b2 +2 2

,

and

b1 = 3 B(3,3) ✓

Line BC Point: Gradient:

)

2 1+4 2

=

b2 +2 2

b2 = 3

BC:

=−

2 3

B(4,7) or C ∵ BC ∥ AD mBC = mAD =

10(ii) Area of rhombus 1 2 3 1 0 2 | = | 2 1 3 4 2 1 1 = (6 + 12 + 2 + 0 − 3 − 3 − 0 − 4)

3 2

−1 3 ( ) 2

=−

2 3

y − y1 = mBC (x − x1 ) 2

y − (7)= (− ) (x − 4) 3

2

8

3 2

3 29

3

3

y−7 =− x+

2 1

= (10)

y

2

=− x+

✓

2

= 5 unit ✓


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164

A math 360 sol (unofficial) 12(ii) Line DE Point:

Ex 6.4 12(iv) Area of quadrilateral ABFD 1

D(−3,3) or E

A (1, ) 2

∵ DE ⊥ BC (y = − x +

Gradient:

3

−1

mDE =

=

mBC

−1 2 3

(− )

=

29 3

Area of ABFD 0 1 1 4 = | 1 7 29 2

3 2

3

3 2

3

2

2

y

3

15

2

2

= x+

= 13(i)

Point E 1

15

3 29

2

At E, DE (y = x + 2

(y = − x + 3

3

x+

2 13 6

15 2

x

6

=1 29

3

3

=9 ⇒ E(1,9) ✓

3

104

unit 2 ✓

3

𝐕𝐚𝐥𝐮𝐞 𝐨𝐟 𝐚 A(3a, 4a + 1), a > 0 B(0,1) =5 1]2

⇒ A(3,5) 13(ii) Line BC Point:

12(iii) Point F 2

At F, BC (y = − x + 3

2

29

3

3

y|x=0 = − (0) + =

)

2

√(3a − + [(4a + 1) − =5 2 2 9a + 16a = 25 2 25a = 25 a2 − 1 =0 (a + 1)(a − 1) =0 a = −1 (rej ∵ a > 0) or a = 1 ✓

3

2

3

0)2

13

y|x=1 = − (1) +

4

+ 0 + (−1) − − 0 − (−29) − 3]

|AB|

29

3

3

3

) intersects BC

)

=− x+ =

x

3 2

116

2

= (

2

9

1

−3 1 1| 3

1 208

y − 3 = (x + 3) 3

3

= [7 +

y − (3)= ( ) [x − (−3)]

y−3 = x+

3

)

y − y1 = mDE (x − x1 )

DE:

29

F (0, ) D(−3,3)

B(4,7)

3

29 3

B(0,1) or C

) cuts y-axis (x = 0) Gradient: ∵ BC ⊥ AB, mBC =

29 3 29

−1 mAB

1

1

(0)−(3)

4 3

= − (1)−(5) = −

=−

3 4

⇒ F (0, ) 3

y − y1 = mBC (x − x1 )

BC:

3

y − 1 = − (x − 0)

Point A A(a1 , a2 ) B(4,7)

29 3

(

2

a1 +0 2

a1

,

=

2

)=(

4+(−3) 2

3

At C, BC (y = − x + 1) cuts x-axis (y = 0) 4

3

4+(−3) 7+3

and

=1

,

2

2

29 a2 +( ) 3

2

a2

4

Point C

For Parallelogram ABFD, MAF = MBD 29

=− x+1✓

y

F (0, ) D(−3,3)

a1 +0 a2 +( 3 )

4 3

− x + 1= 0

) = =

4 3

− x

= −1

x

=

4

7+3 2 1

4 3

4

3

⇒ C ( , 0) 3

1

A (1, ) ✓ 3


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165


Ex 6.4

13(iii) ⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB MAB = (

Point:

13(iv) Area of quadrilateral ABCD

(3)+(0) (5)+(1) 2

,

2

A(3,5)

3

3

C ( , 0)

B(0,1)

4

) = ( , 3)

D(

11 2

, 0)

2

Area of ABCD Gradient: mAB⊥ = mBC = −

3

y − y1 = mAB⊥ (x − x1 )

AB⊥ :

3 4 3

9

4 3

8 33

y

4

8

=− x+

3

0

2

5

1

4

11

3

2

0

0

2

=

3

|

5

1

55

2

2

= (3 + 0 + 0 +

3

y − 3 = − (x − ) y−3 =− x+

1

= |

4

4

− 0 − − 0 − 0) 3

1 175 ( ) 2 6

= 14

7 12

unit 2 ✓

Point D At D, AB⊥ cuts x-axis (y = 0). y =0 3

33

4 3

8

− x+ − x

=−

4

x ⇒ D(

=0

= 11 2

33 8

11 2

, 0) ✓


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166


Rev Ex 6 A2(ii) Point D at k = 2: A(3,1) B(5,3)

Rev Ex 6 A1(i) A(−2,1) B(10,6) C(a, −6) AB

D on x − axis: D(d, 0)

= BC 10]2

√[(−2) − + (1 − 169 169 a2 − 20a + 75 (a − 5)(a − 15) a = 5 or a = 15 ✓

6)2

= √(10 − a)2 + [6 − (−6)]2 = (100 − 20a + a2 ) + 144 = 244 − 20a + a2 =0 =0

BD ⊥ AC: mBD ⋅ mAC = −1 3−0 5−d 3 5−d

Let D be (d1 , d2 ) For ABCD is rhombus, MAC = MBD (−2)+a 1+(−6)

2 (−2)+a 2

d1

,

=

) =(

2 10+d1

and

2

= a − 12

,

)

2 1+(−6) 2

d2

A(k + 1,1) B(2k + 1,3)

3−6 −3 −3

mBD =

= −1 = −1

=

3−0 5−8

tan θ = −1 α = 45°

6+d2 2

= −11

a = 5: D(−7, −11) ✓ a = 15: D(3, −11) ✓

A2(i)

⋅

1−4

Obtuse angle that BD makes with x-axis

10+d1 6+d2 2

⋅

3 =d−5 d =8 ⇒ D(8,0) ✓

A1(ii) A(−2,1) B(10,6) C(a, −6)

(

C(6,4)

∴ obtuse ∡= 180 − 45 = 135° ✓ A3(i)

A(6,7)

B(0,1)

C(9,4)

|BC| = √(0 − 9)2 + (1 − 4)2

C(2k + 2,2k)

= √81 + 9 ∵ A, B, C are collinear, mAB = mBC (1)−(3) (k+1)−(2k+1) −2 −k 2

= √9 × 10 = 3√10 units ✓

(3)−(2k)

= (2k+1)−(2k+2) =

3−2k −1

= 2k − 3

k

= √90

2 = 2k 2 − 3k 2k 2 − 3k − 2 = 0 (2k + 1)(k − 2) = 0 1

k = − or k = 2 ✓ 2

A3(ii) Area of △ ABC 1 6 0 9 6 | = | 2 7 1 4 7 1 = (6 + 0 + 63 − 0 − 9 − 24) 2 1 = (36) 2 = 18 ✓ A3(iii) Area of △ ABC = 18 1 2 1 2

|AF||BC|

|AF|(3√10) = 18

|AF|


sleightofmath.com

= 18

=

36 3√10

=

12 √10

=

12√10 10

=

6√10 5

✓

167


Rev Ex 6

A4(a) y = ax + b ⇒ m1 = a

A5(i)

2x + y = 8 y = −2x + 8 ⇒ m2 = −2

Line BC Point:

B(−1, −3) or C

Gradient:

mBC =

BC:

y − y1

1 2

= mBC (x − x1 ) 1

y − (−3) = ( ) [x − (−1)] 2

y = ax + b ⊥ 2x + y = 8 ⇒ m1 m2 = −1 a(−2) = −1 a

1

= ✓

A5(ii) Line AC Point: Gradient:

2

y = ax + b on y − axis (x = 0): y = a(0) + b =b ⇒ (0, b)

A4(b) AB: y = 3x + 1 BC: y = x − 1 F(3,2)

2

2

y

= x− ✓

−1

=−

mAB

=−

1 −8 ( ) −4

1 2

B

y

C F(3,2)

1

3

2 1

2 13

2

2

1 2

x−

5 2

1

13

2

2

=− x+

1

5

2

2

−1 −3 −1 −3

9 2

=9 1

13

2

2

y|x=9 = − (9) +

Gradient: ∵ AF ⊥ BC, −1

=2 ⇒ C(9,2) ✓

= (1) = −1

A5(iv) Height of triangle ABC A(3,5) B(−1, −3) C(9,2) Equating area of △ ABC, 1

Point A At A, AF intersects AB (y = 3x + 1) −x + 5 = 3x + 1 −4x = −4 x =1

2

|BC||AD|

|BC||AD| √

=4

1

= | 2

=|

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3 5

3 5

9 2

3 | 5 3 | 5

[(−1) − 9]2 −9 − 2 + 45 |AD| = [ ] 2 −(−50) − (−27) − 6 [( ] + −3) − 2

√125|AD| 5√5|AD|

= 60

|AD|

=

= 60

= © Daniel & Samuel A-math tuition 📞9133 9982

✓

At C, AC intersects BC (y = x − )

𝑥

y=x−1

y − y1 = mAF (x − x1 ) y − (2)= (−1)[x − (3)] y − 2 = −x + 3 y = −x + 5

=− x+

A5(iii) Point C

x

mBC

2

y − (5)= (− ) [x − (3)]

y = 3x + 1

−1

1

y − y1 = mAC (x − x1 )

𝑦 A

mAF =

1 (−3)−(5)

((−1)−(3))

=−

y−5 =− x+

Line AF Point: A or F(3,2)

y|x=1 = −(1) + 5 ⇒ A(1,4) ✓

2 5

= x+

A(3,5) or C ∵ ∡BAC = 90°,

AC:

𝑂

AF:

1

2 1

y+3

mAC =

(0, b) lies on x + y + 3 = 0, 0+b+3 =0 b = −3 ✓

1

12 √5 12√5 5

✓

168


Rev Ex 6 𝑦

AD: 3x + 2y = 6

E

C

B AB: 5y + 6 = 3x 𝑥

D

𝑂

A

A6(ii) Line BC ∵ BC ∥ x − axis, BC is y = 6 Point B At B, BC (y = 6)cuts AD (5y + 6 = 3x), 5(6) + 6 = 3x 36 = 3x x = 12 ⇒ B(12,6) ✓ Line CD Pt: C or D(0,3) Grad: ∵ CD ∥ AB,

Point A At A, AD cuts x − axis (y = 0): 3x + 2(0) = 6 3x =6 x =2 ⇒ A(2,0) ✓ Point D At D, AD cuts y − axis (x = 0): 3(0) + 2y = 6 y =3 ⇒ D(0,3) ✓

mCD = mAB = CD: y − y1

5 3

0 =

2

3

2+e1 0+e2

2+e1

2

= x+3 5

Point C At C, CD cuts BC (y = 6),

= MAE ,

= mCD (x − x1 ) 3

y

(0,3) = (

5

y − (3)= [x − (0)]

A6(ii) Point E Let E be (e1 , e2 ) D

3

2

5 3

)

and

e1 = −2 ∴ E(−2,6) ✓

3 =

0+e2

5

2

x+3=6 x

=3

x =5 ⇒ C(5,6) ✓

e2 = 6

Area of trapezium ABCD Area of ABCD 1 2 12 5 0 2 | = | 2 0 6 6 3 0 1 = (12 + 72 + 15 + 0 − 0 − 30 − 0 − 6) 2 1 = (63) 2 = 31.5 unit 2 ✓ B1(a) A(−1,2) B(3,10) C(p, 8) A, B and C lie on the same line: mAB = mBC 2−10 (−1)−3 −8 −4


sleightofmath.com

= =

10−8 3−p 2 3−p 2

2

=

3−p p

=1 =2✓

3−p

169

A math 360 sol (unofficial) B1(b) AC ⊥ BC mAC ⋅ mBC 2−8 10−8 ⋅ (−1)−p 3−p −6 2

⋅

−1−p 3−p 6 2

⋅

Rev Ex 6 B2(iii) 𝐑𝐚𝐭𝐢𝐨 𝐀𝐏: 𝐏𝐐 3

= −1

A ( , 0) P(1, −1) B(0, −3) 2

= −1

By similar triangles (using x-coordinates) [diagram?]

= −1 = −1

p+1 3−p

3

AP: PB = ( − 1) : (1 − 0) 2

12 = (p + 1)(p − 3) 12 = p2 + 2p − 3 2 p − 2p − 15 = 0 (p + 3)(p − 5) = 0 p = −3 or p = 5 ✓ B2(i)

=

:1

2

= 1: 2 ✓ B3(i)

Point P At P, x − y − 2 = 0 intersects 2x − 5y − 7 = 0. x−y−2 =0 y = x − 2 −(1)

Show 𝐚 = 𝟑 A(2,6) B(6, −2) C(a, 8 − 3a) 8 − 3a < 0 −3a < −8 a

2x − 5y − 7 = 0 −(2) sub (1) into (2): 2x − 5(x − 2) − 7 = 0 2x − 5x + 10 − 7 = 0 −3x = −3 x =1

>

8 3

Area of △ ABC

= 10

6 a 2 || = 10 −2 8 − 3a 6 2 6 a 2 | | = 20 6 −2 8 − 3a 6 | − 4 + (48 − 18a) + 6a = 20 −36 − (−2a) − (16 − 6a)| |−8 − 4a| = 20 −8 − 4a = 20 or −8 − 4a = −20 −4a = 28 −4a = −12 a = −7 a =3 8 (rej ∵ a > ) 3 Point C C(3, −1) 1

| | 2

y|x=1 = (1) − 2 = −1 ⇒ P(1, −1) ✓ B2(ii) Line AB Point: P(1, −1) or A or B Gradient: mAB = 2 y − y1 = mAB (x − x1 ) y − (−1) = (2) [x − (1)] y+1 = 2x − 2 y = 2x − 3 ✓ Point A Point B At A, AB cuts x-axis At B where cuts y-axis (x = 0). (y = 0). y =0 y|x=0 = 2(0) − 3 2x − 3 = 0 = −3 3 ⇒ B(0, −3) ✓ x = AB:

1

2 6

B3(ii) Line AB Point:

2

A(2,6) or B(6, −2) (6)−(−2)

mAB =

AB:

y − y1 = mAB (x − x1 ) y − (6)= (−2)[x − (2)] y − 6 = −2x + 4 y = −2x + 10 ✓

(2)−(6)

=

8

Gradient:

−4

= −2

3

⇒ A ( , 0) ✓ 2


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170


Rev Ex 6

B3(iii) Line (through C & ⊥ to AB) Point: C(3, −1)

B4(iv) Line CD Point: C(8,2) or D Gradient: ∵ CD ∥ AB, (4)−(9)

Gradient: ∵ line ⊥ AB (y = −2x + 10) m=−

1 mAB

=−

1 −2

=

mCD = mAB = (2)−(7) =

1

= m (x − x1 )

y − y1

1

y − (−1) = ( ) [x − (3)] 2

1

3

2 1

2 5

2

2

y+1

= x−

y

= x− ✓

B3(iv) Point F (Foot of ⊥) C A

2 5 2

x− x

x

2

1

5

2

2

y|x=3 = (3) − 6 = −3 ⇒ D(3, −3) ✓

= −2x + 10 =

B5(i)

25 2

=5 1

5

2

2

y|x=5 = (5) −

Show △ 𝐏𝐐𝐑 𝐢𝐬 𝐢𝐬𝐨𝐬𝐜𝐞𝐥𝐞𝐬 P(1,0) Q(0,2) R(2,3) |PQ| = √[(1) − (0)]2 + [(0) − (2)]2 = √1 + 4 = √5

=0 ⇒ F(5,0) ✓ B4(i)

= mCD (x − x1 ) = (1) [x − (8)] =x−8 =x−6✓

y − y1 y − (2) y−2 y

B

F

5

=1

B4(v) Point D At D, BM (y = 3x − 12) intersects CD (y = x − 6). 3x − 12 = x − 6 2x =6 x =3

At F, y = x − meets AB (y = −2x + 10). 1

−5

2

CD: Line:

−5

|QR| = √[(0) − (2)]2 + [(2) − (3)]2 = √4 + 1 = √5

Show 𝐀𝐁 = 𝐁𝐂 A(2,4) B(7,9)

C(8,2)

∵ |PQ| = |QR|, PQR is isosceles ✓

AB = √[(2) − (7)]2 − [(4) − (9)]2 = √50 BC = √[(7) − (8)]2 + [(9) − (2)]2 = √50

B5(ii) Area of △ 𝐏𝐐𝐑 1

area of △ PQR = | |

∴ AB = BC [shown] ✓

2

1 0

0 2

2 3

1 || 0

1

= |2 + 0 + 0 − 0 − 4 − 3|

B4(ii) Point M M = MAC = (

(2)+(8) (4)+(2)

,

2

2

2 1

= |−5|

) = (5,3) ✓

= B4(iii) Line BM Point:

2 5 2

unit 2 ✓

B(7,9) or M(5,3) (9)−(3)

6

Gradient: mBM = (7)−(5) = = 3 2

BM:

y − y1 = mBM (x − x1 ) y − (9)= (3) [x − (7)] y − 9 = 3x − 21 y = 3x − 12 ✓


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171


Rev Ex 6

B5(iii) Line QS

B6(ii) Angle that BC makes with the positive x-axis 2−30

mBC = (−12)−12

R Q

tan β = S

6

β ≈ 49.4° ✓

P Point: Q(0,2) or S Gradient: ∵ QS ⊥ PR (using kite property), 1

mQS = −

mPR

=−

1 (0)−(3) ((1)−(2))

=−

1 −3 ( ) −1

=−

y − y1 = mQS (x − x1 )

QS:

7

1 3

B6(iii) Angle ACB ∡ACB = α − β = 69.4 − 49.4 = 20.0° ✓ B6(iv)

𝑦

C(12,30)

1

y − (2)= (− ) [x − (0)] 3

1

y−2 =− x 3 1

y

=− x+2 3

=− x+2

𝛾 Point D D(d1 , d2 ) BC = BD: B = MDC

−(1)

3

(−12,2) = (

x − y= 5 sub (1) into (2):

𝑥

A(−3, −10)

D

B5(iv) Point S At S, QS (x + 3y = 6) intersects x − y = 5. x + 3y = 6 3y = −x + 6 1

α 𝑂

3y = −x + 6 x + 3y = 6 ✓

y

β

B(−12,2)

−(2)

−12 = d1

1

d1 +12 d2 +30 2

d1 +12 2

,

2

)

and

2 =

= −36

d2 +30 2

d2 = −26

x − (− x + 2) = 5 3

4 3 4 3

x−2

=5

x

=7

x

=

∴ D(−36, −26) ✓

B6(v) Line AD Point:

21 4

A(−3, −10) or D(−36, −26)

1 21

y|x=21 = − ( ) + 2 3

4

= ⇒ S( B6(i)

4

tan α =

AD:

y − y1

4

, )✓ 4

(−10)−30 (−3)−12 8

(−10)−(−26) (−3)−(−36)

=

16 33

= mAD (x − x1 ) 16

y − (−10) = ( ) [x − (−3)]

Angle that AC makes with the positive x-axis A(−3, −10) 𝑦 𝐶(12,30) B(−12,2) C(12,30) mAC =

mAD =

1

21 1 4

Gradient:

𝐵(−12,2)

3

α ≈ 69.4° ✓


𝛽

𝛼 𝑂

y + 10

=

y

=

33 16

48

33 16

33 282

33

x+ x−

33

✓

𝑥

𝐴(−3, −10)

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172


Rev Ex 6

B6(vi) Angle ADB mAD = tan γ =

16 33 16 33

γ ≈ 25.9° ∡ADB = β − γ = 29.4 − 25.9 ≈ 23.5° ✓


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173


Ex 7.1 2(e)

Ex 7.1 1(a)

ln 4 + lg 6 ≈ 2.16 ✓

1(b)

3 lg 2 − ln 2 ≈ 0.210 ✓

1(c)

ln 7.1 2 lg 5

1(d)

= log √2 (5 − 2√2) Is base √2 > 0 ? True Is base √2 ≠ 1 ? True Is input 5 − √2 > 0 ? True

≈ 1.40 ✓

lg 9−ln 3 ln(e2 −1)

∴ Yes ✓

≈ −0.0778 ✓ 3(a)

2(a)

log x (5 − 2x)|x=√2

log x (5 − 2x)|x=0.5 = log 0.5 (4)

3−2 log 3

Is base 0.5 > 0 ? True Is base 0.5 ≠ 1 ? True Is input 4 > 0 ? True

3(b)

10n

= 1 9

1 9

= −2 ✓ =5

log10 5 = n lg 5 =n✓

∴ Yes ✓ 3(c) 2(b)

log x (5 − 2x)|x=3 = log 3 (−1)

ex

=4

log e 4 = x ln 4 =x✓

Is base 3 > 0 ? True Is base 3 ≠ 1 ? True Is input −1 > 0 ? False

3(d)

2x

=p

log 2 p = x ✓ ∴ No ✓ 2(c)

3(e)

log x (5 − 2x)|x=2.5 = log 2.5 (0) Is base 2.5 > 0 ? Is base 2.5 ≠ 1 ? Is input 0 > 0 ?

=y

log a y = 3 ✓ True True False

∴ No ✓ 2(d)

a3

4(a)

53 4(b)

log x (5 − 2x)|x=1 = log1 (3) Is base 1 > 0 ? True Is base 1 ≠ 1 ? False Is input 3 > 0 ? True ∴ No ✓

log 5 125 = 3

lg 100 =2 log10 100 = 2 102

4(c)

sleightofmath.com

=x✓

log x 3 = 4 x4


= 100 ✓

ln x = 2 log e x = 2 e2

4(d)

= 125 ✓

=3✓

174


log 4 x = 2 4 x x y

=

log 2 𝑦 = 3 3

=x = 16 16 8

2 y

=y =8

=b

6(a)

log 4 4 −3 log 2 2 = 1 −3(1) = −2 ✓

6(b)

log 2 1 +4 log 5 5 = 0 +4(1) =4✓

6(c)

(3 − log 3 3)3 = [3 − (1)]3 = 23 =8✓

6(d)

(

8(a)

3 logx x+2 2

3(1)+2 2

4−2 log5 1

4−2(0) 5 2

) =[

8(b)

= 6(e)

6(f)

7(a)

16

8(c)

8(d)

9(a)

=x =8✓

ln x = lg 2 log e x = lg 2 =x ≈ 1.35 ✓

lg(3x) = 9 log10 3x = 9 109

= 3x

x

= (109 )

x

≈ 3.33 × 108 ✓

e2x

1 3

=k

log e k = 2x ln k = 2x ✓ 9(b)

10x−4 = 9 log10 9 = x − 4 lg 9 =x−4✓

log 2 x = 3 23 x

= e±√3 ≈ 5.65 or 0.177 ✓

elg 2 x

✓

log 2 (4 − 2 lg 10) = log 2 [4 − 2 log10 10] = log 2 [4 − 2(1)] = log 2 2 =1✓

= 100.61 ≈ 4.07 ✓

(ln x)2 = 3 ln x = ±√3

x x

]

log 2 (6 − 5 log 7 7) = log 2 [6 − 5(1)] = log 2 (1) =0✓

lg x = 0.61 log10 x = 0.61

log e x = ±√3

=( ) 4 25

2

x x

log y b = 2

x1 =a y2 x =a 2 xy = (a)(b) = ab ✓

3

= ✓

x

=2✓

log x a = 1

log 4 8 = x 4x =8 22x = 23 ⇒ 2x = 3

=y✓

2

5(b)

7(c)

log 3 y = n 3n

5(a)

Ex 7.1

9(c)

x4

=2−k

log x (2 − k) = 4 ✓ 7(b)

log x 9 = 2 x2 =9 x = −3 or x = 3 ✓ (rej)


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175


em+5

=x−2

Ex 7.1 12

log 3 p = a p

log e (x − 2) = m + 5 ln(x − 2) = m+5✓

= 3a

log 27 q = b 10(a) log 3 y = n + 1 3n+1

e

33b a−3b

= 3c

13(a) log 2 (2x + 1) = −3

=x−y✓

2x + 1

= 2−3

2x

=−

x

=−

= 4y ✓ 13(b) log 3 (x 2 − 1)

log 4 y = a = 4a

= 21+2a = 21+2a = 1 + 2a = 3b − 1 [shown] ✓

7 8 7 16

✓

=1

x2 − 1 = 31 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 or x = 2 ✓

−(1)

log 8 (2y) = b −(2) sub (1) into (2): log 8 [2(4a )] = b log8 [2(22a )] = b log 8 (21+2a ) = b 8b 23b ⇒ 3b 2a

= 3c

3 = 3c ⇒ c = a − 3b ✓

10(d) log 2 (4y) = p + 1

y

= 3c

27b 3a

=k✓

10√2

11

p q 3a

10(c) lg(x − y) = √2 log10 (x − y) = √2

2p+1

= 27b

=y✓

10(b) ln k = x − 3 log e k = x − 3 x−3

q

13(c)

log x (6x − 8) = 2 x2 = 6x − 8 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x = 2 or x = 4 ✓

13(d) log x 64 = 3 2

3

x2

= 64

3 2

= 26

x

2

x x x


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= (26 )3 = 24 = 16 ✓

176


Ex 7.1

13(e) log 9 √27 = x + 1 9x+1

15(b) log y x = 2

= √27

2x+2

3

= y2

x

3 2

−(1)

=3

⇒ 2x + 2 =

xy = 8 −(2) sub (1) into (2): (y 2 )y = 8 y3 = 23 ⇒ y =2✓

3 2 1

2x

=−

x

=− ✓

2 1 4

14(a) ln 2 × ln(4x)= 3 3

ln(4x)

=

log e (4x)

=

4x

= eln 2

x|y=2 = (2)2

ln 2 3

16

ln 2

=4✓

log 2 (log 3 x) = ln e

3

1

log 2 (log 3 x) = 1 log 3 x = 21

3 ln 2

x

= e

x

≈ 18.9 ✓

4

= 32 =9✓

x x

14(b) lg(x − 2) = (lg 3)2 log10 (x − 2) = (lg 3)2

17(a) For lg(x + 2) to be defined, 2

= 10(lg 3)

x−2

= 2 + 10(lg 3) ≈ 3.69 ✓

x x

x + 2> 0 x > −2

2

17(b) For ln(x 2 − 2x) to be defined, 14(c) ln(4x) = lg 3 × lg 5 log e 4x = lg 3 × lg 5 4x

= elg 3×lg 5

x

= e(lg 3×lg 5)

x

≈ 0.349 ✓

x 2 − 2x x(x − 2) +

1

17(c) For log x (3 − x) to be defined, x > 0, x ≠ 1 and 3 − x > 0 3 >x x <3 ∴ 0 < x < 3, x ≠ 1 ✓

15(a) log x 16 = 4 x4 = 16 x = 2 or x = −2 (rej) ✓ −(1)

y

+ 2

x < 0 or x > 2 ✓

14(d) lg(x − 1) = lg(e2 − 1) ⇒ x − 1 = e2 − 1 x = e2 x ≈ 7.39 ✓

log 2 y = x sub (1) into (2): log 2 y = 2

− 0

4

>0 >0

−(2)

18(a) ln(y + 1) − x = 0 ln(y + 1) =x log e (y + 1) = x y+1 y

= ex = ex − 1 ✓

= 22 =4✓


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177

A math 360 sol (unofficial) 18(b) 2 lg y lg y

Ex 7.1

=x−2

20(ii) log 9 (31 + log x 16) = log 2 √8

x−2

=

log 9 (31 + log x 16) =

2 x−2

log10 y =

log 9 (31 + log x 16) =

2 x−2 2

y

= 10

y

= 102x−1 ✓

= (9)2

31 + log x 16 31 + log x 16 31 + log x 16 log x 16

= (32 )2 = 33 = 27 = −4

x −4 x −4

= 16 = 24

x

= (24 )−4 = 2−1

3

=x =x−4

1

= ln(x − 4) ✓

y

2 3

= log e (x − 4)

2y

2 3

31 + log x 16

1

18(c) e2y + 4 e2y

3

2

1

18(d) ln(x + y) − 4x = 0 ln(x + y) = 4x log e (x + y) = 4x

2

= e4x = e4x − x ✓

x+y y 19

1

= ✓

ln(x 2 + 1 − e3 lg 10 )

21

For equal real roots: b2 − 4ac =0 2 (−4) − 4(1)(log 2 p) = 0 16 = 4 log 2 p log 2 p =4 p = 24 p = 16 ✓

=3

log e (x 2 + 1 − e3 lg 10 ) = 3 x 2 + 1 − e3 lg 10 x 2 + 1 − e3(1) x2

= e3 = e3 = 2e3 − 1

x = ±√2e3 − 1 x = ±6.26 ✓ [textbook answer is wrong] 20(i)

p = log 2 √8 3

p = log 2 22 3

p= ✓ 2

x 2 − 4x + log 2 p = 0 i.e. a = 1, b = −4, c = log 2 p

22(i) (a) (b) (c)

lg 546 ≈ 2.74 ✓ lg 12 458 ≈ 4.10 ✓ lg 464 777 399 ≈ 8.67 ✓

22(ii) Number of digits in integer k from lg k lg k + 1 if lg k ∈ ℤ ={ ✓ ⌈lg k⌉ if lg k ∉ ℤ 22(iii) Number of digits in 342 = ⌈lg 342 ⌉ = ⌈20.03⌉ = 21 ✓ Number of digits in 732 = ⌈lg 732 ⌉ = ⌈27.02⌉ = 27 ✓


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178


Ex 7.1

22(iv) As we use the base-10 numeral system, The ‘index of base 10’ is very indicative of the number of digits. 1 digit: k = 1 to 9 ⇒ lowest k = 1 = 100 2 digits: k = 10 to 99 ⇒ lowest k = 10 = 101 3 digits: k = 100 to 999 ⇒ lowest k = 100 = 102 Observe that the lowest k for each digit range is always 10integer Every integer increase in index results in increase in digits but increase in index smaller than 1 will not result in increase in digits and still be registered as the same number of digits. ∴ for k with lg k is an integer, the no. of digits is lg k + 1 1 digit: k = 100 ⇒ no. of digits = lg 100 + 1 = 0 + 1 = 1 2 digit: k = 10 = 101 ⇒ no. of digits = lg 101 + 1 = 1 + 1 = 2 ∴ for k with lg k is a non-integer, the no. of digits is ⌈lg k⌉ (round to the next higher integer) 1 digit: k = 5 ≈ 100.699 ⇒ no. of digits = ⌈lg 100.699 ⌉ = ⌈0.699⌉ = 1 2 digit: k = 55 ≈ 101.74 ⇒ no. of digits = ⌈lg 101.74 ⌉ = ⌈1.74⌉ = 2


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179


Ex 7.2 4(a)

Ex 7.2 1(a)

1(b)

1

9

9

log 3 27 + log 3 = log 3 (27 ⋅ )

4(b)

log5 4×log2 10 log25 √10

=

= log 6 (6) =1✓

=

3

=

log 5 4 +2 log 5 3 −3 log 5 2 = log 5 4 + log 5 32 − log 5 23 23

=

]

=

9

= log 5 ✓ 2

2(b)

lg

8

=

−2 lg

75

3 5

+4 lg

3

8

3 2

3 4

75

5

2

8 3 4 ( ) 75 2 3 2 ( ) 5

= lg [

5(a)

]

3

= lg ✓ 2

1

+2 log 3 5 − log 3

3(52 ) 3−3

]

5(b)

= log 3 2025 ✓ 3(a)

log 5 7 =

3 ln 3 ln 3

lg 4 lg 10 × lg 5 lg 2 lg √10 lg 25

lg 4 lg 10 lg 5

⋅

lg 22 lg 5 2 lg 2 lg 5 2 lg 2 lg 5 2 lg 5

lg 2

⋅ ⋅ ⋅ ⋅

⋅

1 lg 2 1 lg 2 1 lg 2

lg 25 lg √10

⋅

lg 52 1

lg 102 2 lg 5

⋅1 2

lg 10

2 lg 5

⋅1 2

lg 10

2 lg 5 1 2

4 1 2

y = 100x1.5 lg y = lg(100x1.5 ) = lg 100 + lg x1.5 = 1.5 lg x + lg 102 = 1.5 lg x +2 ✓ m = 1.5 ✓ X = lg x ✓ c =2✓

1 27

= log 3 3 + log 3 52 − log 3 3−3 = log 3 [

ln 3

=8✓

2

= lg ( ) − lg ( ) + lg ( )

2(c)

=

= log 6 ( 2 )

= log 5 [

ln 33

=3✓

log 6 54 −2 log 6 3 = log 6 54 − log 6 32

4(32 )

ln 5

ln 3

=

54

2(a)

ln 27

ln 27

=

1

×

ln 3

=

log 2 8 = log 2 23 =3✓

= log 3 (3) =1✓ 1(c)

ln 5

log 3 5 × log 5 27 =

ln 7 ln 5

y = 0.1(1000)x y = (10−1 )(103 )x y = 103x−1 lg y = 3x − 1 ✓

≈ 1.21 ✓ 3(b)

log 1 5.3 = 2

m = 3✓ X= x✓ c = −1 ✓

ln 5.3 ln

1 2

≈ −2.41 ✓ 6(a)

log a 8 −2 log a 4 = log a 8 − log a 42 8

= log a ( 2 ) 4

1

= log a ✓ 2


sleightofmath.com

180


2 log x 5 −3 log x 2 + log x 4 = log x 52 − log x 23 + log x 4 = log x [ = log x

6(c)

Ex 7.2

52 (4) 23

25 2

8(iv)

log a 3 = 0.477 log a 5 = 0.699 loga 25 loga 3a

= =

loga 52 loga 3+loga a 2 loga 5 (0.477)+1

2

9 5a

log 4 3 = a 3 = 4a

3

= (4a )(4b ) = 4a+3b ✓ 9 2

= log a 9

− log a 5a

= log a 32 −(log a 5 + log a a) = 2 log a 3 − log a 5 − log a a = 2(0.477) −0.699 −1 = −0.745 ✓ log 4 3 = a log 4 5 = b

= 2(lg 3 + lg √x)

− (lg − lg x 2 )

1

4

+

3 4

− (lg − 2 lg x)

= 2 lg 3 + lg x

− lg + 2 lg x

= 3 lg x

+2 lg 3 − lg

2

+ lg 32 − lg + lg

9

= lg x

+ lg

2 9

= lg x

+ lg

2

log 4 20= log 4 × 5) 2 = log 4 2 + log 4 5 =1 +b ✓

4

+

3 4

2 lg 22 +3 lg x 2 2 lg 2+3 lg x 2 2

32

+ lg 2

4 3

32 ×2 4 3

27

✓

2

x = 10p

y = 10q

y2

lg 4+lg x3

3

lg y =q log10 y = 𝑞

100√x

lg 4x3 lg 100

+ lg 2 + lg x

3

lg x =p log10 y = 𝑝

lg (

sleightofmath.com

+ +

3

9

(22

3

4

2


3x2

= 2 (lg 3 + lg x)

10(i) 8(ii)

+ log100 4x 3

− lg

= lg x

log 4 45= log 4 (32 × 5) = log 4 32 + log 4 5 = 2(log 4 3) + log 4 5 = 2a +b ✓

4

2 lg 3√x

= 3 lg x 8(i)

✓

375 = 3 × 53

log a (5a ) = log a 5 + log a a = 0.699 +2 = 2.699 ✓ log a

a+2b

log 4 5 = b 5 = 4b

2(0.699) = 1.477 ≈ 0.947 ✓

7(iii)

2

=

]

= log a 8a ✓

7(ii)

log 4 42 log 4 (3 × 52 ) 2 = log 4 3 + log 4 52 2 = log 4 3 + 2 log 4 5

✓

a6 2

log4 75

=

]

23 (a6 )

log4 16

log 75 16 =

3 log a 2 −4 +2 log a a3 = log a 23 − log a a4 + log a (a3 )2 = log a [

7(i)

8(iii)

) = lg 100 + lg √x − lg y 2 1

=2

+ lg x

−2 lg y

=2

+ p

−2q ✓

2 1 2

181


Ex 7.2

10(ii) lg(y x ) = x lg y = 10p (q) = q(10p ) ✓ 11

14

log 8 5 =

ln K = ln a − ln b + ln c − t − bc

1

ln K = ln (a ⋅ ⋅ c ⋅ e b

K

=

lg(y + 1) = lg ⇒y+1

=

y

=

⇒

√x 100 √x

34

15(ii) lg(√3 − √2) = lg [

(1)−(m) 3(m)

=

1−m 3m

✓

√3+√2 √3+√2

1 √3+√2

]

= lg 1 − lg(√3 + √2)

15(iii) k = log 2 (√9 + √5) log √2 (√9 − √5) = log √2 [(√9 − √5) = log √2 (

9−5 √9+√5

√9+√5 √9+√5

]

)

= (x + 2)3

= log √2 4 − log √2 (√9 + √5)

= 81(x + 2)3

=

3 2

or y = −9(x + 2) ✓ (rej ∵ y > 0)

12(c) 3 + log 2 (x + y) 3 log 2 2 + log 2 (x + y) log 2 8(x + y) ⇒ 8(x + y) 8x + 8y = x − 2y 10y = −7x 7

=

√3+√2

= 3 log 3 (x + 2) = log 3 (x + 2)3

y2

10

3 lg 2

√3 + √2 1 = [shown] ✓

= log 3 (x + 2)3

=−

lg 10−lg 2

=

3−2

=

)

−1✓

34

=

= log 2 (x − 2y) = log 2 (x − 2y) = log 2 (x − 2y) = x − 2y 16(i)

x✓

y = 3, x = 2: log 2 (3 + 1) = 2 log 2 (2) + c log 2 22 =2+c 2 =2+c c =0

log2 22 1 log2 22

2 1 2

−

log2 (√9+√5) 1

log2 22

−

log2 (√9+√5) 1 2

=4

−2 log 2 (√9 + √5)

=4

−2k

= 2(2 − k) ✓

log 2 (y + 1) = 2 log 2 x + c

∴ log 2 (y + 1) log 2 (y + 1) ⇒y+1 y

10 2 lg 23

lg( )

= − lg(√3 + √2) ✓

y2

3 2

=

√3 − √2 = (√3 − √2)

√x 100

y = 9(x + 2)

13

lg 8

100

y2

y

15(i)

t bc

− lg √x − lg √x

12(b) 2 log 3 y −4 log 3 y 2 − log 3 34 log 3

−

[shown] ✓

12(a) lg(y + 1) = 2 lg(y + 1) = lg 102

lg 5

t bc

ln K = ln a − ln b + ln c + ln e ac − t e bc b

m = lg 2

16(ii)

p log a p log a p ⇒p 1 q

= aloga x = log a aloga x = log a x =x✓

3 =3

1 log4 3

=3

1 1 log3 4

= 3log3 4 = 4 ✓

= 2 log 2 x = log 2 x 2 = x2 = x2 − 1 ✓


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182


Ex 7.2

Method 1 (prime factorization) (lg 5)2 + lg 2 lg 50 = (lg 5)2 + lg 2 lg(2 × 52 ) = (lg 5)2 + lg 2 (lg 2 + 2 lg 5) = (lg 5)2 + 2 lg 5 lg 2 + (lg 2)2 = (lg 5 + lg 2)2 = (lg 10)2 =1✓

1 2 18 un = (1 − ) n (ii)(a) 1 2

lg u4 = lg (1 − ) 4

3

= 2 lg [shown] ✓ 4

lg u Method 2 (selective factorization) (lg 5)2 + lg 2 lg 50 = (lg 5)2 + lg 2 lg(5 × 10) = (lg 5)2 + lg 2 (lg 5 + lg 10) = (lg 5)2 + lg 2 (lg 5 + 1) = (lg 5)2 + lg 2 lg 5 + lg 2 = lg 5 (lg 5 + lg 2) + lg 2 = lg 5 lg 10 + lg 2 = lg 5 + lg 2 = lg 10 =1✓ 18(i)

1

2

3

998

2

3

998

999 999

999

lg + lg + lg + ⋯ + lg 1

2

4 3

2 1

3

4

= lg [( ) ( ) ( ) … ( = lg (

)(

+ lg

1000

= lg (1 − = 2 lg (

1 10k

)

2

10k −1 10k

)✓

lg u2 + lg u3 + lg u4 + ⋯ + lg u10k 18 (ii)(b) 1 2 1 2 3 = lg (1 − ) + lg (1 − ) +2 lg 2

3

4

+ ⋯ + 2 lg = 2 lg

1 2 1

= 2(lg

2

+2 lg + lg

2

2 3

+2 lg

+ lg

3

3

3

+ ⋯ + 2 lg

4

+ ⋯ + lg

1

2

3

4 10k −1

2

3

4

10k

= 2 lg [( ) ( ) ( ) … (

10k −1 10k 10k −1 10k

10k −1 10k

)

)]

1 = 2 lg ( k ) 10 = 2 lg(10−k ) = 2(−k) = −2k ✓

999 1000

)]

)

1000 −3 )

= lg(10 = −3 ✓

(10k )

19

1 lg ≈ −0.301 < 0 2 When you multiply both sides of an inequality with a negative number, switch the reverse the inequality sign from 2nd line onwards

20

lg[(−2)(−5)] ≠ lg(−2) + lg(−5) ✓

20(i)

x = −2, y = −5 ✓

20(ii) x = −2, r = 2 ✓


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183


Ex 7.3

Ex 7.3 1(a)

log 5 (x + 1) = log 5 3 ⇒x+1 =3 x = 2✓

1(b)

log 2 (x − 1) ⇒x−1 3x x

1(c)

2(a)

2(b)

3(a)

3(b)

= log 2 (4x − 7) = 4x − 7 =6 = 2✓

3(c)

log 2 (x − 1)2 =2 + log 2 (x + 2) 2 2 (x log 2 − 1) = log 2 2 + log 2 (x + 2) log 2 (x − 1)2 = log 2 4(x + 2) 2 ⇒ (x − 1) = 4(x + 2) 2 x − 2x + 1 = 4x + 8 2 x − 6x − 7 =0 (x − 7)(x + 1) = 0 x = 7 or x = −1 ✓

4(a)

log 3 x = 9 log x 3 log 3 x =

log 3 x + log 3 (x + 2)= 1 log 3 [x(x + 2)] = log 3 3 ⇒ x(x + 2) =3 x2 + 2 =3 2 x + 2x − 3 =0 (x + 3)(x − 1) =0 x = −3 (rej) or x = 1 ✓ log x 25 + log x 5 log x (125) ⇒ 125 x

=3 = log x x 3 = x3 =5✓

9 log3 x

sub u = log 3 x: u=

9 u

u2 = 9 u =3 log 3 x = 3 x x 4(b)

or

u = −3 log 3 x = −3

= 33 = 27 ✓

x

= 3−3

x

=

1 27

✓

log 3 x + 2 = 3 log x 3 3

log 3 x + 2 =

log3 x

3 log x 2 + log x 18 = 2 log x 23 + log x 18 = log x x 2 log x (23 × 18) = log x x 2 log x 144 = log x x 2 ⇒ 144 = x2 x = 12 or x = −12 (rej) ✓

sub u = log 3 x:

lg[(x + 2)(x − 2)] = lg(2x − 1) ⇒ (x + 2)(x − 2) = (2x − 1) x2 − 4 = 2x − 1 2 x − 2x − 3 =0 (x − 3)(x + 1) =0 x = 3 or x = −1 (rej) ✓

x

u+2

=

3 u

u2 + 2u − 3 =0 (u + 3)(u − 1) = 0 u = −3 or u =1 log 3 x = −3 log 3 x = 1

x

= 3−3 =

1 27

x

=3✓

✓

log 2 [(x − 2)(8 − x)] − log 2 (x − 5)= 3 log 2 [ ⇒

(x−2)(8−x)

(x−5) (x−2)(8−x)

]

(x−5)

= log 2 23 = 23

−(x − 2)(x − 8) = 8(x − 5) 2 −(x − 10x + 16) = 8x − 40 x 2 − 2x − 24 =0 (x − 6)(x + 4) =0 x = 6 or x = −4 (rej) ✓


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184


Ex 7.3

ln(3x − y) = ln 36 − ln 9

7(a)

36

ln(3x − y) = ln ( )

2 log 5 x + (

⇒ 3x − y = 4 y = 3x − 4 −(1)

2 log 5 x + (

(ex )2

2 log 5 x + (

9

=e

ey

5

e2x−y = e1 ⇒ 2x − y = 1

2

= 55 2

log 2 x − log 4 (x + 6) = 0 log 2 x = log 4 (x + 6) =[ =[ =[

=2

6

)

x

2

= 5q − p = 5q − 1

log2 (x+6) 2

] ]

= √x + 6

2

=x+6 =0

x = 3 or x = −2 (rej) ✓ 7(c) −(2)

sub (1) into (2): 2 + 3q = 5q − 1 2q =3 3

log 5 (5 − 4x) = log √5 (2 − x) log 5 (5 − 4x) =

log5 (2−x)

log 5 (5 − 4x) =

log5 (2−x)

log 5 (5 − 4x) =

log5 (2−x)

log5 √5 1

log5 52 1 2

log 5 (5 − 4x) = 2 log 5 (2 − x)

= ✓

q

log2 22

]

(x − 3)(x + 2) = 0

=2

1 p

log2 (x+6)

x2 − x − 6

=2

5q−p

log2 4

⇒x

= log 2 2

15q−3p

log2 (x+6)

= log 2 √x + 6

2p+2 = 24 (23q ) 2p+2 = 24+3q ⇒ p + 2 = 4 + 3q p = 2 + 3q −(1) log 2 6 − log 2 (15q − 3p) = 1

⇒

= log 5 55 = (55 )5 = 52 = 25

3

15q−3p 6

= log 5 55

)

2

x

4 [(22 )2q ]

log 2 (

) = log 5 55

= log 5 55

= 16 (4 )

2

log5 52 log5 x

5 2

3 q 2

p+2

) = log 5 55

log5 25 log5 x

log 5 x

⇒x

Put x = 3 into (1): y|x=5 = 3(3) − 4 =5✓ 2

=5

log 5 x 2

7(b)

p+2

log5 x

5

−(2)

sub (1) into (2): 2x − (3x − 4) = 1 −x + 4 =1 −x = −3 x =3✓

6

2 log 5 x + log 25 x

2

log 5 (5 − 4x) = log 5 (2 − x)2 Put q =

3 2

into (1): 3

p|q=3 = 2 + 3 ( ) 2

2

=

13 2

✓

⇒ 5 − 4x

= (2 − x)2

5 − 4x

= 4 − 4x + x 2

−x 2 + 1

=0

x2 − 1

=0

(x + 1)(x − 1) = 0 x = −1 or x = 1 ✓


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185


log 9 y + log 3 y log3 y

+ log 3 y

log3 9 log3 y

+ log 3 y log3 32 1 log 3 y + log 3 y 2 3 log 3 y 2 3

Ex 7.3

= 2 log 3 x +3 log 3 2 = log 3 x

2

9(c)

+ log 3 2

3

= log 3 [(x 2 )(23 )] = log 3 (8x 2 )

1

= log 3 (8x 2 )

log 3 √3x + 1 = log 3 (2x) ⇒ √3x + 1 = 2x 3x + 1 = 4x 2 4x 2 − 3x − 1 = 0 (4x + 1)(x − 1) = 0

2

⇒ y2

= 8x 2

y

= (8x 2 )3 2

=

2 4 (23 )3 x 3

=

4 22 x 3

log 3 (3x + 1) = log 3 (2x)

1

2

x = − (rej) or x = 1 ✓

= (8)3 (x 2 )3

4

9(d)

3 log 4 x − log16 x = 3.75 log 4 x 3 −

4

= 4x 3 ✓ log 4 (6 − x) − log 2 8

log 4 x 3 −

= log 9 3

log 4 x 3 −

1

log 4 (6 − x) − log 2 23 = log 9 92

log4 x

= log 4 43.75

log4 16 log4 x log4

log4 x

= log 4 4

2

1

log 4 x 2

2

5

15

x2

= 44

log 4 (6 − x)

=

6−x

= 42

x

= (22 )2 = 27 = −122 ✓

5

5

15

= (22 ) 4

5

15

x2

= 22 2 15 5 2

= (2 )

x

= 23 =8✓

log 7 (9x + 38) − log 7 (x + 2) = log 9 81 log 7 ( log 7 ( ⇒

9x+38

)

= log 9 92

)

=2

x+2 9x+38 x+2 9x+38

x+2 9x+38

x+2 9x+38 x+2

)

9(e)

log 5 x − log 25 (x + 10) = 0.5 log 5 x −

2

= log 7 7 = 72

log 5 x −

= 49

log 5 x −

9x + 38 −40x

= 49x + 98 = 60

x

=− ✓

15 4

( )

= log 4 4

x2 7

15 4

( )

2 7

=

log 7 (

15 4

( )

1

log 4 (6 − x) −3

15 4

( )

= log 4 4

42

log 4 x 3 − log 4 x 2 = log 4 4

7

9(b)

= log 3 (2x)

log3 32

2

9(a)

= log 3 (2x)

log3 9 log3 (3x+1)

= log 3 (8x 2 )

log 3 y 2

log 9 (3x + 1) = log 3 x + log 3 2 log3 (3x+1)

3

log5 (x+10) log5 25 log5 (x+10) log5 52 log5 (x+10) 2

log 5 x − log 5 √x + 10

3

log 5 (

2

⇒

x

)

√x+10

x √x+10

x2 x+10

= log 5 50.5 = log 5 √5 = log 5 √5 = log 5 √5 = log 5 √5 = √5 =5

x2

= 5x + 50

x 2 − 5x − 50

=0

(x − 10)(x + 5)

=0

x = 10 or x = −5 (rej ∵ x > 0) ✓


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186


Ex 7.3 + log a (7x − 10a)

2 log a x

=1

log a x 2

= log a a + log a (7x − 10a)

1st eqn log 2 (2y − 3x) − log 2 3 = 4 log 4 2

= log a [a(7x − 10a)]

log 2 (

= a(7x − 10a)

log 2 (

log a x ⇒x x

2

2

2

12

2

log 2 (

= 7ax − 10a

x 2 − 7ax + 10a2 = 0

log 2 (

(x − 2a)(x − 5a) = 0

⇒

x = 2a or x = 5a ✓ 11

27 × 3lg x

= 91+lg(x−20)

(33 )(3lg x )

= (32 )1+lg(x−20)

33+lg x

= 32+2 lg(x−20)

⇒ 3 + lg x

= 2 + 2 lg(x − 20)

1 + lg x

= 2 lg(x − 20)

lg 10 + lg x

= lg(x − 20)2

lg(10x)

= lg(x − 20)2

⇒ 10x

= (x − 20)2

2y−3x 3 2y−3x 3 2y−3x 3 2y−3x 3

2y−3x

1

)

= 4 (log 4 42 )

)

= 4( )

)

=2

)

= log 2 22

1 2

= 22

3

2y − 3x 2y

= 12 = 3x + 12

y

= x + 6 −(1)

3 2

2nd eqn log 3 6 + log 3 (x + y) log 3 6 + log 3 (x + y) log 3 [6(x + y)] ⇒ 6(x + y) 6x + 6y

= log 3 (−x) + log 3 (1 − x) = log 3 [(−x)(1 − x)] = log 3 (x 2 − x) = x2 − x = x2 − x −(2)

sub (1) into (2): 3

2

10x

= x − 40x + 400

6x + 6 ( x + 6)

0

= x 2 − 50x + 400

x 2 − 50x + 400

=0

6x + 9x + 36 = x2 − x 15x + 36 = x2 − x x 2 − 16x − 36 =0 (x − 18)(x + 2) =0 x = 18 or x = −2 ✓

= x2 − x

2

(x − 40)(x − 10) = 0 x = 40 or x = 10 (rej) ✓

3

y|x=−2 = (−2) + 6

(rej ∵ −x > 0)

2

=3✓ 12(ii) log 3 6 + log 3 (x + y) = log 3 (x 2 − x) ⇒ x + y > 0 x2 − x > 0 ∴ On top of x = −2, y = 3, the solution includes 3

x = 18, y|x=18 = (18) + 6 = 33 ✓ 2


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187


Ex 7.3

13(i) log 2 (x 3 + 1) − 2 log 2 x = log 2 (x 2 − x + 1) − 2 log 2 (x 3 + 1) − log 2 x 2 = log 2 (x 2 − x + 1) − log 2 22 log 2 ( ⇒

x3 +1 x2

)

= log 2 (

x3 +1

=

x2 (x+1)(x2 −x+1)

=

x2 x+1

=

x2

22

)

x = 2 − 2√2 or (rej ∵ x > 0)

+ log a2 x

x

1 x

loga ( ) 1 x

x2 −x+1

loga ( )

22 x2 −x+1

1 2

+ log a x

+

+ log a x

+

1

7

x 1 2

4

loga x loga a2 loga x 2

+ log a4 x = c + +

loga x loga a4 loga x 4

1 4

log a ( ) x

7

1 2

7

= log a ac

x

4±√16+16 2

=

4±√32 2

=

4±4√2 2

x = 2 + 2√2 ✓

(x 3

7

log a [(x −1 )2 x 4 ]

= log a ac

7

log a [(x −2 )x 4 ]

= log a ac

1

2

(x 2

13(ii) log 2 + 1) − log 2 x = log 2 − x + 1) − 2 3 2 log 2 (x + 1) − 2 log 2 x = log 2 (x − x + 1) − 2 ⇒ x = 2 ± 2√2 ✓

log a (x −4 )

= log a ac

1

⇒ x −4 x 15

= log a ac

+ log a x 4 = log a ac

log a [( ) x 4 ] =

= log a ac

= log a ac

2 log a ( ) + log a x

4

−(−4)±√(−4)2 −4(1)(−4) 2(1)

1

log √a ( ) + log a x loga √a

= x2 =0

4x + 4 x 2 − 4x − 4 x=

x2 −x+1

14

= ac = a−4c ✓ 2 log a b + 4 log b a = 9 2 log a b + 4 (

1 loga b

)=9

sub u = log a b: 2u +

4

=9

u

2u2 − 9u + 4 = 0 (2u − 1)(u − 4) = 0 u=

1 2

sub u = log a b: log a b = b b


sleightofmath.com

1

u=4 sub u = log a b: log a b = 4

2 1 2

=a = √a ✓

b = a4 (rej ∵ a > b > 1)

188


Ex 7.4 3(ii)

Ex 7.4 1(a)

1(b)

4x

= 9(5x )

4 x

( ) 5

5x = 9 lg 5x = lg 9 x lg 5 = lg 9

lg ( ) = lg 9 5

4

x lg ( ) = lg 9 5

lg 9

x

=

x

≈ 1.37 ✓

4e2x

lg 5

= 21

2x

=9

4 x

4(a)

21

e

=

2x

= ln

x

= ln

x

≈ 0.829 ✓

4 21

1

4 21

2

4

lg 9

x

=

x

≈ −0.985 ✓

2x . 3x 6x lg 6x x lg 6

= 10 = 10 = lg 10 =1

x

=

lg

4 5

1 lg 6

≈ 1.29 ✓ 1(c)

4 − 72x −72x 72x lg 72x 2x lg 7

=1 = −3 =3 = lg 3 = lg 3

x

=

x

≈ 0.282 ✓

4(b)

2x+1 = 3x (2x )(21 ) = 3x 2x

=

3x 2 x

( )

lg 3

3

2 lg 7

=

2 x

lg ( )

2

3(i)

3x+1 − 12 3x+1 lg 3x+1 (x + 1) lg 3

=0 = 12 = lg 12 = lg 12

x+1

=

x

=

x

≈ 1.26 ✓

lg 3

2

x lg ( ) = lg 3

x

=

1 2 1 2

1 2 2 lg( ) 3

lg

≈ 1.71 ✓

lg 12 lg 3 lg 12

2 1

= lg

3

2

1(d)

1

4(c) −1

3x+1 . 2x−2 = 21 x x −2 (3 )(3) (2 )(2 ) = 21 32

(3x )(2x )

= (3)(2−2 )

y = 5e

6x lg 6x x lg 6

= 28 = lg 28 = lg 28

y = 12: 5e0.2x = 12

x

=

0.2x

=

0.2x

= ln

x

= 5 ln

x

≈ 4.38 ✓

4

4x 5x 4 x

lg 6

≈ 1.86 ✓

12

e0.2x

x

lg 28

5 12 5

5(i)

9x − 4 = 3x+1 (32 )x − 4 = (3x )(3) (3x )2 − 3(3x ) − 4 = 0 ✓

5(ii)

(3x − 4)(3x + 1) = 0 3x = 4 or 3x = −1 (rej) lg 3x = lg 4 x lg 3 = lg 4

12 5

x)

= 9(5 =9

( ) =9✓ 5


sleightofmath.com

lg 4

x

=

x

≈ 1.26 ✓

lg 3

189


6(ii)

7(a)

Ex 7.4 3

8x + 11(2x )

= 4x+2 − 20

(23 )x + 11(2x ) (2x )3 + 11(2x ) (2x )3 + 11(2x ) (2x )3 − 8(2x )2 + 11(2x ) + 20 sub u = 2x u3 − 8u2 + 11u + 20 (u + 1)(u2 + + 20)

= (22 )x+2 − 20 = (22x )(23 ) − 20 = 8(2x )2 − 20 =0

(u + 1)(u2 − 9u + 20) = 0 (u + 1)(u − 5)(u − 4) = 0 u = −1 or u = 5 or sub u = 2x sub u = 2x x 2 = −1 2x = 5 (rej ∵ 2x > 0) x lg 2 = lg 5

3

=0✓

u =4 sub u = 2x 2x = 22 ⇒ x =2✓

7(d)

=

x

≈ 2.32 ✓

1 2

lg 2

sub y = ex : ex = −

1 2

(NA ∵ ex > 0) ex x

e

2y − 7√y + 3 1

y=

1 4

sub y = ex ex =

8(a)

1 4 1

x

= ln

x

≈ −1.39 ✓

4

x 3 = e6x−1 ln x 3 = ln e6x−1 3 ln x = 6x − 1 ln x = 2x −

12 ex

sub y = ex y

=7−

12

a=2✓

y

b=− ✓

y2 = 7y − 12 2 y − 7y + 12 = 0 (y − 3)(y − 4)= 0 y = 3 or y=4 x sub y = e sub y = ex x e =3 ex = 4 x = ln 3 x = ln 4 x ≈ 1.10 ✓ x ≈ 1.39 ✓


=0 √y = 3 y=9 sub y = ex ex = 9 x = ln 9 x ≈ 2.20 ✓

√y = 2 or

= 7 − 12e−x =7−

2ex = 7√ex − 3

(2√y − 1)(√y − 3) = 0

y=2 sub y = ex : ex = 2 x = ln 2 x ≈ 0.693 ✓

or

y=2 sub y = ex ex = 2 x = ln 2 x ≈ 0.693 ✓

1

lg 5

x

e3x + 2ex = 3e2x (ex )3 + 2ex = 3(ex )2 x sub y = e y 3 + 2y = 3y 2 y 3 − 3y 2 + 2y = 0 y(y 2 − 3y + 2) = 0 y(y − 1)(y − 2) = 0 y=0 or y = 1 or x sub y = e sub y = ex x e = 0 (NA) ex = 1 x =0✓

2ex = 7(ex )2 − 3 sub y = ex 2y = 7√y − 3

2e2x − 3ex =2 x )2 x 2(e − 3e =2 x sub y = e 2y 2 − 3y =2 2y 2 − 3y − 2 = 0 (2y + 1)(y − 2) = 0 y=−

7(b)

7(c)

1 3

1 3

8(b)

xe−x ln xe−x ln x + ln e−x ln x + (−x) ln x

= 2.46 = ln 2.46 = ln 2.46 = ln 2.46 = x + ln 2.46

a=1✓ b = ln 2.46 ≈ 0.9 ✓

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190


Ex 7.4

(xex )2 x 2 e2x x2 ln x 2 2 ln x 2 ln x

= 30e−x = 30e−x = 30e−3x = ln(30e−3x ) = ln 30 + ln e−3x = ln 30 + (−3x)

ln x

= − x + ln 30

3

1

2

2

9(d)

3x × 102x

= 4 × 20x−2

(3)x (102 )x = 4 (20)x (20)−2 (3)x (102 )

x

= 4(20)−2

(20)x

[

3

a=− ✓ 2

1

b = ln 30 2

(3)(102 ) (20)

x

]

= =

lg(15)x

= lg

x lg 15

= lg

x

5x−1 × 3x+2 = 10 (5x )(5−1 ) × (3x )(32 ) = 10 =

15x

=

x lg 15

= lg =

=

100 1 100 1 100

1 100

lg

lg 15

≈ −1.70 ✓

10 (5−1 )(32 ) 50

(5x )(3x )

x

1

(15)x

≈ 1.70 ✓ 9(a)

4 202

9 50 9

50 9

lg

lg 15

≈ 0.633 ✓ 9(b)

22x × 5x+1 =7 2 x x (2 ) × (5) (5) = 7 (4)x × (5)x

=

x

7 5 7

20

=

x lg 20

= lg

x

=

5 7 5

lg

7 5

lg 20

≈ 0.112 ✓ 9(c)

4(32x ) 4(32 )x (32 )

= ex = ex

x

ex 9 x

( ) e

9 x

= =

1 4 1 4

ln ( )

= ln

x ln

= ln

x

e 9

e

=

1 4 1 4

1 4 9 ln e

ln

≈ −1.16 ✓


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191


Ex 7.4

1st eqn 4x+3

= 32(2x+y )

(22 )x+3

= (25 )(2x+y )

22x+6

= 25+x+y

⇒ 2x + 6

=5+x+y

x

=y−1

−(1)

2nd eqn 9x + 3y = 10

−(2)

sub (1) into (2): 9y−1 + 3y = 10 (32 )y−1 32y−2

+3y

= 10

+3y

= 10

(32y )(3−2 ) +3y 1

(3y )2 ( ) +3y

= 10

9

1 9

(3y )2

= 10

+3y −10 = 0

sub u = 3y 1 2 u 9

+ u − 10

=0

u2 + 9u − 90

=0

(u + 15)(u − 6)

=0

u = −15 or

u

=6

3y = −15

3y

=6

(rej)

lg 3y = lg 6 y lg 3 = lg 6 lg 6

y

=

y

≈ 1.631 ✓

lg 3

x|y=1.631 ≈ (1.631) − 1 ≈ 0.631 ✓


sleightofmath.com

192


Ex 7.5

Ex 7.5 1(a)

y = log 2 x ∵ base = 2 > 1, graph slopes up

y O

2(ii)

pH = − lg(7.8 × 10−6 ) ≈ 5.11 ✓ ∴ Blue

3(i)

T = 85(0.96)x T|x=0 = 85(0.96)0 = 85°C ✓

3(ii)

T|x=15 = 85(0.96)15 ≈ 46.1 ✓

3(iii)

T = 30 85(0.96)x = 30

y = log 6 x ∵ base = 6 > 1, graph slopes up y O

x

1

✓ 1(c)

pH = −lg(5.6 × 10−7 ) ≈ 6.25 ✓ ∴ Pink

x

1

✓ 1(b)

2(i)

y = log 0.2 x ∵ 0 < base 0.2 < 1, graph slopes down y O

x

1

✓ 1(d)

6

(0.96)x

=

x lg 0.96

= lg

x

=

x

≈ 25.5 ✓

17 6 17

lg

6 17

lg 0.96

4(i)

N = 100(1.65)t N|t=0 = 100(1.65)0 = 100 ✓

4(ii)

N|t=4 = 100(1.65)4 ≈ 741 ✓

4(iii)

N 100(1.65)t (1.65)t t lg(1.65)

= 400 = 400 =4 = lg 4

t

=

t

≈ 2.77 ✓

y = log 1 x 4 1

∵ 0 < base < 1, graph slopes down 4

y O

1

x ✓

1(e)

y = lg x ∵ base = 10 > 1, graph slopes up y O

1

x

5(i)

R = 100e−0.000427 9t R|t=10 000 = 100e−0.0004279(10 000) = 100(1 − e−4.279 ) ≈ 1.39g ✓

5(ii)

R

✓ 1(f)

y = ln x ∵ base = e > 1, graph slopes up

y

lg 4 lg(1.65)

1

= (100) 2

100e(−0.0004279)t = 50 O

1

x ✓


sleightofmath.com

1

e−(0.0004279)t

=

−0.0004279t

= ln

t

≈ 1620 yrs ✓

2 1 2

193


6(ii)

Ex 7.5

I = 0.87T I|T=1.5 = (0.87)1.5 ≈ 0.811 ✓

9(ii)

I = 0.5 T 0.87 = 0.5 T lg 0.87 = lg 0.5

≈

ln(t + 1)

≈ ln e 6

⇒t+1

≈ e6

t t

≈ e6 − 1 ≈ 7.72

6

13

lg 0.5

13

=

T

≈ 4.98 mm ✓

lg 0.87

∴t= 8✓

y

O

13

ln(t + 1)

13

T

7(i)

S ≈ 62 75 − 6 ln(t + 1) ≈ 62 6 ln(t + 1) ≈ 13

𝑦 = ln 𝑥 𝑦 = lg 𝑥 x

1

9(iii)

S = 75 − 6 ln(t + 1) 6 ln(t + 1) = 75 − S 1

ln(t + 1) = (75 − S)

✓

6

1

(75−S)

7(ii) (a)

Range of values of x: ⇒x>1✓

lg x > 0

Range of values of x: ⇒0
lg x ≤ 0

7(ii) (c)

Range of values of x: ⇒x>1✓

ln x > lg x

8(ii)

1

10(i)

A = 5000(1.04)t A|t=5 = 5000(1.04)5 ≈ $6083 ✓

>

t

𝐴 5000

t 9(i)

1500 1+1499e−0.85(6) 1500 1+1499e−5.1

≥ 0.4(1500) 1500

1+1499e−0.85t 1500 5 3

≥ 1 + 1499e−0.85t ≤ 1499e−0.85t

2 3

≥ e−0.85t

2998

ln ( t

3 2998

)

3 ) 2998

ln(

(1.04)𝑡

−0.85

5000

)✓

S = 75 − 6 ln(t + 1) 0 ≤ t ≤ 12 S|t=4 = 75 − 6 ln(4 + 1) = 75 − 6 ln(5) = 75 − 6 ln 5 ≈ 65.3 ✓


sleightofmath.com

≥ 600 ≥ 1 + 1499e−0.85t

600

lg 1.04

A

−1✓

1+1499e−0.85t

10(ii) y

2

= log1.04 (

(75−S)

1500

y=

lg 1.6

= 5000(1.04) =

= e6

(75−S)

≈ 148 ✓

> 11.98

A

t

=

∴ t = 12 ✓ 8(iii)

= e6

y|t=6 =

A > 8000 t 5000(1.04) > 8000 (1.04)t > 1.6 t lg(1.04) > lg(1.6) t

⇒t+1

1

7(ii) (b)

8(i)

ln(t + 1) = ln e6

≥ −0.85t ≤t

8.13

≤t

t ∴t= 9✓

≥ 8.13

194


t = −2.35 ln (

Ex 7.5

T−24.5 12.4

13(ii)

)

3.5 = Ce

t|T=28.8 = −2.35 ln (

28.8−24.5 12.4

)

3.5

− 12.4

)

2h 58min before 7:29am ⇒ 4: 31am [verified]

log 4 x

=

29 60

4.3

[shown] ✓ 35

k = ln ( ) 43

1

13(iii) T = T0 + Ce−kt T0 = 24.5, C = 12.4, k ≈ 0.426: T = 24.5 + 12.4e−0.426t

2 1 2

x x

3.5

k ≈ 0.426 ✓ t = 0, T = 36.9°C (assumption of normal body temperature of 36.9°C at death in Q11) 36.9 = 24.5 + Ce−0.426k(0) 36.9 = 24.5 + C C = 12.4 ✓

≈ 2.97h ≈ 2h 58min

=1

29 60

=e

e−60k =

7: 29am: T = 28.0°C t|T=28.0 = −2.35 ln (

Ce−kt1

29

2h 29min before 7:00am ⇒ 4: 31am ✓

=4 =2✓

T−24.5

ln (

y

𝑦 = 2 log 4 𝑥 𝑦=1 x

1 2

t 14(i)

✓

12(iii) 2 log x 4 ≤ 1 ⇒0
= e−0.426t

12.4 T−24.5

12(ii) y = 2 log 4 x

O

29 ) 60

− k

4.3

28.0−24.5

−k(t1 +

Ce

=

4.3 3.5

≈ 2.49h ≈ 2h29min

2 log 4 x

−(2)

(2) ÷ (1):

7: 00am: T = 28.8°C

12(i)

29 60

−k(t1 + )

14(ii)

29 60

−k(t1 + )

28.0 = 24.5 + Ce

)

= −0.426t ≈ −2.35 ln (

T−24.5 12.4

)✓

C1 : y = 2 ln x [x > 0] C2 : y = ln x 2 [x 2 > 0] 2 C3 : y = (ln x) [x > 0] All different. C1 ≠ C2 because of the different domain C1 ≠ C3 Evident by algebriac observation ✓ C2 ≠ C3 Evident by algebraic observation

T = T0 + Ce−kt 7: 00am ⇒ T = 28.8°C, T0 , 24.5°C, t = t1 28.8 = 24.5 + Ce−kt1 4.3 = Ce−kt1 −(1) 7: 29am ⇒ T = 28.0°C, T0 , 24.5°C, t = t1 +

12.4

y

29 60

✓

C3 : y = (ln x)2 C2 : y = ln x 2

C1 : y = 2 ln x 𝑥 ✓


sleightofmath.com

195


Rev Ex 7 A2(b) 2 log 5 x

Rev Ex 7 A1(a) log x 27 = 1.5 x1.5 = 27 3

= 33

x2

= (33 ) = 32 =9✓

x x x

A1(b) log 2 x log 8 x log 2 x ( log 2 x (

log2 x

3

) = 12

1 64

A1(c) log 3 (x − 2) log 3 (x − 2) log 3 (x − 2) ⇒x−2

2 log 5 x

=5−(

log5 x log5 52 log5 x 2 log5 x

)

)

)

2

=5−( ) u

2

x

✓ =3 − log 3 (x + 4) 3 = log 3 3 − log 3 (x + 4) = log 3 ( =

= 52 = 25 ✓

x

1 2

=5

= √5 ✓

33

)

A3(a) log 3 (x − 19) x − 19 x x x lg x

x+4

=4 = 34 = 34 + 19 = 81 + 19 = 100 = lg 102 =2✓

33 x+4

A3(b) lg(x + 2) + 7 lg 2

(x − 2)(x + 4) = 33 x 2 + 2x − 8 = 27 2 x + 2x − 35 = 0 (x + 7)(x − 5) = 0 x = −7 or x = 5 ✓ (rej ∵ x − 2 > 0) A2(a) log 3 xy − log 3 (x − 1) log 3 xy − log 3 (x − 1) xy log 3 ⇒

=5−(

= 12

= 2±6

x = 64 or

2 log 5 x

log5 25

2u2 = 5u − 2 2 2u − 5u + 2 = 0 (2u − 1)(u − 2) = 0 1 u =2 u = or 2 log 5 x = 2 1 log 5 x =

(log 2 x)2 = 36 log 2 x = ±6 x

=5−(

2u

) = 12

log2 23

2 log 5 x

sub u = log 5 x:

= 12

log2 8 log2 x

(log2 x)2

2 3

= 5 − log x 25

x−1 xy

x−1

xy y


= log 3 9

+ lg(2x + 1)

lg(x + 2) + lg 27

= log 3 32 + lg(2x + 1)

lg[(x + 2)27 ]

=2

+ lg(2x + 1)

lg[(x + 2)27 ]

= lg 102

+ lg(2x + 1)

lg[128(x + 2)]

= lg[100(2x + 1)]

⇒ 128(x + 2)

= 100(2x + 1)

= log 3 6x − 1 = log 3 6x 2 − log 3 3 = log 3 2x 2

32(x + 2)

= 25(2x + 1)

32x + 64

= 50x + 25

= 2x 2

−18x

= −39

= 2x 2 (x − 1) = 2x(x − 1) ✓

x

=

2

A3(c) e2x + ex − 6 = 0 (ex )2 + ex − 6 = 0 sub u = ex : u2 + u − 6 =0 (u + 3)(u − 2) = 0 u = −3 sub u = ex ex = −3 (NA ∵ ex > 0)

sleightofmath.com

13 6

✓

u=2 ex = 2 sub u = ex : x = ln 2 x ≈ 0.693 ✓ 196


Rev Ex 7 A5(ii) P > 90 000 0.07n 12 000e > 90 000

log 2 x = a = 2a

x

log 8 y = b = 8b

y

x 2 y = (2a )2 (8b ) = 22a+3b ✓ x

=

2a 8b

=

2a 23b

A6

= 2a−3b ✓

=

x

5−3b 2

2a−3b = 2−1 ⇒ a − 3b = −1

5−3b 2

>

n

> 28.78

0.07

log b (xy 2 ) =m log b x + log b y 2 = m log b x +2 log b y = m −(1)

=

2n−m 5

[shown]✓ −(3)

sub (3) into (1): (

2n−m

) + 2 log b y = m

5

2 log b y

7

= ✓ 9

=m+

2 log b y

=

a|b=7 = ✓

log b y

=

Let P be the population of a town after n years from the beginning of 1990 P = 12000e0.07n

log b

4

A5(i)

n

log b x

5 − 3b − 6b = −2 9b =7

9

2

15 2

−(2)

) − 3b = −1

b

ln

15

sub (2) into (1): log b x +2(n − 3 log b x) = m log b x +2n − 6 log b x = m −5 log b x = m − 2n

sub (1) into (2): (

> ln

−(1)

= 0.5

y

0.07n

2

log b (x 3 y) =n log b x 3 + log b y = n 3 log b x + log b y = n log b y = n − 3 log b x −(2)

A4(ii) x 2 y = 32 2a+3b 2 = 25 ⇒ 2a + 3b = 5 a

>

∴ 1990 + ⌊28.78⌋ = 2018 ✓

= 22a (33b )

y

15

e0.07n

3

1990 ⇒ n = 0 2005 ⇒ n = 2005 − 1990 = 15

y x

m−2n 5

6m−2n 5 3m−n 5

✓

= log b y − log b x =

3m−n

=[ =

−

2n−m

5 5 (3m−n)−(2n−m) 5

4m−3n 5

]

✓

1

log b √xy = log b xy P|n=15

2 1

= (log b x + log b y)

0.07(15)

= 12000e ≈ 34 292 ✓

2 1 2n−m

= (

2 5 1 2m+n

= [ =


sleightofmath.com

2 5 2m+n 10

+

3m−n 5

)

]

✓

197


Rev Ex 7 B1(c) log 2 x 2 − log 2 (2x + 5) = 2

A7 y = 2 lg x (i)(a) y|x=4 = 2 lg 4 ≈ 1.20 ✓

log 2 ( ⇒

A7 y = 1.5 (i)(b) 2 lg x = 1.5 lg x = 0.75 log10 x = 0.75

A7 (ii)

x2

= 22

2x+5

=4

2x+5

x2

= 8x + 20

x 2 − 8x − 20

=0

(x − 10)(x + 2) = 0 x = 10 or x = −2 ✓ B1(d) log 2 x

y = 2 lg x y x

1

✓ A7 (iii)

4

x√10 = √10x 1 2

x (10 ) = 10

= log 2 22

)

2x+5

x2

= 100.75 ≈ 5.62 ✓

x x

x2

B2(i)

x 4

= 4 log x 2 1

log 2 x

= 4(

log 2 𝑥

= ±2

x

2±2 = 4 or

log2 x

)

1 4

log 2 x = p log 4 y = q

x 1

= 104−2

x

x

1

4 1

2

lg x

= −

2 lg x

= x−1

log 2 xy = log 2 x + log 2 y = log 2 x + = log 2 x +

2

= log 2 x +

1

∴y= x−1✓ 2

B1(a) log 9 (3𝑥+1 ) 9x

= 3x+1 2

32x = 3x+1 ⇒ 2x 2 =x+1 2 2x − x − 1 =0 (2x + 1)(x − 1) = 0

= = =

2

=

43x + log 2 2 43x −3 43x (22 )3x 26x ⇒ 6x x

=5 =5 =5 =8 = 23 = 23 =3

log2 x

− log 4 y

log2 4 p log2 22 p 2 p

− log 4 y − log 4 y −q ✓

2

B2(iii) log 4y = log2 4y = log2 [4(22q )] x log2 x

=

log2 [22 (22q )]

B2(iv) log 2 x = p

p

p

=

log2 (22+2q ) p

=

2+2q p

✓

log 4 y = q

y = 4q y = 22q x 2 y = (2p )2 (22q ) = 22p+2q ✓ x

1

= ✓


1 2

y

1

8 −3

1

log4 4 2 log4 y

B2(ii) log 4 x = log 4 x − log 4 y

x = − or x = 1 ✓ B1(b) 43x + log (1) 2

log4 2 log4 y

= log 2 x +2 log 4 y =p +2q ✓

= x2

2

log4 y

2

sleightofmath.com

= 2p

198


Rev Ex 7

B3(a) 2x = 128(4y ) 2x = 27 (22y ) 2x = 27+2y ⇒ x = 7 + 2y −(1) ln(4x + y) = ln 40 − 2 ln 2 ln(4x + y) = ln 40 − ln 22 ln(4x + y) = ln 10 ⇒ 4x + y = 10 −(2)

B4(i)

ln 2 = a ln 5 = b 1

3

ln √10e = ln 10e 3 1

= (ln 10 + ln e) 3 1

= [ln(2 × 5) + ln e] 3 1

= [ln 2 + ln 5 +1) 3 1

= (a + b +1)

sub (1) into (2): 4(7 + 2y) + y = 10 (28 + 8y) + y = 10 28 + 9y = 10 9y = −18 y = −2

3 1

= (a + b + 1) ✓ 3

B4(ii) ln x = b−2a = =

x|y=−2 = 7 + 2(−2)

=

=3✓

2 (ln 5)−2(ln 2) 2 ln 5−ln 22 2 ln 5−ln 4 1

2 5

2

4

= ln B3 lg(1 + 2x) − lg x 2 = 1 − lg(2 + 5x) 2 (b)(i) lg(1 + 2x) − lg x = lg 10 − lg(2 + 5x) lg ⇒

1+2x

= lg

x2 1+2x

= lg

x2

5

= ln √

4

10

⇒x =

2+5x 10

√5 ✓ 2

2+5x

(2x + 1)(5x + 2) = 10x 2 10x 2 + 9x + 2 = 10x 2 9x + 2 =0 2

=− ✓

x

9

B3 3y+2 = 5y y 2 (b)(ii) (3) (3 ) = (5)y (3)y

=

(5)y 3 y

( ) 5

=

3 y

lg ( ) 5

1 32 1 9

= lg

3

y lg ( ) = lg 5

1 9 1 9

1 9 3 lg 5

lg

y

=

y

≈ 4.30 ✓


sleightofmath.com

199


Rev Ex 7

y = ln(2x + e2 ) At (0, h), h = ln[2(0) + e2 ] h = ln e2 h =2✓

B6(b) log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 log 4 2 1

log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 log 4 42 1

log 3 5 + log 3 (4 + x) = log 3 (10 − x) + 2 ( ) 2

y = ln(2x + e2 ) At (k, 0), 0 = ln[2(k) + e2 ] 0 = log 𝑒 (2𝑘 + 𝑒 2 ) 0

k

=

1−e2 2

✓

1

y = − eax 2

At (h, k) i.e. (2, 1−e2 2 2

= log 3 (10 − x) + 1 = log 3 (10 − x) + log 3 3 = log 3 [3(10 − x)] = 3(10 − x) = 30 − 3x = 10

x

= ✓

5 4

2

e = 2k + e 1 = 2k + e2 1 − e2 = 2k

1

log 3 5 + log 3 (4 + x) log 3 5 + log 3 (4 + x) log 3 [5(4 + x)] ⇒ 5(4 + x) 20 + 5x 8x

1−e2 2

).

1

= − ea(2)

B6(c) 102x+1 +7(10x ) = 26 x 2 x (10 ) (10) +7(10 ) = 26 x )2 x) 10(10 +7(10 −26 = 0 sub u = 10x : 10u2 +7u −26 =0 (10u − 13)(u + 2) =0 13 u = −2 u= or 10 10x = −2 (rej) 13 x 10 = 10

2

1

x

2

2 1 2 e 2 1 ln ( e2 ) 2 1 1 2

sub x = lg a:

e2a

=

2a

=

a

= ln ( e ) 2 1

2 1

2 1

2

= lg

13

1

− e2 = − e2a

lg a = lg a

=

13 10

10

13 10

✓

= (ln + ln e2 ) = (ln 2−1 + 2) 2 1

= (− ln 2 + 2) 2

1

= 1 − ln 2 ✓ 2

B6 m = 32e−0.02t (a)(i) m|t=20 = 32e−0.02(20) ≈ 21.5 ✓ 1 B6 m = m|t=0 2 (a)(ii) 1 −0.02t 32e = [32e−0.02(0) ] 2

32e−0.02t = 16 1

e−0.02t

=

−0.02t

= ln

t

= −50 ln

t

≈ 34.7 ✓

2 1 2


1 2

sleightofmath.com

200


x 0.2 0.4 0.6 0.8 1.0 1.2 y -1.61 -0.92 -0.51 -0.22 0.00 0.18

Rev Ex 7 B7(ii) xe2x

= √e3 3

xe2x

= e2

ln(xe2x )

=

ln x + 2x = ln x B7(iii)

3 2

3 2 3

= −2x + ✓

y = −2x +

2

3 2

B7(iv) Intersection pt (0.84,0.16) ⇒ x = 0.84 ✓ There is only 1 intersection (and thus only 1 solution) as both curves would not turn around. ✓


sleightofmath.com

201


Ex 8.1 1(a)

Ex 8.1

Method 1 (No action) xy = 3x −5 Y = 3X −5 ⇒m=3✓ ⇒ c = −5 ✓

x

Y = X −2 ⇒m=1✓ ⇒ c = −2 ✓ Method 2 (Divide by 𝐱 𝟐 ) y = x 2 −2x

Method 2 (Divide by x) xy = 3x −5 −

y x2 y

5

y

=3

y

= −5 ( ) + 3

x

xy 1

x

Y = −2X +1 ⇒ m = −2 ✓ ⇒c=1✓

xy

x

Method 3 (Divide by y) y = x 2 −2x 1 =

y 3

xy

x2 y

= −1 =

= −2 ( ) + 1

5

y 3

−

x2

1

y

5y 5 3 1 1 5 y 5 3 1

Y = X−

x2

−

y

=1 +

2x y 2x y

x

= 2 ( ) +1 y

Y = 2X +1 ⇒m=2✓ ⇒c=1✓

= ( )− 5

x

1

3

1 = −

2

−

x2

Method 3 (Divide by 𝐱𝐲) xy = 3x −5

xy 1

=1

1

Y = −5X +3 ⇒ m = −5 ✓ ⇒c=3✓

5

Method 1 (Divide by x) y = x 2 −2x y = x −2

5

3

⇒m= ✓ 5

1

1(c)

⇒c=− ✓

Method 1 (Multiply denominator) y

5

=

3 x−2

xy − 2y = 3 xy = 2y +3 Y = 2X +3 ⇒m=2✓ ⇒c=3✓ Method 2 (Swap y and 𝐱 − 𝟐) y

=

x − 2=

3 x−2 3 y 3

x

=

+2

x

= 3 ( ) +2

y 1 y

Y = 3X ⇒m=3✓ ⇒c=2✓


sleightofmath.com

+2

202


Ex 8.1

Method 1 (Multiply x) 5

3y = 2x

−

3xy= 2x 2

−5

2

xy = x 2

−

3 2

xy = (x 2 ) − 3 2

Y = X− 3

2(a)

x

y 2 = 2x + 3 Y = 2X + 3 𝑦2

5 3 5

3

3

𝑥

𝑂

5

✓

3

2

⇒m= ✓ 3

2(b)

5

⇒c=− ✓ 3

xy = 2y 2 − 5 Y = 2X − 5 𝑥𝑦

Method 2 (Divide by x) 3y

= 2x

−

y

3( )= 2

−

x

y x y

=

2

−

3 5

1

x

= − ( 2)

+

Y

=− X

+

3 x 5

3 5

5

x2 5 3x2 2

✓ 3(a)

3 2 3

Method 1 (Divide by x) xy 2 = 2x +5y y 2 y =2 +5 x

⇒m=− ✓

y

3

2

y

= 5 ( ) +2 x

Y = 5X +2 ✓

Log both sides y = 10 × 2x lg y = lg(10 × 2x ) = lg 10 + lg 2x =1 +x lg 2 = lg 2 (x) +1 ✓ Y = (lg 2)X +1 ⇒ m = lg 2 ✓ ⇒c=1✓

Method 2 (Divide by y) xy 2 = 2x +5y x

xy = 2 ( ) +5 y

+5 ✓

Y = 2X 3(b)

Method 1 (Divide by x) 3xy= 5y −2x y

3y = 5 ( ) −2 y

1(f)

2

⇒c= ✓ 3

1(e)

𝑦2

𝑂 -5

x 5

Log both sides y = 5x 7 lg y = lg(5x 7 ) = lg 5 + lg x 7 = lg 5 +7 lg x = 7 lg x + lg 5 Y = 7X + lg 5 ⇒m=7✓ ⇒ c = lg 5 ✓

x 5 y 3 x 5

3 2

− ✓

Y = X 3

3

Method 2 (Divide by y) 3xy= 5y −2x x

−2 ( )

3x = 5 5

y

2 x

− ( )

x

=

x

=− ( ) +

3

3 y 2 x

5

3 y 2

3 5

Y =− X 3


2

= ( ) −

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+ ✓ 3

203


Ex 8.1

3

y

y

5 Non-linear eqn: (c)(ii)

= 5x 2 3 2

lg y = lg (5x ) lg y = lg 5 + lg x

1

= − (x + y) +2

x

2

y

3 2

y x

3

lg y = lg 5 + lg x 2

1

lg 5 𝑂 5(a) (i)

3 lg x + lg 5 2 lg 𝑥

(0,1) (4,9)

 Gradient:

m=

 Y-intercept:

c=1

1−9 0−4

−8

=

2

1

1

x

2

1

−4

2+x 2x

2

) =

y 5 Points: (d)(i) Gradient:

✓

 Points:

= − x +2

2

y( lg y =

2

1

+ y

2

lg 𝑦

1

2

y ( + ) = − x +2

3

lg y = lg x + lg 5 [shown] ✓ 4(ii)

1

= − x − y +2

x

2 x(4−x) x+2

✓

(0, −2) (4,1) m=

Y − intercept: Linear eqn:

=2

=

4−x

(−2)−(1) (0)−(4)

=

−3 −4

=

3 4

c = −2 Y = mX +c 3

= ( ) X + (−2) 4

3

= X −2 ✓ 4

Linear eqn:

5(a) (ii)

Non-linear eqn:

Y = mX + c = 2X + 1 ✓

 Y − intercept: Linear eqn:

5 Non-linear eqn: (b)(ii) 5  Points: (c)(i)  Gradient:  Y-intercept: Linear eqn:

x

A(−4,5) (0,1) (5)−(1)

m = (−4)−(0) =

4 −4

1

y = −x 3 + ✓

 Point:

(1,3) or (2,2)

 Gradient:

m=

Linear eqn:

Y − Y1 = m (X − X1 ) Y − (3) = (−1)[X − (1)] Y − 3 = −X +1 Y = −X +4

x

0−4

c=2 Y = mX

2 −4

3−2 1−2

=

lg y = lg (

(0,2) (4,0) =

−2x ✓

1

= −1

−1

Non-linear eqn: lg y = − lg x +4 lg y = − lg x + lg 104

xy = −x 3 + 1

2−0

4 3 2 x 4

= −1

c=1 Y = mX +c = (−1)X + 1 = −X +1 ✓

m=

3

= x −2

y=

y = 2x 2 + 1 ✓ 6(a)

5  Points: (b)(i)  Gradient:

y

5 Non-linear eqn: (d)(ii)

=−

⇒y =

1 2

6(b)

+c

104

x 10 000 x

 Point:

(1,3) or (3,4)

 Gradient:

m=

Linear eqn:

Y − Y1

1

3−4 1−3

=

−1 −2

)

✓

=

1 2

= (− ) X +(2) 2

1

=− X 2

+2 ✓

= m (X − X1 ) 1

Y − (3) = ( ) [X − (1)] 2

Non-linear eqn:


sleightofmath.com

1

1

2 1

2 5

2 1

2 5

Y−3

= X−

Y

= X+

x+y x

= x+

x+y

=

y

=

2 2 1 2 5 x + x 2 2 1 2 3 x + x 2 2

✓

204


Ex 8.1

Method 1 (non-linear to linear) Non-linear eqn: y = pqx ln y = ln(pqx ) = ln p + ln qx = ln p +(x ln q) = (ln q)x + ln p

8(i)

 Points:

(1,8) or (4,2)

 Gradient:

m = (1)−(4) =

Linear eqn:

Y − Y1 Y − (8) Y−8 Y (y√x)

= m (X − X1 ) = (−2)[X − (1)] = −2X +2 = −2X +10 = −2(x) + 10

y

= −2√x +

(8)−(2)

Non-linear eqn:

Linear eqn: Y = (ln q)X + ln p 8(ii)

y|x=16 = −2√16 +

Y-intercept = 2 ln p =2 p = e2 ≈ 7.4 ✓

= −2(4) + =− 8(iii)

Gradient = − ln q

=−

7(i)

2 3 2

m=−

10

√16 10 4

✓

C(9, −8) lies on graph of y√x against x

y√x = −8 y(√9) = −8 3y = −8 y

3

8

=− ✓ 3

2

9(i)

3

Y = − X +2 2

Non-linear eqn:

 Points:

(−1, −3) (0,0) (1,3)

 Gradient:

m = (−1)−(0) =

 y-intercept:

c=0

Linear eqn:

Y = mX + c = (3)X + (0) = 3X

Non-linear eqn:

y − x = 3x 3 y = 3x 3 + x ✓

3

ln y = − x + 2 2

3 2

− x+2 3

= e2 (e−2x ) 2

✓

Equate Y(y√x):

Linear eqn: Y = mX +c

=e

√x

3

Method 2 (linear to non-linear) Point: (0,2) Y-intercept: c = 2

y

10

Equate X(x): x=9

=e 2 ≈ 0.2 ✓

Gradient:

2

= −2

3

−

q

11

6 −3

3

= (ee 2 ) (e−2 )

x

9(ii)

(−3)−(0)

−3

=3

−1

y|x=2 = 3(2)3 + (2) = 24 + 2 = 26 ✓

= pqx [given] ⇒ p = e2 ≈ 7.4 ✓ 3

⇒ q = e−2 ≈ 0.2 ✓ 7(ii)

y = pqx = 7.4(0.2)x ✓


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205


Ex 8.1

m + 2n = 18 m = 18 − 2n

11(b) y = ab1−x lg y = lg(ab1−x ) = lg a = lg a = lg a = − lg b (x) lg y = − lg b (x)

−(1)

n2 − m = 17 sub (1) into (2): n2 − (18 − 2n) = 17 n2 + 2n − 18 = 17 2 n + 2n − 35 =0 (n − 5)(n + 7) = 0 n=5 or m|n=5 = 18 − 2(5) =8 ⇒ C(8,5)

−(2)

n = −7 m|n=−7 = 18 − 2(−7) = 32 ⇒ C(32, −7) (rej ∵ graph slopes up)

11(c) aey = b2x ln(aey ) = ln(b2x ) ln a + ln ey = (2x) ln b ln a + y = 2x ln b y = 2x ln b − ln a y = 2 ln b (x) − ln a ✓ 12

Equate X (x 3 ) − coordinate: x3 = 8 x =2✓ Equate Y (y − x) − coordinate: y−x =5 y − (2)= 5 y =7✓ 10(i)

 Gradient:

m=

Linear eqn:

Y − Y1 = m(X − X1 ) Y − (10) = 3 [X − (3)] Y − 10 = 3X − 9 Y = 3X + 1

Non-linear eqn:

y

1

y|x= 1 = 3 ( ) √2

√2

= = = = 11(a) x

=

y−b =

3 √2 3

×

√2 3√2

√2

2 3√2+1 2

−18 −6

=3

1 x

= 3x + x 2 ✓ 1

+( )

2

√2

+ √2

=

= 3( ) + 1

x2

y 10(ii)

⇒m=q✓ ⇒ c = lg p ✓

(3,10) or (9,28) (3)−(9)

y = p(x + 1)q lg y = lg[p(x + 1)q ] = lg p + lg(x + 1)q = q lg(x + 1) + lg p = q lg(x + 1) + lg p Plot lg y against lg(x + 1) ✓

 Point:

(10)−(28)

+ lg b1−x +[(1 − x) lg b] + lg b − lg b (x) + lg a + lg b + lg ab ✓

+ +

1 2 1 2 1 2

✓

a y−b a x a

y

= +b

y

= a( ) + b ✓

x

1 x


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206


Method 1 (non-linear to linear) Non-linear eqn: ax y =

Ex 8.1 14(i)

x−b

yx − yb = ax yx = yb +ax y y = (b) +a x

1

 Point:

A(3,1) or B (8,3 ) 2

 Gradient:

mAB =

Linear eqn:

Y − Y1

Linear eqn: Y = bX +a

Non-linear eqn:

=

1 2

= m (X − X1 ) 3

2 1

2 1

2

2

Y−1

= X−

Y

= X− 1 1

1

2 x 1

2

xy

= ( )−

2xy

= − 1 (shown) ✓ x

2

1

4y

=−

y

=− ✓

2 1 8

−(1)

x

y=x

1

2(2)y = − 1

14(iii) 2xy = 1 − 1 2

−(2)

sub (2) into (1):

Linear eqn: Y = mX +c = 3X +2

1

2x(x 2 )

= −1

2x 3

= −1

x 1 x

2x 4 =1−x 4 2x + x − 1 = 0

Non − linear eqn: y

= 3 ( ) +2 x

y−

3y

let f(x) ≡ 2x 4 + x − 1 f(−1) = 2(−1)4 + (1) − 1 =0 ∴ x = −1 ✓

=2

x 3

y (1 − ) = 2 x

y(

−5

1

14(ii) when x = 2,

Y-intercept = 2 a = 2✓

y

5 2

2

x

Method 2 (linear to non-linear)  Gradient: m = 3  Y-intercept: c = 2

−

1

y

Gradient = 3 b =3

(3)−(8)

=

Y − (1) = ( ) [X − (3)]

= (b) +a

y

1 2

(1)−(3 )

x−3

)

x

y

=2 = =

2x x−3 ax x−b

⇒ a=2✓ ⇒ b=3✓ 13(ii) y =

2x x−3

x=y

−(1) −(2)

sub (2) into (1): (x)

=

2x x−3

x 2 − 3x = 2x x 2 − 5x = 0 x(x − 5) = 0 ∴ x = 0 or x = 5 ✓


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207


Ex 8.1

He connected the dots with a straight which is not necessarily the correct shape as verified by the graphing calculator

𝑦 𝑦 = 𝑥 7 − 14𝑥 5 + 49𝑥 3 − 35𝑥

𝑂

𝑥

y|x=−4 = −5044 ≠ −4 (−4, −4) does not lie on graph ✓ y|x=4 = 5044 ≠ 4 (4,4) does not lie on graph ✓


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Ex 8.2

Ex 8.2 1(i)

Linearization y = ax 2 + b x2 y

1(ii)

1 4 9 16 25 6.2 5.6 4.6 3.2 1.4

Gradient & Y-intercept (X1 , Y1 ) = (0,6.4) (X 2 , Y2 ) = (20,2.4) Gradient ≈

(6.4)−(2.4) (0)−(20) 1

≈− ✓

a

5

Y − intercept ≈ 6.4 b ≈ 6.4 ✓

2(i)

Linearization 1 y

= ax + b x 1 y

2(ii)

1

2

3

4

5

0.40 0.90 1.41 1.89 2.38

Gradient & Y-intercept (X1 , Y1 ) = (0, −0.1) (X 2 , Y2 ) = (4.5,2.15) Gradient ≈ a

(−0.1)−(2.15) (0)−(4.5)

≈ 0.50 ✓

Y − intercept ≈ −0.1 b ≈ −0.1 ✓


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209


Ex 8.2

Linearization y = ax 2 + x − b y − x = ax 2 − b 1 4 9 16 25 x2 y − x 40.5 36 28.5 18 4.5

3(ii)

Gradient & Y-intercept (X1 , Y1 ) = (0,42) (X 2 , Y2 ) = (20,12) Gradient ≈

(42)−(12) (0)−(20)

≈ −1.5 ✓

a

Y − intercept ≈ 42 −b ≈ 42 b ≈ −42 ✓ 4(i)

Linearization a

+

N a

t

=4 =−

N 1

=−

N 1

b

+4

t b at b 1

+

=− ( ) +

N

a

1 t 1 N

4(ii)

b

t

4 a 4 a

1.00 0.50 0.30 0.25 0.20 0.75 0.99 1.09 1.12 1.15

Gradient & Y-intercept (X1 , Y1 ) = (0,1.25) (X 2 , Y2 ) = (0.88,0,80) Gradient ≈ −

b a

(1.25)−(0.80) (0)−(0.88)

≈ −0.51

−(1)

Y − intercept ≈ 1.25 4

≈ 1.25

a

≈ 3.2 ✓

a

−(2)

sub (2) into (1): −b 3.2

b

≈ −0.51 ≈ 1.6 ✓


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Ex 8.2

Linearization xy = h(x + k) xy = hx + hk hk

y

=h+

y

= hk ( ) + h

x 1 x

1 x

5.00 2.50 1.67 1.25 1.00

y 5.25 3.38 2.75 2.44 2.23 5(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.5) (X 2 , Y2 ) = (4,4.5) Gradient ≈ hk

(1.5)−(4.5) (0)−(4)

≈ 0.75

Y − intercept ≈ 1.5 h ≈ 1.5 ✓

−(1)

−(2)

sub (2) into (1): (1.5)k ≈ 0.75 k ≈ 0.5 ✓


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211


Ex 8.2

Linearization y

= ax +

b x

xy = ax 2 + b x 2 0.25 1.00 2.25 4.00 xy 7.3 6.8 6 4.8 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (7.3)−(4.8)

≈ (7.3) − [(0.25)−(4.00)] (0.25) ≈ 7.47 Domain = [0,4] X-interval =

(4)−(0) 10

= 0.4 ⇒ 0.25

X − scale: 1 cm to 0.25 units Range

= [4.8,7.47]

Y-interval = 6(ii)

(7.47)−(4.8) 12

≈ 0.23 ⇒ 0.2

Y-scale: 1 cm to 0.2 units Gradient & Y-intercept (X1 , Y1 ) = (0,7.5) (X 2 , Y2 ) = (3,5.48) Gradient = a

(7.5)−(5.48) (0)−(3)

≈ −0.67 ✓

Y − intercept ≈ 7.5 b ≈ 7.5 ✓ 6(iii) Substitution y ≈ −0.67x +

7.5 x

y|x=1.7 ≈ −0.67(1.7) +

7.5 1.7

≈ 3.3 ✓


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Ex 8.2

Linearization y = ax 2 + bx y = ax + b x

x y x

7(ii)

1

2

3

4

1.6 3.6 5.6 7.6

Gradient & Y-intercept (X1 , Y1 ) = (0, −0.4) (X 2 , Y2 ) = (3.7,7) Gradient ≈

(−0.4)−(7) (0)−(3.7)

≈2✓

a

Y − intercept ≈ −0.4 b ≈ −0.4 ✓ 7(iii)

Intersection 𝑦 Put 𝑎 = 2, 𝑏 = −0.4 into = 𝑎𝑥 + 𝑏, 𝑦 𝑥

𝑥

= 2𝑥 − 0.4

2𝑥 2 − 0.4𝑥 = 10 2𝑥 − 0.4 𝑦

= =

𝑥

x y x

2.0

2.5

10 𝑥 10 𝑥

3.0

3.5

4 .0

5.00 4.00 3.33 2.86 2.50

Intersection Pt (2.35,4.3) ⇒ X ≈ 2.35 𝑥 ≈ 2.35 ✓


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Ex 8.2

Linearization P

=

En R

ln P = ln

En R

ln P = ln E n − ln R ln P = n ln E − ln R ✓ ln E 1.61 2.30 2.89 3.00 3.22 3.40 ln P 0.92 2.30 3.48 3.69 4.14 4.50 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (0.92)−(4.50)

≈ (0.92) − [(1.61)−(3.40)] (1.61) ≈ −2.3 Domain = [0,3.4] X-interval =

(3.4)−(0) 10

= 0.34 ⇒ 0.25

X − scale: 1 cm to 0.25 units Range

= [−2.3,4.5]

Y-interval = 8(ii)

(4.5)−(−2.3) 12

≈ 0.57 ⇒ 0.5

Y-scale: 1 cm to 0.5 units The graph produces a straight line ⇒ true ✓

8(iii) Gradient & Y-intercept (X1 , Y1 ) = (0, −2.3) (X 2 , Y2 ) = (3.25,4.2) Gradient ≈ n

(−2.3)−(4.2) (0)−(3.25)

≈2✓

Y − intercept ≈ −2.3 − ln R ≈ −2.3 R ≈ 10.0 ✓


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214


Linearization f = kλn lg f = lg[kλn ] = lg k + lg λn = lg k +n lg λ = n(lg λ) + lg k lg λ 2.40 2.67 2.81 3.08 3.18 lg f 3.08 2.81 2.67 2.40 2.30

9(ii)

Gradient & Y-intercept (X1 , Y1 ) = (0, 5.48) (X 2 , Y2 ) = (3,2.48) Gradient = n

Ex 8.2

(5.48)−(2.48) (0)−(3)

≈ −1 ✓

Y − intercept = 5.48 lg k = 5.48 k ≈ 301 995 ✓ 9(iii) Intersection kλn = 10−2.5n 𝑓 = 10−2.5𝑛 lg 𝑓 = −2.5𝑛 𝑌 = −2.5𝑛 𝑌 = 2.5 ∵ 𝑛 = −1 Intersection Pt (2.97,2.5) ⇒ X = 2.97 lg λ = 2.97 λ ≈ 933 ✓


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215


Ex 8.2

Linearization y

= Cx +

D x

xy = Cx 2 + D x 2 0.04 0.16 0.36 0.64 xy 1.55 1.69 1.93 2.27 10(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.5) (X 2 , Y2 ) = (0.6,2.22) Gradient ≈ C

(1.5)−(2.22) (0)−(0.6)

≈ 1.2 ✓

Y − intercept ≈ 1.5 D ≈ 1.5 ✓ 10(ii) Intersection Cx 2 + D = 2 𝑌 =2 Intersection Pt (0.42, 2): X ≈ 0.42 x 2 ≈ 0.42 x ≈ ±0.65 ✓


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216


Ex 8.2

Linearization y−x = kx n lg(y − x) = lg(kx n ) = lg k + lg x n = lg k + n lg x = n lg x + lg k lg x 0.30 0.48 0.60 0.70 0.78 lg(y − x) 1.00 1.26 1.45 1.59 1.71 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (1.00)−(1.71)

≈ (1.00) − [(0.30)−(0.78)] (0.30) ≈ 0.56 Domain = [0,0.78] X-interval =

0.78−0 10

= 0.078 ⇒ 0.05

X-scale: 1 cm to 0.05 units = [0.56,1.71]

Range

Y-interval =

(1.71)−(0.56) 12

≈ 0.096 ⇒

0.05 Y-scale: 1 cm to 0.05 units 11(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,0.55) (X 2 , Y2 ) = (0.68,1.56) Gradient = n

(0.55)−(1.56) (0)−(0.68)

≈ 1.5 ✓

Y − intercept ≈ 0.55 lg k ≈ 0.55 k ≈ 3.5 ✓ 11(iii) Substitution 𝑦 − 𝑥 = 3.5𝑥 1.5 −(1) 𝑦 = 𝑥 + 4.5 −(2) 𝑠𝑢𝑏 (2) 𝑖𝑛𝑡𝑜 (1): (𝑥 + 4.5) − 𝑥 = 3.5𝑥 1.5 4.5 = 3.5𝑥 1.5 9 7

𝑥

= 𝑥 1.5 ≈ 1.18


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Ex 8.2

Linearization y = Ae−bx ln y = ln[Ae−bx ] = ln A + ln e−bx = ln A −bx = −bx + ln A x 1 2 3 4 5 6 ln y 2.59 2.29 1.99 1.69 1.39 1.10 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (2.59) − [

(2.59)−(1.10) (1)−(6)

] (1)

≈ 2.89 Domain = [0,6] X-interval =

(6)−(0) 10

= 0.6 ⇒ 0.5

X-scale: 1 cm to 0.5 units = [1.10,2.89]

Range

Y-interval =

(2.89)−(1.10) 12

≈ 0.15 ⇒ 0.1

Y-scale: 1 cm to 0.1 units 12(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,2.88) (X 2 , Y2 ) = (5.8,1.16) Gradient ≈ −b b

(2.88)−(1.16) (0.2)−(5.8)

≈ −0.307 ≈ 0.3 ✓

Y − intercept ≈ 2.88 ln A ≈ 2.88 A ≈ 18 ✓ 12(iii) Substitution y = (18)e−(0.3)x = 18e−0.3x At y = 6: 6 = 18e−0.3x 1

= e−0.3x

3 1

ln = −0.3x 3

x

=

ln

1 3

−0.3

≈ 3.66 ✓


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218


Ex 8.2

13(i) Linearization y=a+

b x

1

= b( ) + a 𝑥

Plot y against 1 x

1 x

2.50 0.59 0.50 0.39 0.31 0.25 0.20

Y -0.5 0.8 1.5 1.6 1.7 1.75 1.8 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (−0.5)−(1.8)

≈ (−0.5) − [(2.50)−(0.20)] (2.50) ≈2 Domain = [0,2.5] X-interval =

(2.5)−(0) (10)

= 0.25 ⇒ 0.25

X-scale: 1 cm to 0.25 units Range

= [−0.5,2]

Y-interval =

(2)−(−0.5) 12

≈ 0.21 ⇒ 0.2

Y-scale: 1 cm to 0.2 units 13 Correction (ii)(a) Abnormal reading: y = 0.8 (2nd data) ✓ Correct reading:y ≈ 1.4 ✓ 13 Gradient & Y-intercept (ii)(b) (X1 , Y1 ) = (0,2) (X 2 , Y2 ) = (2,0) (2)−(0)

gradient ≈ (0)−(2) b

≈ −1 ✓

Y − intercept = 2.0 a ≈2✓


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219


Ex 8.2

Linearization T = kx n ln T = ln(kx n ) = ln k + ln x n = ln k +n ln x = n ln x + ln k ln x ln T

4.06 -1.43

4.68 -0.48

5.43 0.63

6.66 2.47

7.26 3.38

Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (−1.43)−(3.38)

≈ (−1.43) − [ (4.06)−(7.26) ] (4.06) ≈ −7.53 Domain = [0,7.26] X-interval =

(7.26)−(0) 10

≈ 0.73 ⇒ 0.5


= [−7.53,3.38]

Y-interval =

(3.38)−(−7.53) 12

≈ 0.91 ⇒ 0.5

Y-scale: 1 cm to 0.5 units 14(ii) Gradient & Y-intercept (X1 , Y1 ) = (0, −7.5) (X 2 , Y2 ) = (5,0) Gradient ≈ n

(−7.5)−(0) (0)−(5)

≈ 1.5 ✓

Y − intercept ≈ −7.5 ln k ≈ −7.5 k ≈ 5.5 × 10−4 ✓ 14(iii) Substitution Put 𝑛 − 1.5, 𝑘 ≈ 5.5 × 10−4 𝑖𝑛𝑡𝑜 𝑇 = 𝑘𝑥 𝑛 , T = (5.5 × 10−4 )(x)1.5 149.6 × 106 km = 149.6 million km T|149.6 = (5.5 × 10−4 )(149.6)1.5 ≈1✓


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220


Ex 8.2

15(i) The graph shows an upward sloping curve instead of a line.

15(ii) Linearization Plot P against ln t P = a ln t + b [P] = a[ln t] + b ln t 0.00 0.69 1.61 2.30 2.71 3.00 3.40 P 30 42 50 65 73 77 85 Scale (optional) c = 30 Domain = [0,3.4] X-interval =

(3.4)−(0) 10

= 0.34 ⇒ 0.25


= [30,85]

Y-interval =

(85)−(30) 12

≈ 4.58 ⇒ 2.5

Y-scale: 1 cm to 2.5 units 15(ii) Gradient & Y-intercept (X1 , Y1 ) = (0.9, 44) (X 2 , Y2 ) = (3.2, 80) (44)−(80)

Gradient = (0.9)−(3.2) a

≈ 15.7 ✓

Y − intercept ≈ 30 b ≈ 30 ✓ 15(iii) Substitution P = 15.7 ln t + 30 P = 95: (95) = 15.7 ln t + 30 15.7 ln t = 65 ln t ≈ 4.15 t ≈ 63.6s ✓


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221


Ex 8.2

Quadratic eqn y = ax 2 + bx y = ax + b x

Plot

y x

against x

x 2 3 4 5 6 7 y 9 11 13 15 17 19 x

Exponential eqn y = kh−x lg y = lg k + lg h−x = −x lg h + lg k = − lg h (x) + lg k Plot

y x

x lg y

against x 2 3 4 5 6 7 1.26 1.52 1.72 1.88 2.01 2.12

By inspection, Quadratic equation fits better ✓ 16(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,5) (X 2 , Y2 ) = (6.5,18) (5)−(18)

Gradient ≈ (0)−(6.5) a

≈2✓

Y − intercept = 5 b =5✓ 16(iii) Substitution y = 2x 2 + 5x y|x=4.5 = 2(4.5)2 + 5(4.5) = 63 ✓ 16(iv) No. x = 1 is outside the data range and thus the computed value is not reliable. ✓


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222


Ex 8.2

Quadratic Equation It is the only one among the 3 that has a turning point to fit the data given that increases then decreases. ✓


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223


Rev Ex 8 A2

Rev Ex 8 A1(a)  Points:

m=

 Y-intercept: Linear eqn: Non-linear eqn:

a

(0,5) (6,1)

 Gradient:

5−1 0−6

=

4

=−

−6

2 3

b

=

y

=

a

2

b

b

=− X+

3

Gradient = 0.75 a − = 0.75

) =5

y

=− ( )+

Y

2xy

=5 3

b x

Linear eqn:

=5− 2x

2 b 2

3 2 3

y (1 +

a 1

b x a 1

= (− ) +

y

= − xy + 5

3

x

2

y

y+

a

=− +2

y 1

= mX + c =− X+5

2xy

y

y 1

Y

y

b

+ =2

x b

c=5 Y

Method 1 (linear) Non-linear eqn:

b

a

5 2 1+ x 3

15 3+2x

= −0.75b

Y − intercept = −0.5

✓

2

= −0.5

b

A1(b)  Points:  Gradient:

A(−1,1) or B(2,7) mAB =

Linear eqn:

Y − Y1 Y − (1) Y−1 Y Non-linear eqn: (lg y) lg y lg y ⇒y

1−7 (−1)−2

=

−6 (−3)

−(1)

= −4 ✓

b =2

−(2)

sub (2) into (1): a|b=−4 = −0.75(−4) =3✓

= m(X − X1 ) = 2[X − (−1)] = 2X + 2 = 2X + 3 = 2(lg x) + 3 = lg x 2 + lg 103 = lg 1000x 2 = 1000 x 2 ✓

Method 2 (non-linear)  Gradient: m = 0.75  Y − intercept: c = −0.5 Linear eqn: Y = mX +c 0.75X −0.5 Non-linear eqn: 1

1

( )

= 0.75 ( ) −0.5

y

x

1

=

y

− 3 x a x

0.75

− +

x 4 y b y

+

1 y

0.75 x

−0.5

= −0.5 =2 =2

⇒ a=3✓ ⇒ b = −4 ✓ A3

y = √ax + b y 2 = ax + b Plot y 2 against x gradient = a ✓ y 2 − intercept = b ✓


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224


Rev Ex 8

A4(i) Linearization y = Ae−bt ln y = ln(Ae−bt ) = ln A + ln e−bt = ln A + [(−bt) ln e] = −bt + ln A = −bt + ln A t 1 2 3 4 5 ln y 2.50 1.95 1.39 0.83 0.26 A4(ii) Gradient & Y-intercept (X1 , Y1 ) = (0, 3.06) (X 2 , Y2 ) = (4.3, 0.65) Gradient = −b b

(3.06)−(0.65) (0)−(4.3)

≈ −0.56 ≈ 0.56 ✓

Y − intercept ≈ 3.06 ln A ≈ 3.06 A ≈ 21.3 ✓


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225


1 u 1 v

Rev Ex 8

0.050 0.040 0.033 0.025 0.020 0.050 0.059 0.067 0.077 0.080

A5(ii) Linearization 1 u 1 u

1

1

v

f

+ =

=−

1 v

+

1 f

Scale (optional) Y1 = mX1 + c c = Y1 − mX1 (0.05)−(0.08)

≈ (0.05) − [(0.05)−(0.02)] (0.05) ≈ 0.1 Domain = [0,0.05] X-interval =

(0.05)−(0) 10

= 0.005 ⇒ 0.005


= [0.05,0.1]

Y-interval =

(0.1)−(0.05) 12

≈ 0.0042 ⇒

0.0025 Y-scale: 1 cm to 0.0025 units 2 cm to 0.005 units Gradient & Y-intercept Y − intercept = 0.10 1 f

f

= 0.10 ≈ 10 ✓


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226


Rev Ex 8

Linearization xm yn lg(x m y n ) lg x m + lg y n m lg x + n lg y n lg y

= 200 = lg 200 = lg 200 = lg 200 = −m lg x + lg 200

lg y

=−

m n

lg x +

lg 200 n

lg x 0.00 0.48 0.70 0.86 lg y 1.15 0.43 0.10 -0.12 Scale (optional) c = 1.15 Domain = [0,0.86] X-interval =

(0.86)−(0) 10

= 0.086 ⇒ 0.05


= [−0.12,1.15]

Y-interval =

(1.15)−(−0.12) 12

≈ 0.11 ⇒ 0.1

Y-scale: 1 cm to 0.1 units A6(ii) Gradient & Y-intercept (X1 , Y1 ) = (0.1,1.0) (X 2 , Y2 ) = (0.78,0) (1.0)−(0)

Gradient = (0.1)−(0.78) −

m n

≈ −1.47

−(1)

Y − intercept = 1.15 lg 200 n

= 1.15 ≈2✓

n

−(2)

sub (2) into (1): m − ≈ −1.47 2

m ≈3✓ A6(iii) Substitution x 3 y 2 = 200 y = 10: x 3 (10)2 = 200 x3 =2 x 1.26 ✓


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227


Rev Ex 8

Method 1 (linear) Non-linear eqn: y 1 y 1 y

= =

B1(i)

h

Method 2 (non-linear)  Points: A(10,1) B(0, −4) C(r, −2)  Gradient:

2x+k 2x+k

(1)−(−4) (10)−(0)

=

1 2

Y − intercept: c = −4

h 2

k

h

h

= ( )x +

Linear eqn: Y = mX +c 1

Linear eqn: 2

k

h

h

Y = X+

m =

= X −4 2

Non-linear eqn: 1

At A(10,1): 2

1 = (10) 1 =

h 20+k

+

k h

At B(0, −4): k

h k

h

−4

= (0) +

−4

=

y

=

y

=

y

=

2 x

2 x−8

−4

2 2 x−8 4 2x−16 h 2x+k

⇒ h=4✓ ⇒ k = −16 ✓

h

−4h = k

=

y

−(1)

2

=

y 1

h

h = 20 + k

1

= (x) −4

y 1

−(2)

sub (1) into (2): −4(20 + k) = k −80 − 4k =k −80 = 5k k = −16 ✓

B1(ii) C(r, −2) lies on Y = 1 X − 4.

Put k = −16 into (1): h = 20 + (−16) = 4 ✓

B2(a) y =

2

1

(−2) = (r) −4 2

1 2

r

=4✓

r

x y

=2

x px+q

= px + q ✓

B2(b) y = pq−x lg y = lg(pq−x ) = lg p + lg q−x = lg p +(−x) lg q = − lg q (x) + lg p ✓

B2(c) ey = px 2 − qx ey x


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= px − q ✓

228


Rev Ex 8

Linearization axy − b = a(x 2 + bx) axy − b = ax 2 + abx axy − ax 2 = abx + b a(xy − x 2 ) = abx + b xy − x 2

= bx

+

x 1 2 3.00 xy − x Scale (optional) Y1 = mX1 + c c = Y1 − mX1

b a

2 1.00

3 -0.99

4 -3.00

(3)−(−3)

≈ (3) − [ (1)−(4) ] (1) ≈5 Domain = [0,4] (4)−(0)

X-intercept =

10

= 0.4 ⇒ 0.25

X-scale: 1 cm to 0.25 units 2 cm to 0.5 units Range

= [−3,5] (5)−(−3)

Y-intercept =

12

≈ 0.67 ⇒ 0.5

Y-scale: 1 cm to 0.5 units B3 Graphical Reading (ii)(a) x = 1.5 ⇒ X = 1.5 Pt (1.5,2.0) Y xy − x 2 (1.5)y − (1.5)2 y

=2 =2 =2 ≈ 2.83

B3(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,5) (b) (X 2 , Y2 ) = (3.5, −2) (5)−(−2)

Gradient = (0)−(3.5) ≈ −2 ✓

b

−(1)

Y − intercept = 5 b

=5

a

−(2)

sub (1) into (2): (−2) a

a

=5 = −0.4 ✓


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229


Rev Ex 8

Linearization y = Ka−x lg y = lg(Ka−x ) = lg K + (lg a−x ) = lg K + (−x) lg a = − lg a (x) + lg K ✓ x 1 2 3 4 5 6 7 lg y 1.19 0.99 0.79 0.48 0.38 0.18 -0.05

B4 Graphical reading (ii)(a) Flawed reading ⇒ y = 3.0 (4th data) Correct reading Y ≈ 0.59 lg y ≈ 0.59 y ≈ 3.89 ✓ B4 Gradient & Y-intercept (ii)(b) (X1 , Y1 ) = (0,1.4) (X 2 , Y2 ) = (6.9,0) (0)−(1.4)

Gradient = (6.9)−(0) − lg a a

= −0.2 ≈ 1.58 ✓

Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.19) − [

Y − intercept = 1.4 lg K = 1.4 K = 101.4 = 25.1 ✓

(1.19)−(−0.05) (1)−(7)

] (1)

≈ 1.38 Domain = [0,7] X-interval =

(7)−(0) 10

= 0.7 ⇒ 0.5

X-scale: 1 cm to 0.5 units B4 Graphical Reading (ii)(c) y = 10 lg y = 1 Y =1

Range

= [−0.05,1.38]

Y-interval =

(1.38)−(−0.05) 12

≈ 0.12 ⇒ 0.1

Y-scale: 1 cm to 0.1 unit

Point (2,1) ⇒X =2 x =2✓


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230


Rev Ex 8

Linearization y = ln(ax 2 + b) ey = ax 2 + b x2 ey

0.04 0.16 0.36 0.64 1.00 1.38 1.62 2.02 2.58 3.30

B5(ii) Gradient & Y-intercept (X1 , Y1 ) = (0,1.3) (X 2 , Y2 ) = (0.9,3.1) Gradient ≈ a

(1.3)−(3.1) (0)−(0.9)

≈2✓

Y − intercept ≈ 1.3 b ≈ 1.3 ✓ B5 Graphical Reading (iii)(a) y = ln 3 ey = 3 Y =3 Pt (0.85,3) X = 0.85 x ≈ ±0.92 ✓ B5 Graphical Reading (iii)(b) x = 0.1 x 2 = 0.01 X = 0.01 Pt (0.01,1.32) Y ≈ 1.32 y ≈ 0.28 ✓

Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.38) − [ ≈ 1.3 Domain

(0.04)−(1)

] (0.04)

= [0,1]

X-interval =

(1)−(0) 10

= 0.1 ⇒ 0.1

X-scale:

1 cm to 0.1 units

Range

= [1.3,3.3]

Y-interval = Y-scale:


(1.38)−(3.30)

(3.3)−(1.3) 12

≈ 0.17 ⇒ 0.1

1 cm to 0.1 units

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231


Rev Ex 8

B6(i) Linearization P = kc t lg p = lg(kc t ) = lg k + lg c t = lg k +t lg c = lg c (t) + lg k t 1 2 3 4 5 lg P 1.14 1.40 1.67 1.93 2.20 Scale (optional) Y1 = mX1 + c c = Y1 − mX1 ≈ (1.14) − [

(1.14)−(2.20)

] (1)

(1)−(5)

≈ 0.88 = [0,5]

Domain

X-interval =

(5)−(0) 10

= 0.5 ⇒ 0.5

X-scale:

1 cm to 0.5 units

Range

= [0.88,2.2]

Y-interval = Y-scale:

(2.2)−(0.88) 12

≈ 0.11 ⇒ 0.1

1 cm to 0.1 units

B6(ii) Gradient & Y-intercept (x1 , y1 ) = (0,0.87) (x2 , y2 ) = (3.5, 1.8) Gradient = lg c c

(0.87)−(1.8) (0)−(3.5)

= 0.27 ≈ 1.86 ✓

Y − intercept = 0.87 lg k = 0.87 k ≈ 7.41 ✓ B6(iii) Substitution P = (7.41)(1.86)t P|t=10 = (7.41)(1.86)10 ≈ 3.7 × 103 ✓


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232


Ex 9.1 1(c)

Ex 9.1 1(a)

1(b)

x -5 -4 -3 -2 -1 0 y ±3.16 ±2.83 ±2.45 ±2.00 ±1.41 0

y 2 = 4x for 0 ≤ x ≤ 5 x 0 1 2 3 y 0 ±2 ±2.83 ±3.46 Line of symmetry: y = 0

y 2 = −2x for − 5 ≤ x ≤ 0

4 ±4

5 ±4.47

Line of symmetry: y = 0

y 2 = 0.5x for 0 ≤ x ≤ 10 x y

0 1 2 3 4 0 ±0.71 ±1.00 ±1.22 ±1.41

5 6 7 8 9 10 ±1.58 ±1.73 ±1.87 ±2.00 ±2.12 ±2.24 Line of symmetry: y = 0

2(i)


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(2,4) lies on y 2 = 4ax, (4)2 = 4a(2) 16 = 8a a =2✓

233


Ex 9.1

y 2 = 4(2)x = 8x

4(i)

Points A & B At A & B, y = 2x + 4 meets y = x 2 − 4, 2x + 4 = x2 − 4 x 2 − 2x − 8 =0 (x + 2)(x − 4) = 0 x = −2 or x = 4 y|x=−2 = 2(−2) + 4 y|x=4 = 2(4) + 4 =0 = 12 ⇒ A(−2,0) ✓ ⇒ B(4,12) ✓

4(ii)

Length of AB

x 0 1 2 3 4 5 y 0 ±2.83 ±4 ±4.90 ±5.66 ±6.32

|AB| = √[(−2) − 4]2 + (0 − 12)2 = √36 + 144 = √180 = √36 × 5 = 6√5 ✓ 5(i)

Line & Curve y 2 = 2x − 1 y = mx

−(1) −(2)

sub (2) into (1): (mx)2 = 2x − 1 2 2 m x − 2x + 1 = 0

3

Discriminant For line & curve to meet at one point: b2 − 4ac =0 (−2)2 − 4m2 (1) = 0 4 − 4m2 =0 2 m −1 =0 (m + 1)(m − 1) = 0 m = −1 or m = 1 ✓

2y = x + 3 y

1

3

2

2

= x+

−(1)

y 2 = 2x + 3 −(2) sub (1) into (2): 1

3 2

2

2 3

9

2 1

4 3

2

4

( x+ ) 1 2 x 4 1 2 x 4 2

= 2x + 3

+ x+ − x−

= 2x + 3

5(ii)

=0

x − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1 or x = 3 1

3

2

2

y|x=−1 = (−1) + =1 ⇒ (−1,1) ✓


1

3

2

2

y|x=3 = (3) + =3 ⇒ (3,3) ✓

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Point (𝐚𝐭 𝐦 = −𝟏) y 2 = 2x − 1 −(1) y = −x −(2)

Point (𝐚𝐭 𝐦 = 𝟏) y 2 = 2x − 1 −(1) y=x −(2)

sub (2) into (1): (−x)2 = 2x − 1 2 x − 2x + 1 = 0 (x − 1)2 =0 x=1 y|x=1 = −1 ⇒ (1, −1)

sub (2) into (1): x2 = 2x − 1 2 x − 2x + 1 = 0 x 2 − 2x + 1 = 0 (x − 1)2 =0 x=1 y|x=1 = 1 ⇒ (1, −1)

234


y2 = x + 4

Ex 9.1 7(i)

x -4 -3 -2 -1 0 y 0 ±1 ±1.41 ±1.73 ±2

x 5 6 7 8 y ±5.48 ±6.00 ±6.48 ±6.93

x 1 2 3 4 5 6 y ±2.24 ±2.45 ±2.65 ±2.83 ±3 ±3.16

6(ii)

6(iii)

4x 2 − 13x = −5 2 4x − 12x + 9 = x + 4 (2x − 3)2 =x+4

y 2 = 6x x 0 1 2 3 4 y 0 ±2.45 ±3.46 ±4.24 ±4.90

7(ii)

y>3 ⇒ x > 1.5 ✓

7(iii)

4x 2 − 10x + 1 = 0 4x 2 − 4x + 1 = 6x (2x − 1)2 = 6x 2 (2x − 1) = y2 ⇒y = ±(2x − 1) Draw y = 2x − 1:

2

4x − 13x = −5 (2x − 3)2 = x + 4 (2x − 3)2 = y 2 y = ±(2x − 3) Draw y = 2x − 3 or y = −2x + 3 ✓ x = 0.45 or 2.8

⇒ x = 0.1 or 2.4


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235


Ex 9.1

Points A and B

10(i) 2

At A and B, y = x + meets y = 2x − 1: x

x+

2

= 2x − 1

x

x−1−

2

=0

x

x2 − x − 2 =0 (x + 1)(x − 2) = 0 x = −1 or y|x=−1 = 2(−1) − 1 = −3 ⇒ B(−1, −3) ✓ 8(ii)

x=2 y|x=2 = 2(2) − 1 =3 ⇒ A(2,3) ✓

Points A and B At A and B, y 2 = 3x meets y = 6 − x (6 − x)2 = 3x 2 x − 12x + 36 = 3x x 2 − 15x + 36 = 0 (x − 3)(x − 12) = 0 x=3 or x = 12 y|x=3 = 6 − (3) y|x=12 = 6 − (12) =3 = −6 ⇒ A(3,3) ✓ ⇒ B(12, −6) ✓

10(ii) Area of triangle OMB 1

Area of triangle PAB P(5,0) A(2,3) B(−1, −3)

Area of △ OMB = (Area of △ OAB) 2 1 1

0 12 3 0 | 0 −6 3 0 = [0 + 36 + 0 − 0 − (−18) − 0] = ⋅ | 2 2 1

Area of △ PAB 1 5 = | 2 0

2 3

−1 −3

5 | 0

4 1

= (54) 4

1

1

= [15 + (−6) + 0 − 0 − (−3) − (−15)]

= 13 unit 2 ✓

2

2

1

= [9 − (−18)]

11(i)

2

= 9(i)

27 2

unit 2 ✓

Points A & B 1

At A & B, y = meets y = 2x + 1: x

1

= 2x + 1

x

1 = 2x 2 + x 2x 2 + x − 1 =0 (x + 1)(2x − 1) = 0 x = −1

1

or x =

y|x=−1 =

1 (−1)

2

y|x=1 = 2

= −1 ⇒ B(−1, −1) ✓ 9(ii)

1

Points A & B At A & B, y 2 = 12 − 2x meets y = 2 − x (2 − x)2 = 12 − 2x 2 x − 4x + 4 = 12 − 2x 2 x − 2x − 8 =0 (x + 2)(x − 4) = 0 x = −2 or x=4 y|x=−2 = 2 − (−2) y|x=4 = 2 − (4) =4 = −2 ⇒ A(−2,4) ✓ ⇒ B(4, −2) ✓

11(ii) ⊥ bisector of AB

1 ( ) 2

AB⊥ ≡ ⊥ bisector of AB

=2 1

⇒ A ( , 2) ✓ 2

Point:

Point M At M, y = 2x + 1 cuts y-axis (x = 0): y|x=0 = 1 ⇒ M(0,1)

MAB = (

(−2)+4 4+(−2)

Gradient: mAB⊥ = − =−

,

2 1

mAB 1 6 ) −6

(

) = (1,1)

2

=−

1 (4)−(−2) ((−2)−(4))

=1

Ratio 𝐀𝐌: 𝐌𝐁 1

AB⊥ :

Recall A ( , 2) M(0,1) B(−1, −1) 2

By similar triangles (using x-coordinates) 1

AM: MB = ( − 0) : [0 − (−1)]

y − y1 = m (x − x1 ) y − 1 = (1)(x − 1) y−1 =x−1 y=x✓

2

=

1 2

=1

:1 :2✓


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236


Ex 9.1

Points A & B At A & B, y = x 2 + x − 2 meets y = 2x x2 + x − 2 = 2x 2 x −x−2 =0 (x + 1)(x − 2) = 0 x = −1 or x = 2 y|x=−1 = 2(−1) y|x=2 = 2(2) = −2 =4 ⇒ B(−1, −2) ✓ ⇒ A(2,4) ✓

13(i)

A(2, −4) lies on y = x + k: −4 = 2 + k k = −6 Point B At B, y = x − 6 meets y 2 = 8x (x − 6)2 = 8x 2 x − 12x + 36 = 8x x 2 − 20x + 36 = 0 (x − 2)(x − 18) = 0 x = 2 or x = 18 (taken) y|x=18 = (18) − 6 = 12

12(ii) Length of AB |AB| = √[(−1) − 2]2 + [(−2) − 4]2 = √9 + 36 = √45

13(ii) ⊥ bisector of AB

= √9 × 5 = 3√5 units ✓

Recall A(2, −4) B(18,12)

12(iii) ⊥ 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐟𝐫𝐨𝐦 𝐂 𝐭𝐨 𝐀𝐁 Recall A(2,4) B(−1, −2) C(−3,4) |AB| = 3√5

AB⊥ ≡ ⊥ bisector of AB Point:

F ≡ Foot of ⊥ from C to AB |CF| ≡ ⊥ distance from C to AB

MAB = (

2+18 (−4)+12 2

Gradient: mAB⊥ = − Equating Area of △ ABC: 1 2 1 2

=−

(base)(height) = △ area by shoelace formula |AB||CF|

|AB||CF| 3√5 |CF| 3√5 |CF| |CF|

−3 −1 2 | 4 −2 4 2 −3 −1 2 | =| 4 4 −2 4 8 + 6 + (−4) ] =[ −(−12) − (−4) − (−4) 1

= | 2

2 4

AB⊥ :

,

1 mAB 1 −16 ( ) −16

2

=−

) = (10,4) 1

(−4)−(12) ( (2)−(18) )

= −1

y − y1 = mAB⊥ (x − x1 ) y − (4)= (−1) [x − (10)] y − 4 = −x + 10 y = −x + 14 ✓

= 30 = = =

30 3√5 10 √5 10√5 5

= 2√5 ✓


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237


Ex 9.1

Points A & B At A & B, y 2 = 2x − 3 meets 2x + 3y = 7 y 2 = 2x − 3 −(1)

14(iii) ⊥ distance from P to AB F ≡ Foot of ⊥ from P to AB |PF| = ⊥ distance from P to AB

2x + 3y = 7 3y = −2x + 7 y

Area of △ PAB = 1

2

7

3

3

=− x+

2 1

−(2)

sub (2) into (1): (− x + ) 3 4 2 x − 9 4 2 x − 9 2

= 2x − 3

3

28 9 46 9

x+ x+

|AB||PF|

2 1 5√13

7 2

2

(base)(height) =

49

2

2

) |PF|

=

|PF|

= 2x − 3

9 76

(

=

= =

=0

9

2x − 23x + 38 = 0 (x − 2)(2x − 19) = 0 x=2

or 2

7

3

3

y|x=2 = − (2) +

x=

15 19 2 2 19

7

3

3

y|x=19 = − ( ) + 2

=1

= −4

⇒ A(2,1)

⇒ B ( , −4)

2

35 2 35 2 35 2 35 2 14 √13 14 13

units

√13 units ✓

Point B B(b1 , b2 ) lies on y 2 = 2x: (b2 )2 = 2b1 1

= (b2 )2

b1

2

1

⇒ B ( (b2 )2 , b2 ) 2

19 2

Midpoint of AB 1

Length of AB

A(2,6) B ( (b2 )2 , b2 ) 2

19 2

|AB| = √(2 −

2

) + [1 −

1

2+ (b2 )2 6+b2 2

(−4)]2

MAB = (

325

=√

=

4

5√13 2

,

2

)

= (x, y)

4

52 ×13

=√

2

⇒x=

1 2

2+ (b2 )2 2

[shown] ✓ ⇒y =

14(ii) Area of triangle PAB

6+b2 2

2y = 6 + b2 b2 = 2y − 6 −(2) sub (2) into (1):

19

P(6,3) A(2,1) B ( , −4) 2

19

1 6 2 6 2 | Area of △ PAB = | 2 3 1 −4 3 1 57 19 = [6 + (−8) + −6− − 2

2

2

(−24)]

1 2

2+ (2y−6)2

x

=

x

= 1 + (2y − 6)2

x

= 1 + (4y 2 − 24y + 36)

2 1 4 1 4

x = 1 + (y 2 − 6y + 9) x = y 2 − 6y + 10 0 = y 2 − 6y − x + 10 y 2 − 6y − x + 10 = 0 ✓

1

= (35) 2

1

= 17 unit 2 ✓ 2


−(1)

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238


Ex 9.1

Point P P(a, b) lies in y 2 = 8x: b2 = 8a b = √8a or b = −√8a (rej ∵ b > 0)

17(ii) Line AC Point:

⇒ P(a, √8a)

A(0,2at) or C(at 2 , −2at)

Gradient:

mAC =

AC:

y − y1

(2at)−(−2at)

=

(0)−(at2 )

4at −at2

=−

4 t

= m(x − x1 ) 4

y − (2at) = − [x − (0)] Length of PF P(a, √8a) F(2,0)

y 2

PF = √(a − 2)2

+ (√8a)

2

4

(− x + 2at) t 16 2 x − t2 16 2 x − t2 2

= √(a2 − 4a + 4) + 8a =

= − x + 2at t

Another point of intersection At point where AC meets y 2 = 4ax

+ (√8a − 0)

= √(a − 2)2 √a2

t 4

+ 4a + 4 2)2

= √(a + = a + 2 units ✓ 16(ii) Length of PQ PQ = (x − coordinate of P) −(x − coordinate of Q) =a −(−2) =a+2 ⇒ PF = PQ [shown] ✓ 16(iii) Area of triangle PQF P(a, √8a) PQ = a + 2

2

= 4ax

16ax + 4a2 t 2

= 4ax

20ax + 4a2 t 2

=0

16x − 20at 2 x + 4a2 t 4 = 0 4x 2 − 5at 2 x + a2 t 4 =0 (4x − at 2 )(x − at 2 ) =0 x= y|

at2

or

4

at2 x= 4

4 at2

=− ( t

4

x = at 2 (taken)

) + 2at

= −at + 2at = at Another point of intersection (

at2 4

, at) ✓

1

Area of △ PQF = (base)(height) 2 1

= (PQ) (y − coordinate of P) 2 1

= (a + 2)(√8a) 2 1

= (a + 2)(√4 × 2a) 2 1

= (a + 2)2√2a 2

= (a + 2)√2a ✓ 17(i)

𝑦

D(at 2 , 2at) 𝑦 2 = 4𝑎𝑥

A(0,2at) 𝑂 B(0, −2at)

M

𝑥

C(at 2 , −2at)

A(0,2at), B(0, −2at) & C(at 2 , −2at) ✓


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239


Ex 9.1

17(iii) A(0,2at)

18(i)

2

2

M

1

1

l2 : y = √x ∴ l1 ≠ l2 ✓

2 2 1 C(at , −2at) ⃗⃗⃗⃗⃗⃗ = OA ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ OM +AM 2 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = OA + AC

⃗⃗⃗⃗⃗ − OA ⃗⃗⃗⃗⃗ ) + (OC 3 2

⃗⃗⃗⃗⃗ = OA

3

x

3

⃗⃗⃗⃗⃗ + OC 3 2

𝑦2 = 𝑥

2

0 ) + ( at ) 3 −2at 2at

2

=(

𝑦 = √𝑥

2

3 2

⃗⃗⃗⃗⃗ = OA = (

3

at 2 2

− at

✓ 19(i)

)

3

2

2

3

= √(2.2)2 + 12

3

= √5.84 ≈ 2.416609195

Method 2 Let M be (xm , ym )

MC

=

Distance in km = (2.416609195)(93 × 106 )(1.609) ≈ 361 × 106 ≈ 3.61 × 108 ✓

2 1

⇒ x − coordinates xm −xa =2 xc −xm x1 −0

⇒ y − coordinates ym −ya =2 yc −ym y1 −2at

=2

at2 −x1

−2at−y1

=2

3x1

= 2at 2 − 2x1 y1 − 2at = −4at − 2y1 2 3y1 = −2at = 2at

x1

= at 2

x1

2

2

3 2

S(−0.1,0) C(−2.3,1) |SC| = √[(−0.1) − (−2.3)]2 + [(0) − (1)]2

2

⇒ M ( at , − at) ✓

AM

y

⃗⃗⃗⃗⃗ − OA ⃗⃗⃗⃗⃗ + OC

1 3 1

18(ii)

2+1 2

⃗⃗⃗⃗⃗ = OA

l1 : y 2 = x y = ±√x

2

y1

= − at 3

2

⇒ M ( at , − at) ✓ 3

3

17(iv) Line (through M and ⊥ 𝐭𝐨 𝐀𝐂) A(0,2at) B(0, −2at) C(at 2 , −2at) D(at 2 , 2at) 2

2

3

3

Point:

M ( at 2 , − at)

Gradient:

m =− =−

Line:

1

=−

mAC 1 (

4at ) −at2

1 (2at)−(−2at) ( ) (0)−(at2 )

=−

1 −

4 t

=

t 4

= m(x − x1 )

y − y1 2

t

3

4 t

2

1

4 t

6 1

2

4

6

3

y − (− at) = [x − ( at 2 )] 2

3

y + at

= x − at 3

y

= x − at 3 − at ✓

3


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240


Ex 9.1

19(ii) S(−0.1,0) C(c1 , c2 ) C(c1 , c2 ) lies on x = −2.3y 2 c1 = −2.3(c2 )2 ⇒ C(−2.3(c2 )2 , c2 ) |SC| = √[(−0.1) − (−2.3(c2 )2 )]2 + (c2 )2 = √[2.3(c2 )2 − 0.1]2 + (c2 )2 = √[5.29(c2 )4 − 0.46(c2 )2 + 0.01] + (c2 )2 = √5.29(c2 )4 + (0.54)(c2 )2 + 0.01 f(c2 ) = √5.29(c2 )4 + (0.54)(c2 )2 + 0.01 f′(c2 )=

21.16(c2 )3 +(1.08)(c2 ) 2√5.29(c2 )4 +(0.54)(c2 )2 +0.01

Minimum |SC| ⇒ f ′ (c2 ) =0 21.16(c2 )3 +(1.08)(c2 ) 2√5.29(c2 )4 +(0.54)(c2 )2 +0.01 21.16(c2 )3 + (1.08)(c2 ) c2 (21.16c 2 + 1.08)

=0 =0 =0

c2 = 0 Sign Test c2 0− 0 0+ f′(c2 ) sign − 0 + ∴ f(c2 ) is minimum 1 AU = 93 million miles = 93 × 106 miles = 93 × 106 [1.609] km = 93 × 106 [1.609][103 ]m = 93 × 106 [1609] m d = |SC||c

2 =0

≈ √f(0)

× 93 × 106 [1609]

× 93 × 106 [1609]

≈ √[0.01] × 93 × 106 [1609] ≈ 1.49637 × 1010 V=

k √d

=

1.17×1010 √1.49637×1010


≈ 95645ms −1 ✓

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241


Ex 9.2 2(a)

Ex 9.2 1(a)

1(b)

Centre (0,1) Radius: r = 4 Circle: (x − a)2 +(y − b)2 = r 2 (x − 0)2 + (y − 1)2 = 42 ✓

Comparing coefficients: 2g = 2 2f = −10 g=1 f = −5

𝑦

Radius: r = √f 2 + g 2 − c = √(−5)2 + 12 − 1 = 4

2 (3, −2)

Method 2 (standard form) x 2 + y 2 + 2x − 10y + 1 x 2 + 2x +y 2 − 10y (x + 1)2 − 12 +(y − 5)2 − 52 (x + 1)2 +(y − 5)2 2 [x − (−1)] +(y − 5)2

Centre: (3, −2) Radius: r = 2 Circle: (x − a)2 + (y − b)2 = r2 (x − 3)2 + [y − (−2)]2 = 22 (x − 3)2 + (y + 2)2 = 22 ✓ 1(c)

2(b) 3

(−3,4) 𝑥

𝑂

Radius: √f 2 + g 2 − c = √(−3)2 + (−2)2 − 4 = 3 Method 2 (standard form) x 2 + y 2 − 4x − 6y + 4 x 2 − 4x +y 2 − 6y (x − 2)2 − 22 +(y − 3)2 − 32 (x − 2)2 +(y − 3)2 (x − 2)2 +(y − 3)2

(−2,2) 𝑥

Centre: (−2,2) Radius: r = 2 Circle: (x − a)2 +(y − b)2 = r 2 [x − (−2)]2 +[y − (2)]2 = 22 (x + 2)2 +(y − 2)2 = 22 ✓


c=4

Centre: (−g, −f) = (2,3)

𝑦

𝑂

Method 1 (general form) x 2 + y 2 − 4x −6y +4 = 0 x 2 + y 2 + 2gx +2fy +c = 0 Comparing coefficients: 2g = −4 2f = −6 g = −2 f = −3

Centre: (−3,4) Radius: r = 3 Circle: (x − a)2 +(y − b)2 = r 2 2 [x − (−3)] +(y − 4)2 = 32 (x + 3)2 +(y − 4)2 = 32 ✓

2

=0 = −1 = −1 = 25 = 52

Centre: (−1,5) Radius: 5

𝑦

1(d)

c=1

Centre: (−g, −f) = (−1,5)

𝑥

𝑂

Method 1 (general form) x 2 + y 2 + 2x −10y +1 = 0 x 2 + y 2 + 2gx +2fy +c = 0

=0 = −4 = −4 =9 = 32

Centre: (2,3) Radius: 3

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242


Method 1

Ex 9.2 𝑦 A(−8,0) 𝑂 𝑥 B(0, −4) 10

4

r

C(0, −16)

2

= −10 5(i)

centre (−8, −10) radius = 10

=5 = 25 = 20 = 20 ✓

Centre C C(0,0)

1

y

= x−3

−(1)

y − 2x = 0 sub (1) into (2):

−(2)

2

=0

At A(−8,0): (−8)2 + (0)2 + 2g(−8) + 2f(0) + c 64 − 16g + c −(1) At B(0, −4), (0)2 + (−4)2 + 2g(0) + 2f(−4) + c 16 − 8f + c −(2)

1

=0 =0

( x − 3) − 2x = 0 2

3

=0

− x

=3

x

= −2

2

=0 =0

1

y|x=−2 = (−2) − 3 = −4 2

⇒ P(−2, −4) Radius r r = |CP| = √[(−2) − 0]2 + [(−4) − 0]2 = √20 Circle (x − a)2 + (y − b)2

Solving (1), (2) & (3): g = 8, f = 10, c = 64 ∴ x + y + 2(8)x + 2(10)y + 64 x 2 + y 2 + 16x + 20y + 64

− x−3 2 3

At C(0, −16), (0)2 + (−16)2 + 2g(0) + 2f(−16) + c = 0 256 − 32f + c =0 −(3)


=5

Point P on circle At point P on circle, where x − 2y = 6 meets y − 2x = 0, x − 2y = 6 2y =x−6

Circle: (x − a)2 +(y − b)2 = r2 [x − (−8)]2 +[y − (−10)]2 = 102 (x + 8)2 +(y + 10)2 = 102 ✓

2

−c

√5 − c 5−c −c c

(−4)+(−16)

2

+

g2

√12 + (−2)2 − c = 5

y − coordinate of centre

Method 2 x 2 + y 2 + 2gx + 2fy + c

=5

√f 2

x − coordinate of centre = −8

=

C(2, −1) = (−g, −f) Comparing coefficients: g = −2 f=1

(x − 0)2 + (y − 0)2 x + y2 =0 =0

5(ii)

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= r2 2

= (√20) = 20

Check point (2,6) lies on circle (2)2 + (6)2 = 20 40 = 20 [inconsistent] ∴ No, (2,6) does not lie in circle ✓

243


Ex 9.2

Points A & B At A & B, 12x − 5y = −11 cuts x 2 + y 2 − 6x + 2y − 15 = 0 12x − 5y= −11 5y = 12x + 11 y

12x+11

=

−(1)

5

or

12x+11 2

2

5 144x2 +264x+121

)

5

y|x=1 = 2(1) y|x=3 = 2 ( ) 5

5

=2

=0

−6x + 2 (

) − 6x +

25

3

x=

3

−(2)

12x+11

5 24x+22

6

=

⇒ P(1,2) ✓

2

x +(

Points P & Q At P & Q, x 2 + y 2 + 4x − 6y + 3 = 0 & y = 2x meet x 2 + (2x)2 + 4x − 6(2x) + 3 = 0 x 2 + 4x 2 + 4x − 12x + 3 =0 2 5x − 8x + 3 =0 (x − 1)(5x − 3) =0 x=1

x 2 + y 2 − 6x + 2y − 15 = 0 sub (1) into (2): x 2 + y 2 − 6x + 2y − 15 x +(

7(i)

5

3 6

⇒ Q( , ) ✓ 5 5

) − 15 = 0 − 15 = 0

5

7(ii)

⊥ bisector of PQ 3 6

P(1,2) Q ( , ) 5 5

x2 +

144 2 x 25

264

+

25

x

+

PQ ⊥ ≡ ⊥ bisector of PQ

121 25

Point:

−6x +

24 5

x

+

+

234

x−

25

144

13

⇒ A (− 6(ii)

x=

24 13

12(− )+11

24 13

y−

6 13

,−

13

= ) ✓

⇒ B(

6

,

y

6 13

12( )+11 5

13

29 13 29

8(i)

43

13 43

A (−

13

,−

29

) B(

13

6

,

43

) 2

24 6 29 43 |AB| = √[(− ) − ( )] + [(− ) − ( )] 13

13

2

13

mPQ

,

6 5

2

4 8

)= ( , ) 5 5

1

= − (2)−(6 = − )

5 3 (1)−( ) 5

1 4 ( ) 5 2 ( ) 5

=−

1 2

(x − x1 )

=m

8

1

5

2

4

8 5

5

1

2

2 1

5

=− x+

=− x+2✓ 2

x 2 + y 2 − 2x + 8y − 23 = 0 x 2 + y 2 + 2gx +2fy +c =0

c = −23

Centre: (−g, −f) = (1, −4) Radius: √f 2 + g 2 − c = √(4)2 + (−1)2 − (−23)

= √36 =6✓


1

Comparing coefficients 2g = −2 2f = 8 g = −1 f=4

13 13

13

2

2+

) ✓

13 13

Length of AB 24

y − y1

3 5

y − ( ) = (− ) [x − ( )]

y|x= 6 =

5

=−

PQ ⊥ :

=0 =0 or

13

=0 =0

25

24

y|x=−24 =

Gradient: m = −

5

169x +234x −144 (13x + 24)(13x − 6) x=−

MPQ = (

22

−15 169 2 x 25 2

1+

= √40 = √4 × 10 = 2√10

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244


Ex 9.2

Tangent AB

10

P(3, −10) 2√10 AB

⊥ bisector of AB AB⊥ ≡ ⊥ bisector of AB

C(1, −4)

Point:

Point:

P(3, −10)

Gradient: mAB = −

Points on circle A(2,3) B(−1,6)

1 mPC

=−

1 (−10)−(−4)

( (3)−(1) )

=−

1 −6 ) 2

(

1 AB⊥ :

3

y − y1

y−

3

9(i)

1

= x − 11 ✓

y

3 1

⊥ bisector of AB

9(ii)

9(iii)

(−2)+4 2

= 1✓

Centre C At C, y = 6 − 2x meets AB⊥ (x = 1) y|x=1 = 6 − 2(1) =4 ⇒ C(1,4) ✓

2 2

1 (3)−(6)

[(2)−(−1)]

1

= − −3 = 3

= m (x − x1 ) 1

9 2

2

=x−

1 2

=x+4

Radius r r = |AC| = √[2 − (−3)]2 + (3 − 1)2 = √29 Circle (x − a)2

Radius r r = |AC| = √[(−2) − 1]2 + (0 − 4)2 = √25 = 5 Circle (x − a)2 +(y − b)2 = r 2 (x − 1)2 +(y − 4)2 = (5)2 ✓

1 9

)=( , )

Centre C At C, AB⊥ (y = x + 4) meets 2x + 5y = −1 2x + 5(x + 4) = −1 2x + 5x + 20 = −1 7x = −21 x = −3 y|x=−3 = (−3) + 4 =1 ⇒ C(−3,1)

3

Points on circle A(−2,0) B(4,0)

x=

mAB

=−

2

y − (−10) = ( ) [x − (3)]

y

1

2

y − ( ) = (1) [x − ( )]

1

= x−1

y − y1

,

2

9

= mAB (x − x1 )

y + 10

(2)+(−1) (3)+(6)

Gradient: mAB⊥ = −

=

1

AB:

MAB = (

+(y − b)2

= r2

[x − (−3)]2 +(y − 1)2 = (√29) (x + 3)2 +(y − 1)2 = 29 ✓ 11(i)

2

Points A & B At A & B, x 2 + y 2 − 4x + 6y − 12 = 0 cuts x − axis (y = 0). x 2 + (0)2 − 4x + 6(0) − 12 = 0 x 2 − 4x − 12 =0 (x + 2)(x − 6) =0 x = −2 or x = 6 ⇒ A(−2,0) ⇒ B(6,0) Length of AB |AB| = 6 − (−2) = 8 ✓


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245


Ex 9.2

11(ii) Centre C x 2 + y 2 − 4x + 6y − 12 = 0 x 2 + y 2 + 2gx + 2fy + c = 0

12(ii)

𝑦 𝑦=𝑥

𝐶 6

Comparing coefficients: 2g = −4 2f = 6 g = −2 f=3

𝐴

𝑦 = −𝑥

= MAD

(2, −3) = ( 2 =

(−2)+(d1 ) (0)+(d2 ) 2

(−2)+(d1 ) 2

,

and

12(i)

2

−3 =

d1 = 6 ⇒ D(6, −6) ✓

Centre C of 𝐂𝟑 At C, AB⊥ (y = −x) cuts circle (x 2 + y 2 = 36) x 2 + (−x)2 = 36 2x 2 = 36 2 x = 18 x = ±√18 = ±√9 × 2 = ±3√2 x = 3√2 or x = −3√2 y|x=3√2 = −3√2 y|x=−3√2 = 3√2

) (0)+(d2 ) 2

d2 = −6

𝑦 𝑦=𝑥 5 5

O

5

⇒ C(3√2, −3√2)

⇒ C(−3√2, 3√2)

𝑥 Radius r of 𝐂𝟑 r = |BC|

5 Centre: A(−5, −5) or B(5,5) Radius: r = 5 C1 : C2 :

𝐶

Circle with centre (𝟎, 𝟎) & radius 6 (x − 0)2 + (y − 0)2 = 62 x2 + y2 = 36

Point D D(d1 , d2 ) C

6

𝑥

⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 AB⊥ : y = −x

= (−g − f) = (2, −3)

Centre C

𝐵

O

2

2

= √(5 − 3√2) + [5 − (−3√2)]

[x − (−5)]2 + [y − (−5)]2 = 52 ✓ (x − 5)2 + (y − 5)2 = 52 ✓

= √(25 − 30√2 + 18) + (25 + 30√2 + 18) = √96 Circle 𝐂𝟑 C3 :

[x − (3√2)]

or [x − (3√2)]


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2 2

+[y − (−3√2)] +[y − (−3√2)]

2 2

2

= (√96) ✓ 2

= (√96) ✓

246


Ex 9.2

Tangent x + 2y = −5 2y = −x − 5 y

1

5

2

2

=− x−

Normal Point:

14 C(−1,3) P

x + 2y = −5

Normal:

−1 mtan

=

−1 1 2

(− )

=2

y − y1 = mnorm (x − x1 ) [x − (−1)] y − (3)= (2) (x + 1) y−3 =2 y − 3 = 2x + 2 y = 2x + 5

Centre C x 2 + y 2 − 4x − 8y − 5 = 0 x 2 + y 2 + 2gx + 2fy + c = 0

Point P on circle 1

5

2

2

At P, tangent (y = − x − ) meets normal (y = 2x + 5) 1

5

2

2

Comparing coefficients 2g = −4 2f = −8 g = −2 f = −4

− x − = 2x + 5 5 2

x

x

=−

15 2

= −3

Centre C

1

5

2

2

y|x=−3 = − (−3) −

1

Area of △ ABC = | |

13(ii) Centre C(−1,3)

2

Radius r = |CP| = √[(−3) − (−1)]2 + [(−1) − 3]2 = √20

2 −1

5 8

2 4

2 || −1

=

1 16 + 20 + (−2) | | 2 −(−5) − 16 − 8

=

1 |15| 2

= 7.5 units 2 ✓ = r2

[x − (−1)]2 +(y − 3)2 = (√20) (x + 1)2 +(y − 3)2 = 20 ✓


= (−g, −f) = (2,4)

Area of triangle ABC A(2, −1) B(5,8) C(2,4)

= −1 ⇒ P(−3, −1) ✓

Circle (x − a)2 +(y − b)2

=0 =0 =0

10x 2 −70x + 100 = 0 x 2 −7x +10 =0 (x − 2)(x − 5) =0 x=2 or x = 5 y|x=2 = 3(2) − 7 y|x=5 = 3(5) − 7 = −1 =8 ⇒ A(2, −1) ⇒ B(5,8)

C(−1,3) or P

Gradient: mnorm =

Point A & B At A & B, (y = 3x − 7) cuts (x 2 + y 2 − 4x − 8y − 5 = 0) x 2 + (3x − 7)2 −4x −8(3x − 7) −5 2 2 x + (9x − 42x + 49) −4x −24x + 56 −5 (10x 2 − 42x + 49) −28x + 51

2

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247


Ex 9.2

x 2 + y 2 + 4x +6y −12 = 0 x 2 + y 2 + 2gx +2fy +c = 0 Comparing coefficients 2g = 4 2f = 6 g=2 f=3

16

Line & Circle y = mx − 1 x 2 + y 2 − 4x + 3 = 0

−(1) −(2)

sub (1) into (2): x 2 + (mx − 1)2 −4x + 3 = 0 2 2 2 x +(m x − 2mx + 1) −4x + 3 = 0 (1 + m2 )x 2 + (−2m − 4)x + 4 =0

c = −12

Centre C = (−g, −f) = (−2, −3) Radius r = √f 2 + g 2 − c

Discriminant For two distinct points: b2 − 4ac (−2m − 4)2 − 4(1 + m2 )(4) (4m2 + 16m + 16) − 16(1 + m2 ) (4m2 + 16m + 16) − 16 − 16m2 −12m2 + 16m 3m2 − 4m m(3m − 4)

= √32 + 22 − (−12) = √25 = 5 𝑦

(−3, −1)

𝑂

𝑥

(−2, −3)

+

3

= √[(−2) − (−3)]2 + [(−3) − (−1)]2 =

+

+ 4

0

Distance from centre to point (−3, −1) √12

−

>0 >0 >0 >0 >0 <0 <0

4

0 < m < [shown] ✓

(−2)2

3

= √5 < radius ⇒ point lies inside circle

17

∴ any line passing through (-3,-1) cannot be tangent to circle ✓

Centre x 2 + y 2 + 2x − 2y − 3 = 0 x 2 + y 2 + 2gx + 2fy + c = 0 C(−1,1)

Comparing coefficients: 2g = 2 2f = −2 g=1 f = −1

P(1,2) Tangent

Centre: C(−g, −f) = (−1,1) Tangent Point:

P(1,2) or C(−1,1)

Gradient: mtan = =

−1 mnorm

=

−1 (1)−(2)

[(−1)−(1)]

−1 mCP

=

−1 −1 ) −2

(

= −2

Tangent: y − y1 = mtan (x − x1 ) y − (2)= (−2)[x − (1)] y − 2 = −2x + 2 y = −2x + 4 ✓


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248

A math 360 sol (unofficial) 18(i) P(x, y)

Ex 9.2 19(ii) ⊥ bisector of PQ (𝐏𝐐⊥ ) P(2, −2) Q(6,0)

A(−a, 0) B(a, 0)

Point: PA

= 2PB

√[x − (−a)]2 + (y − 0)2

= 2√(x − a)2 + (y − 0)2

√(x + a)2 + y 2

= 2√(x − a)2 + y 2

(x + a)2 + y 2

= 4[(x − a)2 + y 2 ]

(x 2 + 2ax + a2 ) + y 2

= 4[(x 2 − 2ax + a2 ) + y 2 ]

x 2 + 2ax + a2 + y 2

= 4x 2 − 8ax + 4a2 + 4y 2

−3x 2 + 10ax − 3a2 − 3y 2

=0

x2 −

10 3

ax + a2 + y 2 2

5

Gradient: mPQ⊥ = = PQ ⊥ :

19(iii) 2

5

2

3 5

+ a2 + y 2 = 0

2

(x − a) + y 2

=

3

(x − x1

)2

+ (y − y1

)2

16 2 a 9

= r [shown]

=

−1 1 2

= −2

(x − a)2 + (y − b)2 = r2 (x − 6)2 + [y − (−5)]2 = (5)2 (x − 6)2 + (y + 5)2 = 25 ✓ [wrong answer in textbook]

4 3

Circle 𝐂𝟏 Centre: P(2, −2) Point: Q(6,0) Radius: r = |PQ|

20(i)

= √(2 − 6)2 + [(−2) − 0]2 √(−4)2

+

(x − x1 )2 + (y − y1 )2 = r 2 (x − 2)2 − [y − (−2)]2 = (√20) (x − 2)2 + (y + 2)2 = 20 ✓

x 2 + y 2 − 4x − 4y = 1 y=x+c

−(1) −(2)

sub (2) into (1): x 2 + (x + c)2 −4x − 4(x + c) 2 2 2 x + (x + 2cx + c ) −4x − 4x − 4c (2x 2 + 2cx + c 2 ) −8x − 4c 2 2 (2c 2x + − 8)x + c − 4c − 1

(−2)2

= √20


𝑥

5

19(iv) C2 :

radius = a ✓

C1 :

−2 −4

−1 (−2)−(0)

[ (2)−(6) ]

Radius of C2 = 5 Centre of C2 =(6, −5) ✓

2

3

=

−1

=

) = (4, −1)

𝐶2

18(ii) centre (5 a, 0) ✓

19(i)

−1

2

y − y1 = mPQ⊥ (x − x1 ) y − (−1) = (−2) [x − (4)] y+1 = −2x + 8 y = −2x + 7 ✓

𝑂

25 2 a 9

,

mPQ

𝑄(6,0)

3

(x − a) −

2

𝑦

2

[(x − a) − ( a) ] + a + y = 0 3

2+6 (−2)+(0)

=0

2

5

MPQ = (

2

x 2 + (c − 4)x +

c2 −4c−1 2

=1 =1 =1 =0 =0

[shown] ✓

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249


Ex 9.2

20(ii) Sum of roots Roots: x1 & x2

21 b

Sum of roots = x1 + x2 = −

a

=−

(c−4) 1

(2)]2

=4−c

P(9,2) C(2, −1)

By Pythagoras’ Theorem, |PT|2 = |PC|2

Midpoint of PQ

=( =( =( =(

2 x1 +x2 2 x1 +x2 2 4−c 2 4−c 2

2 x1 +x2 +2c

,

2 x1 +x2 +2c

,

2 (4−c)+2c

,

2 4+c

,

2

(x, y) = MPQ ( ⇒x = 2x c

2

4−c 4+c 2

,

2

)

= 72 + 32 = 58 = 45

) ) PT

)

22

)

y=

= √45

Method 1 (graphical inspection) 𝑦 x = (y − k)2 + h

4−c 2

(h, k)

4+c 2

𝑂

𝑥

∴ two y − intercepts

−(2)

Method 2 (solution) x = a(y − k)2 + h

sub (1) into (2) y=

−13 −13

= 3√5

=4−c = 4 − 2x −(1)

⇒y=

2

= √9 × 5

) ✓

20(iii) Curve traced by midpoint

−|CT|2

= [√(9 − 2)2 + [2 − (−1)]2 ] −(√13)

(x1 +c)+(x2 +c)

,

2

= (√13)

T

Q(x2 , y2 ) lies on y = x + c: y2 = x2 + c ⇒ Q(x2 , x2 + c)

x1 +x2

= 13

(−1)]2

[x − + [y − Centre: C(2, −1) Radius: r = √13

Points P & Q P(x1 , y1 ) lies on y = x + c: y1 = x1 + c ⇒ P(x1 , x1 + c)

MPQ = (

Circle (x − 2)2 + (y + 1)2

4+(4−2x) 2 8−2x

At y − axis (x = 0): 0 = a(y − k)2 + h 2 a(y − k) = −h > 0 ∵h<0

2

y = −x + 4 ✓

h

(y − k)2

=−

(y − k)

= ±√−

y

= k ± √−

a

>0

∵a>0

h a h a

∴ two y − intercepts ✓


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250


Ex 9.2

Method 3 (discriminant) a(y − k)2 + h =0 a(y 2 − 2ky + k 2 ) +h = 0 ay 2 −2aky +ak 2 + h = 0 i.e. A = a, B = −2ak, C = ak 2 + h

23

9 4

Grad: mAB⊥ = AB⊥ :

−1 mAB

8

4

8

x

= =

31 8 31 18

31

y|x=31 = 2 ( ) 18

18

=

⇒ D(

31

9 31 31 18

,

9

)

31

2

31

= √[(3) − ( )] + [(1) − ( )] 18

=√

7

) = ( , 3)

2

2

9

2465 324

2

−1

=

(1)−(5)

[(3)−(4)]

=−

1

Circle (x − x1 )2 + (y − y1 )2

4

y − y1 = mAB⊥ (x − x1 ) 1

7

4

2

(x −

y − (3)= (− ) [x − ( )] 1

7

4 1

8

y−3 =− x+ y

7

Radius r r = |AD|

3+4 1+5 2

1

x

⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐁 AB⊥ ≡ ⊥ bisector of AB ,

7

4

− x + 3 = 2x

Points A(3,1) B(4,5) C(−1,3)

Point: MAB = (

1

At D, AB⊥ (y = − x + 3 ) meets AC⊥ (y = 2x).

Discriminant B 2 − 4AC = (−2ak)2 − 4(a)(ak 2 + h) = 4a2 k 2 − 4a2 k 2 − 4ah = −4ah >0 ∵ a > 0, h < 0 ∴ two y-intercepts 23

Centre D

=− x+3 4

31 2

) + (y −

18

31 2 9

)

= r2 2

= (√

2465 324

) ✓

7 8

⊥ 𝐛𝐢𝐬𝐞𝐜𝐭𝐨𝐫 𝐨𝐟 𝐀𝐂 AC⊥ ≡ ⊥ bisector of AC Point: MAC = ( Grad: mAC⊥ = AC⊥ :

3+(−1) 1+3 2 −1 mAC

,

=

2

) = (1,2)

−1 (1)−(3) [(3)−(−1)]

=2

y − y2 = mAC⊥ (x − x2 ) y − (2)= (2) (x − 1) y − 2 = 2x − 2 y = 2x


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251


Rev Ex 9 A2

Rev Ex 9 A1(i)

Line AC A(4,4) C(6,0) Point: A(4,4) or C(6,0) (4)−(0)

Gradient: mAC = (4)−(6) =

4 −2

−(1) −(2)

sub (1) into (2): (mx + 1)2 =x 2 2 m x + 2mx + 1 =x 2 2 m x + (2m − 1)x + 1 = 0

= −2

y − y1 = m (x − x1 ) y − (0)= (−2)[x − (6)] y = −2x + 12

AC:

Line & Curve y = mx + 1, m > 0 y2 = x

Point B At B, AC meets y 2 = 4x (−2x + 12)2 = 4x 2 4x − 48x + 144 = 4x 4x 2 − 52x + 144 = 0 x 2 − 13x + 36 =0 (x − 4)(x − 9) =0 x=4 or x = 9 (taken) y|x=9 = −2(9) + 12 = −6 ⇒ B(9, −6) ✓

Discriminant For two distinct points: b2 − 4ac (2m − 1)2 − 4(m2 )(1) (4m2 − 4m + 1) − 4m2 −4m + 1 −4m

>0 >0 >0 >0 > −1

m

<

1 4

Combine inequalities: m > 0 and m <

1 4

1

⇒ 0 < m < [shown] ✓ 4

A1(ii) A(4,4) B(9, −6) C(6,0) |AC| = √(4 − 6)2 + (4 − 0)2 =

√(−2)2

+

A3(i)

𝑦 𝑃(10,18)

42

= √4 + 16

13

= √20 = √4 × 5 = 2√5 units ✓

13

𝑥

Radius: r = 13 Point: (10,18) y − coordinate of centre C = 13

|CB| = √(6 − 9)2 + [0 − (−6)]2 C1 : (x − a)2 + (y − b)2 = r2 2 2 (x − a) + (y − 13) = 132 [(10) − a]2 + [(18) − 13]2 = 169 (10 − a)2 + 52 = 169 2 (10 − a) = 144 a2 − 20a + 100 = 144 2 a − 20 − 44 =0 (a + 2)(a − 22) =0 a = −2 or a = 22 ⇒ C(−2,13) ⇒ C(22,13)

= √(−3)2 + (−6)2 = √9 + 36 = √45 = √9 × 5 = 3√5 ✓ A1(iii) AC: CB

= 2√5: 3√5 = 2: 3 ✓

∴ [x − (−2)]2 + (y − 13)2 = 132 ✓ or (x − 22)2 + (y − 13)2 = 132 ✓


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252


Rev Ex 9

A3(ii) Centre: (−2, −13) or (22, −13) Radius: r = 13 [x − (−2)]2 + [y − (−13)]2 = 132 ✓ C2 : or (x − 22)2 + [y − (−13)]2 = 132 ✓

A4(iii) Radius 1

r = |AB| 2

1

= √[(−2) − 6]2 + [4 − (−2)]2 2 1

A4(i)

= √100 = 5 2

Points on circle A(−2,4) B(6, −2)

Circle (x − x1 )2 + (y − y1 )2 = r 2 [x − (2)]2 + [y − (1)]2 = (5)2 (x − 2)2 + (y − 1)2 = 25

Centre C C = MAB = (

(−2)+6 4+(−2) 2

,

) = (2,1) ✓

2

A4(ii) Line DE Point: C(2,1) Gradient: ∵ DE ⊥ AB, mDE = DE:

−1 mAB

Points D & E =

−1 (4)−(−2)

[(−2)−(6)]

=

−1 [

6 ] −8

=

4 3

y − y1 = mDE (x − x1 ) y − (1)=

4 3 4

[x − (2)]

y−1 = x− y

3 4

3 5

3

3

5

3

3

+ [( x − ) − 1]

(x − 2)2

+( x− )

25 2 x 9 25 2 x 9 2

= x− ✓

4

(x − 2)2

− −

100 9 100 9

3 3 16 2 64 +( x − x 9 9 100

x+ x−

= 25 = 25

+

64 9

)= 25 = 25 =0

9

=0 =0 or x = 5

4

5

3

3

y|x=−1 = (−1) − = −3 ⇒ E(−1, −3) ✓

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3

9 125

x − 4x − 5 (x + 1)(x − 5) x = −1


5

3 2

8 2

4

(x 2 − 4x + 4)

8

4

At D & E, circle meets DE (y = x − )

4

5

3

3

y|x=5 = (5) − =5 ⇒ D(5,5) ✓

253


Rev Ex 9

Circle & Line x 2 + y 2 + 6x − 8y = 0 y = mx −

B1(i) −(1)

1

−(2)

3

sub (2) into (1): 1 2

1

x 2 + (mx − )

+6x − 8 (mx − ) = 0

3

2

x +

(m2 2

3

2

1

8

3 2

9 1

3 8

3

9

x − mx + ) +6x − 8mx +

(1 + m2 )x 2 − mx + (1 + m2 )x 2 + (6 −

26 3

+(6 − 8m)x +

m) x +

3

25

=0 =0 =0

9

Discriminant b2 − 4ac = (6 −

26 3

m)

B1(ii) Length of AB A(−9, −6) B(−4,4)

2

25

− 4(1 + m2 ) ( )

|AB| = √[(−9) − (−4)]2 + [(−6) − 4]2

9

= (36 − 104m + = (36 − 104m +

676 9 676 9

m2 ) −

100

2)

100

m

−

9 9

= √125

(1 + m2 ) −

100 9

= √25 × 5 = 5√5 units ✓

2

m

224

= 64m2 − 104m +

B1(iii) Area of triangle CAB A(−9, −6) B(−4,4) C(−4, −4)

9

For line to intersect circle at two distinct points: b2 − 4ac >0 224

Area of △ CAB

7

4

1 −4 −4 −9 −4 | | 2 −4 4 −6 −4 1 = [(−16) + 24 + 36 − 16 − (−36) − 24] 2 1 = (40) 2

24

3

= 20 unit 2 ✓

64m2 − 104m +

9

>0

=

576m2 − 936m + 224 > 0 72m2 − 117m + 28 >0 (24m − 7)(3m − 4) >0 +

m<

−

7

+

4

24

or m > ✓ 3

A5(ii) For line to be tangent to circle: b2 − 4ac =0 64m2 − 104m + m=

Points A & B At A & B, y 2 = −4x meets y = 2x + 12 (2x + 12)2 = −4x 2 4x + 48x + 144 = −4x 4x 2 + 52x + 144 = 0 x 2 + 13x + 36 =0 (x + 9)(x + 4) =0 x = −9 or x = −4 y|x=−9 = 2(−9) + 12 y|x=−4 = 2(−4) + 12 = −6 =4 ⇒ A(−9, −6) ✓ ⇒ B(−4,4) ✓

7 24

224 9

B1(iv) ⊥ distance from C to AB F ≡ Foot of ⊥ from C to AB |CF| = ⊥ distance from C to AB

=0

Equate area of △ CAB:

4

1

3

2 1

or m = ✓

2

A5(iii) For line to not meet circle: b2 − 4ac <0 64m2 − 104m +

224 9

|AB||CF|

= 20

(5√5)|CF| = 20

CF

=

40 5√5

=

8 √5

=

8√5 5

units ✓

<0

(24m − 7)(3m − 4) < 0 +

7 24

−

+

7

4

24

3

4 3

✓


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254


Line & circle y = mx x 2 + y 2 − 4mx + 3 = 0

Rev Ex 9

−(1) −(2)

B2(iii) Line & Circle ∵ line is parallel to y = x, y=x+c

x 2 + (x + c)2 − 4x + 3 = 0 −(2) sub (1) into (2): x 2 + (x + c)2 −4x + 3 = 0 2 2 2 x + (x + 2cx + c ) −4x + 3 = 0 (2x 2 + 2cx + c 2 ) −4x + 3 = 0 2x 2 + (2c − 4)x + (c 2 + 3) =0

sub (1) into (2): x 2 + (mx)2 − 4mx + 3 = 0 x 2 + m2 x 2 − 4mx + 3 = 0 (1 + m2 )x 2 − 4mx + 3 = 0 Discriminant For no intersection: b2 − 4ac (−4m)2 − 4(1 + m2 )(3) 16m2 − 12(1 + m2 ) 4m2 − 3(1 + m2 ) 4m2 − 3 − 3m2 m2 − 3 (m + √3)(m − √3) +

−

−(1)

<0 <0 <0 <0 <0 <0 <0

Discriminant For line to be tangent to circle: b2 − 4ac =0 2 2 (2c − 4) − 4(2)(c + 3) =0 2 2 (4c − 16c + 16) − 8(c + 3) = 0 (4c 2 − 16c + 16) − 8c 2 − 24 = 0 −4c 2 − 16c − 8 =0 2 c + 4c + 2 =0

+

c=

−√3 √3 −√3 < m < √3 ∵ m is positive integer, m=1✓

=

−(4)±√(4)2 −4(1)(2) 2(1) −4±2√2 2

=

−4±√8 2

= −2 ± √2

Line with 𝐜 = −𝟐 ± √𝟐 y = x + (−2 ± √2) = x − 2 ± √2 ✓


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255


Rev Ex 9

Method 1 (algebraic) x 2 + y 2 + 2gx + 2fy + c = 0 (13,0): 132 + 2g(13) + c = 0 169 + 26g + c = 0 (27,0): 272 + 2g(27) + c = 0 729 + 54g + c = 0

B3

Method 2 (graphical) ⊥ bisector of PQ PQ ⊥ ≡ ⊥ bisector of PQ PQ ⊥ : x =

−(1)

(13)+(27) 2

= 20

𝑅(0,9) O

⊥ bisector of PR Point:

−(2)

MPR = ( =(

(0,9):

𝑦

02 + 92 + 2f(9) + c = 0 81 + 18f + c = 0 −(3)

Gradient: mPR⊥ = PR ⊥ :

(1) − (2): −560 − 28g = 0 −28g = 560 g = −20

13+0 0+9 2 13 9 2

2

𝑥

)

, ) 2

−1

=

mPR

y − y1

,

𝑄(27,0) 𝑃(13,0)

−1

=

[

−9 ] 13

=

13 9

(x − x1 )

=m

9

−1 (0)−(9)

[(13)−(0)]

13

13

y − ( ) = ( ) [x − ( )] 2

y−

9 2

y

Put g = −20 into (1): 169 + 26(−20) + c =0 c = 351

= =

9 13

169

9 13

18 44

9

2

x− x−

9

Centre C At C, PQ ⊥ (x = 20) meets PR ⊥ (y =

Put g = −20 & c = 351 into (3): 81 + 18f + 351 = 0 f = −24

y|x=20 =

13 9

(20) −

13 9

x−

44 9

)

44 9

= 24 ⇒ C(20,24) Radius r C(20,24) P(13,0) Q(27,0) R(0,9) r = |CP| = √(20 − 13)2 + (24 − 0)2 = √625 = 25 Circle (x − x1 )2 + (y − y1 )2 = r 2 (x − 20)2 + (y − 24)2 = 252 ✓ B4(i)

x 2 + y 2 − 6x − 2y − 15 = 0 x 2 + y 2 + 2gx + 2fy + c = 0 Comparing coefficients: 2g = −6, 2f = −2, g = −3 f = −1 Centre:


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c = −15

Z(3,1) ✓

256

A math 360 sol (unofficial) B4(ii)

Rev Ex 9 B4(iii)

𝐵

𝐵

𝑀(2,3) 𝐴

𝑍(3,1) 𝐴

AB:

𝑍(3,1)

𝐶

Chord AB Point: M(2,3) Grad: mAB =

−1 mZM

=

−1 (1)−(3)

[(3)−(2)]

=

−1 −2 ) 1

(

=

1 2

Line CD Point: Z(3,1) Gradient: ∵ CD ∥ AB

y − y1 = mAB (x − x1 )

mCD = mAB =

1

y − (3)= [x − (2)] 2 1

1 2

y − y1 = m(x − x1 )

CD:

1

y−3 = x−1 y

𝐷

𝑀

y − (1)= ( ) [x − (3)]

2 1

2

1

= x+2✓

y − 1= x −

2

2

3 2

1

1

2

2

= x− ✓

y

−(1)

Points C & D 1

1

2

2

At C & D, CD (y = x − ) meets circle (x 2 + y 2 − 6x − 2y − 15 = 0) 1 2

1

x2 + ( x − ) 2 1 2 x + ( x2 4 5 2 1

2

1

1

2 1

4

− x+ )

( x − x+ ) 4

5 2 x 4 2

−

2 15 2

x−

4 55

2

=0

−6x − x + 1 − 15

=0

−7x − 14

=0

=0

−(−6)±√(−6)2 −4(1)(−11) 2(1)

y|x=3−2√5

=

6±4√5 2

= 3 ± 2√5

y|x=3+2√5

1

1

2

2

= (3 − 2√5) −

1

2

=0

4

5x − 30x − 55 x 2 − 6x − 11 x=

1

−6x − 2 ( x − ) − 15

1

1

2

2

= (3 + 2√5) −

3 1 3 1 = ( + √5) − = ( − √5) − 2 2 2 2 = 1 + √5 = 1 − √5 ⇒ C(3 − 2√5, 1 − √5) ✓ ⇒ D(3 + 2√5, 1 + √5) ✓


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A math 360 sol (unofficial) B5 Circle & line x 2 + y 2 − 4x − 2y + 1 x 2 − 4x + y 2 − 2y + 1 x 2 − 4x + (y − 1)2 − 12 + 1 x 2 − 4x + (y − 1)2

Rev Ex 9 B5(i) =0 =0 =0 =0

For two distinct points: b2 − 4ac >0 4k(3k + 4) > 0 +

−

−(1)

− 4

+

0

3 4

k < − or k > 0 ✓

kx − y = k + 1 −y = −kx + k + 1 y = kx − k − 1

3

B5(ii) For intersection at one exact point: b2 − 4ac =0 4k(3k + 4) = 0

−(2)

4

sub (2) into (1): x 2 − 4x + [(kx − k − 1) − 1]2 x 2 − 4x +(kx − k − 2)2 x 2 − 4x +[kx − (k + 2)]2 x 2 − 4x +[k 2 x 2 − 2k(k + 2)x + (k + 2)2 ] (1 + k 2 )x 2 +[−4 − 2k(k + 2)]x +(k + 2)2 (1 + k 2 )x 2 +(−4 − 2k 2 − 4k)x +(k + 2)2 (1 + k 2 )x 2 +(−2k 2 − 4k − 4)x +(k + 2)2

k = 0 or k = − ✓ 3

=0 =0 =0 =0 =0 =0 =0

B5(iii) For no points of intersection: b2 − 4ac <0 4k(3k + 4) < 0 + −

− 4 3

+ 0

4

−
Discriminant b2 − 4ac = (−2k 2 − 4k − 4)2 −4(1 + k 2 )(k + 2)2 = (−2)2 (k 2 + 2k + 2)2 −4(k 2 + 1)(k 2 + 4k + 4) = 4(k 2 + 2k + 2)2 −4(k 2 + 1)(k 2 + 4k + 4) k 2 (k 2 + 2k + 2) k 2 (k 2 + 4k + 4) ] = 4 [+2k(k 2 + 2k + 2)] −4 [ +1(k 2 + 4k + 4) 2 +2(k + 2k + 2) = 4(k 4

−4(k

+2k 3 +2k 3 4

= 4(k 4 −4(k 4

+4k

3

+4k 3 +4k 3

= 4k 4 +16k 3 −4k 4 −16k 3 = 12k 2 + 16k = 4k(3k + 4)

+2k 2 +4k 2 +2k 2 +4k 2 +k 2

+4k +4k

+4)

+4k

+4)

+8k 2 +5k 2

+8k +4k

+4) +4)

+32k 2 −20k 2

+32k −16k

+16 −16


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Ex 10.1

Ex 10.1 1(i)

1(ii)

AD = CD (S) ∡BDA = ∡BDC (A) BD = BD (S) △ ABD ≅ △ CBD ✓

(D is mid − pt of AC) (BD ⊥ AC) (common side) (SAS congruency)

B

A

D

(corr. ∡s of ≅ △ s)

∡ABD = ∡CBD i. e. BD bisects ∡ABC ✓

B

A

2(i)

(given) AP = QE AP + PQ = QE + PQ AQ = EP✓

D

AQ = EP (S) ∡BAQ = ∡DEA (A) ∡BQA = ∡DPE (A)

D C

(Proven) (Given) (iso.△ CPQ with CP = CQ) (AAS congruency)

△ ABQ ≅ △ EDP ✓ 2(iii)

∡QBA = ∡PDE ✓


C

B

A

2(ii)

C

P

Q

E

B

D C A

P

E

Q

B

D C A

3(i)

△ BOA ~ △ BYT ⇒

AB BO

=

BT BY

✓

(given) (corr. ∡s of ≅ △ s)

P

Q

T

A B

Y

O

3(ii)

(SS)

AB BO

=

BT BY

(A) ∡ABT = ∡OBY △ BAT ~ △ BOY ✓

(proven)

∡ABC = ∡AMN (A) ∡ACB = ∡ANM (A) △ ABC ~ △ AMN ✓


T

A

(opp. ∡) (SAS similarity)

B

O

4(i)

E

Y

(corr. ∡) (corr. ∡) (AA similarity)

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AC

=

BC

Ex 10.1

AN


MN

A

AC × MN = AN × BC NC × AC = AN × BC ✓ (MN = NC; equal sides of iso.△ MNC with ∡NMC = ∡NCM)

M

N

B

5(i)

∡CAB = ∡BCD (A) ∡ACB = ∡CBD (A) △ ABC ~ △ CDB ✓

(90°) (alt. ∡) (AA similarity)

C

C

A

B 5(ii)

AC

=

CB

CB

D


BD

AC × BD = CB 2 ✓ 6

AC ∥ GF ⇒ AC ∥ GE ✓ GE ∥ BD ∴ AC ∥ BD✓

(BG = GA, BF = FC, mid − pt thm)

C

A

(given) E G

F

D

B

7(i)

DE ∥ BC ✓

AD = DB, AE = EC, mid − pt thm

A

E

D

C

B

7(ii)

1

DE = BC 2 1

= (2x) 2

(AD = DB, AE = EC, mid − pt thm)

A

(BC = 2x)

=x ✓

D

𝑥

E

2𝑥 C

B

8(i)

∡SDM = ∡TCM (A) ∡DMS = ∡CMT (A) DM = CM (S) △ SDM ≅ △ TCM ✓

(given) (opp. ∡) (M is mid-pt of DC) (AAS congruency)

D A S

B

T

M C


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9

Ex 10.1

In △ DSB & △ CTA, ∡DSB = ∡CTA(A) DS = CT (S) ∡SDM = ∡TCM 2∡SDB = 2TCA ∡SDB = ∡TCA(A)

(△ SDM ≅ △ TCM) (△ SDM ≅ △ TCM) (△ SDM ≅ △ TCM) (DB bisects ∡SDM & CA bisects ∡TCM)

△ DSB ≅△ CTA ⇒ SB = TA ✓

(AAS congruency) (corr. sides of ≅△ s)

D A S

B

C

Prove: △ ANT ≅ △ ELR

G

A

10

(A) (iso.△ AGE with AG = EG) (given) (given)

E

R I

(S) (SAS congruency)

Z

J

11(i)

T

ZX = ZQ (S) (given) ∡JXZ = ∡FQZ (A) (iso.△ QZX with QZ = XZ) JQ = XF JQ + QX = XF + QX (given) JX = FQ (S) △ JZX ≅ △ FZQ ✓

L

N

AR = ET (given) AR − (TR) = ET − (TR) AT = ER (S) ∡NAT = ∡LER AG = EG NG = LG Subtracting sides, AG − NG = EG − LG AN = EL △ ANT ≅ △ ELR ✓

T

M

(SAS congruency)

Q

F

X

JX = FQ

∡BAE = ∡DAC (A)

(AE bisects ∡BAD)

∡AEB = ∡DEC = ∡DCE = ∡ACD (A)

(opp. ∡) (iso.△ CDE with DE = DC) (common ∡)

△ ABE ~ △ ADC ✓

(AA similarity)

A D

11(ii)

AB AE

=

AD AC

E B

C

(△ ABE ~ △ ADC)

AB × AC = AD × AE ✓


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∡DEB = ∡FDE (A)

(alt. ∡)

∡BDE = ∡CEF = ∡EFD (A)

(90°) (alt. ∡)

△ BDE ~ △ EFD ✓

(AA similarity)

12(ii) ∡EFD = ∡CEF (A)

13(i)

Ex 10.1 A

B

A

∡FED = ∡DBE = ∡ECF (A)

(△ BDE ~ △ EFD) (iso.△ ABC with AB = AC)

△ EFD ~ △ CEF ✓

(AA similarity)

13(ii) ∡FAC = ∡EBC (A) ∡FCA = ∡ECB (A) △ AFC ~ △ BEC ✓

C

E

(alt. ∡s)

∡AFC = 180° − ∡ACF − ∡CAF 180°) = 180° − (∡BCE) − (∡EAF) = 180° − ∡BCE − (∡EBF) = ∡BEC ✓ 180°)

F

D

F

D

B

C

E

(∡s in △ AFC =

B

(common ∡) (∡EAF = ∡EBF) (∡s in △ BEC =

F

A

C

E

(∡EAF = ∡EBF) (common ∡ ) (AA similarity)

B

F

A

13(iii)

CF

=

CA

CE CB

C

E

(△ AFC ~ △ BEC)

CF × CB = CE × CA ✓

14(i)

In △ DEC & △ ABC, ∡EDC = ∡BAC (A) ∡DCE = ∡ACB (A) △ DEC ~ △ ABC CD CA

= =

A

(right angles) (common ∡) (AA similarity) (corr. sides of ~ △ s)

CE CB 1

D

C

E B

(E is mid-pt of BC)

2

⇒ D is the mid-point of AC ✓ 14(ii) In △ ADE & △ CDE, DA = DC (S) ∡ADE = ∡CDE (A) DE = DE (S) △ ADE ≅ △ CDE ⇒ AE = CE ⇒△ AEC is isosceles ✓ © Daniel & Samuel A-math tuition 📞9133 9982

A

(D is mid-point of AC) (90°) (common side) (SAS congruency) (corr. sides of ≅△ s) (△ with 2 equal ∡s)

D

C

E B

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∡DBG = ∡ABC = ∡ACB = ∡DGB ⇒△ DBG is isosceles ⇒ DG = BD = CE ✓

15(ii) DG = EC (S) ∡DFG = ∡EFC (A) ∡DGF = ∡ECF (A) △ DGF ≅ △ ECF ✓

Ex 10.1 A

(iso.△ ABC with AB = AC) (corr. ∡) (△ with 2 equal ∡s) (equal sides of iso.△ DBG) (BD = CE)

D

B

C

F

G

E

(proven) (opp. ∡) (alt. ∡) (AAS congruency)

A

D

B

C

F

G

E

15(iii) DF = FE ✓

(corr. sides of ≅ △ s)

A

D

B

C

F

G

E

16(i)

BP = AC AB = QC

(S) (S)

(Given) (Given)

∡FBP = 90° − ∡FPB = 90° − ∡EPC = ∡ECP ⇒ ∡ABP = ∡QCA (A)

(⊿FBP) (opp. ∡) (⊿ECP)

Q

A F E P

B C

△ ABP ≅ △ QCA ✓ 16(ii) ∡BAP = ∡CQA AP = QA ⇒△ QAP is isosceles ⇒ ∡CQA = ∡APQ ⇒ ∡BAP = ∡APQ ✓

(SAS congruency) (corr. ∡s of ≅ △ s) (corr. sides of ≅ △ s) (△ with 2 equal sides) (equal ∡s of iso.△)

Q

A F E P

B C

16(iii) ∡QAP = ∡FAQ + ∡BAP = ∡FAQ + (∡CQA) (△ ABP ≅ △ QCA) = ∡FAQ + (90° − ∡FAQ) (⊿FAQ) = 90° ⇒ AP ⊥ AQ ✓

Q

A F E P

B C


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In △ FEC & △ BAC ∡FCE = ∡BCA (A) ∡FEC = ∡BAC (A) △ FEC ~ △ BAC CF CB

= =

CE CA 1

Ex 10.1 B

D

(Common ∡) (corr. ∡) (AA similarity) (corr. sides of ~ △ s)

F

A

C

E

2

⇒ BF: BC = 1: 2 ✓ 17(ii) AB = 2EF

(CE = EA, CF = FB, midpt thm)

B

D

F

In △ DCA & △ FEA, ∡DAC = ∡FAE (A) ∡DCA = ∡FEA (A) △ DCA ~ △ FEA CD EF

=

CD =

AC AE AC AE

(common ∡) (corr. ∡) (AA similarity) (corr. sides of~ △ s)

A

C

E

EF

= 2EF ∴ AB: CD = 1: 1 18(i)

AC = 3EC (Given) AE + EC = 3EC = 2EC = 2(2EF) (F is mid − pt of EC) = 4EF ✓

A

E G B

18(ii) BE ∥ DF

AG AD

=

AG =

AE AF AE AF 4 5

AG GD

A

E G B

F D

C

AD

= AD ⇒

(common ∡) (corr. ∡) (AA similarity) (corr. sides of ~ △)

C

D

(CD = DB, CF = FE, mid − pt thm)

In △ GAE & △ DAF, ∡GAE = ∡DAF (A) ∡AGE = ∡ADF (A) △ GAE ~ △ DAF

F

(AE = 4EF)

=4


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264


Ex 10.1

1st pair of similar triangles In △ EFC & △ ABC, ∡EFC = ∡ABC (A) ∡ECF = ∡ACB (A) △ EFC ~ △ ABC

A AC = BC = 1cm

E EF FC

= =

AB BC AB AB

(AB = BC)

=1 EF = FC

B −(1)

C

DF BD = DC = 0.5cm

2nd pair of similar triangles In △ BFE & △ ABD, ∡BFE = ∡ABD (A) (90°) ∡EBF = ∡ABD − ∡ABE = 90° −∡ABE (∡ABD = 90°) = ∡BAD (A) (⊿BAD) △ BFE ~ △ ABD (AA similarity) BF FE

=

AB BD AB

=1 2

BC

AB

=1 2

AB

(corr. sides of ~ △ s) (D is midpoint of BC) (AB = BC)

=2 BF = 2EF

−(2)

Sum of sides BF + FC =1 sub (1) & (2) into (3): (2EF) + (EF) =1 3EF =1 EF

=

−(3)

1 3

20(i)

∡ADB = ∡CDA (A) ∡ABD = 90° − ∡BAD = ∡CAD (A) △ ABD ~ △ CAD ✓

20(ii)

BD AB

=

BD = BD DC AD AB AD AB

= = =

AD

(corr. sides of ~ △ s)

CA AD2 CD AD2 CD2 CD CA AB AC

(90°) (⊿ BAD) (∡BAC = 90°) (AA similarity)

A

B

D

C

−(1) (corr. sides of ~ △ s) −(2)

sub (2) into (1): BD DC

=

AB2 AC2


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265


Ex 10.2

Ex 10.2 1(i)

1(ii)

∡AEB = ∡ECD (A) AE = CE (S) BE = DE (S) △ ABE ≅ △ CDE ✓

(opp. ∡) (AC and BD bisects at E) (AC and BD bisects at E) (SAS congruency)

A

D

E B

ABCD is a parallelogram ✓ (diagonals bisect each other)

C

A

D

E B

2(i)

BF = DE (S) ∡AFB = ∡CED (A) AF = CE (S) △ ABF ≅ △ CDE ✓

(Given) (90°) (Equal sides of square) (SAS congruency)

A

B

2(ii)

AB = CD

(= sides)

BF = ED FC = EA

∡ABD = ∡CDB (A) ∡ADB = ∡CBD (A) BD = DB (S) △ ABD ≅ △ CDB ✓

D

C

△ ABF ≅ △ CDE

(2 pairs of equal opp. sides)

(alt. ∡) (alt. ∡) (common side) (AAS congruency)

A

B

3(ii)

(△ ABD ≅ △ CDB) AB = CD (= sides) (△ ABD ≅ △ CDB) AD = CB (= sides) ⇒ ABCD has 2 pairs of equal sides ✓

4(i)

∡A + ∡B + ∡C + ∡D ∡A + ∡B + (∡A) + (∡B) 2(∡A + ∡B) ∡A + ∡B ✓

4(ii)

∡A = ∡C (= ∡𝐬) (Given) ∡B = ∡D (= ∡𝐬) (Given) ABCD is a parallelogram ✓ (2 pairs of equal opp. ∡s)


F

E

(given) (equal sides of square AFCE)

Adding sides, BF + FC = ED + EA BC = DA (= sides) ABCD is a parallelogram ✓

3(i)

C

D

C

= 360° (∡s of quad = 360°) = 360° (∡A = ∡C, ∡B = ∡D) = 360° = 180°

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Ex 10.2

A

X

D

Y

C

B

In △ ABC & △ CDA, ∡BAC = ∡DCA AC = CA

(A) (S)

(given) (common side)

∡XBA = ∡YDC 180° − ∡XBA = 180° − ∡YDC ∡ABC = ∡CDA (A) ∴△ ABC ≅ △ CDA ⇒ BA = DC (= 𝐬𝐢𝐝𝐞𝐬) ⇒ BC = DA (= 𝐬𝐢𝐝𝐞𝐬) ABCD is a parallelogram ✓

(given) (∡s in a line = 180°) (AAS congruency) (corr. sides of ≅△ s) (corr. sides of ≅△ s) (2 pairs of equal opp. sides)

(∥ 𝐬𝐢𝐝𝐞𝐬) (alt. ∡, ∡BAC = ∡ACD) BA ∥ CD (corr. ∡, BA ∥ CD) ∡XBA = ∡XCD ∡XBA = ∡YDC (given) ⇒ ∡XCD = ∡YDC (∥ 𝐬𝐢𝐝𝐞𝐬) (alt. ∡) ⇒ BC ∥ AD (2 pairs of ∥ sides) ABCD is a parallelogram ✓ 6)

(∥ sides of ▱ABCD)

AD ∥ BC ⇒ DX ∥ BY AD AX = CY

X

(∥ 𝐬𝐢𝐝𝐞𝐬) = CB

AD + (AX) = CB + (CY) DX = BY (= 𝐬𝐢𝐝𝐞𝐬)

A

(Equal opp. sides of ▱ABCD) (given) (AX = CY)

B C

D Y

XBYD is a parallelogram ✓ (1 pair of equal & ∥ opp. sides) 7(i)

BE = DF AB = CD

(S) (S)

∡EBA = ∡FDC (A) △ ABE ≅ △ CDF ✓

7(ii)

(given ) (Equal opp sides of ▱ ABCD) (alt. ∡) (SAS congruency)

A

D F E

B

C

(= 𝐬𝐢𝐝𝐞𝐬) AE = CF (△ ABE ≅ △ CDF) ∡AEB = ∡CFD (△ ABE ≅ △ CDF) 180° − ∡𝐴𝐸𝐵 = 180° − ∡𝐶𝐹𝐷

A

D F E

(∥ 𝐬𝐢𝐝𝐞𝐬) ⇒ AE ∥ CF (alt. ∡) ∴ AECF is a parallelogram ✓ (1 pair of equal & ∥ opp. sides)


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B

C

267


AE = CF = 7cm

Ex 10.2

(△ ABE ≅ △ CDF) (CF = 7cm)

A

D F E

B

8(i)

AC = AC (S) AB = AD (S) BC = DC (S) △ ABC ≅ △ ADC ✓

C

(common side) (equal adj. sides of kite) (equal adj. sides of kite) (SSS congruency)

A

E B

D C

8(ii)

CE = CE (S) ∡BEC = ∡DEC (A) BE = DE (S) △ BCE ≅ △ DCE ✓

(common side) (Diagonals are ⊥) (Longer diagonal bisects shorter diagonal) (SAS congruency)

A

E B

D C

8(iii)

BD ⊥ AC ✓

(Diagonals of kite are ⊥)

A

E B

D C

9?

(incomplete) Diagonals are only equal if ∡x = ∡y = ∡z = 90°

x A

1)

Statements ∡x = ∡z

2) 3) 4) 5)

∡BAD = ∡x ∡ABC = ∡y ∡BCD = ∡z ∡BAD + ∡ABC = 180°

6)

AD ∥ BC

7)

∡ABC + ∡BCD = 180°

8)

AB ∥ DC


Reasons ∡x + ∡y = ∡y + ∡z ∡x = ∡z opp. ∡ opp. ∡ opp. ∡ ∡BAD + ∡ABC = ∡x + ∡y = 180°

y

D

B

C

z

Supplementary int. ∡s ∡ABC + ∡BCD ∡y + ∡z = 180° Supplementary int. ∡s

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EH ∥ BD GF ∥ BD ⇒ EH ∥ GF ✓

Ex 10.2

(AE = ED, AH = HD, Mid − pt thm) (CF = FB, CG = GD, Mid − pt thm)

D H A G E I B

∡EID = 90° ∡EFG = 90° ✓

C

F

(BD ⊥ DF) (corr. ∡)

D H A G E I B

10(ii) EH ∥ GF ∡EFG = 90° 1

(proven) [∥ 𝐬𝐢𝐝𝐞𝐬] (proven) [𝐫𝐢𝐠𝐡𝐭 ∡]

A

(AE = ED, AH = HD, mid − pt thm)

GF = BD

(CF = FB, CG = GD, mid − pt thm)

2

∴ EFGH is a rectangle ✓

In △ EBF & △ ABD, ∡EBF = ∡ABD (A) ∡EFB = ∡ADB (A) △ EFB ~ △ ADB BF

= =

I C

F

(∥ & equal opp. sides & 1 right ∡)

EF ∥ BC

BD

G E

B

[𝐞𝐪. 𝐬𝐢𝐝𝐞𝐬]

⇒ EH = GF

11(i)

D H

EH = BD 2 1

C

F

BE BA 1

(AE = EB, AF = FC, midpt thm)

A F

E

(common ∡) (corr. ∡) (AA similarity) (corr. ∡s of ~ △ s)

D

G

B

C

2

⇒F is midpt of BD ✓ 11(ii) In △ FDG & ∡BDC, ∡FDG = ∡BDC ∡DFG = ∡DBC △ DFG ~ △ DBC DG DC

= =

DF DB 1 2

A

(common ∡) (corr. ∡) (AA similarity) (corr. ∡s of ~ △ s)

E

B

D F G

C

(F is midpt of BD)

⇒ G is midpt of CD ✓ 11(iii) AF = FC (Given) BF = FD (F is midpt of BD) ABCD is a parallelogram ✓ (Diagonals bisect each other)


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Ex 10.2

In △ BAD & △ CDA, BA = CD (S) AD = DA (S) Draw line through A ∥ to DC ∡ABC = ∡AFB = ∡DCB ⇒ ∡BAD = ∡CDA △ BAD ≅△ CDA

A

(given) (common side)

(A)

(equal ∡s of iso. △ ABF) (corr. ∡) (supple. int. ∡s) (SAS congruency)

D E

B

C A

D E F

B (△ BAD ≅ △ CDA) 12(ii) ∡ADB = ∡DAC ⇒△ ADE is isosceles (△ with 2 equal ∡s) (equal sides of iso.△) ⇒ AE = DE ✓

C

A

D E

B 12(iii) BD = CA BD − (DE) = CA − (AE) BE = CE ⇒△ BEC is isosceles ✓

C

(△ BAD ≅ △ CDA) (AE = DE)

A

D E

(△ with 2 equal sides)

B 13(i)

AE = AD (S) EB = DC (S) ∡AED = ∡ADE ∡AED + 90° = ∡ADE + 90° ∡AEB = ∡ADC (A) △ ABE ≅ △ ACD ✓

13(ii) AE = AD

(S)

(given) (equal opp. sides of rect.) (iso.△ AED) (∡ of rect. ) (SAS congruency)

A

E

D F

C

A

(△ ABE ~ △ ACD)

∡AED = ∡ADE ⇒ ∡AEF = ∡ADG (A)

(iso.△ AED)

△ AEF ≅△ ADG ✓

(AAS congruency)

E

D F

=

2EF 1

C

A

(BF = 2EF)

(1) E


D F

30° (2) 60°

2

∡EBF = 30° ∡ABC = 60°

G

B

13(iii) sin ∡EBF = EF =

G

B

(given)

∡EAB = ∡DAC ⇒ ∡EAF = ∡DAG (A)

BF EF

C

B

G

60° C

(Complementary ∡)

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270


Ex 10.2

AB = AC ⇒ iso.△ ABC ⇒ ∡ABC = ∡ACB = 60° ⇒△ ABC is equilateral ✓ 14(i)

AE ∥ XY ✓

(△ ABE ≅ △ ACD) (△ with 2 equal sides) (equal ∡s of iso.△) (△ with 2 ∡s = 60°)

(CX = XA, CY = YE, mid − pt thm)

B

A X C D Y

F E

14(ii) AE ∥ XY XY ∥ BF ⇒ AE ∥ BF ✓

(Proven) (DX = XB, DY = YF,mid-pt thm)

B

A X C D Y

F E

14(iii) AB ∥ DC EF ∥ DC ⇒ AB ∥ DC (∥ 𝐬𝐢𝐝𝐞𝐬) (∥ 𝐬𝐢𝐝𝐞𝐬) AE ∥ BF ∴ BAEF is a parallelogram ✓

(ABCD is a ▱ ) (CDEF is a ▱ )

B

A X

(proven) (∥ opp. sides )

C D Y F E

15

In △ ADE & △ CBF, AE = CF (S) ∡EAD = ∡FCB (A)

given alt. ∡

H

D

JH ∥ GI [= 𝐬𝐢𝐝𝐞𝐬] (given) ED ∥ BF [= 𝐬𝐢𝐝𝐞𝐬] (∥ opp. sides of ▱ EBFD) ⇒ JG ∥ IH HJGI is a parallelogram ✓

F

J

AD = CB (S) (equal opp. sides of ▱ ABCD) △ ADE ≅ △ CBF (SAS congruency) ⇒ DE = BF [= 𝐬𝐢𝐝𝐞𝐬] ⇒ ∡AED = ∡CFB ⇒ ∡DEF = ∡BFE (∡s in a line = 180°) ⇒ DE ∥ BF [∥ 𝐬𝐢𝐝𝐞𝐬] (alt. ∡) EBFD is a parallelogram ✓ (1 pair of equal & ∥ opp. sides)

I

E A

B

G

H

D

(∥ opp. sides)

J

1)

C F

2)

2) E A


C

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1) G

I B

271


Ex 10.3

Ex 10.3 1

∡ABC + ∡ABD = 90° + ∡ABD (∡ in semicircle) = 90° + 90° (∡ in semicircle) = 180° i.e. C, B and D lie on a straight line ✓

D B C

A

2

3(i)

∡APB = ∡ABP = 180° − ∡ABC = ∡ADC ⇒ ∡CDP = ∡CPD ⇒△ CDP is isosceles ⇒ CD = CP ✓

(iso △ APB with AB = AP) (∡s in a line = 180°) (∡s in opp segment)

C B P A

(△ with 2 equal ∡s) (equal sides of iso.△)

∡AEF + ∡CDA = ∡AEF + (180° − CBA) = ∡AEF + (ABF) = ∡AEF + (180° − ∡AEF) = 180°

D

(∡s in opp. segment)

C B

(∡s in a line st. line) (∡s in opp. segment)

D F A E

3(ii)

∡AEF + ∡CDA = 180° ⇒ CD ∥ FE ✓

(Proven) (Supplementary int. ∡s)

C B

D F A E

4

PB = PA = PC

(Equal tangents from ext pt) (Equal tangents from ext pt)

P

C B A

5(i)

∡ABC = ∡ACB = ∡BAT

(iso.△ ABC with AB = AC) (alt. segment thm)

C

B O

S


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A

T

272


∡ABC = ∡BAT ⇒ CB ∥ ST ✓

Ex 10.3

(proven) (alt. ∡s)

C

B O

S

6(i)

OA = OD OD = OC

(S) (S)

T

A

(radius) (radius)

B O

∡AOD = ∡OBC = ∡OCB = ∡DOC (A)

(corr. ∡) (iso.△ OCB with OC = OB) (alt. ∡)

C A D

△ AOD ≅ △ DOC ⇒ AD = CD ✓ 6(ii)

E

(SAS congruency) (corr. sides of ≅△ s) B

O C A D

E

∡CDE = 180° − ∡ODC − ∡ODA = 180° − ∡ODC − (∡OCD) = ∡COD ✓ 7(i)

(∡s in a line = 180°) (△ AOD ≅△ DOC) ∡s in △= 180°

∡EDC = ∡DAC + ∡DCA (ext. ∡ = sum of int. opp. ∡s) = ∡DAC + (iso.△ ADC with AD = DC) (∡DAC) = 2∡DAC ✓

E D A C

O

B

7(ii)

∡ABD = ∡ACD = ∡CAD = ∡CBD

(∡s in same segment) (iso.△ ADC with DA = DC) (∡s in same segment)

E D A C

O

B


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273


∡APB = 90° = ∡PQA

Ex 10.3

(A)

(∡ in semicircle) (Given)

B P

∡PBA = ∡QAP (A) △ ABP ~ △ PAQ ✓

(alt. segment thm) (AA similarity)

O

Q

A

8(ii)

1)

AP

=

AB

PQ

△ ABP ~ △ PAQ

PA

(AP)2 = AB × QP = √AB × QP ✓

AP 9

∡BFD = ∡BAF + ∡ABF (ext. ∡ = sum of int. opp. ∡s) = ∡BAF + (∡BAF) (iso.△ AFB with FA = FB) = 2∡BAF (alt. segment thm) = 2∡ACB ✓

C

B O

S

10(i)

∡APC = ∡BPA (A) ∡PAC = ∡PBA (A) △ APC ~ △ BPA ✓

(common ∡) (alt. segment thm) (AA similarity)

A

D

F

B

C A

10(ii)

PC AP

=

PA BP

P

(△ APC ~ △ BPA)

BP × CP = AP 2 ✓ 11

In △ BAN & △ BPL, ∡BAN = 90° = ∡BPL (A) ∡BNA

= ∡ALN = ∡BLP (A) ∴△ BAN ~ △ BPL ∡ABN = ∡PBL ⇒ ∡ABL = ∡PBL ✓

12(i)

∡PBL = ∡ABN (A) ∡BPL = 90° = ∡BAN (A) △ BPL ~ △ BAN ✓

B

(tan ⊥ rad) (∡ in semicircle)

P L O

(iso.△ ALN with AN = AL) (opp. ∡) (AA similarity) (corr. ∡ of ~ △ s)

(BN bisects ∡ABP) (∡ in semicircle) (tan ⊥ rad) (AA similarity)

A

Q

N

B P L

O

M A


sleightofmath.com

N

Q

274


Ex 10.3

12(ii) ∡BPL = 90° = ∡BMA (A) = ∡AMN (A)

(∡ in semicircle) (∡ in semicircle) (∡s in a line = 180°)

B P L

O

13

14(i)

∡PBL = ∡ABM = ∡MAN (A)

(BN bisects ∡ABP) (alt. segment thm)

△ BPL ~ △ AMN ✓

(AA similarity)

M A

PQ + QR + RP = PQ + QT + TR +RP = PQ + (QU) + TR +RP (Equal tangents QT = QU) = PQ + QU + (RS) +RP (Equal tangents TR = RS) = (PQ + QU) +(RS + RP) = PU +PS = PU +PU (Equal tangents PS = PU) = 2PU ✓ ∡BAC = ∡CAP (A)

(common ∡)

∡ABC = ∡CAX = ∡ACP (A)

(alt. segment thm) (alt. ∡)

△ ABC ~ △ ACP ✓

(AA similarity)

R

S

T P Q

U

B

C

P

X

14(ii)

AC AB AC2 AB

=

16(i)

Y

A

(corr. sides of ~ △ s)

AC

= AP

PA = 15

AP

Q

N

AC2 AB

✓

∡PCB = ∡PCA = ∡CAX = ∡PBC ⇒△ PBC is isosceles ⇒ PB = PC ✓ ∡BEC = 90° = ∡DAC

(PC bisects ∡BCA) (alt. ∡) (alt. segment thm) (△ with 2 equal ∡s) (equal sides of iso.△)

B

C X

CE ⊥ AB (A) (∡ in semicircle)

Y

A

C

(∡s in same segment) ∡CBA = ∡CDA ⇒ ∡CBE = ∡CDA (A) (AA similarity) △ EBC ~ △ ADC ✓


P

O A B

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E D

275


Ex 10.3

16(ii) area of △ ABC

C

1

= (AB)(CE) 2

EC BC

=

EC =

AC

O

(△ EBC ~ △ ADC)

DC BC×AC

A E

B

DC

D

area of △ ABC 1

= (AB)(CE) 2 1

BC×AC

2 1

DC BC×AC

= (AB) ( = (AB) ( = = 17(i)

1) 2) 3) 4)

5) 5)

6)

2 2r AB×BC×AC 2(2r) AB×BC×AC 4r

) )

(diameter DC = 2r)

✓

AP × BR = PQ2 PQ2 = PC × PA (AP) × BR = PC × (PA) RB = PC AB = AC

given Tangent-secant thm (1,2)

AQ is line of symmetry of iso △ APR [Trigonometric proof] [geometric] 2 2 2 Pythagoras thm AQ = AP − PQ 2 2 [ ] AQ = AP − PC × PA tangent −secant thm = AP[AP − PC] = AP[AC] 2 (4) AQ = AP × AB ✓

17(ii) (work in progress) (S) AQ = AQ (A) ∡AQP = ∡AQR (S) QP = QR △ AQP ≅ △ AQR ✓

P

R

Q

C

B

A

P

(common side) (90°) (AQ is line of symmetry) [Trigonometric proof] SAS(1,2,3) R

17(ii) (R) ∡ABQ = ∡ACQ (H) AQ = AQ (S) AB = AC △ ABQ ≅ △ ACQ ✓

rt ∡ in semicircle common side AQ is line of symmetry [trigonometric proof] RHS(1,2,3)

C

B

A

P

R


Q

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Q

C

B

A

276

A math 360 sol (unofficial) 17(iii) AP = AR AP =1 AR AC = AB AC AB AP AR

=1 =

AC AB

=1

Ex 10.3 AQ is line of symmetry [trigonometric proof] AQ is line of symmetry [trigonometric proof] (1,2)

AP × AB = AR × AC ✓


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277


Ex 10.3

C

B

F α

β

A

D

E

Proposition Given AE is tangent to diameter CD AB = EF Then CD is line of symmetry Proof 0 < α < 90° cos α = tan α =

AB AD 2r AD

⇒ AD = ⇒ AD =

AB cos α 2r tan α

Equate AD: AB 2r = cos α tan α (AB) tan α = 2r cos α (AB) sin α = 2r cos 2 α (AB) sin α = 2r[1 − sin2 α] (AB) sin α = 2r − 2r sin2 α 2r sin2 α + (AB) sin α − 2r = 0 (2r)[sin α]2 + (AB)[sin α] − 2r = 0 sin α =

−(AB)±√(AB)2 −4(2r)(−2r) 2(2r)

=

sin α =

−(AB) ± √(AB)2 + 16r 2 4r

−(AB)+√(AB)2 +16r2 4r

or sin α =

−(AB)−√(AB)2 +16r2 4r

<0

[rej ∵ sin α > 0 for 0 < α < 90°] There could only be 1 angle α for a certain fixed length AB Similarly, there could be only 1 angle β for the same fixed length EF ∵ AB = EF ⇒α=β △ CAD ≅ △ CED CD is line of symmetry Alternate proof Increasing function of sine when ∡ is increasing from 0° to 90° and adjacent is increasing


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278


Ex 10.3

area of △ ABC = area of △ ABO +△ BCO +△ CAO 1

1

1

2 1

2

2

= (AB)r + (BC)r + (CA)r = r(AB + BC + CA) 2 1

= r(AB + BC + CA) 2 1

= r(perimeter of △ ABC) 2 1

= r(2s) 2

(S is semi-perimeter of △ ABC)

= rs ✓


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279


Rev Ex 10

Rev Ex 10 A1(i)

Prove: △ GHI ≅ △ ABC

H I

HI = BC ∡GHI = ∡ABC

(S) (A)

(Given) (alt. ∡)

AH = GB AH − (AG) = GB − (AG) GH = AB (S) △ GHI ≅ △ ABC ✓ A1(ii) Prove: CA ∥ GI

(Given)

G A C B

(SAS congruency) H I

∡HGI = ∡BAC (△ GHI ≅ △ ABC) ⇒ ∡AGI = ∡GAC (∡s in a line = 180°) (alt. ∡) ⇒ CA ∥ GI ✓

G A C B

A2(i)

Prove: △ APQ ~ △ ABC ∡APQ = ∡ABC (A) ∡PAQ = ∡BAC (A) △ APQ ~ △ ABC ✓

A2(ii) Prove: AQ = AQ AP

=

AQ = ( =( = A3(i)

AB AC

AB AC2

2AB

(Given) (Common ∡) AA similarity

P

Q

C

AC2 2AB

AC AB AC

A

B

(△ APQ ~ △ ABC) ) AP AC

)( ) 2

(AP =

AC 2

)

✓

Prove: △ AEF ~ △ DEG (A) ∡AEF = ∡DEG (A) ∡EAF = ∡EDG △ AEF ~ △ DEG ✓

A F

(opp. ∡) (alt. ∡) (AA similarity)

E B


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D

G C

280


Rev Ex 10

A3(ii) Prove: BF = 4AF

A

(1) DG AF

=

F

DE AE

(4)

1

(AE = AD)

=2

B

In △ FCB & △ GCD, ∡FCB = ∡GCD (A) ∡FBC = ∡GDC (A) △ FCB ~ △ GCD BF

=

G

(2)

3

DG = 2AF

GD

E

C

D

(common ∡) (corr. ∡) (AA similarity)

BC DC

=2 BF = 2DG = 2(2AF) = 4AF ✓ A3(iii) Find: |AB|

(D is midpt of BC) (DG = 2AF) A

1 cm

A4(i)

AB = AF + BF = AF + (4AF) = 5AF = 5cm ✓ Prove: ∠DPC = ∠BQA In △ PDC & △ QBA, PD = QB (S) ∡PDC = ∡QBA (A) DC = BA

(S)

F

BF = 4AF

4 cm

AF = 1cm

B

G C

D A

B Q

P

(given) (equal opp. ∡s of ▱ABCD) (equal opp. sides of ▱ABCD) (SAS congruency)

D

△ PDC ≅ △ QBA ⇒ ∡DPC = ∡BQA ✓ A4(ii) Prove: AQCP is parallelogram ∡DPC = ∡BQA ∡DPC = ∡PCQ ⇒ ∡PCA = ∡BQA PC ∥ QA (∥ 𝐬𝐢𝐝𝐞𝐬) PC = QA (= 𝐬𝐢𝐝𝐞𝐬) AQCP is a parallelogram ✓


E

C

A

B Q

(proven) (alt. ∡)

P D

C

(corr. ∡) (1 pair of ∥ & eq. sides)

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281


Rev Ex 10

3

Prove: TU = PQ

S

There is only 1 straight line connecting T & U.

T

4

If V is the midpoint of PR, TV ∥ SR (PT = TS, PG = GR, mid − pt thm) VU ∥ PQ (RU = RQ, RG = GP, mid − pt thm)

R

U V Q

P

TU = TV + VU ⇒ TU ∥ SR ∥ PQ (SR ∥ PQ) TU = TV

+VU

1

= ( SR) +VU 2

1

(PT = TS, TV = VR, midpt thm)

1

+ ( PQ) (RV = VR = RU =

= SR 2

2

UQ, midpt thm) 1

1

2

2 1

= ( PW) + PQ 1 1

(equal height of trapezium)

= ( PQ) + PQ 2 2 1

2 1

= PQ

+ PQ

4 3

2

= PQ 4

A5(ii) TU = 3 PQ 4 3

8 = PQ PQ = A6(i)

4 32 3

✓

Prove: CE ∥ DG C

In △ ABD & △ ACD, AB = AC (H) AD = AD (S) ∡ADB = 90° = ∡ADC (R) △ ABD ≅ △ ACD BD = CD CE ∥ DG ✓

(given) (common side) (∡ in semicircle) (∡s in a line = 180°) (RHS congruency) (corr. sides of ≅△ s)

D F A

(BD = CD, BG = GE, mid − pt thm)

A6(ii) Prove: AF = 1 AD

C

2

(A) ∡FAE = ∡DAG (A) ∡AFE = ∡ADG △ FAE ~ △ DAG AF AD

= =

B

G

E

AE AG 1 2 1

(common ∡) (corr. ∡) (AA similarity) (corr. sides of~ △ s)

D F A

E

G

B

(AE = AG)

= AD ✓ 2


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282


Rev Ex 10

C D A

O

F

P

B

E

Prove: △ PDF ~ △ POC ∡DPF = ∡OPC

(A) (common ∡)

∡PDF = 180° − ∡CDF (∡s in line = 180°) = ∡CAE (∡s in opp. segments) = 2OAC (OA bisects ∡CAE ∵ arc AC = arc AE) = ∡OAC (iso.△ OCA with 2 sides as radius) + ∡OCA = ∡POC (A) (ext. ∡ = sum of int. opp. ∡s) (AA similarity) △ PDF ~ △ POC ✓ A7(ii) Prove: PD × PC = PF × PO PD

=

PF

PO

(△ PDF ~ △ POC)

PC

PD × PC = PF × PO ✓ A7(iii) Prove: PB × PA = PF × PO In △ PBD & △ PCA, ∡BPD = ∡CPA (A) ∡PDB = 180° − ∡BDC = ∡PAC (A) △ PBD ~ △ PCA ⇒

PB PD

=

D

(common ∡ ) (∡s in a line) (∡s in opp. segments) (AA similarity)

A

O

F

B

P

E

PC PA

PB × PA = PD × PC = PF × PO✓ A8(i)

C

from (ii)

P F D

A

Q O

C

E

B

∡ADQ + ∡QDC + ∡CDF = ∡ADQ + (90°) + ∡CDF = ∡ADQ + 90° + (∡CFD) = ∡ADQ + 90° + (∡OFA) = ∡ADQ + 90° + (∡OAF) = ∡ADQ + 90° + (90° − ∡ADQ) = 180° i.e. A,D and F lie on straight line ✓


(tan ⊥ rad) (iso.△ CDF with CD = CF) (Common ∡) (iso.△ OAF with OA = OF as radius) (⊿ADQ)

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283


Rev Ex 10

A8(ii) Prove: AD × AF = AQ × AB [

AD AQ

=

AB AF

] or

AD AB

In △ ADQ & △ ABF, ∡DAQ = ∡BAF (A) △ ADQ = 90° = ∡AFB (𝐀) △ ADQ ~ △ ABF ⇒ B1(i)

AD

=

AQ

=

P F

AQ AF

D

(common ∡) (given) (∡ in semicircle) (AA similarity)

A

Q O

B

E

AB AF

AD × AF = AQ × AB ✓ Prove: ∠ABF = ∠EDF

C

(given) ∡ABD = ∡EDB (given) ∡FBD = ∡FDB Subtracting angles, ∡ABD − ∡FBD = ∡EDB − ∡FDB ∡ABF = ∡EDF ✓ B1(ii) Prove: △ ABF ≅ △ EDF

B

D

A

E

F C

(proven) (iso. △ ACE with CA = CE) (iso. △ DBF with ∡FBD = ∡FDB) (AAS congruency) △ ABF ≅ △ EDF ✓ B1(iii) Prove: F is the mid − point of AE ∡ABF = ∡EDF(A) ∡BAF = ∡DEF(A) BF = DF (S)

B

D

A

E

F C

AF = EF (△ ABF ≅ △ EDF ) ⇒ F is the mid-point of AE ✓

B

D

A

B2(i)

C

E

F

Prove: △ YDC ~ △ XYB ∡YCD = ∡XBY(A) ∡YDC = ∡XYB(A) △ YDC ~ △ XYB ✓

X Y

A

(corr. ∡) (corr. ∡) (AA similarity)

B

D

C

B2(ii) Prove: YC = 2XB YC XB

= = =

DC YB AB YB (2YB)

X A

(△ YDC ~ △ XYB) (DC = AB; equal opp. sides of ▱ABCD) (AB = 2YB)

YB

(1)

Y

B

(2) D

C

=2 ∴ YC = 2XB✓


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284


Rev Ex 10

Prove: CE = GE

A D

In △ CAE & △ GAE, ∡CAE = ∡GAE (A)

(AE bisects ∡BAC)

∡ACB = 90° = ∡ADF = ∡AGE (A)

(given) (AB ⊥ CD) (corr.∡)

F

G B

AE = AE (S) △ CAE ≅ △ GAE ⇒ CE = GE ✓ B3(ii) Prove: CF = CE

B4(i)

C

(Common side) (AAS congruency) (corr. sides of ≅△ s) A

∡CEA = 90° − ∡CAE = 90° − ∡DAF = ∡DFA = ∡CFE ⇒ iso.△ CEF ⇒ CF = CE ✓ Prove: SM = 2TU

(⊿ ACE) (AE bisects ∡BAC) (⊿ AFD) (opp. ∡) (△ with 2 equal ∡s) (equal sides of iso.△)

SM ∥ QT

(PS = SQ, PM = MT, mid − pt thm)

In △ RSM & △ RUT, ∡SRM = ∡URT (A) ∡RSM = ∡RUT (A) △ RSM ~ △ RUT SM RM

=

UT RT

=2 SM = 2TU ✓ B4(ii) Prove: 4TU = QT QT = 2SM = 2(2TU) = 4TU ⇒ 4TU = QT✓

E

D F

G B

C

E

P

common ∡ corr. ∡ (AA similarity) (corr. sides of ~ △ s)

S

(2)

M

(1) U Q

R

(RM = 2RT)

P (PS = SQ, PM = MT, mid − pt thm) (SM = 2TU)

M

S

T U Q

R

B4(iii) Find: area of △ TQR

P

area of △ TQR = (QT)(height) = (4TU)(height) 2

1

= 4 [ (TU)(height)]

M

S

1 2 1

T

QT = 4TU

U Q

𝑥

T R

2

= 4(area of △ UTR) = 4x ✓


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285


Rev Ex 10

Prove: BXDY is a parallelogram EX ∥ BY ⇒ XD ∥ BY

(AE = EB, AX = XY, mid − pt thm)

FY ∥ BX ⇒ BX ∥ YD

(CF = FB, CY = YX, mid − pt thm)

A X E Y B

A

(S)

E Y B

C

F

(equal opp. sides of ▱BXDY)

(SAS congruency) △ AXB ≅ △ CYD ✓ B5(iii) Prove: ABCD is a parallelogram AB = CD (= 𝐬𝐢𝐝𝐞𝐬) ∡CAB = ∡ACD ⇒ AB ∥ CD(∥ 𝐬𝐢𝐝𝐞𝐬) ABCD is a ▱ ✓

A

(△ AXB ≅ △ CYD) (△ AXB ≅ △ CYD) (alt. ∡) (1 pair of equal & ∥ opp. sides)

B6

Prove: ∠BCD = 2∠BDA

B7

∡BCD = ∡BCA + ∡DCA = (∡BDA) + ∡DCA (∡s in same segment) = ∡BDA + (∡DBA) (∡s in same segment) = ∡BDA + (∡BDA) (iso.△ ABD with AB = AD) = 2∡BDA ✓ Prove: BC ∥ QR

D X

E Y B

C

F C

D B

A

R ∡QRP = ∡PQC = ∡PBC ⇒ BC ∥ QR ✓

D X

(S) (given)

∡YXB = ∡XYD (alt. ∡) ⇒ ∡AXB = ∡CYD (A) (∡s in a line = 180°) XB = YD

C

F

(∥ opp. sides) BXDY is a ▱ ✓ B5(ii) Prove: △ AXB ~ △ CYD AX = CY

D

C P

(alt. segment thm) (∡s in same segment) (alt. ∡)

B Q


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Rev Ex 10

B8(i)

P

P

Q

Q

A

A C

B

B

H

O

C O

R

H R

Prove: PC 2 = PA × PB − AC × BC In △ APQ & △ QPB, ∡APQ = ∡QPB ∡PQA = ∡PBA △ APQ ~ △ QPB PQ AP

=

(common ∡) (alt. segment thm) (AA similarity)

PB QP

PQ2 = PA × PB PC 2 = PH 2 + HC 2 = (PQ2 − QH 2 ) + HC 2 = PQ2 + (HC 2 − QH 2 ) = PQ2 + (HC + QH)(HC − QH) = PQ2 + (CH + HR)(−QC) = PQ2 − QC × RC = PQ2 − AC × BC = PA × PB − AC × BC ✓

B8(ii) Prove: PC 2

1

1

PC

= (

1

2 PA

+

1 PB

(pythagoras ′ thm) (pythagoras ′ thm) [a2 − b2 = (a + b)(a − b)] QH = HR (QC × RC = AC × BC) (PQ2 = PA × PB) (tangent − secant thm) (not in syllabus)

)

= PA × PB − AC × BC = PA × PB − (PC − PA) × (PB − PC) = PA × PB + (PA − PC) × (PB − PC) = PA × PB + (PA)(PB) − (PA + PB)(PC) + PC 2 = 2(PA)(PB) − (PA + PB)(PC) + PC 2

PC(PA + PB) = 2(PA)(PB) PC 1 PC

= =

2(PA)(PB) PA+PB PA+PB 2(PA)(PB) 1 1 1

= (

2 PA

+

PB

)✓


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Ex 11.1 4(b)

Ex 11.1 1(a)

sin 45° cos 30°+sin 60°

1(b)

√2 2 √3 √3 + 2 2

=

=

√2 2

√3

=

√2 2√3

=

√6 6

✓

1st quadrant ✓ α = 390 − 360 = 30° ✓

1

√3

=1 +1 =2✓ π

sin cos = =

π

3 6 √3 √3 ( )( ) 2 2 3 4 5

+ cos + +

4(c)

π 3

2 1

𝑂

2

4

2

π tan 3 π π π sin tan +cos 3 4 6

Method 1 (graphical) 𝑦

1

= ✓ 1(d)

=

√3 (

√3 √3 )(1)+( ) 2 2

=

√3 √3

1 2

=1✓

∵ sin θ =

2

4(d)

Method 1 (graphical) 𝑦 𝑂

2

𝑥 θ = −100°

α

= ✓ 4

= 2(2)

+

1 tan A 1 2

Method 2 (non-graphical) θ = −100 ≡ 260° ✓ 3rd quadrant α = θ − 180 = 260 − 180 = 80° ✓

3rd quadrant ✓ α = 180 − 100 = 80° ✓

+ tan(90° − A)

= 2 tan A +

𝑥 θ = −60°

1

1

2 tan A

α

Method 2 (non-graphical) θ = −60 ≡ 300° ✓ 4th quadrant α = 360 − θ = 360 − 300 = 60° ✓

4th quadrant ✓ α = 60° ✓

sin θ cos(90° − θ) = sin θ sin θ = (sin θ)2 =( )

3

Method 2 (non-graphical) θ = 390 ≡ 30° 1st quadrant ✓ α= θ = 30° ✓

tan 45° + tan 30° tan 60° = (1) + ( ) (√3)

1(c)

Method 1 (graphical) 𝑦 θ = 390° α 𝑥 𝑂

∵ tan A = 2

1

=4 ✓ 2

4(a)

Method 1 (graphical) 𝑦

α

θ = 250° 𝑥 𝑂

3rd quadrant ✓ α = 250 − 180 = 70° ✓

Method 2 (non-graphical) θ = 250° 3rd quadrant ✓ α = θ − 180 = 250 − 180 = 70° ✓

5(a)

α = 20° θ = α, 180 − α, 180 + α, 360 − α = 20°, 160°, 200°, 340° ✓

5(b)

α = 70° θ = α, 180 − α, 180 + α, 360 − α = 70°, 110°, 250°, 290° ✓

5(c)

α = 35° θ = α, 180 − α, 180 + α, 360 − α = 35°, 145°, 225°, 315° ✓

6(a)

α=

π 3

θ = α, π − α, π + α, 2π − α π 2π 4π 5π = , , , ✓ 3

6(b)

α=

3

3

3

π 5

θ = α, π − α, π + α, 2π − α π 4π 6π 9π = , , , ✓ 5


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5

5

288


7(i)

Ex 11.1

α = 0.3 θ = α, π − α, π + α, 2π − α = 0.3, 2.84, 3.44, 5.98 ✓ R=

2(v0 )2 g

9(i)

Right angle triangle sin A =

√3

By Pythagoras’ Thm,

sin θ cos θ

2

adj = √(√3) − 12

If v0 = 10, g = 10, θ = 30°: 2(10)2

R =

√3 1 A adj

1

10

= √3 − 1 = √2

sin 30° cos 30° 1

√3 2

⇒ cos A =

= 20 ( ) ( ) 2

√2 √3

, tan A =

1 √2

= 5√3 ✓ Trigonometric ratio 7(ii)

If v0 = 7, g = 9.8, θ = 45°: 2(7)2

R = =

√2

sin 45° cos 45°

9.8 98 √2 ( ) 9.8 2 2

√2 ( ) 2

= 9(ii)

= 10 ( ) 4

1 2

5 A 𝑜𝑝𝑝 3

3 5

By Pythagoras’ Thm, √52

opp =

−

32

2

4

5

3

= √2 ✓ 10(i)

𝑦

π

+ tan ( − A) 2

4

1

3

4 3

4

3

= + = + 3

√3

3

4

= tan A +

8(ii)

sin(90° − A)tan 60° = cos A (√3) = (√ )

Trigonometric ratio

=2

9(iii)

=4

⇒ sin A = , tan A =

tan A

1

√2

= ✓

Right angle triangle cos A =

1

√2

tan A sin 45° = ( ) ( )

=5✓ 8(i)

√3 2 √3 √2 ✓ 2

cos A cos 30° = ( ) ( )

1

α

A = 230° 𝑥 𝑂

tan A

α = 230 − 180 = 50° ✓

4

1 12

✓

4 sin A

For 0° < B < 360° B = α, 180 − α, 180 + α, 360 − α = 50°, 130°, 230°, 310° ✓

π

+ cos ( − A) 2

= 4 sin A + sin A = 5 sin A

10(ii)

𝑦

4

= 5( )

𝑂

5

=4✓

α

𝑥 A = 320°

α = 360 − 320 = 40° ✓ For 0° < B < 360° B = α, 180 − α, 180 + α, 360 − α = 40°, 140°, 220°, 320° ✓


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Ex 11.1

Right angle triangle opp sin A = k =

1

12

A 𝑘 adj

hyp

Right angle triangle cos 35° = k By Pythagoras’ Thm,

By Pythagoras’ Thm,

opp = √1 − k 2

adj = √1 − k 2

tan 35° = Trigonometric ratio cos A = √1 − k 2 ✓

=

1

= √1 − k 2 ( ) √3

1−k2

11(iii)

3

π 2 cos 6

tan(90°−A)

✓

=

√3 2( ) 2 1 tan A

= (√3) tan A = √3 ( = k√

k

√1−k2 k

Trigonometric ratio tan 35° + tan 45° + tan 55°

11(ii) sin(90° − A)tan 30° = cos A tan 30°

=√

1 opp 35° k

) 2

√1−k2 k

+1

=1

+

=1

+

=1

+

=1

+

=1

+

+ tan(90° − 35°)

√1−k2

+

k √1−k2

+√

1−k2 k

k √1−k2 k 1−k2 +k2 k√1−k2 1 k√1−k2

1 tan 35° 1

+

k √1−k2

✓

√1−k 3

1−k2


✓

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Ex 11.2 3(a)

Ex 11.2

cos 330° = cos 30° =

1

Coordinates x = −1

√3 2

S

A 330°

✓

T C α = 360 − 330 = 30°

y = −√3 3(b)

2

sin 225° = − sin 45°

r = √(−1)2 + (−√3) = 2

=

√2 − 2

cos 240° = tan 240° = sec 240° = csc 240° = cot 240° = 2(a)

y r x r y x

=

T C α = 360 − 225 = 45°

(−√3)

= =

(2) (−1) (2)

1 sin 240° 1 tan 240°

√3 2

✓ 3(c)

1

= = =

sin 230° < 0 ✓

cos 350° > 0 ✓

tan 340° < 0 ✓

sin 160° > 0 ✓

cos

3π 4

1 1 2

(− ) 1 √3 (− ) 2

1 (√3)

= −2 ✓ =− =

S

S

S

S

2 √3

1 √3

𝑦

9π 8

6

T

𝑦

𝑦

S

S

5π 6

3(e)

cos(−150°) = − cos 30°

=

π 6

✓ =

3(f)

A 350° 𝑥 C A 340° 𝑥 C

3

π

) = tan ( ) 3

sin θ = −0.2 ⇒ 3rd or 4th quadrant α ≈ 11.5°

S A T C

2π

−

3 2π 3

S

A

α T

α C

tan θ = −3 ⇒ 2nd or 4th quadrant α ≈ 71.6

S α

A

0° < θ < 360°

T

0° < θ < 360°

=

π 3

θ = 180 + α, 360 − α ≈ 191.5°, 348.5° ✓

4(b)

A

3π

C

2π

✓

S A −150° T C α = 180 − 150 = 30°

α=π− 4(a)

A 160° 𝑥 C

tan (−

√3 − 2

= √3 ✓

𝑥

α C

θ = 180 − α, 360 − α ≈ 108.4°, 288.4° ✓

𝑦

>0✓

C

tan(−45°) = − tan 45° S A = −1 ✓ −45° T C α = 45°

𝑦

<0✓

A

5π

3(d)

✓

A 230° 𝑥 C

𝑦

T tan

S

6

α=π−

4

2(f)

π

= √3 ✓

T 2(e)

= sin 2

T 2(d)

6

= ✓

T 2(c)

5π

1

T 2(b)

sin

2

(−1)

cos 240°

=−

=− ✓

(−√3)

1

A 225°

✓

Trigonometric ratio sin 240° =

S

A 9π 8

T © Daniel & Samuel A-math tuition 📞9133 9982

𝑥

C sleightofmath.com

291


cos θ =

Ex 11.2

1

5(a)

√2

⇒ 1st or 4th quadrant π α= 4

S T

A α α C

tan A = −

8 15

< 0 ⇒ 2nd or 4th quadrant

90° < A < 180° ⇒ 2nd quadrant ∴ 2nd quadrant

0 < θ < 2π θ = α, 2π − α π 7π = , ✓ 4

Quadrant

Coordinates tan A = −

4

8 15

=

y

r A

8

x

−15 4(d)

cos θ = −0.7 ⇒ 2nd or 3rd quadrant α ≈ 0.795

S α α T

x = −15, y = 8,

A

r = √82 + (15)2 = 17

C

Trigonometric ratio

0 < θ < 2π θ = π − α, π + α ≈ 2.35,3.94 ✓ 4(e)

tan2 θ = 3 tan θ = ±√3 ⇒ 1st, 2nd, 3rd or 4th quadrant α = 60°

5(b) S α α T

A α α C

(17) 8

r

17

=−

15 17

✓

2 √5

< 0 ⇒ 3rd or 4th quadrant

90° < B < 270° ⇒ 2nd or 3rd quadrant ∴ 3rd quadrant Coordinates sin B = −

2 √5

=

r

3 sin θ = 2 2

sin θ

=

sin θ

= ±√

3 2

S α α T

A α α C

B

x

y

−2

y = −2, r = √5,

2

✓

Quadrant sin B = −

θ = α, 180 − α, 180 + α, 360 − α = 60°, 120°, 240°, 300° ✓

2

(−15)

r y

sin A = =

0° < θ < 360°

4(f)

x

cos A = =

√5

2

x = −√(√5) − (−2)2 = −√5 − 4 = −1

3

Trigonometric ratio

⇒ 1st, 2nd, 3rd or 4th quadrant α ≈ 0.955

x

cos B = = r

0 < θ < 2π

cot B =

θ = α, π − α, π + α, 2π − α = 0.955 2.19, 4.10, 5.33 ✓


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1 tan B

−1 √5

=

✓ 1 y x

=

1 (−2) (−1)

1

= ✓ 2

292


Ex 11.2

Quadrant

8(i)

1

cos A = > 0

Coordinates sin 20° = k =

2

⇒ 1st or 4th quadrant

y r

1

y = k, r = 1,

cos A & sin A have same sign ⇒ 1st or 3rd quadrant

k

20° x

x = √1 − k 2

∴ 1st quadrant Trigonometric ratio sin 200° = − sin 20° = −k ✓

Coordinates 1

x

2

r

cos A = =

2

A 𝑦 8(ii)

1

x = 1, r = 2, y = √22 − 12 = √3

8(iii)

Trigonometric ratio y

sin(−A) = − sin A = − = r

y

tan A = = x

7(i)

√3 1

√3 − 2

8(iv)

✓

5 12

9(a)

<0

=

12

y x

9(b) −5

r

+

(−5)2

y

5

r

13

⇒ sin A = = −

= 13 x

12

r

13

, cos A = =

cos(−A) = cos A =

7(iii) 7(iv)

12

=

y x

=−

1

=

cos(90°−70°)

k √1−k2 1 cos 20°

✓ =

1 √1−k2

100 + 71 sec x = 0 71 sec x = −100 sec x

=−

cos x

=−

13

π

5

2

13

π

1

2

tan A

=


71 71

1 5 12

(− )

5

S α α T

100

A

C

= 180 − α, 180 + α ≈ 135.2°, 224.8° ✓

5 cot x + 3 = 2 + 3 cot x 2 cot x = −1 =−

1 2

0° < x < 360°

✓

cos ( − A) = sin A = −

100

x = 180 − α, 360 − α = 116.6°, 296.6° ✓

tan(−A) = − tan A = − (−

tan ( − A) =

1 sin 70°

tan x = −2 α ≈ 63.4° ⇒ 2nd or 4th quadrant

Trigonometric ratio

7(ii)

csc 70° =

cot x

y = −5, x = 12, r=

tan(−20°) = − tan 20° = −

x

Coordinates

√122

= √1 − k 2 ✓

0° < x < 360°

∴ 4th quadrant

12

1

α ≈ 44.8° ⇒ 2nd or 3rd quadrant

tan A & cos A have opp. signs ⇒ 3rd or 4th quadrant

5

r

✓ [typo in book]

⇒ 2nd or 4th quadrant

tan A = −

√1−k2

= √3 ✓

Quadrant tan A = −

x

cos 20° = =

) =

12

5 12

S α T

A α C

✓

✓ =−

12 5

✓

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293


Ex 11.2 10(a) 2 sin x − √3 = 0

5 cos x = √3 sin 60° √3 2

5 cos x = √3 ( ) 5 cos x = cos x

=

sin x

3

α=

2 3 10

S

0° < x < 360°

T

A α α C

2 sin(−x) = 0.3 −2 sin x = 0.3 sin x = −0.15 α ≈ 8.6° ⇒ 3rd or 4th quadrant

S

A

α T

α C

3

0 < x < 2π

α ≈ 72.5° ⇒ 1st or 4th quadrant

0° < x < 360°

x = α, π − α π 2π = , ✓ 3

10(b)

3

cos x−1 0.5−cos x

=2

cos x − 1 = 1 − 2 cos x 3 cos x = 2 cos x S

A

α T

α C

2

=

3

α ≈ 0.841 ⇒ 1st or 4th quadrant 0 < x < 2π

x = 180 + α, 360 − α ≈ 188.6°, 351.4° ✓ 9(e)

π

√3 2

⇒ 1st or 2nd quadrant

x = α, 360 − α = 72.5°, 287.5° ✓ 9(d)

=

S T

A α α C

x = α, 2π − α ≈ 0.841,5.44 ✓

2 cos x = sec x 2 cos x = cos 2 x = cos x

1

10(c)

cos x 1

=7

4 = 7 − 14 tan2 x 14 tan2 x = 3

2

=±

4 1−2 tan2 x

1 √2

α = 45° ⇒ 1st, 2nd, 3rd or 4th quadrant

S α α T

0° < x < 360°

A α α C

3

tan2 x

=

tan x

= ±√

14 3 14

α ≈ 0.434 ⇒ 1st, 2nd, 3rd or 4th quadrant

x = α, 180 − α, 180 + α, 360 − α = 45°, 135°, 225 °, 315° ✓

0 < x < 2π 9(f)

3 sin x + 2 = cot 15° 3 sin x = cot 15° − 2 1

sin x

= tan 15°

−2

0° < x < 360° x = α, 180 − α ≈ 35.3°, 144.7° ✓


A α α C

x = α, π − α, π + α, 2π − α ≈ 0.434, 2.71, 3.58, 5.85 ✓

3

sin x ≈ 0.57735 α ≈ 35.3 ⇒ 1st or 2nd quadrant

S α α T

11(i)

π

π

2

2 π

sin (x − ) = sin [− ( − x)] = − sin ( − x) 2

S

A

α T

α C

= − cos x [shown] ✓

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Ex 11.2

11(ii) √3 sin (x − π) + 3 = √3 cos x 2

13

√3(− cos x) + 3 = √3 cos x 3 − √3 cos x = √3 cos x 2√3 cos x =3 cos x α=

=

π

√3 2

3 sin B + 2 = 0 3 sin B = −2 S

A α α C

6


T

6

12

6

=− <0 3

⇒ 3rd or 4th quadrant ∵ A & B are in different quadrant, Case 1: A in 1st quadrant, B in 3rd quadrant Case 2: A in 1st quadrant, B in 4th quadrant Case 3: A in 3rd quadrant, B in 4th quadrant Coordinates of A 1st quadrant: y x1 = 1 tan A = 2 = x in 1st or 3rd quadrant y1 = 2

x = α, 2π − α π 11π

2

sin B

0 < x < 2π = ,

Quadrant tan A = 2 > 0 ⇒ 1st or 3rd quadrant

✓

Quadrant tan θ = 7 > 0 ⇒ 1st or 3rd quadrant 0° < θ < 180° ⇒ 1st or 2nd quadrant ⇒ 1st quadrant

r1 = √12 + 22 = √5 r1

Coordinates y tan θ = 7 =

𝑟 θ 1

x

y = 7, x = 1,

2

−1 −2

7

3rd quadrant: x2 = −1 y2 = −2

1 r2

r2 = √12 + 22 = √5 3rd quadrant: y3 = −2 r3 = 3

Coordinates of B

r = √12 + 72

2

y

3

r

sin B = − =

= √50 = √25 × 2 = 5√2

in 3rd or 4th quadrant

x3 = −√32 − (−2)2 = −√5

Value v= =

770 sin 135° sin θ 385√2 7 5√2

=

770 sin 45°

= 385√2 ⋅

y r

5√2 7

=

=

770(

√2 ) 2

x3

7 5√2

385(5)(2) 7

−2 = 550 ✓

x4

4th quadrant: −2 y4 = −2 r4 = 3

3 3

x4 = √32 − (−2)2 = √5 Values Case 1: 3 sin A − tan B y

y

r1 2

x3 −2

√5

−√5

= 3 ( 1) − ( 3) = 3( ) − (

) =

4 √5

✓

Case 2: 3 sin A − tan B y

y

r1 2

x4 −2

8

√5

√5

√5

= 3 ( 1) − ( 4) =3 ( )−( ) =

✓

Case 3: 3 sin A − tan B y

y

r2 −2

x4 −2

4

√5

√5

√5

= 3 ( 2) − ( 4)

=3 ( )−( ) =−


sleightofmath.com

✓

295


Ex 11.2

Coordinates y sin 130° = k =

15(i)

r

1 130°

k x

y = k, r = 1, x = − √1 − k 2

k 1

14(ii) Coordinates y sin 50° = k = r

1 𝑘 50° 𝑥

y = k, r = 1, x = √1 − k 2

1

1

=

y ( ) x

1

=

(

k √1−k2

)

=

√1−k2 k

15(ii)

✓

y = √1 − r=

√k 2

k

k2, x

=

𝑟

= k,

=0

40° √1 − k

=

1

x r

A

α α T C x ≠ 180 + α, 360 − α ≠ 185.7°, 354.3° Combine answers: x = 161.6°, 198.4° ✓

cos(−220°) −( )

1

S

Trigonometric ratio

1

α ≈ 18.4° ⇒ 3rd or 4th quadrant

0° < x < 360°

= √1 − k 2 + k 2 =1

sec(−220°) =

1 √10

10 sin x + 1 ≠ 0

k2

2 k2)

+ (√1 −

3 √10

≠−

sin x ≠− <0 ⇒x= 10 α ≈ 5.7° 161.6°, 198.4° ✓ ⇒ 3rd or 4th quadrant

y x

3+√10 cos x 10 sin x+1

14(iii) Coordinates √1−k2

sin x

Combine answers: x = 161.6° ✓

Trigonometric ratio tan 40° = cot(90° − 40°) tan 50°

=−

√10 sin x + 1 ≠ 0

α ≈ 18.4 ⇒ 2nd or 3rd quadrant 0° < x < 360° 0° < x < 360° S A S A α α α T C α x ≠ 180 + α, 360 − α T C ≠ 198.4°, 341.6° x = 180 − α, 180 + α ≈ 161.6°, 198.4°

sin 50° = sin(180° − 130°) = = k ✓

tan 40° =

=0 √10 sin x + 1 3 + √10 cos x = 0 = −3 √10 cos x cos x

Trigonometric ratio

=

3 + √10 cos x

=

=

1 −

k 1

1 cos(220°) 1

=− ✓ k

=

1 − cos(40°)

16(i) (a)

3 sin x − 2 = 0 sin x

=

2 3

α ≈ 41.8° ⇒ 1st or 2nd quadrant 0° < x < 360° x = α, 180 − α ≈ 41.8°, 138.2° ✓


sleightofmath.com

S

A

α T

α C

296

A math 360 sol (unofficial) 16(i) (b)

tan x − 4 = 0 tan x =4 α ≈ 76.0° ⇒ 1st or 3rd quadrant 0° < x < 360° x = α, 180 + α ≈ 76.0°, 256° ✓


Ex 11.2 16(ii) 3 sin x tan x − 12 sin x −2 tan x + 8 = 0 3 sin x (tan x − 4) −2(tan x − 4) = 0 (tan x − 4)(3 sin x − 2) =0 S

α T

A α

tan x = 4 or sin x =

2 3

⇒ x = 41.8°, 76.0°, 138.2°, 256.0° ✓

C

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Ex 11.3 1(e)

Ex 11.3 1(a)

(sin x − 1)(sin x + 1) = 0 sin x = 1 or sin x = −1 0° ≤ x ≤ 360°

0° ≤ x ≤ 360°

𝑦 1

90° 180° 270° 360°

1(b)

90° 180° 270° 360°

−1

−1

x = 90° ✓

x = 270° ✓

(3 sin x − 1)(tan x + 1) = 0 1 tan x = −1 sin x = or 3 α = 45° α ≈ 19.5° ⇒ 2nd or 4th quadrant ⇒ 1st or 2nd quadrant 0° ≤ x ≤ 360° 0° ≤ x ≤ 360° S A S A α α α α T C T C x = 180 − α, 360 − α x = α, 180 − α = 135°, 315° ✓ = 19.5°, 160° ✓

2(a)

−1 ≤ sin x ≤1 −7 ≤ 7 sin x ≤7 −10 ≤ 7 sin x − 3 ≤ 4

1

90° 180° 270° 360°

𝑥

90° 180° 270° 360° −1

x = 0°, 360° ✓

x = 180° ✓

𝑥

sin x (2 cos x − 3) = 0 3 sin x = 0 or cos x = 2

0° ≤ x ≤ 360°

(no solution)

𝑦 1

90° 180° 270° 360°

𝑥

For y = 7 sin x − 3, min = −10 ⇒ sin x = −1 0° ≤ x ≤ 360° x = 270° ✓

−1

x = 0°, 180°, 360° ✓ 1(d)

A

1(f)

𝑦

−1

S α

−1

0° ≤ x ≤ 360°

1

1(c)

𝑥

(cos x − 1)(cos x + 1) = 0 cos x = 1 or cos x = −1 𝑦

𝑥

α T C x = 0°, 180°, 360° ✓ x = 180 − α, 360 − α ≈ 104.0°, 284.0° ✓ 90° 180° 270° 360°

𝑥

0° ≤ x ≤ 360°

0° ≤ x ≤ 360°

𝑦 1

𝑦

1

sin2 x (tan x + 4) = 0 tan x = −4 sin2 x = 0 or α ≈ 76.0° sin x = 0 ⇒ 2nd or 4th quadrant 0° ≤ x ≤ 360°

tan x (2 cos x − 1) = 0 tan x = 0 or cos x = 1

𝑦 1

90° 180° 270° 360°

𝑥

−1

2

0° ≤ x ≤ 360° 𝑦

α = 60° ⇒ 1st or 4th quadrant

max = 4 ⇒ sin x = 1 x = 90° ✓

0° ≤ x ≤ 360° 90° 180° 270° 360°

𝑥

A α α x = 0°, 180°, 360° T C ✓ x = α, 360 − α = 60°, 300° ✓


S

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−1 −5 −3

Ex 11.3

≤ cos x ≤1 ≤ 5 cos x ≤5 ≤ 5 cos x + 2 ≤ 7

For y = 5 cos x + 2, min = −3 ⇒ cos x = −1 0° ≤ x ≤ 360° x = 180°

𝑦 1

90° 180° 270° 360°

𝑥

−1

max = 7 ⇒ cos x = 1 x = 0°, 360° ✓ 2(c)

−1 ≤ sin x ≤1 −3 ≤ −3 sin x ≤3 1 ≤ 4 − 3 sin x ≤ 7 For y = 4 − 3 sin x, min = 1, ⇒ sin x = 1 0° ≤ x ≤ 360° x = 90° ✓

𝑦 1

90° 180° 270° 360°

𝑥

−1

max = 7 ⇒ sin x = −1 x = 270° ✓


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Ex 11.3

3(a) y = 3 cos(2x) + 2 i.e. a = 3, b = 2, c = 2

𝑦 y = 3 cos(2x) + 2 5

Amplitude = |a| = |3| = 3 ✓ Period =

360 b

=

360

= 180° ✓

2

2 𝑂 −1

180°

360°𝑥 ✓ Workings

Domain 0° ≤ x ≤ 360° Axis with y=2±3 Amplitude Shape +cos 360−0

Cycle

180

=2

3(b) y = 5 − sin 2x = − sin 2x + 5 i.e. a = −1, b = 2, c = 5 Amplitude = |a| = |−1| = 1 ✓ Period

=

2π b

=

2π 2

𝑦 6

y = 5 − sin 2x

5 4

=π✓

𝑂

𝜋

𝑥

2𝜋

✓

Workings Domain 0 ≤ x ≤ 2π Axis with y=5±1 Amplitude Shape −sin 2π−0

Cycle

3(c)

π

=2

y = 4 sin 8x i.e. a = 4, b = 8, c = 0 Amplitude = |a| = |4| = 4 ✓ Period

=

360 b

=

360 8

𝑦 4 𝑂

= 45° ✓

y = 4 sin(8x) 45°

-4

90°

𝑥 ✓

Workings

Domain 0° ≤ x ≤ 90° Axis with y=0±4 Amplitude Shape +sin Cycle

90−0 45

=2


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Ex 11.3

3(d) y = 6 cos (x) − 4

𝑦

2

𝑥

1

i.e. a = 6, b = , c = −4

2π b

=

2π 1 2

4𝜋

−4 𝑂

Amplitude = |a| = |6| = 6 ✓ =

2

2

2

Period

𝑦 = 6 cos ( ) − 4 𝑥

−10

= 4π ✓

✓

Workings Domain 0 ≤ x ≤ 4π Axis with y = −4 ± 6 Amplitude Shape +cos 4π−0

Cycle

4π

=1

3(e) y = 2 sin(3x) − 1 i.e. a = 2, b = 3, c = −1 Amplitude = |a| = |2| = 2 ✓ Period

=

360 b

=

360 3

= 120° ✓

𝑦 1 −1𝑂

𝑦 = 2 sin(3𝑥) 120° 180° 𝑥

−3

✓

Workings Domain 0° ≤ x ≤ 180° Axis with y = −1 ± 2 Amplitude Shape +sin 180−0

Cycle

3(f)

120

= 1.5

y = 2(2 − cos 5x) = 4 − 2 cos 5x = −2 cos 5x + 4 i.e. a = −2, b = 5, c = 4

𝑦 y = 2(2 − cos 5x) 6 4

Amplitude = |a| = |−2| = 2 ✓ Period

=

360 b

=

360 5

= 72° ✓

2 𝑂

72°

144° 180° 𝑥 ✓

Workings Domain 0° ≤ x ≤ 180° Axis with y=4±2 Amplitude Shape −cos Cycle

180−0 72


= 2.5

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Ex 11.3

3(g) y = 1 − 2 sin x

𝑦

3

1

1

y = −2 sin ( x) + 1

= −2 sin ( x) + 1 3

i.e. a = −2, b = , c = 1

1

3

=

360 b

=

360 1 3

𝑥

810°

-1 O

Amplitude = |a| = |−2| = 2 ✓ Period

3

3

1

✓

= 1080° ✓ Workings

Domain 0° ≤ x ≤ 810° Axis with 𝑦 =1±2 Amplitude Shape −𝑠𝑖𝑛 810−0 3 Cycle = 1080

4

3(h) y = 1.3 + 1.2 cos 2t = 1.2 cos 2t + 1.3 i.e. a = 1.2, b = 2, c = 1.3

𝑦 y = 1.3 + 1.2 cos 2t 2.5

Amplitude = |a| = |1.2| = 1.2 Period

=

2π

=

b

2π 2

=π

1.3 0.1 𝑂

π

2π

𝑥 ✓

Workings Domain 0 ≤ 𝑡 ≤ 2𝜋 Axis with 𝑦 = 1.3 ± 1.2 Amplitude Shape +𝑐𝑜𝑠 2𝜋−0 Cycle =2 𝜋

4(ii) y|t=4 = 1.2 cos[2(4)] + 1.3 = 1.2 cos 8 + 1.3 = 1.13m ✓ 5(i)

x

y = p cos ( ) + q 2

1

i.e. a = p, b = , c = q 2

Period

=

2π b

=

2π 1 2

= 4π ✓


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Ex 11.3

5(ii) q = min+max = (−5)+(3) = −1 ✓ 2

|p| =

2

max−min 2

=

(3)−(−5)

=4

2

p = 4 or p = −4 ✓ 5(iii) p < q ⇒ p = −4, q = −1

𝑦 1

y = −4 cos ( x) − 1 2

3 x

𝑂

y = −4 cos ( ) − 1 2

4𝜋

-1

Amplitude = 4 Period = 4π

𝑥

-5

✓

Workings Domain 0 ≤ x ≤ 4π Axis with y = −1 ± 4 Amplitude Shape −cos 4π−0

Cycle

6(i)

4π

=1

y1 = 4 sin 2x i.e. a = 4, b = 2, c = 0 Amplitude = |a| = |4| = 4 Period

=

360 b

=

360 2

= 180°

y2 = 2 cos x − 1 i.e. a = 2, b = 1, c = −1 Amplitude = |a| = |2| = 2 Period

=

360 b

=

360 1

𝑦 4 1 𝑂 −3 −4

180°

𝑦2 = 2 cos 𝑥 − 1 𝑥 360° 𝑦1 = 4 sin 2𝑥 ✓

= 360°

Workings 𝑦1 = 4 sin 2𝑥 Domain 0° < x ≤ 360° Axis with y=0±4 Amplitude Shape +sin Cycle

360−0 180

=2

𝑦2 = 2 cos 𝑥 − 1 Domain 0° < x ≤ 360° Axis with y = −1 ± 2 Amplitude Shape +cos Cycle

360−0 360


=1

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Ex 11.3

6(ii) 4 sin(2x) + 1 = 2 cos x 4 sin 2x = 2 cos x − 1 y1 = y2 4 Intersections ⇒ 4 real roots ✓ 7(a) y = 5 tan x i.e. a = 5, b = 1, c = 0 Period

=

180 b

=

180 1

y

= 180° 𝑂

y = 5 tan 𝑥 x 360°

180°

𝑥 = 90°

𝑥 = 270° Workings

Domain 0° ≤ x ≤ 360° Axis with Amplitude Shape +𝑡𝑎𝑛 360−0 Cycle =2 180

7(b)

y = −4 tan 3x i.e. a = −4, b = 3, c = 0 180 180 Period = = = 60° b

y

𝑦 = −4 tan 3𝑥

3

𝑂

60°

𝑥 = 30°

120°

𝑥 = 90°

180°

x

𝑥 = 120°

Workings Domain 0° ≤ x ≤ 180° Axis with Amplitude Shape −𝑡𝑎𝑛 180−0 Cycle =3 60


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Ex 11.3

y = 3 tan 2x i.e. a = 3, b = 2, c = 0 Period

=

π b

=

𝑦 𝑦 = 3 tan 2𝑥

π 2

𝑂

𝜋 2 𝑥=

𝜋 4

𝑥=

3𝜋 4

𝑥

3𝜋 2

𝜋

𝑥=

5𝜋 4

2𝜋

𝑥=

7𝜋 4 ✓

Workings Domain 0 ≤ x ≤ 2π Axis with Amplitude Shape +tan 2π−0

Cycle

π 2

=4

7(d) y = −1.5 tan 4x i.e. a = −1.5, b = 4, c = 0 Period

=

π b

=

𝑦 𝑦 = −1.5 tan 4𝑥

π 4

𝑂

𝑥=

𝜋 8

π

π

3𝜋

4

2

4

𝑥=

3𝜋 8

𝑥=

5𝜋 8

𝜋

𝑥=

7𝜋 8

𝑥

✓

Workings Domain 0 ≤ x ≤ 2π Axis with Amplitude Shape −𝑡𝑎𝑛 𝜋−0 Cycle =4 𝜋 4


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Ex 11.3 10(d) A = max−min = (3.5)−(0.5) = 1.5 ✓

1

sin−1 ( ) = α 2

2

= 30° ✓ 8(b) cos −1 (− 1) = 180° − α 2

= 180° − 60° = 120° ✓ 8(c)

tan−1 (−√3) = −α 3

1 √2

b= c=

2π

π 4

1

A = sin−1 (− )

⇒ 4th quadrant

3

B = cos −1 h, h < 0 ⇒ 2nd quadrant C = tan−1 k ⇒ 1st or 4th quadrant A and B cannot be in the same quadrant 9(ii) For A and C to be in the same quadrant, they have to be in the 4th quadrant. ⇒ k<0✓ 10(a) A = max−min = (5)−(−1) = 3 ✓ 2

2

c= 10(b) A =

1

= ✓ =

2 (3.5)+(0.5) 2

=2✓

11(ii) −1 ≤ sin 4t ≤ 1 −20 ≤ 20 sin 4t ≤ 20 ⇒ max = 20 ✓ 20 sin 4t = 20 sin 4t = 1 π ⇒ 4t = t

=

2 π 8

≈ 0.393 ✓ 11(iii) I = 20 sin 4t i.e. a = 20, b = 4, c = 0 Amplitude = |a| = |20| = 20 ✓ =

2π b

=

2π 4

1

= π 2

𝐼

a = −3 360

2π

2

Period

T = 90° ✓

b=

=

T 4π max+min

) = −α =− ✓

9(i)

a = 1.5 ✓

11(i) I = 20 sin 4t i.e. a = 20, b = 4, c = 0 Amplitude = |a| = |20| = 20 ✓

π

=− ✓ 8(d) sin−1 (−

2

T = 4π✓

=

360

T 90 max+min

=

2 max−min

=

2

𝐼 = 20 sin 4𝑡

= 4✓

20

(5)+(−1) 2 1−(−3) 2

=2✓

0 -20

= 2✓

𝜋

𝜋

2

𝑡 ✓

T= π✓ Workings a=2 b= c=

2π

=

2π

T π max+min

Domain 0 ≤ x ≤ 2π Axis with 𝑦 = 0 ± 20 Amplitude Shape +𝑠𝑖𝑛

= 2✓

2

=

(1)+(−3) 2

= −1 ✓

10(c) A = max−min = (4)−(−2) = 3 ✓ 2

Cycle

2

T = 180° ✓

𝜋−0 1 𝜋 2

=2

a = −3 ✓ b= c=

2π

=

360

T 180 max+min 2

=2✓ =

(4)+(−2) 2


=1✓

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Ex 11.3

12(a) y = |6 sin 3x + 2| i.e. a = 6, b = 3, c = 2

𝑦

Amplitude = |a| = |6| = 6 Period

=

2π b

=

8 4 2 𝑂

2π

y = |6 sin 3x + 2|

𝜋

𝑥

3

✓

Workings Domain 0≤x≤π Axis with 𝑦 =2±6 Amplitude Shape +𝑠𝑖𝑛 𝜋−0 3 Cycle 2𝜋 = 2

3

12(b) y = |4 cos (x) − 1|

𝑦

2

1

i.e. a = 4, b = , c = −1 2

3 𝑥


=

2π b

=

2π 1 2

✓

= 4π

Workings Domain 0≤x≤π Axis with 𝑦 = −1 ± 4 Amplitude Shape +𝑐𝑜𝑠 𝜋−0 1 Cycle = 4𝜋


4

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Ex 11.3

12(c) y = |2 tan 3x| i.e. a = 2, b = 3, c = 0 Period

=

π b

=

y

y = |2 tan 3x|

π 3

𝑂

𝜋 3 𝑥=

𝜋 6

𝑥

2𝜋 3 𝑥=

𝜋 2

𝜋

𝑥=

5𝜋 6

✓

Workings Domain 0≤x≤π Axis with Amplitude Shape +𝑡𝑎𝑛 𝜋−0 Cycle =3 𝜋 3

12(d) y = |5 + 3 sin x| = |3 sin x + 5| i.e. a = 3, b = 1, c = 5

𝑦

8 Amplitude = |a| = |3| = 3 Period

=

2π b

=

2π 1

𝑦 = |5 + 3 sin 𝑥|

5

= 2π

𝜋

𝑥 ✓ Workings

Domain 0≤x≤π Axis with 𝑦 =5±3 Amplitude Shape +𝑠𝑖𝑛 𝜋−0 1 Cycle = 2𝜋


2

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Ex 11.3

12(e) y = |sin 2x + 1| for 0 ≤ x ≤ π. i.e. a = 1, b = 2, c = 1

𝑦 𝑦 = |sin 2𝑥 + 1|

2 Amplitude = |a| = |1| = 1 Period

=

2π b

=

2π 2

1

=π

O

𝜋

𝑥 ✓

Workings Domain 0≤x≤π Axis with 𝑦 =1±1 Amplitude Shape +𝑠𝑖𝑛 𝜋−0 Cycle =1 𝜋

12(f) y = |1.5 cos 4x − 2| for 0 ≤ x ≤ π. i.e. a = 1.5, b = 4, c = −2 Amplitude = |a| = |1.5| = 1.5 Period

=

2π 4

1

= π 2

y

𝑦 = |1.5 cos 4𝑥 − 2|

3.5 2 0.5 𝑂

1 𝜋 2

𝜋

x

-2 ✓ Workings Domain 0≤x≤π Axis with y = −2 ± 1.5 Amplitude Shape +cos π−0

Cycle

1 π 2

=2

13(i) Amplitude = 4 ⇒a=4 Period 360 b

b

= 120 = 120 =3

c = max − amplitude =2−4 = −2 ✓


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A math 360 sol (unofficial) 13(ii) y = 4 sin 3x − 2 13(iii) Amplitude = 4 Period = 120° y = b + c cos ax = 3 − 2 cos 4x = −2 cos 4x + 3 Amplitude = 2 Period

=

360 4

Ex 11.3 y 5 2 1 𝑂 -2

𝑦 = 3 − 2 cos 4𝑥 90°

180°

𝑥

𝑦 = 4 sin 3𝑥 − 2

-6 ✓

= 90° Workings

𝑦 = 4 𝑠𝑖𝑛 3𝑥 − 2 Domain 0° ≤ x ≤ 180° Axis with 𝑦 = −2 ± 4 Amplitude Shape +𝑠𝑖𝑛 180−0 Cycle = 1.5 120

y = b + c cos ax = 3 − 2 cos 4x = −2 cos 4x + 3 Domain 0° ≤ x ≤ 180° Axis with 𝑦 =3±2 Amplitude Shape −𝑐𝑜𝑠 180−0 Cycle =2 90


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Ex 11.3

14(i) y = a sin (x) + c

16(i) Vertical component ⇒ r sin θ

b

Amplitude = 4 ⇒ a=4✓ Period 2π 1 b

( )

2πb b 14(ii) c

a=

40

= 20

2 2π

= 4π

b=

= 4π

c = 25

= 4π =2

16(ii)

(𝑟 𝑐𝑜𝑠 𝜃, 𝑟 sin 𝜃)

360

=

π 180

y y = 20 sin

45

= min + amplitude =1+4 =5✓

π 180

t + 25

25 5

1

14(iii) y = 4 sin ( x) + 5

x

𝑂

✓

2

Amplitude = 4 Period = 4π

16(iii) y 20 sin

𝑦 9 5

(𝜋, 9)

sin

(5𝜋, 9)

π 180

π 180

= 40 t + 25 = 40

t

=

3 4

α ≈ 0.848 1

y = 4 sin ( x) + 5 2

1

(3𝜋, 1)

𝑥

𝑂

6𝜋 Workings

0≤ π 180 π 180

t

π 180

t ≤ 2π

t = α, π − α t = 0.848,2.29 = 48.6s, 131. s ✓

Domain 0 ≤ x ≤ 6π Axis with 𝑦 =5±4 Amplitude Shape +𝑠𝑖𝑛 6𝜋−0

Cycle

15

4𝜋

=

3 2

y = a sin bt Period 2π b

b

=4 =4 =

π 2

Max = 0.55 ⇒ a = 0.55 π ∴ y = 0.55 sin t ✓ 2


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Ex 11.3

y1 = 2 cos 2x i.e. a = 2, b = 2, c = 0

y


=

2π b

=

2π 2

=π

y2 = 1 + sin x = sin x + 1 i.e. a = 1, b = 1, c = 1

𝑦1 = 2 cos 2𝑥

2 1 O

𝜋

2𝜋

𝑦2 = sin 𝑥 + 1 x

-2


=

2π b

=

2π 1

= 2π

Workings 𝑦1 = 2 𝑐𝑜𝑠 2𝑥 Domain 0 ≤ x ≤ 2π Axis with 𝑦 =0±2 Amplitude Shape +𝑐𝑜𝑠 Cycle

2𝜋−0 𝜋

=2

𝑦2 = 𝑠𝑖𝑛 𝑥 + 1 [0,2𝜋] Domain Axis with 𝑦 =1±1 Amplitude Shape +𝑠𝑖𝑛 Cycle

2𝜋−0 2𝜋

=1

17(i) 2 cos 2x = sin x + 1 y1 = y2 4 intersections ⇒ 4 distinct values of x 17(ii)

y 𝑦 = 2 cos 2𝑥

2 O

𝜋

2𝜋

𝑦 = sin 𝑥 + 1 x

-2 ✓ 7 intersections

⇒ 7 distinct values of x


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Ex 11.3

18(i) y1 = sin x i.e. a = 1, b = 1, c = 0 Amplitude = |a| = |1| = 1 Period

=

2π b

=

2π

= 2π

1

y2 = 2 cos x i.e. a = 2, b = 1, c = 0 Amplitude = |a| = |2| = 2 Period

=

2π b

=

2π 1

y 2 1

𝑂 𝜋 -1 2

𝑦2 = 2 cos 𝑥

𝜋

x 2𝜋 𝑦1 = sin 𝑥

-2 ✓

= 2π

18(ii) 2 intersection ⇒ 2 roots ✓ π 2

Workings 𝑦1 = 𝑠𝑖𝑛 𝑥 Domain 0 ≤ 𝑥 ≤ 2𝜋 Axis with 𝑦 =0±1 Amplitude Shape +𝑠𝑖𝑛 Cycle

2𝜋−0 2𝜋

=1

𝑦2 = 2 𝑐𝑜𝑠 𝑥 Domain 0 ≤ 𝑥 ≤ 2𝜋 Axis with 𝑦 =0±2 Amplitude Shape +𝑐𝑜𝑠 Cycle

2𝜋−0 2𝜋


=1

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Ex 11.3

y1 = tan 3x i.e. a = 1, b = 3, c = 0 π π Period = = b

𝑦

𝑥=

𝜋

𝑥=

4

4

3

5

5

y2 = − tan 2x 2

b

=

𝑂

π 2

2 tan 3x

= −5 tan 2x

tan 3x

= − tan 2x

2

2

i.e. a = − , b = 2, c = 0 2 π

𝑦2 = − tan 2𝑥

𝜋

5

Period =

3𝜋

𝜋

2𝜋

3

3

𝜋

𝑥

𝑦1 = tan 3𝑥

5 2

𝑥=

y1 = y2 6 intersections ⇒ 6 solutions ✓

𝜋 6

𝑥=

𝜋 2

𝑥=

5𝜋 6

✓

Workings 𝑦1 = 𝑡𝑎𝑛 3𝑥 Domain 0≤x≤π Axis with Amplitude Shape +𝑡𝑎𝑛 𝜋−0

Cycle

𝜋 3

=3

5

y2 = − tan 2x 2

Domain 0≤x≤π Axis with Amplitude Shape −𝑡𝑎𝑛 Cycle

𝜋−0 𝜋 2

=2

20(a) tan [cos −1 (1)] = tan(α) 2 π = tan ( ) 3 = √3 ✓ 20(b) sin[tan−1 (−1)] = sin(−α) π

= sin (− ) 4

π

= − sin (− ) 4

=

√2 − 2

✓

20(c) cos [sin−1 (− √3)] = cos(−α) 2 π = cos (− ) 3 π = cos 3 1 = ✓ 2


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A math 360 sol (unofficial) 20(d) cos −1 [cos 4π]

Ex 11.3 π

𝑦

= cos −1 (− cos )

3

3

S

1

= cos −1 (− ) 2

=π−α π =π−

π 3

A 𝑂

T

𝑥

C

3

=

2π 3

✓

21(i) y = 3 cos x + 2 i.e. a = 3, b = 1, c = 2 Amplitude = |a| = |3| = 3 Period

=

2π b

=

2π 1

= 2π

21(ii) ∵ x = π is line of symmetry α+β 2

y 𝑦 = 3 cos 𝑥 + 2

5 2

𝑂

-1

𝜋

2𝜋

x

=π

β = 2π − α ✓

Workings Domain 0 ≤ x ≤ 2π Axis with 𝑦 =2±3 Amplitude Shape +𝑐𝑜𝑠

Cycle

2𝜋−0 2𝜋


=1

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Ex 11.3

22(i) y1 = 2 + cos x = cos x + 2 i.e. a = 1, b = 1, c = 2 Amplitude = |a| = |1| = 1 Period

=

2π

2π

=

b

1

= 2π

y 4 3

𝑦 = 2 + cos 𝑥

1

1

y2 = 4 sin x 2

1

i.e. a = 4, b = , c = 0

O

2

𝜋

1 𝑦 = 4 sin 𝑥 2 x 2𝜋

✓


=

2π b

=

2π 1 2

= 4π

Workings 𝑦1 = 2 + 𝑐𝑜𝑠 𝑥 Domain 0 ≤ x ≤ 2π Axis with 𝑦 =2±1 Amplitude Shape +𝑐𝑜𝑠 Cycle

2𝜋−0 2𝜋

=1

1 𝑦2 = 𝑠𝑖𝑛 𝑥 2 Domain 0 ≤ x ≤ 2π Axis with 𝑦 =0±4 Amplitude Shape +𝑠𝑖𝑛 Cycle

2𝜋−0 4𝜋

=

1 2

22(ii) 2 sin 1 x − k cos x = 1 2

−1 − k cos x

1

= −2 sin x 2

1

2 + 2k cos x

= 4 sin x

y1

= y2

2 + cos x

2

1

= 4 sin x 2

Compare coefficient of cos x ⇒ 2k = 1 1

k= ✓ 2

22(iii) ∵ x = π is line of symmetry a+c 2

c

=π = 2π − a ✓


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Ex 11.3

y1 = −3.1 tan 2x i.e. a = −3.1, b = 2, c = 0 Period

180

=

=

b

180 2

𝑦

= 90°

𝑦1 = −3.1 tan 2𝑥 √2

y2 = √2 cos 4x i.e. a = √2, b = 4, c = 0 Amplitude = |a| = |√2| = √2 Period

=

360

=

b

360 4

𝑂 −√2

90°

𝑦2 = √2 cos 4𝑥 𝑥 180°

= 90° 𝑥 = 45°

𝑥 = 135°

✓

Workings 𝑦1 = −3.1 𝑡𝑎𝑛 2𝑥 Domain 0° ≤ x ≤ 180° Axis with Amplitude Shape −𝑡𝑎𝑛 180−0

Cycle

90

=2

𝑦2 = √2 𝑐𝑜𝑠 4𝑥 Domain 0° ≤ x ≤ 180° Axis with 𝑦 = 0 ± √2 Amplitude Shape +𝑐𝑜𝑠 180−0

Cycle

90

=2

T = 90° 45° < α < 90° √2 cos 4x + 3.1 tan 2x = 0 3.1 tan 2x = −√2 cos 4x −3.1 tan 2x = √2 cos 4x y1 = y2 ∵ Period = 90° ⇒ x = α, α + 90°, α + 180°, α + 270° ✓ 24(a) Let A = cos −1 (− 1) 5

⇒ 2nd quadrant 1

x

5

r

cos A = − =

𝑦 5 −1

x = −1, r = 5 y = √52 − (−1)2 = √24

= 2√6

1

y

5

r

sin [cos −1 (− )] = sin A=


=

2√6 5

✓

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Ex 11.3

24(b) Let B = sin−1 (− 2)

𝑥

3

⇒ 4th quadrant 2

y

3

r

sin B = − =

3

−2

y = −2, r = 3, x = √32 − (−2)2 = √5 2

y

3

x

tan [sin−1 (− )] = tan B =

=

−2 √5

✓

25(a) y = cos x tan x Familiar y = sin x ✓ 25(b) y = sin2 x y = cos 2 x Complementary shape ✓ 26(a) [check for error] sin−1 (sin x) = x Dgf = Df = ℝ True ✓ 26(b) sin(sin−1 x) = x π π Dgf = Df = [− , ] ≠ ℝ 2 2 False ✓ 26(c) cos −1 (cos x) = x Dgf = Df = ℝ False ✓ 26(d) tan(tan−1 x) = x π π Dgf = Df = (− , ) ≠ ℝ 2 2 False ✓ 26(e) − π < x < π 2 2 −1 (tan tan x) = x π π π π Dgf = Df = (− , ) = (− , ) 2 2 2 2 True ✓ 26(f) If cos x = 0.2 then x = cos −1 0.2 Not necessarily true. ✓ x can assume other values depending on the domain restriction


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Rev Ex 11 A2(i) Quadrant

Rev Ex 11

3

tan A = > 0 A1(a) 2 cos x − 1 = 4 cos x 2 cos x = −1 cos x

=−

3

cos B = − < 0 ⇒ 2nd or 3rd quadrant 5

Same quadrant ⇒ 3rd quadrant

1 2

α = 60 ⇒ 2nd or 3rd quadrant

S α α T

0° ≤ x ≤ 360° x = 180 − α, 180 + α = 120°, 240° ✓

Coordinates of A

A

=

sin x α=

x

-3

r = √(−3)2 + (−4)2 = 5 Coordinates of B 3

x2

5

r2

cos B = − = √3 2

S α α T

3

⇒ 1st, 2nd, 3rd, 4th quadrant

-3 y2

5

x2 = −3, r2 = 5

A α α C

y2 = −√52 − (−3)2 = −4 Trigonometric ratio

0 ≤ x ≤ 2π x = α, π − α, π + α, 2π − α π 2π 4π 5π = , , , ✓ 3

-4 r

y = −3, x = −4,

4

=±

3

y

4

C

3

π

3

3

tan A = =

A1(b) 4 sin2 x = 3 sin2 x

⇒ 1st or 3rd quadrant

4

y

−3

r

5

sin A = =

3

✓

A2(ii) tan(−B) = − tan B = − (y2 ) = − (−4) = − 4 ✓ x2

A1(c) (2 sin x + 1)(2 cos x + 5) = 0 sin x = −

1 2

or

cos x = −

5

A2(iii) sec A tan B =

2

(no solution) α = 30° ⇒ 3rd or 4th quadrant 0° ≤ x ≤ 360° S

= =

A

π

1 2

or

x r

( ) 1 −4 5

( ) x2

−4

( ) −3

3

A3(i) y = a cos bx + c

cot x = − tan x = −

3

Amplitude = 3 |a| =3 a = 3 or a = −3 (rej ∵ a is a positive integer)

3 2 2 3

⇒ 1st or 4th quadrant α ≈ 0.588 ⇒ 2nd or 4th quadrant 0 ≤ x ≤ 2π S

A α α T C x = α, 2π − α π 5π = , ✓ 3

tan B

cos A 1 y2

=− ✓

A1(d) (2 cos x − 1)(2 cot x + 3) = 0 α=

3

5

α α T C x = 180 + α, 360 − α = 210°, 330° ✓

cos x =

1

−3

3


0 ≤ x ≤ 2π

S

Period = 60° 360 b

b

A α

α T C x = π − α, 2π − α = 2.55,5.70 ✓

= 60 =

360 6

=6✓ A3(ii) c (a)

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= min + |a| =4 +3 = 7✓

319

A math 360 sol (unofficial) A3(ii) (b)

Rev Ex 11 A4(ii) 3 intersections ⇒ 3 sol ✓

𝑦 𝑦 = 3 cos 6𝑥 + 7 10 7

A5(a) 4 tan−1 1 − cos −1 (− √3) 2

4 𝑂

60°

120°

180°

π

𝑥

= 4( )

−(π − α)

=π

− (π − )

4

✓

=π

Workings

π 6

−

5π

π

6

= ✓

Domain 0° ≤ x ≤ 180° Axis with y=7±3 Amplitude Shape +cos 180−0 Cycle =3

6

A5(b) cos [sin−1 (−

1 √2

)] = cos(−α) π

= cos (− )

60

= cos =

1 √2

π

4

4

✓

A4(ii) y = 5 tan 2x i.e. a = 5, b = 2, c = 0 Period

=

π b

=

π 2

𝑦

𝑦= 𝑂

𝜋 2

3𝜋

𝜋

2

𝑥 𝜋

𝑥 2𝜋 𝑦 = 5 tan 2𝑥 ✓

Workings Domain 0 ≤ x ≤ 2\pi Axis with Amplitude Shape +tan 2π−0 Cycle =4 π 2


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Rev Ex 11

A6(i) y1 = cos 2x i.e. a = 1, b = 2, c = 0 Amplitude = |a| = |1| = 1 Period

=

2π b

=

2π 2

=π

y2 = 2 sin x i.e. a = 2, b = 1, c = 0 Amplitude = |a| = |2| = 2 Period

=

2π b

=

2π 1

= 2π

y 2 1

O

𝑦1 = cos 2𝑥

𝜋

-1

2𝜋

x

𝑦2 = 2 sin 𝑥

-2

Workings y1 = cos 2x Domain 0 ≤ x ≤ 2π Axis with y=0±1 Amplitude Shape +cos 2π−0 Cycle =2 π

y2 = 2 sin x Domain 0 ≤ x ≤ 2π Axis with y=0±2 Amplitude Shape +sin 2π−0 Cycle =1 2π

A6(ii) cos 2x = 0.6 for 0 ≤ x ≤ 2π (a) y1 = 0.6 4 intersections ⇒ 4 sol ✓

𝑦 1 O

𝑦1 = cos 2𝑥 𝑦 = 0.6 𝜋

2𝜋

𝑥

-1 A6(ii) cos 2x = 2 sin x for 0 ≤ x ≤ 2π (b) y1 = y2 2 intersections ⇒ 2 sol ✓

𝑦 2 1 O

𝑦1 = cos 2𝑥 𝜋

-1

⇒ 4 sol ✓

𝑥

𝑦2 = 2 sin 𝑥

-2 A6(ii) cos 2x − 2 sin x = 1 for 0 ≤ x ≤ 2π (c) cos 2x = 2 sin x + 1 y1 = 2 sin x + 1

2𝜋

y 3 1 O

𝑦 = 2 sin 𝑥 + 1 𝑦1 = cos 2𝑥 𝜋

2𝜋

x

-1


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Rev Ex 11

A6(ii) 2|sec 2x| = |csc x| for 0 ≤ x ≤ 2π 2 1 (d) = |cos 2x| |cos 2x| 2

𝑦

|sin x|

= |sin x| 𝜋

𝑂

|cos 2x| = |2 sin x| |y1 | = |y2 |

𝑦 = 2|sin 𝑥| 𝑦 = |cos 2𝑥| 𝑥 2𝜋

⇒ 4 sol ✓ B1(a) (2 sin2 x + sin x)(tan x − 3 cos 20°) = 0 sin x (2 sin x + 1)(tan x − 2.82) =0 1 sin x = 0 or sin x = − or 2 0° ≤ x ≤ 360° α = 30° 𝑦 ⇒ 3rd or 4th quadrant 1 90° 180° 270° 360°

𝑥

−1

x = 0°, 180°, 360° ✓


0° ≤ x ≤ 360° S

A

α α T C x = 180 + α, 360 − α = 210°, 330° ✓

tan x = 2.82 α ≈ 70.5° ⇒ 1st or 3rd quadrant 0° ≤ x ≤ 360° S

A α

α T C x = α, 180 + α = 70.5°, 250.5° ✓

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322


Rev Ex 11

B1(b) |3 tan x − 1| = 2 3 tan x − 1 = 2 3 tan x =3 tan x =1 α = 45° ⇒ 1st or 3rd quadrant 0° ≤ x ≤ 360° S

tan x

3

0 < x < 2π

π

α=

𝑦

A

1

4

−2

𝑟

r = √12 + (−2)2 = √5 Trigonometric ratio

1 tan A

y

−2

2

r

√5

√5

4

=

1 −2

✓

1

=− ✓ 2

B3(iii) sin (π − A) cos 150° 2

(− cos 30°)

= cos A

0 < x < 2π

4

1

y = −2, x = 1

B3(ii) cot A =

√2

S A α α T C x = π − α, π + α 3 5 = π, π ✓

x=π✓

Coordinates y tan A = −2 =

sin(−A) = − sin A = − = − ( ) =

⇒ 2nd or 3rd quadrant 𝑥

B3(i) Quadrant A = tan−1 (−2) ⇒ 4th quadrant

x

α T C x = 180 − α, 360 − α = 161.6°, 341.6° ✓

B1(c) tan x (√2 cos x + 1) = 0 tan x = 0 cos x = −

2

1

α ≈ 18.4° ⇒ 2nd or 4th quadrant

S α

α T C x = α, 180 + α = 45°, 225° ✓

B1(d)

=−

0° ≤ x ≤ 360°

A α

90° 180° 270° 360°

B2(iv) q = −6 p = 2π − 8π = −6π ✓

3 tan x − 1 = −2 3 tan x = −1

= = =

1 √5 1 √5 √5 5

=−

⋅

√5 √5

√15 10

(−

√3 ) 2

(−

√3 ) 2

(−

√3 ) 2

✓

=3

sec x−1

sec x − 1 = sec x

=

cos x

=

2 3 5 3 3 5

α ≈ 0.927 ⇒ 1st or 4th quadrant

S T

A α α C

0 < x < 2π x = α, 2π − α = 0.927,5.36 ✓ B2(i) A = max−min = (2)−(−6) = 4 ✓ 2 2 B2(ii) T = 8π ✓ a = −4 ✓ b= c=

2π

=

2π

T 8π max+min 2

1

= ✓ =

4 2+(−6) 2

= −2 ✓

B2(iii) (6π + 8π, 2) ⇒ (14π, 2) ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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323


Rev Ex 11

B4(i) y1 = |2 cos x − 1| i.e. a = 2, b = 1, c = −1

B5(ii) y = √3 tan x + 1 i.e. a = √3, b = 1, c = 1 180 180 Period = = = 180° b 1 𝑦


=

2π b

=

2π 1

= 2π

𝑦 3 1

𝑦1 = |2 cos 𝑥 − 1| 2𝜋

𝑂

y = √3 tan x + 1

𝑥

y2 = −

2x 3π

1

+1

𝑂

✓

150°

330°

𝑥 = 90°

Workings Domain 0 < x < 2π Axis with y = −1 ± 2 Amplitude Shape +cos 2π−0 Cycle =1

360°

180°

𝑥

𝑥 = 270°

✓

Workings Domain 0° ≤ x ≤ 360° Axis with y=1 Amplitude Shape +tan 360−0 Cycle =2

2π

180

B4(ii) −2x + 3π = 3π|2 cos x − 1| −

2x 3π

+1

= |2 cos x − 1| B5(iii) √3 tan x + 1 ≥ 0

y2 = y1 ⇒ 3 sol ✓

0° ≤ x < 90° or 150° ≤ x < 270° or 330° ≤ x ≤ 360° ✓

B5(i) √3 tan x + 1 = 0 tan x

=−

1 √3

α = 30° ⇒ 2nd or 4th quadrant 0° ≤ x ≤ 360° x = 180 − α, 360 − α = 150°, 330° ✓


S α T

A α C

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324


Rev Ex 11

B6(i) y = cos 2x i.e. a = 1, b = 2, c = 0

y


=

2π b

=

2π 2

=π

2 1 −1 𝑂 −2

𝜋

2𝜋

3

3

𝑦 = cos 2𝑥 x 𝜋 𝑦 = 2 sin 6𝑥

y = 2 sin 6x i.e. a = 2, b = 6, c = 0 Amplitude = |a| = |2| = 2 Period

=

2π b

=

2π 6

=

π 3

Workings y1 = cos 2x Domain 0≤x≤π Axis with y=0±2 Amplitude Shape +sin π−0 Cycle =3 π 3

y2 = 2 sin 6x Domain 0≤x≤π Axis with y=0±1 Amplitude Shape +cos π−0 Cycle =1 π

B6(ii) cos 2x = 2 sin 6x [Symmetry] π x = is vertical line of symmetry for y = cos 2x 2

π

( , 0) is point of symmetry for y = cos 2x 2

π

⇒x=α+( )✓ 2

Period of collective graph = π ⇒ x = α + (π) ✓


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325


Ex 12.1 2(i)

Ex 12.1 1(a)

sec x cos x =

1 cos x

1(d)

sin x cot x = sin x (

cos x sin x

3rd quadrant: y = −1, r = 3,

)

x

x = −√32 − (−1)2 = −√8 4th quadrant: y2 = −1, r2 = 3,

sin x (1−cos x)(1+cos x)

sec θ =

=

=

sin x 1−cos2 x sin x

=

sin2 x 1

=

sin x

sin θ = −

2(ii)

cos θ = 1 − sin2 θ

=±

sec θ = ±

tan θ =

1

x r

1

2 √1−(−1) 3

=±

1 8 √ 9

1

= ± √8 = ± 3

3 √8

✓

3(a)

3 √8 3 √8

or

1 √8 3

3 √8

✓

=

= sin θ ⋅ sec θ 1

3

3 1

√8

√8

x −1 −√8 1

=(

cos θ

= (− ) ⋅ (±

y −1

or

√8

1 cos x

√8

✓

LHS = sec x

sin θ

=±

or

−√8 3

=±

√1−sin2 θ 1

tan θ = =

cos θ = ±√1 − sin2 θ cos θ

-1

3

2

1

3

1

=±

1

3

cos θ 1

=− Method 1 (non-graphical)

-1

x2

x2 = √32 — 12 = √8 Values

= csc x ✓

2(ii)

r

(sec x + 1)(sec x − 1) = sec 2 x −1 = (tan2 x + 1) −1 = tan2 x ✓

=

2(i)

y

3

⇒ 3rd or 4th quadrant

= cos x ✓ 1(c)

1

sin θ = − = < 0

cos x

=1✓ 1(b)

Method 2 (graphical) Coordinates

sin x ) sin x

sin x cos x

= tan x

)

= RHS [proven] ✓

✓ 3(b)

LHS = cos x tan x = cos x (

sin x cos x

)

= sin x = RHS [proven] ✓ 3(c)

LHS = =

cos2 x+sin2 x 1−cos2 x 1 sin2 x

= csc 2 x = RHS [proven] ✓


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3(e)

Ex 12.1

LHS = 2 + 3 sin2 x = 2 + 3(1 − cos 2 x) = 2 + (3 − 3 cos 2 x) = 5 − 3 cos 2 x = RHS [proven] ✓

6(ii)

=

1 cos2 x

)

sin2 x cos2 x 2

=

1 sin2 x 1

(

sin x

sin x

5

)

sin θ (

−

7(c)

2+2 cos θ −2+2 cos θ

1 cos2 x

)

LHS

= (1 + =(

)

sin x 1 cos x 1 ) × (1 + ) − + cos x cos x sin x sin x

cos x+sin x−1 cos x

)

×(

sin x+cos x+1 sin x

=

(sin x + cos x)2 − 12 sin x cos x

= 4 cot θ [shown] ✓

=

sin2 x + 2 sin x cos x + cos 2 x − 1 sin x cos x

x = 3 cos θ x cos θ =

=

= sin θ ( =

sin2 θ

)

4 cos θ sin θ

3

y

= 2 tan θ

y

=2

−(1)

sin θ

1 2

−(2)

=2 = RHS [proven] ✓

sub (1) into (2): 1

x

2 xy

3

)

2 sin x cos x + sin2 x + cos 2 x − 1 sin x cos x 2 sin x cos x + (1) −1 = sin x cos x 2 sin x cos x = sin x cos x

cos θ

sin θ = y cos θ

6

= (1 + tan x − sec x) × (1 + cot x + csc x)

)

1−cos2 θ

xy

= tan2 x = RHS [proven] ✓

1−cos θ 1+cos θ 2(1+cos θ)−2(1−cos θ)

= sin θ (

6(i)

2

)

LHS = sin2 x + tan2 x sin2 x = sin2 x (1 + tan2 x) = sin2 x (sec 2 x) = sin2 x (

LHS = (sin θ + cos θ)2 = sin2 θ + 2 sin θ cos θ + cos 2 θ = sin2 θ + cos 2 θ +2 sin θ cos θ =1 +2 sin θ cos θ = RHS [shown] ✓ 2

6

7(b)

= csc x cot x = RHS [proven] ✓ 4

xy 2

LHS = sin4 x − cos 4 x = (sin2 x)2 − (cos 2 x)2 = (sin2 x + cos 2 x) (sin2 x − cos 2 x) (sin2 x − cos 2 x) =1 = sin2 x − cos 2 x = RHS [proven] ✓

) cos x cos x

+ (6 ⋅

7(a)

LHS = (1 + cot 2 x) cos x = (csc 2 x) cos x =(

= 4x 2

= 4(9 cos 2 θ) +36 sin2 θ = 36 cos 2 θ +36 sin2 θ = 36(sin2 θ + cos 2 θ) = 36 ✓

= tan x = RHS [proven] ✓ 3(f)

+x 2 y 2 +(xy)2

= 4(3 cos θ)2 +(6 sin θ)2 ∵ x = 3 cos θ , sin θ =

LHS = sin2 x (1 + tan2 x) = sin2 x (sec 2 x) = sin2 x (

4x 2 = 4x 2

sin θ = y ( ) sin θ =

6

✓


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327


Ex 12.1

LHS =

8(ii)

LHS = csc 2 x + sec 2 x

3 − 6 cos 2 x sin x − cos x

= =

3(sin2 x + cos 2 x) − 6 cos 2 x = sin x − cos x 2

2

=

2

=

3 sin x + 3 cos x − 6 cos x sin x − cos x

=

3 sin2 x − 3 cos 2 x sin x − cos x

1 sin2 x

+

1 cos2 x

cos2 x+sin2 x sin2 x cos2 x 1 sin2 x cos2 x

= csc 2 x sec 2 x = RHS [proven] ✓ 8(iii)

3(sin2 x − cos 2 x) = sin x − cos x 3(sin x + cos x)(sin x − cos x) = sin x − cos x

LHS = (cot x + tan x)2 = (csc x sec x)2 = csc 2 x sec 2 x = csc 2 x + sec 2 x

= 3(sin x + cos x)

= RHS [deduced] ✓

= RHS [proven] ✓ 9 7(f)

LHS 1

= 1−

1

1 1−

1−

1 1− − tan2 x =

cot x)

1 1 − (tan2 x + 1)

1

=

1

1

1−

1 1− 1 − sec 2 x

=

1

=

1−

LHS = (sec x + tan x)(sec x − tan x) +(csc x + cot x)(csc x −

1 1 + cot 2 x

= sec 2 x − tan2 x

+ csc 2 x − cot 2 x

= (tan2 x + 1) − tan2 x

+(cot 2 x + 1) − cot 2 x

=1

+1

=2 1

1 1− csc 2 x 1 1 = = = sec 2 x = RHS 2 1 − sin x cos 2 x

= RHS [proven] ✓ 10

LHS = (sin α + cos β)2 − (sin α + cos β)(sin α − cos β) = (sin α + cos β)[(sin α + cos β) − (sin α − cos β)]

[proven] ✓

= (sin α + cos β)(2 cos β) 8(i)

LHS = cot x + tan x = = =

cos x sin x

+

= 2 cos β (sin α + cos β)

sin x

= RHS [proven] ✓

cos x

cos2 x+sin2 x sin x cos x 1 sin x cos x

= csc x sec x = RHS [proven] ✓


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328


Ex 12.1

sin x + sin y = a sin y = a − sin x

13 −(1)

cos x + cos y = a cos y = a − cos x

1−sin x

LHS = √

1+sin x 1−sin x

=√

1+sin x

−(2)

(1)2 + (2)2 : sin2 y + cos 2 y = (a − sin x)2 +(a − cos x)2 2 2 1 = (a − 2a sin x + sin x) +(a2 − 2a cos x + cos 2 x) 1 = 2a2 − 2a(sin x + cos x) + sin2 + cos 2 x 1 = 2a2 − 2a(sin x + cos x) +1 2a(sin x + cos x) = 2a2 sin x + cos x = a ✓ k= 1 k

= = = = =

1−sin x 1−sin x

(1−sin x)2

=√

12(i)

×

1−sin2 x

(1−sin x)2

=√ =

cos2 x

1−sin x cos x

∵ (1 − sin x) > 0 [ ] cos x > 0 for acute x

= sec x − tan x = RHS [proven] ✓ Different restriction results in different output of trigo f(x) ✓

1+sin x cos x

13(i)

cos x 1+sin x cos x 1+sin x

⋅

√

1−sin x 1+sin x

1−sin x 1−sin x

cos x(1−sin x)

=√

1 − sin x 1 − sin x × 1 + sin x 1 − sin x

=√

(1 − sin x)2 1 − sin2 x

=√

(1 − sin x)2 cos 2 x

1−sin2 x cos x(1−sin x) cos2 x 1−sin x cos x

[proven] ✓

12(ii) k + 1 = 1+sin x + 1−sin x k k2 +1 k

=

cos x =

cos x 2

cos x 2k k2 +1

Put cos x = 2k2 k2 +1

2k k2 +1

into k =

= 1 + sin x

= = =

=

✓

− sin x = 1 − sin x

1 − sin x − cos x ∵ x is obtuse, (1 − sin x) > 0, cos x < 0 = − sec x + tan x = tan x − sec x ✓

cos x

1+sin x cos x

, 13(ii) √

2k2 k2 +1

2k2 k2 +1 2k2 k2 +1 k2 −1 k2 +1

={

−1 −

k2 +1

1 − sin x 1 + sin x

14

k2 +1

✓

sec x − tan x , for x in 1st or 4th quadrant ✓ tan x − sec x for x in 2nd or 3rd quadrant

4 sin A cos A +(cos 2 A − 10 sin A cos A + 25 sin2 A) +9 cos 2 A + 6 sin A cos A + sin2 A = 10 cos 2 A + 26 sin2 A = 10 cos 2 A + 26(1 − cos 2 A) = 26 − 16 cos 2 A = p cos 2 A + q ⇒ p = 26 ✓ ⇒ q = −16 ✓


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329


Ex 12.2 2(b)

Ex 12.2 1(a)

5 cos x + 2 sin x = 0 2 sin x = −5 cos x tan x

=−

1

5 2

90° 180° 270° 360°

S α

=

3

S α T

0° < x < 360° x = α, 180 + α ≈ 53.1°, 233.1° ✓

A α C

=−

3(b)

5

S α T

0° < x < 360° x = 180 − α, 360 − α = 149.0°, 329.0° ✓

A α C

𝑥

S

A

α α T C x = 180 + α, 360 − α = 191.5°, 348.5° ✓

A

90° 180° 270° 360°

−1

x = 270° ✓

2 tan2 x − 3 tan x − 2 = 0 (2 tan x + 1)(tan x − 2) = 0 1 tan x = − or tan x = 2 2 α ≈ 63.4° α ≈ 26.6° ⇒ 1st or 3rd quadrant ⇒ 2nd or 4th quadrant 0° ≤ x ≤ 360° S α

sin x − sin x tan x = 0 sin x (1 − tan x) = 0 sin x = 0 or tan x = 1 𝑦 α = 45° 1 ⇒ 1st or 3rd quadrant 90° 180° 270° 360°

0° < x < 360°

α α T C x = α, 180 − α = 30°, 150° ✓

3



2 sin2 x + sin x − 1 =0 (2 sin x − 1)(sin x + 1) = 0 1 sin x = or sin x = −1 2 0° ≤ x ≤ 360° α = 30° 𝑦 ⇒ 1st or 2nd quadrant 1 0° ≤ x ≤ 360° 𝑥 S

3 cos x + 7 sin x = 2 sin x 5 sin x = −3 cos x tan x

2(a)

3(a)

4

α ≈ 53.1° ⇒ 1st or 3rd quadrant

1(c)

0° < x < 360° x = 180° ✓

α C

3(sin x − cos x) = cos x 3 sin x − 3 cos x = cos x 3 sin x = 4 cos x

𝑥

−1

A

T

0° < x < 360° x = 180 − α, 360 − α ≈ 111.8°, 291.8° ✓

tan x

5

𝑦


1(b)

tan x + 5 tan x sin x = 0 tan x (1 + 5 sin x) = 0 tan x = 0 or sin x = − 1

A

α T C x = 180 − α, 360 − α = 153.4°, 333.4° ✓

0° ≤ x ≤ 360° S

A α

α T C x = α, 180 + α = 63.4°, 243.4° ✓

0° < x < 360°

−1

0° < x < 360° x = 180° ✓

S

A α

α T C x = α, 180 + α = 45°, 225° ✓


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330


Ex 12.2

6 cos 2 x − 5 cos x − 1 =0 (6 cos x + 1)(cos x − 1) = 0 1 cos x = − or cos x = 1

4(a)

4 sin x cos x − 3 sin x = 0 sin x (4 cos x − 3) = 0 3 sin x = 0 or cos x = 4

6

0 ≤ x ≤ 2π

α ≈ 80.4° 0° ≤ x ≤ 360° ⇒ 2nd or 3rd quadrant


𝑦

𝑦

1

0 ≤ x ≤ 2π

1

0° ≤ x ≤ 360° S A α α T C x = 180 − α, 180 + α = 99.6°, 260.4° ✓ 3(d)

90° 180° 270° 360°

𝑥

90° 180° 270° 360°

1 3

−1

x = 0, π, 2π ✓

x = 0°, 360° ✓

4(b)

or sec x = 2

cos x = 3 (no solution)

S

−1

3 sec 2 x − 7 sec x + 2 = 0 (3 sec x − 1)(sec x − 2) = 0 sec x =

𝑥

cos x =

1

3 sin2 x + sin x cos x sin x (3 sin x + cos x) sin x = 0 or 0 ≤ x ≤ 2π 𝑦

2

1


90° 180° 270° 360°

𝑥

−1

x = 0, π, 2π ✓

0° ≤ x ≤ 360° S

A α α T C x = α, 360 − α = 60°, 300° ✓ 3(e)

𝑥

−1

x = 0°, 180°, 360° ✓


4(c)

0° ≤ x ≤ 360° S

=0 =0 3 sin x = − cos x tan x = −

1 3

α ≈ 0.322 ⇒ 2nd or 4th quadrant 0 ≤ x ≤ 2π S α

A

α T C x = π − α, 2π − α = 2.82,5.96 ✓

7 sin3 x + sin2 x = 0 sin2 x (7 sin x + 1) = 0 sin x = 0 or sin x = − 1 7 0° ≤ x ≤ 360° α ≈ 8.2° 𝑦 ⇒ 3rd or 4th quadrant 1 90° 180° 270° 360°

A α α T C x = α, 2π − α = 0.723,5.56 ✓

2 csc 2 x − 7 csc x − 4 = 0 (2 csc x + 1)(csc x − 4) = 0 1 csc x = 4 csc x = − or 2

sin x = −2 (no solution)

sin x =

1 4

α ≈ 0.253 ⇒ 1st or 2nd quadrant 0 ≤ x ≤ 2π

A

S

α α T C x = 180 + α, 360 − α ≈ 188.2°, 351.8° ✓

A

α α T C x = α, π − α = 0.253, 2.89 ✓

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331


Ex 12.2

2 cos 2 x − cos x − 1 = 0 (2 cos x + 1)(cos x − 1) = 0 1 cos x = − or cos x = 1 α=

π 3

0 ≤ x ≤ 2π 1

90° 180° 270° 360°

S A α α T C x = π − α, π + α

4(e)

2π 4π 3

,

3

S

A α

α T

C

x − 60° = α, 180 + α = 30, 210 x = 90°, 270° ✓

x = 0, 2π ✓ 5(c)

✓

3 sin 2x + 2 = 0 sin 2x

2

=−

2 3


2 cos x − 3 cos x sin x = 0 cos x (2 − 3 sin2 x) = 0 cos x = 0 or 3 sin2 x = 2 0 ≤ x ≤ 2π

0° < x < 360° 0° < 2x < 720°

2 3 2

S

A

α T

α C

sin x = ±√

𝑦

3

1

90° 180° 270° 360° −1

π 3π

x= , 2

5(a)

𝑥

−1

sin2 x =

1 √3

0° < x < 360° −60° < x − 60° < 300°

𝑦

0 ≤ x ≤ 2π

tan(x − 60°) =

α = 30° ⇒ 1st or 3rd quadrant

2

⇒ 2nd or 3rd quadrant

=

5(b)

2

✓

𝑥

2x = 180 + α, 360 − α, 540 + α, 720 − α = 221.8°, 318.2°, 581.8°, 678.2° x = 110.9°, 159.1°, 290.9°, 339.1° ✓

α ≈ 0.955 ⇒ 1st, 2nd, 3rd or 4th quad. 0 ≤ x ≤ 2π

S A α α α α T C x = α, π − α, π + α, 2π − α = 0.955,2.19,4.10,5.33 ✓

cos 2x = 0.5 α = 60° ⇒ 1st or 4th quadrant

S

5(d)

S

0° < x < 360° −40° < 2x − 40° < 680°

T

A α α C

2x − 40 = −α, α, 360 − α, 360 + α, 720 − α = −36.9°, 36.9°, 323.1°, 406.9°, 683.1° x ≈ 1.6°, 38.4°, 181.6°, 218.4° ✓

A α α C

T 0° < x < 360° 0° < 2x < 720° 2x = α, 360 − α, 360 + α, 720 − α = 60°, 300°, 420°, 660° x = 30°, 150°, 210°, 330° ✓

cos(2x − 40°) = 0.8 α ≈ 36.9° ⇒ 1st or 4th quadrant

6(a)

10 sin 2x = 3 sin 2x

=

3 10

α ≈ 0.305 ⇒ 1st or 2nd quadrant 0 < x < 2π 0 < 2x < 4π

S

A

α T

α C

2x = α, π − α, 2π + α, 3π − α = 0.305, 2.84, 6.59, 9.12 x = 0.152, 1.42, 3.29, 4.56 ✓


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332


Ex 12.2

cot 2x = −2 tan 2x = −

7

2

R = 55 64 sin 2θ = 55

α ≈ 0.463 ⇒ 2nd or 4th quadrant

S α

0 < x < 2π 0 < 2x < 4π

sin 2θ

A α C

T

1

3

2

2

tan ( x) =

α ≈ 0.983 ⇒ 1st or 3rd quadrant S 0 < x < 2π

α T

1

0< x<π 2

1 2

x 6(d)

A α

=−

A

α T

α C

3 sin x + 2 cos x = 2(sin x + 3 cos x) 3 sin x + 2 cos x = 2 sin x + 6 cos x sin x = 4 cos x tan x =4 α ≈ 76.0° ⇒ 1st or 3rd quadrant S A α α 0° < x < 360° T C x = α, 180 + α ≈ 76.0°, 256.0° ✓

8(b)

sin x−3 cos x

C

5 4 4 5

S α α T

2x = π − α, π + α, ≈ 2.50, 3.79 x ≈ 1.25, 1.89


2 sin x+cos x

=2

sin x − 3 cos x = 4 sin x + 2 cos x 3 sin x = −5 cos x

α ≈ 0.644 ⇒ 2nd or 3rd quadrant 0 < x < 2π 0 < 2x < 4π

S

8(a)

4 sec 2x + 5 = 0 cos 2x

64

2θ = α, 180 − α ≈ 59.2°, 120.8° θ ≈ 29.6°, 60.4° ✓

≈ 0.983 ≈ 1.97 ✓

=−

55

0 < θ < 90° 0 < 2θ < 180°

x = α, π + α

sec 2x

=

α ≈ 59.2° ⇒ 1st or 2nd quadrant

2x = π − α, 2π − α, 3π − α, 4π − α = 2.68, 5.81, 8.96, 12.1 x = 1.34,2.91, 4.48, 6.05 ✓ 6(c)

R = 64 sin 2θ

1

tan x

A

=−

5 3


C

0° < x < 360° x = 180 − α, 360 − α ≈ 121.0°, 301.0° ✓

3π − α, 3π + α 8.78, 10.1 4.39, 5.03 ✓

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S α T

A α C

333


2 sin x

Ex 12.2

= tan x

2 sin x

9(d)

sin x

=

cos x

2 sin x cos x = sin x sin x (2 cos x − 1) = 0 sin x = 0 or cos x = 1 2

0° < x < 360° 𝑦

6 − 2 cos 2 x 6 − 2(1 − sin2 x) 6 − 2 + 2 sin2 x 2 sin2 x − 9 sin x + 4 (2 sin x − 1)(sin x − 4) sin x =


= 9 sin x = 9 sin x = 9 sin x =0 =0 or sin x = 4

1 2

(no solution) α = 30° ⇒ 1st or 2nd quadrant

1

90° 180° 270° 360°

𝑥

S α

−1

A

S

α T C x = α, 360 − α = 60°, 300° ✓

x = 180° ✓

9(b)

0° < x < 360°

0° < x < 360°

sec 2 x = 4 sec x − 3 2 sec x − 4 sec x + 3 =0 (sec x − 1)(sec x − 3) = 0 sec x = 1 or sec x = 3 1 cos x = 1 cos x =

α α T C x = α, 180 − α = 30°, 150° ✓ 9(e)

𝑦

90° 180° 270° 360°

𝑥

−1

⇒ no solution

9(c)


S α

cos x = −

1 2

=

1 cos2 x

1

or

2

cos x = 3

0° < x < 360°

α T C x = α, 360 − α ≈ 70.5°, 289.5° ✓

S

A α α T C x = α, 360 − α = 60°, 300° ✓ 9(f)

2 tan x

= 3 + 2 cot x

2 tan x

= 3 + 2(

1 tan x

)

2

2 tan x = 3 tan x + 2 2 2 tan x − 3 tan x − 2 =0 (2 tan x + 1)(tan x − 2) = 0

or cos x = 3

α = 60° (no solution) ⇒ 2nd or 3rd quadrant

tan x = −

1 2

or tan x = 2

α ≈ 26.6° α ≈ 63.4° ⇒ 2nd or 4th quadrant ⇒ 1st or 3rd quadrant

0° < x < 360°

S A α α T C x = 180 − α, 180 + α = 120°, 240° ✓


7 cos x − 3

2 sec2 x 2

(no solution) α = 60° ⇒ 1st or 4th quadrant

A

2 sin2 x + 5 cos x + 1 =0 2 2(1 − cos x) + 5 cos x + 1 = 0 2 − 2 cos 2 x + 5 cos x + 1 = 0 −2 cos 2 x + 5 cos x + 3 =0 2 2 cos x − 5 cos x − 3 =0 (2 cos x + 1)(cos x − 3) = 0

=

cos x =

0° < x < 360°

1

7 cos x − 3

7 cos x − 3 = 2 cos 2 x 2 2 cos x − 7 cos x + 3 = 0 (2 cos x − 1)(cos x − 3) = 0

3

0° < x < 360°

A

0° < x < 360° S α

A

α T C x = 180 − α, 360 − α ≈ 153.4°, 333.4° ✓

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0° < x < 360° S

A α

α T C x = α, 180 + α ≈ 63.4°, 243.4° ✓

334


Ex 12.2

10(a) 2 tan2 x − 1 = 5 sec x 2 2(sec x − 1) −1 = 5 sec x (2 sec 2 x − 2) −1 = 5 sec x 2 2 sec x − 5 sec x − 3 = 0 (2 sec x + 1)(sec x − 3) = 0 sec x = −

1 2

11(b) |2 cos x + 3 sin x| = sin x 2 cos x + 3 sin x = sin x or 2 sin x = −2 cos x tan x = −1 α ≈ 45° ⇒ 2nd or 4th quadrant

or sec x = 3

cos x = −2 (no solution)

cos x =

1

0° < x < 180°

0 ≤ x ≤ 2π

α T C x = 180 − α, 360 − α ≈ 135° ✓

S α

A α α T C x = α, 2π − α = 1.23, 5.05 ✓

⇒ 2nd or 4th quadrant 0 ≤ x ≤ 2π

0 ≤ x ≤ 2π S

A α

α T C x = α, π + α = 0.785, 3.93


S

S α

A α C

0° < x < 360° 10 ° < 2x + 10° < 730° 0 2x + 10° = 180 − α, 360 − α, 540 − α, 720 − α ≈ 116.6°, 296.6°, 476.6°, 656.6° x

S α α T

A α α C

T

T

2

0° < x < 180° x = α, 180 − α, 180 + α, 360 − α = 55°, 125° ✓

A

α T C x = 180 − α, 360 − α = 153.4° ✓

12(b) cot(2x + 10°) = −0.5 tan(2x + 10°) = −2 α ≈ 63.4° ⇒ 2nd or 4th quadrant

1−sin 20°

α = 55° ⇒ 1st, 2nd, 3rd or 4th quadrant

S α

2x − 20 = −α, α, 360 − α, 360 + α, 720 − α = −18.2°, 18.2°, 341.8°, 378.2°, 701.8° x = 0.9°, 19.1°, 180.9°, 360.9° ✓

11(a) 2 cos 2 x + sin 20° = 1 2 cos 2 x = 1 − sin 20° = ±√

2

0° < x < 180°

A

4

cos x

1


0° < x < 360° −20° < 2x − 20° < 700°


α T C x = π − α, 2π − α = 2.82, 5.96 ✓

=−

12(a) cos(2x − 20°) = 0.95 α ≈ 18.2° ⇒ 1st or 4th quadrant

10(b) csc 2 x +2 cot x = 4 2 (cot x + 1) + 2 cot x = 4 cot 2 x + 2 cot x − 3 = 0 (cot x + 3)(cot x − 1) = 0 cot x = −3 or cot x = 1 1 tan x = 1 tan x = − 3 π α= α ≈ 0.322

A

tan x

3


S

S α

2 cos x + 3 sin x = − sin x 4 sin x = −2 cos x

≈ 53.3°, 143.3°,

233.3°, 323.3° ✓

A α α C

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335


Ex 12.2

12(c) csc(2x + 60°) = 4 sin(2x + 60°) =

13(b) sin (3x + 5) = √3 2

1

α ≈ 14.5° ⇒ 1st or 2nd quadrant 0° < x < 360° 60° < 2x + 60° < 780°

6

S

A

α T

α C

0

5

<

6

0.833 < 2x + 60° = α, 180 − α, 360 + α, 540 − α, 720 + α, 900 − α = 14.5, 165.5, 374.5, 525.5 734.5°, 885.5° x ≈ 52.8°, 157.24°, 232.8°, 337.2° ✓

3x 2

x

S α T

0° < x < 360° −60° < 2x − 60° < 660° 2x − 60° = −α, 180 − α, 360 − α,

6

2

+ +

6 5 6

< 3π +

A

α T

α C

5 6

< 10.258

= α, π − α, 2π + α, 3π − α,

4π + α,

= 0.448,2.69, 6.73, 8.97, = 1.24, 3.93, 5.43 ✓

13.0

=−

A

sin(2x − 3)

=−

α C

α ≈ 0.216 ⇒ 3rd or 4th quadrant

14 3 3 14

0
495, 675

S

A

α T

α C

2x − 3 = −2π − α, −π + α, −α, 3π + α, 4π − α,

277.5° ✓

x = 7.5°, 97.5°, 187.5° 13(a) cot(3x + 0.5) = 3

π + α, 2π − α,

= −6.50 −2.93, −0.216, 3.36, 6.06, 9.64, 12.4

1 3

α ≈ 0.22 ⇒ 1st or 3rd quadrant 0
S α T

A α

x

C

3x + 0.5 = α, π + α, 2π + α, 3π + α, 6π + α

x ≈ 0.988, 6.22 ✓

5

2 3x

< 2π 5

csc(2x − 3)

540 − α, 720 − α,

= −45, 135,315,

≈ 0.322,3.46, 19.2

+

3x

S

13(c) 3 csc(2x − 3) + 15 = 1

12(d) tan(2x − 60°) = −1 α = 45° ⇒ 2nd or 4th quadrant

tan(3x + 0.5) =

4

α ≈ 0.448 ⇒ 1st or 2nd quadrant

4

14(a) sin(3x + 70°) = 0.2 α ≈ 11.5° ⇒ 1st or 2nd quadrant

4π + α, 5π + α,

6.60, 9.75,

12.88, 16.03,

2.04, 3.08,

4.13, 5.17


= 0.0372, 1.39, 3.18, 4.53 ✓

sleightofmath.com

0° < x < 180° 70° < 3x + 70° < 610° 3x + 70° = α, 180 − α, = 11.5, 168.5, x = 32.8°,

S

A

α T

α C

360 + α, 540 − α 371.5, 528.5 100.5°, 152.8° ✓

336


Ex 12.2

14(b) 8 csc 2x cot 2x 8(

1 sin 2x

)(

cos 2x sin 2x

=3 )

=3

8 cos 2x 8 cos 2x 8 cos 2x 3 cos 2 2x + 8 cos 2x − 3 (3 cos 2x − 1)(cos 2x + 3) cos 2x =

1 3

16

= 3 sin2 2x = 3(1 − cos 2 2x) = 3 − 3 cos 2 2x =0 =0

1 cos θ

or cos 2x = −3

2 3

1

y

=4

cos θ

2 2

cos θ =

3

1 4

θ is obtuse ⇒ 2nd quadrant (rej ∵ cos θ < 0 in 2nd quadrant)

θ is obtuse ⇒ 2nd quadrant

0° < x < 180° 0° < 2x < 360°

3 -2

S

A α α T C 2x = α, 360 − α = 70.5, 289.5 x = 35.3°, 144.7° ✓

x = −2, r = 3, y = √(3)2 − (−2)2 = √5 y

tan θ = = x

2 sin x cos x − cos x +4 sin x − 2 (2 = cos x sin x − 1) +2(2 sin x − 1) = (2 sin x − 1)(cos x + 2) ✓

15(ii) 2 sin x cos x − 2 = cos x − 4 sin x 2 sin x cos x − cos x + 4 sin x − 2 = 0 (2 sin x − 1)(cos x + 2) =0 1 cos x = −2 sin x = or 2 (no solution) α = 30° ⇒ 1st or 2nd quadrant −360° ≤ x ≤ 360° S

=−

cos θ = −

(no solution) α ≈ 70.5° ⇒ 1st or 4th quadrant

15(i)

2 tan2 θ = 5 sec θ + 10 2 2(sec θ − 1) = 5 sec θ + 10 2 sec 2 θ − 2 = 5 sec θ + 10 2 2 sec θ − 5 sec θ − 12 = 0 (2 sec θ + 3)(sec θ − 4) = 0 3 sec θ = 4 sec θ = − or

√5 −2

=−

√5 2

✓

17(a) 3 cos 2 2x + 4 sin 2x 3(1 − sin2 2x) + 4 sin 2x 3 − 3 sin2 2x + 4 sin 2x 3 sin2 2x − 4 sin 2x − 2

A

sin 2x = = sin 2x =

=1 =1 =1 =0

4±√(−4)2 −4(3)(−2) 2(3) 4±2√10 6 2+√10 3

≈ 1.72 (no solution)

=

=

4±√40 6

=

4±√4×10 6

2±√10 3

sin 2x =

2−√10 3

≈ −0.387

α ≈ 22.8° ⇒ 3rd or 4th quadrant 0° < x < 360° 0° < 2x < 720°

α α T C x = −360 + α, −180 − α, α, 180 − α

S

A

α α T C 2x = 180 + α, 360 − α, 540 + α, 720 − α

= −330°, −210°, 30°, 150° ✓

≈ 202.8°, 337.2°, 562.8°, 697.2° x


sleightofmath.com

≈ 101.4°, 168.6°, 281.4°, 348.6° ✓

337

A math 360 sol (unofficial) 17(b) sec x (tan x − 2) (

1 cos x

Ex 12.2 = 2 csc x

) (tan x − 2)

=

sin x cos x

= 2(

(tan x − 2)

= =

sin x 2

(

cos θ

)

sin x ) cos x

tan x

2±2√3 2

S α

=

2±√12 2

=

2±√4×3 2

( tan x = 1 + √3 α ≈ 69.9° ⇒ 1st or 3rd quadrant 0° < x < 360°

A

S

α T C x = 180 − α, 360 − α = 143.8°, 323.8° ✓

𝑥

π 3π

3x = , 2

2π, x

3π 2

2

,

π 2

+

+ 2π π π

5π

6 2

6

= ,

✓

−(2)

F

20−10 sin A 2 F

) +(

−(3)

10 cos A 2

)

F

+

=1

100 cos2 A

=1

F2

400 − 400 sin A + 100 sin2 A + 100 cos 2 A = F 2

α T C x = α, 180 + α = 69.9°, 249.9° ✓

S α

20−10 sin A

F2

400 − 400 sin A + 100(sin2 A + cos 2 A)

= F2

400 − 400 sin A + 100

= F2

F2

= 500 − 400 sin A

F = √500 − 400 sin A or F = −√500 − 400 sin A ✓ (NA) 18(ii)

tan θ =

sin θ cos θ

=

20−10 sin A ) F 10 cos A ( ) F

(

=

20−10 sin A 10 cos A

=

2−sin A cos A

[shown] ✓ 18(iii) Put A = 30° in F = √500 − 400 sin A,

0≤x≤π 0 ≤ 3x ≤ 3π

−1

=

400−400 sin A+100 sin2 A

A α

17(c) sin 3x (4 sin 3x − 3 cos 3x) =4 2 4 sin 3x − 3 sin 3x cos 3x =4 2 4(1 − cos 3x) − 3 sin 3x cos 3x = 4 (4 − 4 cos 2 3x) − 3 sin 3x cos 3x = 4 −4 cos 2 3x − 3 sin 3x cos 3x =0 cos 3x (4 cos 3x + 3 sin 3x) =0 cos 3x = 0 3 sin 3x = −4 cos 3x 4 0≤x≤π tan 3x = − 3 0 ≤ 3x ≤ 3π α ≈ 0.927 𝑦 ⇒ 2nd or 4th quadrant 1 90° 180° 270° 360°

−(1)

sin2 θ + cos 2 θ = 1 sub (1) & (2) into (3):

=

= 1 ± √3

tan x = 1 − √3 α ≈ 36.2° ⇒ 2nd or 4th quadrant 0° < x < 360°

10 cos A

sin θ

tan x − 2 tan x =2 2 tan x − 2 tan x − 2 = 0 2(1)

=

10 sin A + F sin θ = 20 F sin θ = 20 − 10 sin A

2

2±√(−2)2 −4(1)(−2)

= cos θ

cos A

2

tan x =

10 cos A = F cos θ 10

(tan x − 2)

tan x − 2

18(i)

2

F|A=30° = √500 − 400 sin 30° 1 = √500 − 400 ( ) 2 = √500 − 200 = √300 = 10√3 ✓ Put A = 30° in tan θ = tan θ =

A

α T C 3x = π − α, 2π − α, 3π − α

2−sin 30° cos 30°

=

1 2 √3 2

2−

2−sin A cos A

=

3 2 √3 2

= √3

⇒ θ = 60°

= 2.21, 5.36, 8.50 x


= 0.738, 1.79, 2.83 ✓

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338


Rev Ex 12 A2(a) 5 sin2 x − 8 sin x cos x = 0 sin x (5 sin x − 8 cos x) = 0 sin x = 0 or 5 sin x = 8 cos x

Rev Ex 12 A1(a) x = a sin θ −(1) (i) a2 − x 2 −(2) sub (1) into (2): = a2 − (a sin θ)2 = a2 − a2 sin2 θ = a2 (1 − sin2 θ) = a2 cos 2 θ ✓

𝑦

tan x =

1

90° 180° 270° 360°

𝑥

−1

0° ≤ x ≤ 360° x = 0°, 180°, 360° ✓

3

= (1 −

A2(b) 5 cot 2 x

)

5(

3 sin2 θ)2

= = cos 3 θ ✓

A1(b) Coordinates of 𝛉 (i) √2 cos θ = − sin θ y tan θ = −√2 = x

θ is obtuse ⇒ 2nd quadrant y = −√2, x = 1, 2

=

1 −1 ( ) √3

cos2 x sin2 x 2

S

-1

+7

= 11 csc x

) +7

=

11 sin x

= −√3 ✓

A

α α T C x = α, 180 − α = 30°, 150° ✓

Trigonometric ratio tan(−θ) = − tan θ = −(−√2) = √2 ✓ x r

A α

0° ≤ x ≤ 360°

𝑟

√2

r = √(√2) + (−1)2 = √3

1

S

5 cos x +7 sin2 x = 11 sin x 2 2 5(1 − sin x) +7 sin x= 11 sin x 5 − 5 sin2 x +7 sin2 x = 11 sin x 2 sin2 x − 11 sin x + 5 = 0 (2 sin x − 1)(sin x − 5) = 0 1 sin x = 5 sin x = or 2 (no solution) α = 30° ⇒ 1st or 2nd quadrant

3 (cos 2 θ)2

A1(b) sec θ = 1 = cos θ (ii)

0° ≤ x ≤ 360°

α T C x = α, 180 + α ≈ 58.0°, 238.0° ✓

(a sin θ)2 2 a2

5


3 A1(a) x2 2 (1 − 2 ) −(3) (ii) a sub (1) into (3):

= (1 −

8

A2(c) 1 + 2 cos (3 x + 75°) = 0 2

3

cos ( x + 75°) 2

1 1 A1(b) csc(90° − θ) = = = −√3 ✓ sin(90°−θ) cos θ (iii)

=−

1 2

α = 60° ⇒ 2nd or 3rd quadrant 0° ≤ x

≤ 360°

S α α T

A C

3

75° ≤ x + 75° ≤ 615° 2

3 2

x


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x + 75° = 180 − α, 180 + α, 540 − α, 540 + α = 120, 240, = 30°, 110°,

480, 600 270°, 350° ✓

339


Rev Ex 12

A3(a) 10 cos 3x + 8 = 0 cos 3x

=−

A4(b) Prove:

4

S α α T

LHS =

A5(i)

C =

2

= =

3x−2 4 3x−2 4

x

< 14.5

1 cos θ 1

⋅ (

1 1

sin θ cos θ 1 sin θ 1 sin θ

⋅ ⋅

−

sin θ

1−cot x 1+cot x

S

A

α T

α C

= π + α, 2π − α, 3π + α, 4π − α,

5π + α

≈ 3.87, 5.55, ≈ 16.4 ✓

16.4

sin θ

− cos θ)

1−cos2 θ cos θ sin2 θ cos θ

sin θ cos θ

= tan θ

10.2, 11.8,

S

A α

0° < x < 360° 0° < 2x < 720°

α T

2x = α, 180 + α, = 45, 225,

360 + α, 540 + α 405, 585

C

x = 22.5°, 112.5°, 202.5°, 292.5° A4(a) Prove:

LHS =

1−csc x cot x

1−csc x cot x

= RHS

cos θ

A5(ii) sec 2x csc 2x − cot 2x = 1 tan 2x =1 α ≈ 45° ⇒ 1st or 3rd quadrant

3

< 20

=

= RHS [proven] ✓

α ≈ 0.730 ⇒ 3rd or 4th quadrant x

=

Prove: sec θ csc θ − cot θ = tan θ

=

A3(c) 3 sin (3x−2) = −2 =−

tan x+1

tan x 1 − tan x tan x tan x 1 + tan x tan x

LHS = sec θ csc θ − cot θ

A

2x − 70° = 180 − α, 180 + α, 1260 − α, 1260 + α, 1620 − α, 1620 + α = 110,250, 1190, 1330, 1550, 1690 x = 700° ✓

)

tan x−1

LHS = ln(sec x + tan x)(sec x − tan x) = ln(sec 2 x − tan2 x) = ln[(tan2 x + 1) − tan2 x] = ln 1 =0 = RHS [proven] ✓

=

4

1+cot x

A4(c) Prove: ln(sec x + tan x) + ln(sec x − tan x) = 0

C

A3(b) cos(2x − 70°) = sin 200° = − sin 20° = − cos 70° α = 70° S α ⇒ 2nd or 3rd quadrant α T x < 800° 2x − 70° < 1530°

sin (

1−cot x

[proven] ✓

A

3x = π − α, π + α, 3π − α, 3π + α, 5π − α, 5π + α, 7π − α = 2.50, 3.79, 8.78, 10.1, 15.1, 16.4, 21.3 x ≈ 7.12 ✓

4 3x−2

=

tan x+1

5

α ≈ 0.644 ⇒ 2nd or 3rd quadrant x >6 3x > 18

tan x−1

✓

= tan x − sec x

=

1 sin x cos x sin x

1−

=

sin x−1 sin x cos x sin x

=

sin x−1 cos x

= tan x −

sec x = RHS [proven] ✓


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340


Rev Ex 12

1st curve y = 4 cos 3x Point P y|x=0 = 4 cos[3(0)] ⇒ P(0,4) ✓

=4

Intersection point At intersection point, y = 4 cos 3x cuts x − axis (y = 0) 4 cos 3x = 0 cos 3x = 0

x

𝑦

π

3x =

2 π

=

1

6

90° 180° 270° 360°

π

⇒ ( , 0)

𝑥

−1

6

2nd curve y = 2 sin x + k π

At ( , 0), 6

π

2 sin ( ) + k = 0 6

1

2( ) + k = 0 2

k

= −1

∴ y = 2 sin x − 1 Point Q π At max, x = y|

π x= 2

2 π

= 2 sin − 1 2

= 2(1) − 1 =1 π ⇒ Q ( , 1) 2


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341


Rev Ex 12

A6(ii) y1 = 4 cos 3x i.e. a = 4, b = 3, c = 0

𝑦


=

360 b

=

360 3

= 120°

y2 = 2 sin x − 1 i.e. a = 2, b = 1, c = −1

4 1𝑂

180° 𝑥 𝑦 = 2 sin 𝑥 − 1 𝑦 = cos 3𝑥

−3 −4

✓


=

360 b

=

360 1

= 360° Workings

𝑦1 = 4 cos 3𝑥 Domain 0° ≤ 𝑥 ≤ 180° Axis with 𝑦 =0±4 Amplitude Shape +𝑐𝑜𝑠 180−0 Cycle = 1.5 120

𝑦2 = 2 sin 𝑥 − 1 Domain 0° ≤ 𝑥 ≤ 180° Axis with 𝑦 = −1 ± 2 Amplitude Shape +𝑠𝑖𝑛 180−0 1 Cycle = 360


2

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342


P = 105 − 20 cos (

Rev Ex 12

16πt

P

7

B1(b) 4 sin x cos x

)

4 sin x cos x

= 100

105 − 20 cos ( −20 cos ( cos (

16πt 7

16πt 7

16πt 7

) = 100

)

= −5

)

=

16πt 7 16πt 7 16πt 7

t

S

> 10 >

160 7

π

sin x = 0 0 ≤ x ≤ 2π

90° 180° 270° 360°

x = 0, π,2π ✓

B1(a) sec 2x = 2 1

cos 2x = 2 π α= 3 ⇒ 1st or 4th quadrant A α α C

T

π 5π

x

3 3 π 5π

= , 6

6

, ,

2π + α, 4π − α 7π 11π

,

3 3 7π 11π 6

,


6

✓

=±

π

1 2

3

⇒ 1st, 2nd, 3rd or 4th quadrant

π 2π 4π 5π

= , 3

3

,

3

,

3

✓

B1(c) (tan x + 1)(2 tan x − 1) = 5 2 tan2 x + tan x − 1 =5 2 2 tan x + tan x − 6 =0 (2 tan x − 3)(tan x + 2) = 0 3 2

tan x = −2

α ≈ 0.983 α ≈ 1.11 ⇒ 1st or 3rd quadrant⇒ 2nd or 4th quadrant 0 ≤ x ≤ 2π S

= ,

α=

4

S A α α α α T C x = α, π − α, π + α, 2π − α

tan x =

S

1 cos x 1

0 ≤ x ≤ 2π

= α, 2π − α, … 22π + α, 24π − α

2x = α, 2π − α,

𝑥

−1

≈ 1.32,4.97, … 70.4,74.1 ≈ 10.3 ✓

) =0

cos x

> 71.8

0 ≤ x ≤ 2π 0 ≤ 2x ≤ 4π

cos x

cos 2 x =

𝑦

4

sin x cos x 1

4 cos x =

1

A α α C

T

=

sin x (4 cos x −

1

α ≈ 1.318 ⇒ 1st or 4th quadrant t

= tan x

A α

α T C x = α, π + α = 0.983,4.12 ✓

sleightofmath.com

0 ≤ x ≤ 2π S α

A

α T C x = π − α, 2π − α = 2.03,5.18 ✓

343


Rev Ex 12

B2(i) x = 4 sin θ + 3 cos θ y = 4 cos θ − 3 sin θ

B3(b) Prove: LHS =

y = 2x 4 cos θ − 3 sin θ = 2(4 sin θ + 3 cos θ) 4 cos θ − 3 sin θ = 8 sin θ + 6 cos θ −11 sin θ = 2 cos θ tan θ

=−

2 11

α ≈ 10.3° ⇒ 2nd or 4th quadrant obtuse ⇒ 90° < θ < 180°

S α T

cos x 1−sin x

A α C

θ = 180 − α, 360 − α = 169.7°, 349.7° ✓ B2(ii) x 2 + y 2 = (4 sin θ + 3 cos θ)2 + (4 cos θ − 3 sin θ)2

−

cos x 1−sin x

1 cos x

−

= tan x

1 cos x

=

cos 2 x − (1 − sin x) (1 − sin x) cos x

=

cos 2 x − 1 + sin x (1 − sin x) cos x

=

(1 − sin2 x) − 1 + sin x (1 − sin x) cos x

=

sin x − sin2 x (1 − sin x) cos x

=

sin x (1 − sin x) (1 − sin x) cos x

=

sin x cos x

= tan x = RHS [proven] ✓

2

2

= (16 sin θ + 24 sin θ cos θ + 9 cos θ) +(16 cos 2 θ − 24 sin θ cos θ + 9 sin2 θ) = 25 sin2 θ + 25 cos 2 θ = 25(sin2 θ + cos 2 θ) = 25 [shown] ✓ B3(a) LHS 2

=

1 − tan x 1 + tan2 x

=

1 − (sec 2 x − 1) sec 2 x

2 − sec 2 x sec 2 x 2 = −1 sec 2 x 2 = −1 1 cos 2 x =

B3(c) LHS = ln(1 + cos x) + ln(1 − cos x) = ln(1 − cos 2 x) = ln(sin2 x) = ln[(sin x)2 ] = 2 ln(sin x) = RHS [proven] ✓ B4(a) tan 5x = 0.6 2

α ≈ 0.540 ⇒ 1st or 3rd quadrant x 5x 2 5x 2

x

>4 > 10

S α T

A α C

= α, π − α 2π + α, 3π − α, 4π + α ≈ 0.540,2.60, 6.82, 8.88, ≈ 5.24 ✓

13.11

= 2 cos 2 x −1 = RHS [proven] ✓


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344


Rev Ex 12

B4(b) csc 2x = 6 sin 2x =

1 6

α ≈ 0.167 ⇒ 1st or 2nd quadrant x <5 2x < 10

2x

α C

α ≈ 63.4° α ≈ 18.4° α = 26.6° ⇒ 2nd or 4th quad.⇒ 2nd or 4th quad.⇒ 1st or 3rd quad.

cos ( )

=−

0° < θ < 180° S α

4 3 3

2x

> 666.67°

3 2x

A C

α, 900 + α ≈ 138.6,221.4, 858.6°, 941.4 x ≈ 1287.9° ✓ 3

0° < θ < 180° S α

A

α T C θ = 180 − α, 360 − α = 161.6° ✓

or tan θ =

2

0° < θ < 180° S

A α

α T C θ = α, 180 + α = 26.6° ✓

=

1 cos 2x − cos 2x 1 + sin 2x

=

(1 + sin 2x) − cos 2 2x cos 2x (1 + sin 2x)

(1 − cos 2 2x) + sin 2x = cos 2x (1 + sin 2x)

= 180° − α, 180° + α, 540 − α, 540 + α,

3

or tan θ = − 3

B6(i) f(x)

4

S α α T

> 1000°

A

α T C θ = 180 − α, 360 − α = 116.6° ✓

α ≈ 41.4° ⇒ 2nd or 3rd quadrant x

1

α T

=−

3

1

tan θ = −2

sec ( ) 3 2x

=2

A

= −4

3

6 tan3 θ + 11 tan2 θ − 3 tan θ

S

B4(c) 3 sec (2x) + 4 = 0 3 sec ( )

= 2 cot θ

6(tan θ)3 + 11(tan θ)2 − 3(tan θ) − 2 = 0

2x = α, π − α, 2π + α, 3π − α, 4π + α = 0.167,2.974, 6.451,9.257 12.734 x ≈ 4.63 ✓ 3 2x

B5(ii) 6 tan2 θ + 11 tan θ − 3

900 − =

sin2 2x + sin 2x cos 2x (1 + sin 2x)

=

sin 2x (sin 2x + 1) cos 2x (1 + sin 2x)

=

sin 2x cos 2x

498.6,581.4

2

B5(i) 6x + 11x − 3x − 2 =0 2 (x + 2)(6x + + ⬚) (x + 2)(6x 2 + − 1) (x + 2)(6x 2 − x − 1) =0 (x + 2)(3x + 1)(2x − 1) = 0 1

1

3

2

= tan 2x [shown] ✓

x = −2 or x = − or x = ✓


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345


Rev Ex 12

B6(ii) y = f(x) = tan 2x i.e. a = 1, b = 2, c = 0 Period

=

180 b

=

B7(i)

πt 2 π i.e. a = 1.6, b = , c = 0 v = 1.6 sin

2

T=

180

2π

= 90°

2

b

=

B7(ii) v

𝑦 𝑦 = 𝑓(𝑥)

90°

𝑂

180° 270°

✓

πt 2

Workings

= 4s ✓

πt 2

πt 2

= 1.3 =

13 16


𝑥

x = 45° x = 135° x = 225°

π 2

= 1.3

1.6 sin sin

2π

t

= α, π − α ≈ 0.948,2.193 ≈ 0.604, 1.40 ✓

S

A

α T

α C

Domain 0° ≤ x ≤ 270° Axis with y=0 Amplitude Shape +tan Cycle

270−0 90


=3

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346


Ex 13.1

Ex 13.1 1(a)

1(b)

1(c)

LHS = sin(90° + θ) = sin 90° cos θ + cos 90° sin θ = (1) cos θ +(0) sin θ = cos θ = RHS [proven] ✓ LHS = cos(90° + θ) = cos 90° cos θ − sin 90° sin θ = (0) cos θ −(1) sin θ = − sin θ = RHS [proven] ✓ LHS = sin (

3π 2 3π

3(a)

sin 15° cos 75° + cos 15° sin 75° = sin(15° + 75°) = sin 90° =1✓

3(b)

sin 50° cos 20° − cos 50° sin 20° = sin(50° − 20°) = sin 30° 1

= ✓ 2

3(c)

= − θ) 3π

= sin ( )

cos θ − cos ( ) sin θ

2

π

2 π

2

2

3(d)

= [− sin ( )] cos θ − cos ( ) sin θ = (−1) cos θ −(0) sin θ = − cos θ = RHS [proven] ✓ 1(d)

=

3(e)

2(c)

2(d)

✓

sin 15° sin 30° − cos 15° cos 30° = −(cos 15° cos 30° − sin 15° sin 30°) = − cos(45°) =−

√2 2

✓

tan 35°+tan 10° 1−tan 35° tan 10°

1+tan 2π tan θ (0)−tan θ 1+(0) tan θ

LHS = sin θ cos 2θ + cos θ sin 2θ = sin(θ + 2θ) = sin 3θ = RHS [proven] ✓ LHS = sin 2θ cos θ − cos 2θ sin θ = sin(2θ − θ) = sin θ = RHS [proven] ✓ Prove: LHS = cos 3θ cos 2θ + sin 3θ sin 2θ = cos(3θ − 2θ) = cos θ = RHS [proven] ✓ Prove: LHS =

tan 3θ−tan 4θ 1+tan 3θ tan 4θ

= tan(3θ − 4θ) = tan(−θ) = − tan θ = RHS [proven] ✓


= tan(35° + 10°) = tan 45° =1✓

tan 2π−tan θ

3(f)

= − tan θ = RHS [proven] ✓

2(b)

√3 2

LHS = tan(2π − θ) =

2(a)

cos 70° cos 40° + sin 70° sin 40° = cos(70° − 40°) = cos 30°

tan 30°−tan 75° 1+tan 30° tan 75°

= tan(30° − 75°) = tan(−45°) = − tan 45° = −1 ✓

4(a)

cos 75° = cos(30° + 45°) = cos 30° cos 45° − sin 30° sin 45° √2 √3 2 2 √6−√2 ✓ 4

= ( )( ) = 4(b)

2

√2 2

sin 15° = sin(45° − 30°) = sin 45° cos 30° − cos 45° sin 30° √2 √3 2 2 √6−√2 ✓ 4

= ( )( ) = 5(a)

1

−( )( )

√2 2

1

−( )( ) 2

LHS = sin(A + B) + sin(A − B) = sin A cos B + cos A sin B + sin A cos B − cos A sin B = 2 sin A cos B = RHS [proven] ✓

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347


Ex 13.1

5(b) LHS = cos(A + B) − cos(A − B) = cos A cos B − sin A sin B −(cos A cos B + sin A sin B) = cos A cos B − sin A sin B − cos A cos B − sin A sin B = −2 sin A sin B = RHS [proven] ✓

6(d) LHS

6(a) LHS

=

1 cos(α − β)

1 1 cos α cos β = = cos α cos β + sin α sin β cos α cos β + sin α sin β cos α cos β sec α sec β 1 + tan α tan β

= RHS [proven] ✓

=

sin(A + B) − sin(A − B) cos(A + B) − cos(A − B)

=

(sin A cos B + cos A sin B) − (sin A cos B − cos A sin B) (cos A cos B − sin A sin B) − (cos A cos B + sin A sin B)

sin A cos B + cos A sin B = cos A cos B − sin A sin B 2 cos A sin B = −2 sin A sin B

=

7(a)

LHS = tan(A + 45°) tan(A − 45°) =

− sin A cos B + cos A sin B − cos A cos B − sin A sin B

= =

= − cot A

7(b)

sin(A + B) = cos(A − B)

tan A−(1) 1+tan A(1)

tan2 A−1 1−tan2 A

LHS = tan A + tan B =

sin A cos B + cos A sin B cos A cos B + sin A sin B sin A cos B + cos A sin B cos A cos B = cos A cos B + sin A sin B cos A cos B tan A + tan B = 1 + tan A tan B

7(c) LHS

= RHS [proven] ✓

= cot(x + y)

=

=

=

sin A cos A

+

sin A cos B

sin A cos B +sin A cos A cos A cos B sin(A+B) cos A cos B

= RHS [proven] ✓

=

6(c) LHS

1 tan(x + y)

1 = tan x + tan y 1 − tan x tany

sin(A + B) = sin(A − B) sin A cos B + cos A sin B sin A cos B + cos A sin B cos A cos B = = sin A cos B − cos A sin B sin A cos B − cos A sin B cos A cos B tan A + tan B = tan A − tan B


⋅

tan A−tan 45° 1+tan A tan 45°

= RHS [proven] ✓

LHS

= RHS [proven] ✓

tan A+(1) 1−tan A(1)

⋅

= −1

= RHS [proven] ✓ 6(b)

tan A+tan 45° 1−tan A tan 45°

1 − tan x tan y 1 − tan x tan y cot x cot y − 1 tan x tan y = = tan x + tan y = tan x + tan y cot y + cot x tan x tan y =

cot x cot y − 1 cot x + cot y

= RHS [proven] ✓

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348


Ex 13.1

7(d) LHS

9(i)

7

− cos A sin B =

10

− cos A sin B

9(ii)

= RHS [proven] ✓

√2 √2 √2 = 2 [sin A ( ) + cos A ( )] [cos A ( ) − 2 2 2 √2 sin A ( )] 2 2 √2 = 2 ( ) (cos A + sin A) (cos A − sin A) 2 2 2 2

7

∵ sin A cos B =

2

10

1 5

1

= ✓ 5

7

1

10

5

∵ sin A cos B = = 9(iii)

tan A tan B

= = = =

2

= cos A − (1 − cos A) = 2 cos 2 A − 1 = RHS [proven] ✓

9 10

7 10

, cos A sin B =

1 5

✓

sin A cos A sin B cos B

sin A cos A

×

cos B sin B

sin A cos B sin B cos A 7 10 1 5

∵ sin A cos B =

7 10

, cos A sin B =

1 5

7

= ✓ 2

10

LHS = cos(A + B + C) = cos[(A + B) + C]

sin(A+B) sin A cos B+cos A sin B

= cos A cos B cos C − sin A sin B cos C − sin A cos B sin C − cos A sin B sin C = RHS [proven] ✓

2

= cos A − sin A cos A − sin2 cos A − cos A sin2 A

sin A cos B

= cos A −3(1 − cos A) cos A

= −5

cos A sin B

= −5 ✓

tan A cot B tan(A − B) tan A−tan B 1+tan A tan B 3 −tan B 4 3 1+ tan B 4

= = =

25 24 25 24 25

∵ tan A =

24

18 − 24 tan B = 25 + 4

= cos A −3 sin A cos A 2

3

= −5 cos A sin B

sin A cos B

2

3

2

3 sin A cos B + 3 cos A sin B = 2 sin A cos B − 2 cos A sin B

171 3

3

=

sin A cos B−cos A sin B

11

LHS = cos 3A = cos A cos A cos A − sin A sin A cos A − sin A cos A sin A − cos A sin A sin A

2

=

sin(A−B)

= cos(A + B) cos C − sin(A + B) sin C = (cos A cos B − sin A sin B) cos C −(sin A cos B + cos A sin B) sin C

3

2 1

sin(A + B) = sin A cos B + cos A sin B

= 2 ( ) (cos A − sin A)

8(ii)

2 1

=( )+( )

7(e) LHS = 2 sin(A + 45°) cos(A + 45°) = 2[sin A cos 45° + cos A sin 45°] [cos A cos 45° − sin A sin 45°]

8(i)

1

=−

cos A sin B

sin y cos x − sin x cos y sin(y − x) = sin y cos x + cos y sin x sin(y + x)

4 2

=

sin A cos B − cos A sin B =

sin y sin x sin y cos x − sin x cos y tan y − tan x cos y − cos x cos x cos y = = = sin y sin y cos x + cos y sin x sin x tan y + tan x + cos y cos x cos x cos y =

sin(A − B)

tan B

tan B

75 4

3 4

tan B

= −7 =−

28 171

✓

= cos 3 A −3 cos A + 3 cos 3 A = 4 cos 3 A − 3 cos A = RHS [proven] ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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349


Ex 13.1

Quadrant

13(i)

4

tan A = > 0

4


3

Coordinates of A cos A = − for 90° < A < 180° 5

3 5

4

-4

sin B =

y

4

x

3

r

5

r

5

3

x

4

12 13

for 0° < B < 90°

13 B 12 𝑥

x = √132 − 122 = 5

3

B

x

5

y

5

y = 12, r = 13,

Coordinates of B sin B = − in 3rd quadrant

3

r

Coordinates of B

𝑟

r = √(−4)2 + (−3)2 = 5 sin A = = − , cos A = = −

y

⇒ sin A = = , tan A = = −

A

-3

y = −4, x = −3,

-4

y = √52 − (−4)2 = 3

Coordinates of A 3

-3

x

5

r

13

⇒ cos B = =

5

,

y

12

x

5

tan B = =

Trigonometric ratio sin(A + B) = sin A cos B + cos A sin B

y = −3, r = 5, x = −√(5)2 − (−3)2 = −4

3

5

−4

12

33

5

13

5

13

65

= ( )( ) + ( )( ) = − x

4

y

3

r

5

x

−4

cos B = = − , tan B = = −

=

tan A+tan 45°

=

1−tan A tan 45°

4

13(ii)

4 ( )+(1) 3 4 1−( )(1) 3

−4

−4

−3

5

5

5

tan(A + B) =

13(iii) sec(A − B)= =

= ( )( ) +( )( ) =

5 24 25

tan A+tan B 1−tan A tan B

= −7 ✓

12(ii) cos(A − B) = cos A cos B + sin A sin B −3

✓

3

Trigonometric ratio tan(A + 45°) =

A

x = −4, r = 5,

same quadrant ⇒ 3rd quadrant

tan A = in 3rd quadrant

5

𝑦

sin B = − < 0 ⇒ 3rd or 4th quadrant

=

✓

=

3 12 )+( ) −4 5 3 12 1−( )( ) −4 5

(

=

33 56

✓

1 cos(A−B) 1 cos A cos B+sin A sin B 1 −4 5 3 12 )( )+( )( ) 5 13 5 13

(

=

65 16

✓

12(iii) sin(B − A) = sin B cos A − cos B sin A −3

−3

5

5

= ( )( ) =

7 25

−4

−4

5

5

−( )( )

✓


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350


Ex 13.1

Right angle triangle hyp =

√(24)2

sin θ =

7 25

+

72

, cos θ =

24 θ hyp

= √625 = 25

24 25

7

, tan θ =

= cos(x − 50°)

cos x cos 10° − sin x sin 10° = cos x cos 50° + sin x sin 50°

7

24

15(c) cos(x + 10°)

cos x cos 10° − cos x cos 50° = sin x sin 10° + sin x sin 50° Trigonometric ratio sin A = sin(θ + 60°) = sin θ cos 60° + cos θ sin 60° 7

1

24

= ( )( ) =

25 2 7+24√3 50

cos x (cos 10° − cos 50°) cos 10°−cos 50°

= tan x

sin 10°+sin 50°

√3 2

+( )( ) 25

= sin x (sin 10° + sin 50°)

tan x

=

✓ =

cos A = cos(θ + 60°) = cos θ cos 60° − sin θ sin 60° 24

1

7

= ( )( ) =

25 2 24−7√3 50

√3 2

✓

tan A = tan(θ + 60°) =

tan θ+tan 60° 1−tan θ tan 60°

=

7 24

( )+(√3) 7 1−( )(√3) 24

=

7+24√3 24−7√3

✓

15(a) sin(x + 30°) = 2 cos x sin x cos 30° + cos x sin 30° = 2 cos x √3 sin x ( ) 2 √3 sin x ( ) 2

1

+ cos x ( ) 2

= 2 cos x

α = 20° ⇒ 1st or 3rd quadrant

S

= √3

cos x cos 60°−sin x sin 60° 2

α T

0° < x < 360°

= 5 sec(x − 20°)

2

2

1 2

cos x( )−sin x(

= =

√3 ) 2

5 cos(x−20°) 5 cos x cos 20°+sin x sin 20° 5 cos x cos 20°+sin x sin 20° 5

5 2

5

√3 sin x + 2 sin x sin 20°

= cos x − 2 cos x cos 20° 2

5

5

sin x ( √3 + 2 sin 20°) 2

x = α, 180 + α = 60°, 240° ✓

= cos x ( − 2 cos 20°) 2

tan x

=5 2

√2 2

√2 2

= 3 sin x = 3 sin x

tan x α ≈ 7.1° ⇒ 1st or 3rd quadrant

2 [cos x ( ) − sin x ( )]

= 3 sin x

√2 cos x − √2 sin x √2 cos x

= 3 sin x = (3 + √2) sin x

tan x

=


S

0° < x < 360°

α T

5

2 cos x cos 20° + 2 sin x sin 20° = cos x − √3 sin x 2 2

C

15(b) 2 cos(x + 45°) 2(cos x cos 45° − sin x sin 45°)

)

C

=

= ( ) cos x

2

A α

α T

0° < x < 360° x = α, 180 + α = 20°, 200° ✓

2

A α

10°−50°

= tan 20°

cos(x+60°)

S

10°+50° 10°−50° ) sin( ) 2 2 10°+50° 10°−50° 2 sin( ) cos( ) 2 2

−2 sin(

tan x

15(d) 2 sec(x + 60°)

3

tan x α = 60° ⇒ 1st or 3rd quadrant

sin 10°+sin 50°

= − tan (

−( )( ) 25

cos 10°−cos 50°

√2 3+√2

0° < x < 360° x = α, 180 + α = 7.1°, 187.1°

5 −2 cos 20° 2

√3+2 sin 20°

≈ 0.124 S α T

A α C

A α C

x = α, 180 + α = 17.8°, 197.8° ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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351


tan x+tan 15° 1−tan x tan 15°

Ex 13.1 17(a) sin x cos 1 − cos x sin 1 = 0.2 sin(x − 1) = 0.2 α ≈ 0.201 S A ⇒ 1st or 2nd quadrant α α T C 0
=2

tan(x + 15°) = 2 α ≈ 63.4° ⇒ 1st or 3rd quadrant 0° < x < 360° 15° < x + 15° < 375°

S

A α

α T

C

x − 1 = α, π − α = 0.201,2.94 x = 1.20, 3.94 ✓

x + 15° = α, 180 + α ≈ 63.4°, 243.4° x ≈ 48.4°, 228.4° ✓ 17(b)

16(b) 2 tan x + 3 tan(x − 45°) = 0 2 tan x + 3 ( 2 tan x + 3 [ 2 tan x +

tan x−tan 45° 1+tan x tan 45° tan x−(1)

1+tan x(1) 3 tan x−3

]

1+tan x

2 tan x

S

A α

α T C x = α, 180 + α = 26.6°, 206.6° ✓

=

tan(2x − 1) =

)= 0

1 2 1 2

α ≈ 0.464 ⇒ 1st or 3rd quadrant

=0 =0 =

3−3 tan x

0
1+tan x

2 tan x + 2 tan2 x = 3 − 3 tan x 2 tan2 x + 5 tan x − 3 =0 (2 tan x − 1)(tan x + 3) = 0 1 tan x = −3 tan x = 2 α ≈ 71.6° α ≈ 26.6° ⇒ 2nd or 4th quadrant ⇒ 1st or 3rd quadrant 0° < x < 360°

tan 2x−tan 1 1+tan 2x tan 1

2x − 1 = α, π + α, = 0.464, 3.61, x = 0.732,2.30,

A

α T C x = 180 − α, 360 − α = 108.4°, 288.4° ✓

16(c) tan 5x sin 5x cos 5x

sin 5x cos 2x sin 5x cos 2x − sin 2x cos 5x sin(5x − 2x) sin 3x

C

2π + α, 3π + α 6.75, 9.89 3.87, 5.44 ✓

sin 2x cos 2x

cos 0.4

tan x

=

tan x α ≈ 0.339 ⇒ 1st or 3rd quadrant

≈ 0.353

0 < x < 2π x = α, π + α ≈ 0.339, 3.48 ✓

= tan 2x =

α T

A α

17(c) cos(x − 0.4) = 3 sin x cos x cos 0.4 + sin x sin 0.4 = 3 sin x cos x cos 0.4 = sin x (3 − sin 0.4)

0° < x < 360° S α

S

S α T

3−sin 0.4

A α C

= sin 2x cos 5x =0 =0 𝑦 =0 1

0° < x < 360° 0° < 3x < 1080°

90° 180° 270° 360°

𝑥

−1

3x = 0,180, 360,540 720,900, 1080 x = 60°, 120°, 180°, 240°, 300° ✓


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352

A math 360 sol (unofficial) 17(d) 3 cos x 3 cos x 3 cos x 3 cos x + 4 cos x sin 2 cos x (3 + 4 sin 2) tan x

Ex 13.1 = 4 sin(x − 2) = 4(sin x cos 2 − cos x sin 2) = 4 sin x cos 2 − 4 cos x sin 2 = 4 sin x cos 2 = 4 sin x cos 2 =

√5

5

cos(A + B) = −

cos(A + B) = −

√5 A 2

𝑦

<0

5 13

=

y

1

x

2

Recall

r

y

12

r

13

4

=

√10

y = 3, r = √10,

B 𝑥

2

x = √(√10) − 32 = 1

3

5

1+tan B tan A

5

13

13

, cos(A + B) = −

3

( )

13 56 65

−5

4

13

5

−( )

5

✓

19(ii) Coordinates of C acute C ⇒ 1st quadrant sin C =

12 13

13 C 12 𝑥

y

= , r

x = √132 − 122 = 5 x

5

r

13

⇒ cos C = =

=3✓ tan B−tan A

12

y = 12, r = 13,

Trigonometric ratio

tan(B − A) =

-5

cos A = ,

5

12

and B is acute

1

,

A+B

3

sin A = ,

=( )

3

13

𝑦

sin B = sin[(A + B) − A] = sin(A + B) cos A − cos(A + B) sin A

18(ii) Coordinates

tan B = =

⇒ Q2 or Q3

x

sin(A + B) =

tan A = = [shown] ✓

18(iii)

13

⇒ Q1 or Q2

∴ Q2

Trigonometric ratio

x

5

⇒ sin(A + B) = =

2

y

5

y = √(13)2 − (−5)2 = 12

y = √(√5) − 22 = 1

3

4

r

Coordinates of (𝐀 + 𝐁) A and B are acute

and A is acute

√10

y

x = −5, r = 13,

x = 2, r = √5,

sin B =

5 𝑦 A 3

y = √52 − 32 = 4

4 cos 2

18(i) Coordinates 2

3

cos A = for acute A

⇒ sin A = = ,

3+4 sin 2

tan x ≈ −3.99 α ≈ 1.33 S A ⇒ 2nd or 4th quadrant α α T C 0 < x < 2π x = π − α, 2π − α ≈ 1.82, 4.96 ✓

cos A =

19(i) Coordinates of A

=

3 1

1 2 3 1 1+( )( ) 1 2

( )−( )

=1✓

Recall

sin C =

12

,

13

cos C =

sin(A + B) =

⇒ B − A = 45° ✓

5 13

,

12

5

13

13

, cos(A + B) = −

sin(A + B + C) = sin[(A + B) + C] = sin(A + B) cos C + cos(A + B) sin C =

12 13

5

( ) 13

−5

12

13

13

+( )

=0✓ 19(iii) sin(A + B + C) = 0 0° < A + B + C < 270° A + B + C = 180° ∴ Yes, A, B and C are angles of triangle ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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353


Ex 13.1

Show: sin(A + B) sin(A − B) = sin2 A − sin2 B LHS = (sin A cos B + cos A sin B)(sin A cos B − cos A sin B) = sin2 A cos 2 B − cos 2 A sin2 B 2 (1 2 = sin A − sin B) −(1 − sin2 A) sin2 B = sin2 A − sin2 A sin2 B − sin2 B + sin2 A sin2 B = sin2 A − sin2 B = RHS [shown] ✓

x

tan(α + β) tan α+tan β

90° 180° 270° 360° 90° 180° 270° 360°

𝑥

5 x

( )+tan β

5+x tan β

=

x−5 tan β

tan β (x + tan β (

20(ii) LHS = sin 15° sin 75° = sin 75° sin 15° = sin(45° + 30°) sin(45° − 30°) = sin2 45° − sin2 30°

= =

2 4 1

2

1 2

−( ) 2

−

1 4

4

10 x 10 x 10

∵ tan α =

x

5 x

10 x

= 10 −

50 x

tan β

50 x

) = 10 − 5

x2 +50 x

) =5

tan β

=

5x x2 +50

✓

21(ii) tan(2α + β) = tan[α + (α + β)] =

−1

2x = α, 180 − α, 4x = 0,180, = 0,180, 360,540, 720 x = 0°, 45°, x = 0°, 45°, 90°, 135°, 180°

√2 2

=

5 1−( ) tan β x

𝑥

−1

=( )

=

1−tan α tan β

1

1

=

5 + x tan β

20(i) sin2 3x = sin2 x sin2 3x − sin2 x =0 sin(3x + x) sin(3x − x) = 0 sin 4x sin 2x =0 sin 4x = 0 sin 2x = 0 0° ≤ x ≤ 180° 0° ≤ x ≤ 180° 0 ≤ 2x ≤ 360° 𝑦 0° ≤ 4x ≤ 720° 𝑦

21(i) tan α = 5

=

360 + α 360 180° ✓

tan α+tan(α+β) 1−tan α tan(α+β) 5 x

10 x 5 10 1−( )( ) x x

( )+( )

5

10

x

x

∵ tan α = , tan(α + β) = = =

15 x x2 −50 x2

15x x2 −50

0° < 2α + β < 90° ⇒ tan(2α + β) > 0 15x x2 −50

>0 x

>0

(x+√50)(x−√50) x

>0

(x+5√2)(x−5√2)

−

= RHS [shown] ✓

−5√2

−

+ 0

+ 5√2

−5√2 < x < 0 or x > 5√2 ✓?


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354

A math 360 sol (unofficial) 21(ii) tan(2α + β)

= =

Ex 13.1

tan 2α+tan β

0° < 2α + β < 90° tan(2α + β)

1−tan 2α tan β 2 tan α ( )+tanβ 1−tan2 α 2 tan α 1−( ) tan β 1−tan2 α

5

5x

x

x2 +50

∵ tan α = , tan β =

>0

15x[x2 +25]

>0

x2 −25x2 +25

x

25+√(−25)2 −4(1)(25) 25−√(−25)2 −4(1)(25) [x− ][x− ] 2(1) 2(1)

,

tan(2α + β)

x

>0

25+√525 25−√525 [x− ][x− ] 2 2

5 2( ) x ] + ( 5x ) [ x 2 + 50 5 2 1−( ) x = 5 2( ) x ] ( 5x ) 1−[ 5 2 x 2 + 50 1−( ) x

−

+ 0

−

25−√525 2

22 A+B+C tan(A + B + C)

10 5x ) [ 2 x ]+( 2 x − 25 x + 50 x2 = 10 5x ) 1−[ 2 x ]( 2 x − 25 x + 50 x2

tan(A+B)+tan C 1−tan(A+B) tan C

⇒ tan(A + B) + tan C tan A+tan B 1−tan A tan B

+

25−√525 25+√525 2

0
>0

+ tan C

2

or x >

25+√525 2

✓

= 180° = tan(180°) =0 =0 =0

tan A + tan B + tan C − tan A tan B tan C = 0 tan A + tan B + tan C = tan A tan B tan C [shown] ✓

10x 5x [ 2 ]+( 2 ) x − 25 x + 50 = 10x 5x ]( ) 1−[ 2 x − 25 x 2 + 50 10x 5x (x 2 − 25)(x 2 + 50) 2 − 25] + (x 2 + 50) x = × 2 10x 5x (x − 25)(x 2 + 50) ]( ) 1−[ 2 x − 25 x 2 + 50 [

=

10x(x 2 + 50) + 5x(x 2 − 25) (x 2 − 25)(x 2 + 50) − 50x 2

=

10x 3 + 500x + 5x 3 − 125x x 4 + 25x 2 + 25 − 50x 2

=

15x 3 + 375x x 4 − 25x 2 + 25

15x[x 2 + 25] = 2 x − 25x 2 + 25


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355


Ex 13.2 2(b)

Ex 13.2 1(a)

Coordinates 1

A = cos −1 (− ) for obtuse A 1

1

2

2

2

cos 2 22 ° − sin2 22 ° = cos(45°) =

√2 2

cos 𝐴 = −

1 2

2

y

✓

-1

y = √(2)2 − (−1)2 = √3 1(b)

π

π

8

8

2 sin cos

π

= sin ( ) =

1(c)

π 2 cos 2 12

√2 2

𝑦 𝑟

√3 , 2

=

cos 𝐴 =

𝑥 𝑟

=−

1 2

✓ Trigonometric ratio sin 2A = 2 sin A cos A

π

− 1 = cos ( ) 6

√3 2

= 1(d)

sin 𝐴 =

4

(cos 75° + sin 75°)2 = cos 2 75° + 2 sin 75° cos 75° + sin2 75° = (sin2 75° + cos 2 75°) +2 sin 75° cos 75° = (1) + sin 150° (1) = + sin 30° =1

−1

√3 2

= 2( )( )

✓

+

=−

2

√3 2

✓

−1 2

√3 2

cos 2A = cos 2 A − sin2 A = ( ) − ( ) 2

2

=−

1 2

✓

tan 2A =

1

2 tan A 1−tan2 A

2(

=

√3 ) −1

1−(

2

3

√3 ) −1

2

=

−2√3 1−3

= √3 ✓

= ✓ 2

1(e)

3(i)

sin2 67.5° = = = =

1(f)

2 tan 15° 1−tan2 15°

1−cos(135°)

3(

2 1−(− cos 45°)

3

2

2

1−(−

4

2

1 √3

3(ii)

✓

sin 2x

3

= ✓ 9

3(iii) 5

3

5

cos 8x = cos[2(4𝑥)] = 2 cos 2 4x −1 1 2

= 2( )

𝑥 x = √52 − 32 = 4 = , 5

=−

tan 𝐴 =

𝑦 𝑥

=

3 4

4(a)

Trigonometric ratio 3

4

24

5

5

25

sin 2A = 2 sin A cos A= 2 ( ) ( ) =

LHS = =

✓

3 2

7

5

5

25

✓

4(b)

LHS = =

2 tan A 1−tan2 A

=

3 4 3 2 1−( ) 4

2( )

=

24 7

✓

9 79

81

−1

✓

cos 2A cos A+sin A cos2 A−sin2 A cos A+sin A

=

(cos A+sin A)(cos A−sin A) cos A+sin A

=

cos A − sin A = RHS [proven] ✓

4 2

cos 2A = cos 2 A − sin2 A = ( ) − ( ) =

tan 2A =

3

1

sin 𝐴 = for acute A

𝑟

2

= ✓

2 2

3

cos 𝐴 =

=1

cos 4x = cos[2(2𝑥)] = 1 − 2 sin2 2x

✓

4

=1

= 1 − 2( )

Coordinates

𝑥

)

sin 2x

√2 ) 2

2 2+√2

sin 2x

= tan 30° =

2(a)

3 sin x cos x = 1

4(c)

LHS =

1−cos 2A sin 2A sin A cos A

=

1−(1−2 sin2 A) 2 sin A cos A

=

2 sin2 A 2 sin A cos A

= tan A = RHS [proven] ✓

1−cos 2A 1+cos 2A 2

=

1−(1−2 sin2 A) 1+(2 cos2 A−1)

=

2 sin2 A 2 cos2 A

= tan A = RHS [proven] ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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356


Ex 13.2

LHS = 4 sin A cos 3 A − 4 sin3 A cos A = 4 sin A cos A (cos 2 A − sin2 A) 1

6(i)

(cos 2 A − sin2 A)

= 4 ( sin 2𝐴) 2

sin A =

= 2 sin 2𝐴 cos 2𝐴 = sin 4A = RHS [proven] ✓ 5(a)

1 cos A

)(

LHS = =

1 sin A

)=

1 sin A cos A

=1 2

1 sin 2A

=

sin 2A

5(c)

LHS =

x

Trigonometric ratio cos 2A = cos 2 A − sin2 A

6(ii)

=

1

=− ✓ 2

cos 4A = 2 cos 2A − 1 1 2

1

=− ✓

= 2 (− ) − 1 2

)(

cos A sin A

)=

1 sin A cos A

=1 2

1 sin 2A

=

A sec 2 2

=

1 cos2

A 2

=

1 1+cos A 2

=

2 1+cos A

6(iii)

sin 4A = √1 −

1 2

6(iv)

=

cos A cos

21 2

= 2 cos

2

=

2 2 tan A ) 1−tan2 A

(

=

A =

1+cos A 1−cos A

=

1 2 1 1−[1−2 sin2 ( A)] 2

1+[2 cos2 ( A)−1]

1 cot 2 ( A) 2

=

2

√3 2

✓

A−1

2

1

1+cos A

2

2

1−tan2 A tan A

=

2 1 2

1+cos A

cos A = √ tan 2A

2

cos 2 4𝐴

= √1 − (− )

= 2 csc 2A = RHS [proven] ✓

LHS = 2 cot 2A =

LHS =

2

2

=√

1 2

1+(− ) 2

= cot A − tan A = RHS [proven] ✓ 5(e)

√3 2

2

RHS [proven] ✓ 5(d)

y

2

1 2

1 ) cos2 A sin A ( ) cos A

2

1

r

= (− ) − ( )

tan A

1

x

2 sin 2A

(

cos2 A

𝑥

cos A = = − , tan A = = −√3

sec2 A

=(

2 A

√3 2

= 2 csc 2A = RHS [proven] ✓ 5(b)

√3 2

x = −√(2)2 − (√3) = −1

LHS = sec A csc A =(

Coordinates obtuse A ⇒ 2𝑛𝑑 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡

1

1+cos A

2

2

or cos A = −√

1

(

rej ∵ cos A > 0 2 1

for 45° < A < 90°

)

2

=√

1 2 1 2 sin2 ( A) 2

2 cos2 ( A)

1 2

2

1

=√

4

= RHS [proven] ✓

1

= ✓ 2


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357


Ex 13.2

Coordinates 2 tan A = 3 3π π
A

-3 -2

8(i)

Coordinates 𝑎𝑐𝑢𝑡𝑒 𝜃 ⇒ 1𝑠𝑡 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡 1

𝑦

k

𝑥

cot θ = k⇒ tan θ = =

𝑟

𝑟 θ 𝑘

𝑦 = 1, 𝑥 = 𝑘,

2

1

r = √12 + k 2 = √1 + 𝑘 2 r=

√(−2)2

sin 𝐴 =

𝑦 𝑟

(−3)2

+

=−

2 √13

= √13 𝑥

, cos 𝐴 =

𝑟

=−

𝑦

sin 𝜃 =

3

𝑟

=

1 √1+𝑘 2

, cos 𝜃 =

𝑥

𝑘

=

𝑟

√1+𝑘 2

√13

Trigonometric ratio Trigonometric ratio tan 2A = 7(ii) 7(iii)

tan 3A =

2 tan A 1−tan2 A

=

2 2( ) 3 2 2 1−( ) 3

tan 2A+tan A 1−tan 2A tan A

=

tan A

sin 2θ = 2 sin θ cos θ = 2 ( =

1 2 1 1−tan2 A 2

12 5

2 3 12 2 1−( )( ) 5 3

=

1

6 tan A

tan A = 2 1

−3−√13

2

2

8(ii) =−

46 9

1−tan θ tan 45° tan θ+tan 45°

8(iii)

2

sec 2θ = =

1 2 tan2 A 2

0 =

0

or tan A =

8(iv)

−3±√13 2

2

7(iv)

Recall sin 𝐴 = − csc 2A = =

1 sin 2A 1 12 13

=

2 √13

= 13 12

rej ∵ tan A < 0 π

2 1

2

2

for < A <

, cos 𝐴 = − 1

2 sin A cos A

=

3π

=

) ✓

1 k

1−( )(1) 1 k

( )+(1)

1

=

cos 2θ

1

2k 1+k2

1 cos2 θ−sin2 θ

=

1+k2 k2 −1

1

=

2

k 1 ( ) −( ) 2 √1 + k √1 + k 2 1

✓

1+k

1

k2 −1 1+k2

=

tan θ+tan 45° 1−tan θ tan 45°

k−1

=

2

k2 1 + k2

−

1 1 + k2

✓

tan 2θ+tan θ 1−tan 2θ tan θ 2 tan θ )+tan θ 1−tan2 θ 2 tan θ 1−( ) tan θ 1−tan2 θ

(

=

2 tan θ+(tan θ−tan3 θ) (1−tan2 θ)−2 tan2 θ

4

3 √13

=

1 −2 −3 2( )( ) √13 √13

✓ = 9(a)

3 tan θ−tan3 θ

3 1 − k k3 3 1− 2 k

LHS = 1 2 1 cos A 2

9(b)

9(c)

sleightofmath.com

=

3k2 −1

=

k3 −3k

sin A 1+cos A

1 3 k 1 2 1−3( ) k 1 k

3( )−( )

=

3k2 −1

✓

k(k2 −3) 1 2

1 2

2 sin A cos A 1 2

1+(2 cos2 A−1)

=

1 2

1 2

2 sin A cos A 1 2

2 cos2 A

=

1

= tan A = RHS [proven] ✓

LHS = =

=

1−3 tan2 θ

sin A


) =

2 1

(

k √1+k2

tan 3θ = tan(2𝜃 + 𝜃) =

−3+√13

=

tan(θ+45°)

=

3

1

1

cot(θ + 45°) =

✓

3

=

2(1)

)(

✓

2

=2−

2 1 21 2 tan A + 6 tan A − 2 = 2 2 1 21 tan A + 3 tan A − 1 = 2 2 1 −3±√(3)2 −4(1)(−1)

tan A =

✓

5

( )+( )

=

2 tan A

12

1 √1+k2

2

1+tan2 A 1−tan2 A

cos2 A+sin2 A cos2 A−sin2 A

=

=

sin2 A cos2 A sin2 A 1− 2 cos A

1+

1 cos 2A

=

cos2 A+sin2 A cos2 A cos2 A−sin2 A cos2 A

= sec 2A = RHS [proven]✓

LHS = cos 4 A − sin4 A = (cos 2 A + sin2 A) (cos 2 A − sin2 A) (cos 2A) =1 = cos 2A = RHS [proven] ✓ 358


9(e)

LHS = cos 4A = cos(2 ⋅ 2𝐴) = 1 − 2 sin2 2A = 1 − 2 (sin 2A)2 = 1 − 2(2 sin A cos A)2 = 1 − 2(4 sin2 A cos 2 A) = 1 − 8 sin2 A cos 2 A = 1 − 8 sin2 A (1 − sin2 A) = 1 − 8 sin2 A + 8 sin4 A = 8 sin4 A − 8 sin2 A + 1 = RHS [proven] ✓

Ex 13.2 10(a) 4 sin 2x 4(2 sin x cos x) 8 sin x cos x sin x (8 cos x − 1)

= = =

0° < x < 360°


𝑦 1

90° 180° 270° 360°

0° < x < 360°

𝑥

S

A α α T C x = α, 360 − α = 82.8°, 277.2° ✓

−1

x = 180° ✓

1−cos 2A 2

)

2 1−2 cos 2A+cos2 2A 4

10(b) 4 sin x cos x = 1 4(

4 2−4 cos 2A+1+cos 4A

2 sin 2x

=1

sin 2x

=

8

= (3 − 4 cos 2A + cos 4A) = RHS [proven] ✓ 1+cos 2A+sin 2A

1 + (2 cos 2 A − 1) + (2 sin A cos A) 1 − (1 − 2 sin2 A) + (2 sin A cos A)

=

2 cos 2 A + 2 sin A cos A 2 sin2 A + 2 sin A cos A

=

2 cos A (cos A + sin A) 2 sin A (sin A + cos A)

= cot A

)

=1 1 2

0° < x < 360° 0° < 2x < 720°

1−cos 2A+sin 2A

=

2

α = 30° ⇒ 1st or 2nd quadrant

8

LHS =

sin 2x

1+cos 4A 1−2 cos 2A+( ) 2

1

9(f)

1

or cos x = 8

sin x = 0

LHS = (sin2 A)2 =(

= sin x = sin x = sin x =0

S

A

α T

α C

2x = α, 180 − α, = 30, 150, x = 15°, 75°,

360 + α, 540 − α 390, 510 195°, 255° ✓

10(c) (sin x − cos x)2 =2 sin2 x − 2 sin x cos x + cos 2 x = 2 (sin2 x + cos 2 x) − 2 sin x cos x = 2 1 − sin 2x =2 sin 2x = −1 𝑦

0° < x < 360° 0° < 2x < 720°

= RHS [proven] ✓

1

90° 180° 270° 360°

𝑥

−1

2x = 270, x = 135°,


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270 + 360 315° ✓

359


Ex 13.2

10(d) cos 2x − 3 cos x + 2 =0 (2 cos 2 x − 1) − 3 cos x + 2 = 0 2 cos 2 x − 3 cos x + 1 =0 (2 cos x − 1)(cos x − 1) =0 1 cos x = or cos x = 1 2 0° < x < 360 α = 60° 𝑦 ⇒ 1st or 4th quad. 1 0° < x < 360°

90° 180° 270° 360°

S

A α α T C x = α, 360 − α = 60°, 300° ✓

𝑦 1

1

𝑥

90° 180° 270° 360°

⇒ 𝑛𝑜 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

⇒ x = 0, π ✓

⇒ x = 0, 2π ✓

cos x [5 (

sin 2𝑥 2

)

5

4

0 ≤ x ≤ 2π

α ≈ 0.927 ⇒ 3rd or 4th quadrant

𝑦 1

2 90° 180° 270° 360°

(no solution)

𝑥

−1

π 3π

x= , 2

2

2 tan x

1−tan2 x 2

11(c) cos 2x 1 − 2 sin2 x 2 sin2 x − sin x − 1 (2 sin x + 1)(sin x − 1)

=3

) tan x = 3 =3 = 3 − 3 tan2 x =3

tan2 x

=

sin x = − α=

3

= ±√

3 5

α ≈ 37.8° ⇒ 1st, 2nd, 3rd or 4th quadrant

2

or

S α α T

A α α C

α T

α C

= − sin x = − sin x =0 =0

sin x = 1

,

𝑦 1

𝑥

−1

7π 11π 6

3π + α, 4π − α 10.4, 11.6 5.18, 5.82 ✓

90° 180° 270° 360°

A

α α T C x = π + α, 2π − α =

sleightofmath.com

A

0 ≤ x ≤ 2π

6

S

x = α, 180 − α, 180 + α, 360 − α = 37.8°, 142.2°, 217.8°, 322.2° ✓


π

1

⇒ 3rd or 4th quad. 0 ≤ x ≤ 2π

5

S

2x = π + α, 2π − α, ≈ 4.07,5.36, x ≈ 2.03, 2.68,

𝑥

2 tan x 5 tan2 x

tan x

0 ≤ x ≤ 2π 0 ≤ 2x ≤ 4π

✓

⇒ x = 180° ✓

1−tan2 x 2 tan2 x

=0

or sin 2x = − 5

cos x = 0

3

=0 =0

=0

2

−1

(

+ 2]

cos x ( sin 2x + 2)

𝑦

tan 2x tan x

𝑥

−1

sin x = −

or

90° 180° 270° 360°

−1

1

10(f)

𝑥

−1

= 3 sin x = 3 sin x = 3 sin x = 2 sin2 𝑥 + 3 sin 𝑥 =0 =0

0° < x < 360°

90° 180° 270° 360°

𝑦

11(b) 5 sin x cos 2 x + 2 cos x cos x (5 sin x cos x + 2)

10(e) cos 2x − 1 (1 − 2 sin2 x) − 1 −2 sin2 x 0 2 sin2 x + 3 sin x sin x (2 sin x + 3) sin x = 0

11(a) 2 sin x = sin 2x 2 sin x = 2 sin x cos x sin x = sin x cos x sin x (cos x − 1) = 0 sin x = 0 or cos x = 1 0 ≤ x ≤ 2π 0 ≤ x ≤ 2π

6

π

⇒x = ✓ 2

✓

360


Ex 13.2

11(d) 2 cos 2x = 11 cos x + 1 2 2(2 cos x − 1) = 11 cos x + 1 4 cos 2 x − 2 = 11 cos x + 1 2 4 cos x − 11 cos x − 3 = 0 (4 cos x + 1)(cos x − 3) = 0 1 cos x = 3 cos x = − 4 (no solution) α ≈ 1.32 ⇒ 2nd or 3rd quadrant

12(i)

At x-axis, y =0 2 3 sin 2x + 2 cos x =0 3(2 sin x cos x) + 2 cos 2 x = 0 6 sin x cos x + 2 cos 2 x =0 2 3 sin x cos x + cos x =0 cos x (3 sin x + cos x) =0 cos x = 0 3 sin x = − cos x

0 ≤ x ≤ 2π

𝑦

S A α α T C x = π − α, π + α = 1.82, 4.46 ✓ 11(e) tan 2x

90° 180° 270° 360°

𝑥

−1

0° < x < 360°

=−

2 tan x = 3 tan x − 3 tan3 x 3 tan3 x − tan x =0 2 tan x (3 tan x − 1) = 0 tan x = 0 or tan2 x = 1 𝑦

tan x = ±

1

𝑥

−1

α≈

π

3 1

√3

6

0 ≤ x ≤ 2π

⇒ 1st, 2nd, 3rd or 4th quadrant

x = 0, π, 2π ✓

0 ≤ x ≤ 2π S α α T

A α α C


6

S α

A

12(ii) y =2 2 3 sin 2x + 2 cos x =2 2 3(2 sin x cos x) + 2(1 − sin x) = 2 6 sin x cos x + (2 − 2 sin2 x) = 2 6 sin x cos x − 2 sin2 x =0 2 3 sin x cos x − sin x =0 sin x (3 cos x − sin x) =0 sin x = 0 3 cos x = sin x 𝑦 tan x = 3 1 α ≈ 71.6° 𝑥 ⇒ 1st or 3rd quadrant 90° 180° 270° 360° −1

π 5π 7π 11π 6

3


x = 180 − α, 360 − α = 161.6°, 341.6° ✓

0° < x < 360°

x = α, π − α, π + α, 2π − α = ,

1

α T C x = α, 360 − α = 90°, 270° ✓ 0° < x < 360°

= 3 tan x

1−tan2 x

tan x

1

= 3 tan x

2 tan x

90° 180° 270° 360°

y = 3 sin 2x + 2 cos 2 x

,

6

,

6

0° < x < 360° S

x = 180° ✓

✓

sleightofmath.com

A α

α T C x = α, 180 + α = 71.6°, 251.6° ✓

361


Ex 13.2 14(ii) 8 cos 3 x 8 cos 3 x − 6 cos x 2(4 cos 3 x − 3 cos x) 2 cos 3x

a = cos x + sin x b = cos x − sin x a + b = 2 cos x

a+b

⇒ cos x =

a − b = 2 sin x

⇒ sin x =

2

cos 3x

=

a+b 2 2

)

−(

a2 +2ab+b2 4

=−

α = 60° ⇒ 2nd or 3rd quadrant

a−b 2

0° < x < 180° 0° < 3x < 540°

cos 2x = cos 2 x − sin2 x =(

= 6 cos x − 1 = −1 = −1 = −1

a−b 2

1 2

S α α T

A C

)

−

2 a2 −2ab+b2

3x = 180 − α, 180 + α, = 120, 240, x = 40°, 80°,

4

4ab 4 = ab ✓ =

540 − α 480 160° ✓

sin 2x = 2 sin x cos x = 2(

a−b 2

)(

a+b 2

) = 2(

a2 −b2 4

) =

a2 −b2 2

✓

13(ii) a2 +b2 = (cos x + sin x)2 +(cos x − sin x)2 = (cos 2 x + 2 sin x cos x + sin2 x) +(cos 2 x − 2 sin x cos x + sin2 x) = 2 cos 2 𝑥 + 2 sin2 𝑥 = 2(cos 2 𝑥 + sin2 𝑥) = 2 [proven] ✓ 14

14(i)

LHS = cos 3x = cos(2x + x) = cos 2x cos x = (2 cos 2 x − 1) cos x = 2 cos 3 x − cos x = 2 cos 3 x − cos x = 2 cos 3 x − cos x = 4 cos 3 x − 3 cos x = RHS [shown] ✓

− sin 2x sin x −(2 sin x cos x) sin x −2 cos x (sin2 x) −2 cos x (1 − cos 2 x) −2 cos x + 2 cos 3 x

8 cos 3 10° − 6 cos 10° 3 = 2(4 cos 10° − 3 cos 10°) = 2 cos 30° √3 2

= 2( ) = √3 ✓


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362


Ex 13.2

Prove: tan x = cot x − 2 cot 2x

16 Prove: cos 6 θ + sin6 θ =

RHS = cot x − 2 cot 2x 2 = cot x − tan 2x 2 = cot x − 2 tan x = cot x −

1−tan2 x 1−tan2 x

tan x

= −(sin θ cos θ)2 + (

tan x = cot x − 2 cot 2x 2 cot 2x = cot x − tan x =

8

LHS = cos 6 θ + sin6 θ = (cos 2 θ)3 + (sin2 θ)3 = [cos 2 θ + sin2 θ] [(cos 2 θ)2 − (cos 2 θ)(sin2 θ) + (sin2 θ)2 ] [(cos 2 θ)2 − sin2 θ cos 2 θ + =1 (sin2 θ)2 ] = − sin2 θ cos 2 θ +(cos 2 θ)2 +(sin2 θ)2

= cot x − cot x + tan x = tan x = LHS [proven] ✓

cot 2x

5+3 cos 4θ

1

= − ( sin 2θ)

cot x−tan x

2

2

1

2

2

= − sin 2θ

+

4

Show: 2 tan 20° + 4 tan 40° + 8 tan 80° = 9(cot 10° − tan 10°)

1 1−cos 4θ

=− ( =

LHS = 2 tan 20° +4 tan 40° +8 tan 80° = 2(cot 20° − 2 cot 40°) +4(cot 40° − 2 cot 80°) +8(cot 80° − 2 cot 160°)

= = =

4 2 cos 4θ−1

+

)

8 cos 4θ−1 8 cos 4θ−1 8 5+3 cos 4θ

+ + + +

1+cos 2θ 2 2

)

1+2 cos 2θ+cos2 2θ 4 2+2 cos2 2θ

+( +

1−cos 2θ 2

)

2

1−2 cos 2θ+cos2 2θ 4

4 2+2(

1+cos 4θ ) 2

4 2+1+cos 4θ 4 3+cos 4θ 4 6+2 cos 4θ 8

8

= RHS [proven] ✓

= 2 cot 20° −4 cot 40° +4 cot 40° −8 cot 80° +8 cot 80° −16 cot 160° = 2 cot 20° −16 cot 160° = 2 cot 20° − = 2 cot 20° −

16 tan 160° 16 − tan 20°

= 2 cot 20° +16 cot 20° = 18 cot 20° = 18 (

cot 10°−tan 10° 2

)

∵ cot 2x =

cot x−tan x 2

= 9(cot 10° − tan 10°) = RHS [shown] ✓


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363


Ex 13.2

3 +4 cos 2 x = 3 +4 (

1+cos 2x 2

)

= 3 +2 + 2 cos 2x = 2 cos 2x + 5 ✓ 17(ii) y = 3 + 4 cos 2 x = 2 cos 2x + 5

𝑦 7 5.5 5 4 3 2.5

i.e. 𝑎 = 2, 𝑏 = 2, 𝑐 = 5 Amplitude = |𝑎| = |2| = 2 Period

=

2𝜋 𝑏

=

2𝜋 2

=𝜋

𝑂

𝑦 = 3 + 4 cos 2 𝑥

𝑦 = 4 − 3 sin 𝑥 cos 𝑥 𝜋

𝑥

2𝜋

17(iii) y = 4 − 3 sin x cos x 1

= 4 − 3 ( sin 2x) 2

3

= − sin 2x + 4 2

3

i.e. 𝑎 = − , 𝑏 = 2, 𝑐 = 4 2

3

3

2

2

Amplitude = |𝑎| = |− | = Period

=

2𝜋 𝑏

=

2𝜋 2

=𝜋

2

17(iv) 4 cos 𝑥 + 3 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 − 1 = 0 4 𝑐𝑜𝑠 2 𝑥 = 1 − 3 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 2 3 + 4 𝑐𝑜𝑠 𝑥 = 4 − 3 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 4 intersections ⇒ 4 sol ✓ Workings 𝑦1 = 2 cos 2𝑥 + 5 Domain 0 ≤ x ≤ 2π Axis with y=5±2 Amplitude Shape +cos 2π−0 Cycle =2 π

𝑦2 = −4 − 3 sin 𝑥 cos 𝑥 Domain 0 ≤ x ≤ 2π 3 Axis with y=4± 2 Amplitude Shape −sin 2π−0 Cycle =2 π


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364


R=

1 2 v sin θ cos θ 4.9

R|v=16 1 4.9

(16)2

2

19

x Show: √2 + √2 + 2 cos x = 2 sin

4

LHS = √2 + √2 + 2 cos x

= 25 sin θ cos θ = 25

sin θ cos θ 1

Ex 13.2

sin 2𝜃

sin 2θ

= = =

1

= √2 + √2 + 2 [2 cos 2 ( x) − 1]

245

2

512 245 512 490

1

= √2 + √2 + 4 cos 2 ( x) − 2 2

512

α ≈ 73.1° ⇒ 1st or 2nd quadrant 0° < θ < 90° 0° < 2θ < 180°

1

S

A

α T

α C

= √2 + √4 cos 2 ( x) 2

1

= √2 + 2√cos 2 ( x) 2

2θ = α, 180 − α ≈ 73.1,106.9 θ ≈ 36.6°, 53.4° ✓

1

= √2 + 2 [− cos ( x)] 2

1

1

2

2

∵ cos x < 0 for 135° < 𝑥 < 180° 1

= √2 − 2 cos ( x) 2

1

= √2 − 2 [1 − 2 sin2 ( x)] 4

1

= √2 − 2 + 4 sin2 ( x) 4

1

= √4 sin2 ( x) 4

1

= 2√sin2 ( x) 4

= 2 sin

x 4 1

1

4

4

∵ sin ( 𝑥) > 0 𝑓𝑜𝑟 67.5° < 𝑥 < 90° = RHS [shown] ✓


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365


Ex 13.3 4(a)

Ex 13.3 1(a)

 R = √32 + 42 = 5 4  α = tan−1 ( ) ≈ 53.1° 3 ∴ y = 5 sin(θ − 53.1°)

cos θ + sin θ  

R = √12 + 12 = √2 1 α = tan−1 ( ) = 45° 1

max = 5 ✓ ⇒ sin(θ − 53.1°) = 1 0° < θ < 360° −53.1 < θ − 53.1° < 306.9° θ − 53.1° = 90° θ ≈ 143.1° ✓

∴ cos θ + sin θ = √2 cos(θ − 45°) ✓ 1(b)

√3 cos θ − sin θ 2



R = √(√3) + 12 = √3 + 1 = 2



α = tan−1 ( ) = 30°

1

√3

∴ √3 cos θ − sin θ = 2 cos(θ + 30°) ✓ 1(c)

R = √32 + 42 = 5 (4) α = tan−1 [(3)] ≈ 53.1°

∴ 3 sin θ + 4 cos θ = 5 sin(θ + 53.1°) ✓ 1(d)

4(b)



2

R = √12 + (√2) = √1 + 2 = √3 α=

√2 tan−1 ( ) 1

 3(i)

α≈

1

≈ 1.18 ✓

𝑦 1

90° 180° 270° 360°

𝑥

−1

I = 15 sin(120πt) − 8 cos(120πt)  

Min = −√10 ✓ θ + 71.6° ≈ 180° θ ≈ 108.4° ✓

R = √152 + 82 = 17 8 α = tan−1 ( ) ≈ 0.489 96 15

∴ I = 17 sin(120πt − 0.489 96) ✓ 3(ii)

−1

R = √12 + 32 = √10 3 α = tan−1 ( ) ≈ 71.6°

Max = √10 ✓ ⇒ cos(θ + 71.6°) = 1 0° < θ < 360° 71.6° < θ + 71.6° < 431.6° θ + 71.6° ≈ 0°, 360° θ ≈ 288.4° ✓

R = √52 + 122 = 13 ✓ 12 tan−1 ( ) 5

𝑥

= 54.7°

5 sin θ − 12 cos θ = R sin(θ − α) 

90° 180° 270° 360°

∴ y = √10 cos(θ + 71.6°)

∴ sin θ − √2 cos θ = √3 sin(θ − 54.7°) ✓ 2

𝑦 = sin 𝑥

y = cos θ − 3 sin θ  

sin θ − √2 cos θ 

𝑦 1

min = −5 ✓ ⇒ sin(θ − 53.1°) = −1 θ − 53.1° = 270° θ ≈ 323.1° ✓

3 sin θ + 4 cos θ  

y = 3 sin θ − 4 cos θ

A = 17 ✓


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366


Ex 13.3

y = 6 cos θ + 5 sin θ  

√62

5(a)

3 cos x − 4 sin x = 1  R = √32 + 42 = 5 4  α = tan−1 ( ) ≈ 53.1° 3 ∴ 5 cos(x + 53.1°)= 1

52

R= + = √61 −1 (5) α = tan ≈ 39.8° 6

∴ y = √61 cos(θ − 39.8°)

cos(x + 53.1°) max = √61 ✓ ⇒ cos(θ − 39.8°) = 1 0° < θ < 360° −39.8° < θ − 39.8° < 320.1° θ − 39.8° ≈ 0° θ ≈ 39.8° ✓

𝑦

90° 180° 270° 360°

𝑥

T

A α α C

x + 53.1° ≈ α, 360 − α ≈ 78.5,281.5 x ≈ 25.3°, 228.4° ✓ 5(b)

√3 sin x − cos x = 1 2



R = √(√3) + 12 = √3 + 1 = 2



α = tan−1 ( ) = 30°

1

√3

∴ 2 sin(θ − 30°) = 1

3

sin(θ − 30°)

∴ y = 3√5 sin(θ + 63.4°)

min = −3√5 ✓ ⇒ sin(θ + 63.4°) = −1 x + 63.4° ≈ 270° x ≈ 206.6° ✓

S

0° < x < 360° 53.1° < x + 53.1° < 413.1°

−1

R = √32 + 62 = √45 = √9 × 5 = 3√5 6 α = tan−1 ( ) ≈ 63.4°

max = 3√5 ✓ ⇒ sin(θ + 63.4°) = 1 0° < θ < 360° 63.4° < θ + 63.4° < 423.4° x + 63.4° ≈ 90° x ≈ 26.6° ✓

5

1

y = 3 sin θ + 6 cos θ  

1


min = −√61 ✓ cos(θ − 39.8°) = −1 θ − 39.8° ≈ 180° θ ≈ 219.8° ✓ 4(d)

=

1 2

α = 30° ⇒ 1st or 2nd quadrant

𝑦 1

=

𝑦 = sin 𝑥 90° 180° 270° 360°

0° < x < 360° −30° < x − 30° < 330°

𝑥

−1

S

A

α T

α C

x − 30° = α, 180 − α = 30,150 x = 60°, 180° ✓ 5(c)

6 cos x − 2 sin x = 3.5  

R = √62 + 22 = √40 = √4 × 10 = 2√10 2 α = tan−1 ( ) ≈ 18.4° 6

∴ √10 cos(x + 18.4°) = 3.5 cos(x + 18.4°)

=

3.5 2√10

α ≈ 56.4° ⇒ 1st or 4th quadrant 0° < x < 360° 18.4° < x + 18.4° < 378.4°

S T

A α α C

x + 18.4° ≈ α, 360 − α ≈ 56.4,303.6 x ≈ 38.0°, 285.2° ✓


sleightofmath.com

367


Ex 13.3 6(i)

sin x + 2 cos x = √2  

√12

R= + = √5 −1 (2) α = tan ≈ 63.4° 1

=√

R = √82 + 62 = 10 6 α = tan−1 ( ) ≈ 0.643 50 8

∴ G(t) = 10 cos(4t − 0.643 50) ✓

∴ √5 sin(x + 63.4°) = √2 sin(x + 63.4°)

G(t) = 8 cos 4t + 6 sin 4t  

22

2

6(ii)

5

α ≈ 39.2° ⇒ 1st or 2nd quadrant 0° < x < 360° 63.4° < x + 63.4° < 423.4°

S

A

α T

α C

−1 ≤ cos(4t − 0.643 50) ≤1 −10 ≤ 10 cos(4t − 0.643 50) ≤ 10 min = −10 ✓ 10 cos(4t − 0.643 50) = −10 cos(4t − 0.643 50) = −1

x + 63.4° ≈ α, 180 − α, 360 + α ≈ 39.2,140.8, 399.2 x ≈ 77.3°, 335.8° ✓

𝑦

t >0 4t − 0.643 50 > −0.643 50

1

90° 180° 270° 360°

𝑥

−1

5(e)

4t − 0.643 50 = π t ≈ 0.946 ✓

2.1 cos x − sin x = 1.6 

R = √(2.1)2 + (1)2 = 1

 ∴

α = tan−1 ( ) 2.1

1

1 10

√541

≈ 25.5°

7(i)

√541 cos(x + 25.5°) = 1.6 10

cos(x + 25.5°)

=

16

4

√541


S

0° < x < 360° 25.5° < x + 25.5° < 385.5°

= 5(1 + 0) = 5 coulombs ✓

A α α C

T

7(ii)

16

R = √12 + ( ) = √



α = tan−1 ( 4 ) ≈ 0.245

=

√17 4

1

=

 R= + ≈ 17.26 −1 ( e )  α = tan ≈ 40.9° π ∴ 17.26 cos(x − 40.9°) ≈ 2 ≈

17

4



∴ q=

e2

cos(x − 40.9°)

1 2

1

π cos x + e sin x = 2 √π2

1

q= 5e−10t (cos 60t + sin 60t) 4

x + 25.5° ≈ α, 360 − α ≈ 46.5,313.5 x ≈ 21.1°, 288.0° ✓ 5(f)

1 q = 5e−10t (cos 60t + sin 60t) 4 1 −10(0) (cos q|t=0 = 5e 0 + sin 0)

√17 5e−10t [ cos(60t 4 5√17 −10t

⇒A =

e

4 5√17 4

− 0.245)]

cos(60t − 0.245)

✓

⇒ B = 0.245 ✓ 2

7(iii)

17.26

α ≈ 61.2° ⇒ 1st or 4th quadrant 0°
S T

A α α C

t →∞ −10t e →0 q → 0✓

x − 40.9° ≈ α, 360 − α ≈ 61.2,298.8 x ≈ 102.1°, 339.6° ✓


sleightofmath.com

368


Ex 13.3

y = 4 sin θ + 3 cos θ − 2

8(c)

 R= + =5 −1 (3)  α = tan ≈ 36.9° 4 ∴ y = 5 sin(θ + 36.9°) − 2 √42

32

−1 ≤ sin(θ + 36.9°) −5 ≤ 5 sin(θ + 36.9°) −7 ≤ 5 sin(θ + 36.9°) − 2

≤1 ≤5 ≤3

max = 3 ✓ ⇒ sin(θ + 36.9°) = 1 0° < θ < 360° 36.9° < θ + 36.9° < 399.9° θ + 36.9° ≈ 90° θ ≈ 53.1° ✓

R = √12 + (√2) = √3



α = tan−1 ( )

√2 1

≈ 54.7°

−1 ≤ sin(θ − 54.7°) ≤1 −√3 ≤ √3 sin(θ − 54.7°) ≤ √3 3√3 ≤ √3 sin(θ − 54.7°) ≤ 5√3 𝑦

max = 5√3 ✓ ⇒ sin(θ − 54.7°) = 1 0° < θ < 360° −54.7° < θ − 54.7° < 305.3° θ − 54.7° ≈ 90° θ ≈ 144.7° ✓ 𝑦

𝑦 = sin 𝑥

1

90° 180° 270° 360°

𝑥

−1

𝑦 = sin 𝑥

1

min = 3√3 ✓ ⇒ sin(θ − 54.7°) = −1 θ − 54.7° ≈ 270° θ ≈ 324.7° ✓

y = 3 − 7 cos θ + 24 sin θ = 3 + (24 sin θ − 7 cos θ)  R = √242 + 72 = 25 7  α = tan−1 ( ) ≈ 16.3° 24 ∴ y = 3 + 25 sin(θ − 16.3°)

8(d)

y = 7 + 15 cos 2θ + 8 sin 2θ = 7 + (15 cos 2θ + 8 sin 2θ)

max = 24 ✓ ⇒ cos(2θ − 28.1°) = 1

𝑦 1

90° 180° 270° 360°

𝑦 = sin 𝑥 90° 180° 270° 360°

𝑥

−1 ≤ cos(2θ − 28.1°) ≤1 −17 ≤ 17 cos(2θ − 28.1°) ≤ 17 −10 ≤ 17 cos(2θ − 28.1°) + 7≤ 24

max = 28 ✓ ⇒ sin(θ − 16.3°) = 1 0° < θ < 360° −16.3° < θ − 16.3° < 343.7° θ − 16.3° ≈ 90° 𝑦 θ ≈ 106.3° ✓ 1

90° 180° 270° 360° −1

 R = √152 + 82 = 17 8  α = tan−1 ( ) ≈ 28.1° 15 ∴ y = 7 + 17 cos(2θ − 28.1°)

−1 ≤ sin(θ − 16.3°) ≤1 −25 ≤ 25 sin(θ − 16.3°) ≤ 25 −22 ≤ 3 + 25 sin(θ − 16.3°) ≤ 28

min = −22 ✓ sin(θ − 16.3°) = −1 θ − 16.3° ≈ 270° θ ≈ 286.3° ✓

2



∴ y = √3 sin(θ − 54.7°) + 4√3

min = −7 ✓ ⇒ sin(θ + 36.9°) = −1 θ + 36.9° ≈ 270° θ ≈ 233.1° ✓ 8(b)

y = sin θ − √2 cos θ + 4√3 = (sin θ − √2 cos θ) + 4√3

𝑥

−1

𝑥

0° < θ < 360° −28.1° < 2θ − 28.1° < 691.9° 2θ − 28.1° = 0°, 360° θ = 14.0°, 194.0° ✓

−1

min = −10 ✓ ⇒ cos(2θ − 28.1°) = −1 2θ − 28.1° ≈ 180°, 180° + 360° θ ≈ 104.0°, 284.0° ✓


sleightofmath.com

369


Ex 13.3

πx

πx

6

6

10(a)

h = 1.6 + 0.5 sin ( ) − 1.2 cos ( )  

∴ h = 1.6 + 1.3 sin ( −1

≤ sin (

πx 6

6

≤ 1.3 sin (

πx 6 πx 6

 

− 1.18)

− 1.18)

−1.3 ≤ 1.3 sin ( 0.3

πx

≤ 1.3

sin (

πx 6

πx 6

=

4 13

α ≈ 0.313 ⇒ 1st or 2nd quadrant 0
S

A

α T

α C

 

α C

2

∴√5 cos(2x − 26.6°) = −2 =−

2 √5

α ≈ 26.6° ⇒ 2nd or 3rd quadrant

− 1.18

≈ α, π − α, 2π + α, 3π − α ≈ 0.313,2.83, 6.60,9.11 x ≈ 2.84h, 7.64h, 14.8h, 19.6h ≈ 2h 51 min, 7h 39 min 14h 51min, 19h 39min ⇒ 02: 51, 07: 39, 14: 51, 19: 39 ✓


α T

R = √22 + 12 = √5 1 α = tan−1 ( ) ≈ 26.6°

cos(2x − 26.6°)

6

A

10(b) 2 cos 2x + sin 2x + 2 = 0 2 cos 2x + sin 2x = −2

6

πx

S

x − 56.3° ≈ α, 180 − α ≈ 56.3,123.7 x ≈ 112.6°, 180° ✓

− 1.18) = 2

− 1.18)

3 √13

0°
1.6 + 1.3 sin (

=


− 1.18) + 1.6 ≤ 2.9

h

2

sin(x − 56.3°)

⇒ max. = 2.9 ✓ ⇒ min. = 0.3 ✓ 9(ii)

R = √22 + 32 = √13 3 α = tan−1 ( ) ≈ 56.3°

∴√13 sin(x − 56.3°) = 3

≤1

− 1.18)

=2

sin x+cos x

4 sin x − cos x − 3 = 2 sin x + 2 cos x 2 sin x − 3 cos x = 3

R = √0.52 + 1.22 = √1.69 = 1.3 1.2 α = tan−1 ( ) ≈ 1.18 0.5

4 sin x−cos x−3

sleightofmath.com

0°
S α α T

A C

540 − α, 540 + α 513.4, 566.6 270°, 296.6° ✓

370


Ex 13.3

10(c) sin 1 x cos 1 x +2 cos x = 1 2

1 2

10(e) 2 sin x(2 sin x − cos x) = 1 4 sin2 x −2 sin x cos x = 1

2

sin x

+2 cos x = 1

4(

1 2

17

2

4



R = √( ) + 22 = √



α = tan−1 ( 1 ) ≈ 76.0°

=

√17 2

2

∴

√17 sin(x 2

+ 76.0°) = 1

sin(x + 76.0°)

=

√17

x + 76.0° ≈ α, 180 − α, ≈ 29.0,151.0 x ≈ 75.0°,

cos x sin x

S

A

= 3 + 2(

α T

α C

1 sin x

x ≈ 45°, ✓ 11(i)

=

1 √5

S

A

α T

α C

360 + α, 540 − α,

720 +

386.6, 513.4,

746.6

161.6°, 225°,

341.6°

V = 70 sin t + 85 cos t 

R = √702 + 852 ≈ 110.11 ✓ 85

 α = tan−1 ( ) ≈ 0.881 87 ✓ 70 ∴ V = 110.11 sin(t + 0.881 87)

2

∴ √13 cos(x + 56.3°) = 2 =

1

2x + 63.4° ≈ α, 180 − α, α ≈ 26.6, 153.4,

)

R = √22 + 32 = √13 3 α = tan−1 ( ) ≈ 56.3°

cos(x + 56.3°)

R = √12 + 22 = √5 2 α = tan−1 ( ) ≈ 63.4°

0° < x < 360° 63.4° < 2x + 63.4° < 783.4°

360 + α, 540 − α 389.0, 511.0 313.1° ✓

2 cos x = 3 sin x + 2 2 cos x − 3 sin x = 2  

=1 =1


= 3 + 2 csc x )

2 − 2 cos 2x − sin 2x sin 2x + 2 cos 2x

sin(2x + 63.4°)

0° < x < 360° 76.0° < x + 76.0° < 436.0°

2(

=1

2

∴ √5 sin(2x + 63.4°) = 1

2


10(d) 2 cot x

) − sin 2x

 

2

1−cos 2x

2 √13

α ≈ 56.3° ⇒ 1st or 4th quadrant 0° < x < 360° 56.3° < x + 56.3° < 416.3°

S T

A α α C

11(ii) −1 ≤ sin t ≤1 −110.11 ≤ 110.11 sin(t + 0.881 87) ≤ 110.11 ⇒ max = 110.11 ✓ 110.11 sin(t + 0.881 87) = 110.11 sin(t + 0.881 87) =1

x + 56.3° ≈ α, 360 − α, 360 + α ≈ 56.3,303.7 416.3 x ≈ 247.4° ✓

t ≥0 t + 0.881 87 ≥ 0.881 87

𝑦 1

𝑦 = sin 𝑥 90° 180° 270° 360°

t + 0.881 87 ≈ t


sleightofmath.com

π

𝑥

−1

2

≈ 0.689 ms −1 ✓

371


Ex 13.3

11(iii) V = 50 110.11 sin(t + 0.881 87) = 50 sin(t + 0.881 87)

=

α ≈ 0.471 ⇒ 1st or 2nd quadrant t ≥0 t + 0.881 87 ≥ 0.881 87

13(i)

In △ QRS, cos θ° =

110.11

S

A

α T

α C

θ

SR 8

T

8

S

θ

R

𝑂

R = √22 + 82 = √68 = √4 × 17 = 2√17 8

α = tan−1 ( ) ≈ 76.0° 2

∴ OR = 2√17 sin(θ° + 76.0°) max = 2√17 ⇒ sin(θ° + 76.0°) = 1 0° < θ < 90° 76.0° < θ + 76.0° < 166.0° θ° + 76.0° = 90° θ° ≈ 14.0° ✓

10

∴ x = √149 cos(8t − 0.611) ✓ 12(ii) −1 ≤ cos t ≤1 −√149 ≤ √149 cos(8t − 0.611) ≤ √149

𝑦 1

90° 180° 270° 360°

𝑥

−1

13(ii) OR =6 2√17 sin(θ° + 76.0°) = 6

max. = √149 ✓

sin(θ ° + 76.0°) x

= √149 √149 cos(8t − 0.611) = √149 cos(8t − 0.611) =1 t ≥0 8t − 0.611 ≥ −0.611

=

3 √17

α ≈ 46.7° ⇒ 1st or 2nd quadrant 𝑦 1

90° 180° 270° 360°

0° < θ° < 90° 76.0° < θ° + 76.0°< 166.0°

𝑥

S

A

α T

α C

−1

θ° + 76.0° ≈ α, 180 − α ≈ 46.7,133.3 θ° ≈ 57.4° ✓

8t − 0.611 ≈ 0 t ≈ 0.0763 ✓ 12(iii) x

=9 √149 cos(8t − 0.611) = 9 =

14(i) 9

0
3 sin θ + 4 cos(60° − θ) =2 [ ] 3 sin θ + 4 cos 60° cos θ + sin 60° sin θ = 2 1

√149



P

Q

OR = OS +SR = PT +SR = (2 sin θ° +8 cos θ°) cm [shown] ✓

R = √102 + 72 = √149 7 α = tan−1 ( ) ≈ 0.611

cos(8t − 0.611)

2

2

⇒ SR = 8 cos θ°

x = 10 cos 8t + 7 sin 8t  

PT

⇒ PT = 2 sin θ°

50

t + 0.881 87 ≈ α, π − α ≈ 0.471,2.67 t ≈ 1.79 ms ✓ 12(i)

In △ PQT, sin θ° =

=2

3 sin θ + 2 cos θ + 2√3 sin θ (3 + 2√3) sin θ + 2 cos θ

=2 =2✓

2

S T

A α α C

√3 2

3 sin θ + 4 [( ) cos θ + ( ) sin θ]

2π + α 7.02 0.954 ✓

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372


Ex 13.3

14(ii) For (3 + 2√3) sin θ + 2 cos θ 

15(ii) OM

=4 √29 cos(θ − 68.2°) = 4

2

R = √(3 + 2√3) + 22

cos(θ − 68.2°)

= √(9 + 12√3 + 12) + 4

α = tan−1 (

2 3+2√3

∴ √25 + 12√3 sin(θ + 17.2°) = 2

15

=


S

A

α T

α C

tan θ = AC

16(i)

360 + α 377.2 360° ✓

= 2 cot θ

sin θ

=

sin θ

=

𝑀

𝑂

(5,2) 2θ 5 𝐶 𝐴

𝑥

OM OA OM 5+2 cot θ

cos θ sin θ

)

18 16

tan x

𝑦 1

90° 180° 270° 360°

𝑥

−1

= tan x =

9 8


R = √22 + 52 = √29 5 α = tan−1 ( ) ≈ 68.2°

0° < x < 90° x = α, 180 − α ≈ 48.4°, 131.6° ✓

2

∴ OM = √29 cos(θ − 68.2°) max = √29 ✓ ⇒ cos(θ − 68.2°) = 1 0° < θ < 90° −68.2° < θ − 68.2° < 28.8° θ − 68.2° = 0° θ = 68.2° ✓

2

16 area of △ COD = 2(area of △ AOB) (ii)(b) 18 cos x = 2(8 sin x) 18 cos x = 16 sin x

For 2 cos θ + 5 sin θ


+18 cos x [shown] ✓

Max = 2√97 ✓ ⇒ sin(x + 66.0°) = 1 0° < θ < 90° 66.0° < θ + 66.0° < 156.0° x + 66.0° = 90 x ≈ 24.0° ✓

= 5 sin θ +2 cos θ = 2 cos θ+5 sin θ [shown] ✓

 

= 8 sin x

8

sin θ (5 + 2 cot θ) = OM OM = sin θ (5 + 2 cot θ)

15(i)

+ (6)(6) sin(90° − x)

∴ S = 2√97 sin(x + 66.0°)

𝐵

= sin θ (5 + 2

1

= (4)(4) sin x

16 For 8 sin x + 18 cos x, (ii)(a)  R = √82 + 182 = 2√97 18  α = tan−1 ( ) ≈ 66.0°

𝑦

OA = OC +CA =5 +2 cot θ [shown] ✓

S = area of △ AOB +area of △ COD 2

2

T

θ − 68.2° ≈ −α, α, 360 − α ≈ −42.0 42.0, 318.0 Θ ≈ 26.2° ✓

√25+12√3

1

AC

A α α C

2

sin(θ + 17.2°)

θ + 17.2° ≈ α, 180 − α, ≈ 17.2,162.8, θ ≈ 0°, 146°,

S

0° <θ < 90° −68.2° < θ − 68.2° < 28.8°

) ≈ 17.2°

0° ≤ θ ≤ 360° 17.2° ≤ θ + 17.2° ≤ 377.2°

√29


= √25 + 12√3 

4

=

S α T

A α C

𝑦 1

90° 180° 270° 360° −1

𝑥

17(i) p = A sin(ωt) +B sin(ωt + c) = A sin(ωt) +B[sin(ωt) cos(c) + cos(ωt) sin(c)] = A sin(ωt) +B cos(c) sin(ωt) + B sin(c) cos(ωt) = [A + B cos(c)] sin(ωt) +B sin(c) cos(ωt) = B sin (c) cos(ωt) +[A + B cos (c)] sin(ωt) ✓

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373


Ex 13.3

17(ii) Why? R2 = [Bsin (c)]2

18(v) In △ ABC,

+[A + B cos(c)]2

= B 2 sin2 (c)

In △ OBN,

+A2 + 2AB cos(c)

cos θ =

ON

AM = AC = AC = 2 cos θ = 6 sin θ

A

6

θ C

2 cm

sin θ =

6

C

B

M

N

θ

+CM +BN +6 sin θ +2 cos θ ✓

⇒ AM = R sin(θ + α) ✓ 18(vi) △ OAM area = 1 (base)

M

2 1

N

(height)

= OM

⋅ AM

2 1

⇒ ON = 6 cos θ In ⊿ ABC,

BN

2 cm

B

θ

O

sin θ =

θ

O

= A2 + B 2 + 2AB cos c [shown] ✓ In ⊿ OBN,

A

2

⇒ BN = 6 sin θ

= B 2 [sin2 (c) + cos 2 (c)] +A2 + 2AB cos(c)

18(i)

AC

⇒ AC = 2 cos θ

+A2 + 2AB cos(c) +B 2 cos 2 (c)

= B 2 [1]

cos θ =

= R cos(θ + α)⋅ R sin(θ + α)

BC

2 1

1

= R2 ( ) [2 sin(θ + α) cos(θ + α)]

2

2 1

⇒ BC = 2 sin θ

2

2

= R sin 2(θ + α) 4

1 2 (√40) sin 2(θ + α) 4 = 10 sin 2(θ + α) [shown] ✓

OM = ON −MN = ON −BC = 6 cos θ −2 sin θ [shown] ✓ 18(ii)  

=

R = √62 + 22 = √40 = √4 × 10 = 2√10 2 α = tan−1 ( )≈ 18.4°

18(vii) −1 ≤ sin 2(θ + α) ≤1 −10 ≤ 10 sin 2(θ + α) ≤ 10

6

∴ OM = 2√10 cos(θ + 18.4°)

⇒ max 10 sin 2(θ + α) sin 2(θ + α)

18(iii) OM =5 2√10 cos(θ + 18.4°) = 5 cos(θ + 18.4°)

=

0° < θ < 90° 36.9° < 2(θ + 18.43) < 216.9°

5 2√10

α ≈ 37.8° ⇒ 1st or 4th quadrant 0° < θ < 90° 18.4° < θ + 18.4° < 108.4°

= 10 = 10 =1

S T

A α α C

𝑦 1

90° 180° 270° 360°

2(θ + 18.43) = 90° θ ≈ 26.6° ✓

𝑥

−1

θ + 18.4° ≈ α, 360 − α ≈ 37.8°, 322.2° θ ≈ 19.3° ✓ 18(iv) OA = R cm [pythagoras thm] ✓ ∡AOB = α ✓


sleightofmath.com

374


Ex 13.3

P = (k − 1) sin 4t + (2 + k) cos 4t R = √(k − 1)2 + (2 + k)2 = √(k 2 − 2k + 1) + (4 + 4k + k 2 ) = √2k 2 + 2k + 5 ✓ ⇒ max = √2k 2 + 2k + 5 = √2(k 2 + 1) + 5 1 2

1 2

2

2

= √2 [(k + ) − ( ) ] + 5 1 2

1

2

2

= √2 (k + ) − + 5 1 2

9

2

2

= √2 (k + ) + ≥√

9

2

≥

1 2

∵ (k + ) ≥ 0 2

3 √2 3

≥ √2 2 3

∴ max cannot be less than √2 ✓ 2


sleightofmath.com

375


Rev Ex 13 A1(b) y = 2 cos (3π − 2x) + 3

Rev Ex 13

2

A1(a) cos 105° = cos(60° + 45°) = cos 60° cos 45° 1

√2 ( ) 2 2 1−√3 √2 ( )( ) 2 2 √2−√6 ✓ 4

=( ) = =

− sin 60° sin 45° √3 2

−( )

√2 2

= 2(− sin 2𝑥) + 3 = −2 sin 2𝑥 + 3 i.e. 𝑎 = −2, 𝑏 = 2, 𝑐 = 3

( ) Amplitude

= |𝑎| = |−2| = 2

Period

=

2

5 3π 2 3π

− 2x) + sin ( ) sin(2x)

= cos ( ) cos(2x)

+ [− sin ( )] sin 2x

= (0) cos 2𝑥 = − sin 2x = RHS [shown] ✓

+(−1) sin 2𝑥

2

2𝜋 2

=𝜋

𝑦 = −2 sin 2𝑥 + 3

2

π

1 𝑂

𝜋

𝑥 ✓

2

Workings Domain 0≤x≤π Axis with y=3±2 Amplitude Shape −sin

Cycle


=

3 3π

= cos ( ) cos(2x) 2 π

𝑏

𝑦

A1(b) Show: cos (3π − 2x) = − sin 2x LHS = cos (

2𝜋

sleightofmath.com

π−0 π

=1

376


Rev Ex 13

Coordinates of 𝜶 90° < α < 180° ⇒ 2nd quad. 4

𝑥

5

𝑟

cos α = − = 𝑥 = −4, y=

√52

−

5 13

𝑦 = −5,

=

𝛼 -4

(−4)2

=3

sin x = −

Coordinates of 𝜷 180° < β < 270° ⇒ 3rd quad. sin β = −

5

𝑦

𝑟=5

A3(a) 4 cos 2x +2 sin x 4(1 − 2 sin2 x) +2 sin x 4 − 8 sin2 x +2 sin x 8 sin2 x − 2 sin x − 1 (4 sin x + 1)(2 sin x − 1)

𝑦

13

𝑟 = 13

0° < x < 360° S

tan α−tan 45°

tan(α − 45°) =

1+tan α tan 45°

=

−4

24

5

5

25

3 )−(1) −4 3 1+( )(1) −4

(

=

✓

cos 2β = 1 − 2 sin2 β = 1 − 2 (−

1

cos 2 β =

1 2 cos 2 ( β) 2 1+cos β

2

7( 5 2

119

13

169

) =

✓

1 2

S

A

sin 2x 2

sin 2x

)

1+cos β 2

α α T C x = α, 180 − α = 30°, 150° ✓

=2 =

4 7

−1

2

= −√

A


1

cos β

2

A3(b) 7 sin x cos x = 2

3

= 2( )( ) = −

=

1

0° < x < 360°

α α T C x = 180 + α, 360 − α ≈ 194.5°, 345.5° ✓

−7 ✓ A2(ii) sin 2α = 2 sin α cos α

A2(iv) cos β

4

sin x =

-5

𝑟

x = −√132 − (−5)2 = −12 Trigonometric ratio

A2(iii)

or

α ≈ 14.5° α = 30° ⇒ 3rd or 4th quadrant ⇒ 1st or 2nd quadrant

𝛽

x

1

=3 =3 =3 =0 =0

(

∵ cos β < 0 2 1

for 90° < β < 135°

)

0° < x < 360° 0° < 2x < 720°

S

A

α T

α C

2

= −√

−12 ) 13

1+(

2 1

= −√ 13 2

= −√ =−

1

26 1

√26

✓

2x ≈ α, 180 − α, ≈ 34.8,145.2, 𝑥 ≈ 17.4°, 72.6°,

360 + α, 540 − α 394.9, 505.2 197.4°, 252.6° ✓

A3(c) sin 2x = sin2 x 2 sin x cos x = sin2 x sin x (2 cos x − sin x) = 0 sin x = 0 or 2 cos x = sin x 𝑦 tan x = 2 1 α ≈ 63.4° 𝑥 ⇒ 1st or 3rd quadrant 90° 180° 270° 360° −1

0° < x < 360° x = 180° ✓

0° < x < 360° S

A α

α T C x = α, 180 + α = 63.4°, 243.4° ✓


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377

A math 360 sol (unofficial) A3(d) cot 2x

Rev Ex 13

= 1 + cot x

1

A5(i) 7 cos x + 24 sin x − 12

= 25 cos(x − 73.7°) − 12

= 1 + cot x

tan 2x 1

=1+

2 tan x 1−tan2 x 1−tan2 x

=

2 tan x

−1 ≤ cos(x − 73.7°) ≤1 −25 ≤ 25 cos(x − 73.7°) ≤ 25 −37 ≤ 25 cos(x − 73.7°) − 12 ≤ 13

1 tan x

tan x+1 tan x

1 − tan2 x = 2 tan x + 2 2 tan x + 2 tan x + 1 = 0 (tan x + 1)2 =0 tan x = −1 α = 45° S A ⇒ 2nd or 4th quadrant α α T C 0° < x < 360°

max = 13 ✓ ⇒ cos(x − 73.7°) = 1 x − 73.7° ≈ 0° x ≈ 73.7° ✓

=

sin A cos A

sin A cos A

cos(x − 73.7°)

(2 cos 2 A − 1)

1

= (cos 2 2A − sin2 2A)

−1

2 5

x > 0° x − 73.7° > −73.7°

sec 2 2A

= cos(2 ⋅ 2A)

=


= tan A (cos 2A) = cos 2A tan A = RHS [proven] ✓ A4(b) LHS = cos 4A

𝑥

A5(ii) 7 cos x = 10 − 24 sin x 7 cos x + 24 sin x = 10 25 cos(x − 73.7°) = 10

− tan A

= 2 sin A cos A −

90° 180° 270° 360°

min = −37 ✓ ⇒ cos(x − 73.7°) = −1 x − 73.7° ≈ 180° x ≈ 253.7° ✓

x = 180 − α, 360 − α = 135°, 315° ✓ A4(a) LHS = sin 2A

𝑦 1

S

T

A α α C

x − 73.7° ≈ −360 + α, −α, ≈ −293.6, −66.4, x ≈ 7.3° ✓

cos2 2A 1 cos2 2A

2

= 1 − tan 2A = RHS [proven] ✓

α, 360 − α 66.4,293.6

A6(i) LHS = 1−cos θ+sin θ 1+cos θ+sin θ

A4(c) LHS =

sin(A+B) sin A cos B

=

sin A cos B+cos A sin B sin A cos B

=1+

cot A tan B = RHS [proven] ✓ A5(i) 7 cos x + 24 sin x = R cos(x − α) R = √72 + 242 = 25 ✓ 24

α = tan−1 ( ) ≈ 73.7° ✓ 7

=

θ 2

θ θ 2 2 θ θ θ 1+(2 cos2 −1)+(2 sin cos ) 2 2 2

=

θ θ θ 2 2 2 θ θ θ 2 cos2 +2 sin cos 2 2 2

=

θ θ θ 2 2 2 θ θ θ 2 cos (cos +sin ) 2 2 2

1−(1−2 sin2 )+(2 sin cos )

2 sin2 +2 sin cos

2 sin (sin +cos )

= tan

θ 2

= RHS [proven] ✓


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378


Rev Ex 13

A6(ii) 1 + sin θ = 2 cos θ sin θ − 2 cos θ = −1  

B1(i)

=

cos(A+B) cos A cos B+sin A sin B

R = √12 + 22 = √5 2 α = tan−1 ( ) ≈ 1.11

7 5 7 5

5 cos A cos B + 5 sin A sin B = 7 cos A cos B −7 sin A sin B

1

=−

=

cos A cos B−sin A sin B

⇒ √5 sin(θ − 1.11) = −1 sin(θ − 1.11)

cos(A−B)

1 √5

α ≈ 0.464 ⇒ 3rd or 4th quadrant 0 <θ <π −1.11 < θ − 1.11 < 2.03

S

A

α T

α C

2 cos A cos B

= 12 sin A sin B

cos A cos B

= 6 sin A sin B [proven] ✓

tan A tan B

= ✓

1

B1(ii) tan(𝐴 + 𝐵) θ − 1.11 ≈ −π − α, −α, ≈ −2.68, −0.464, θ ≈ 0.644 ✓

π + α, 2π − α 3.61,5.82

= tan(45°)

tan 𝐴+tan 𝐵

=1

1−tan 𝐴 tan 𝐵 tan 𝐴+tan 𝐵 1−

−(1)

6

=1

1 6

5

tan 𝐴 + tan 𝐵 = ✓ 6

A7(i) 2.5

1 θ A m θ B

1.5

cos 2 A = θ

𝑥 = −1, y=

4 1

1

𝑥

4

2

𝑟

𝑦 𝑟

=

∴ √7.25 cos(θ − 68.2°) = 1.5

θ − 68.2° ≈ −360 + α, −α, ≈ −303.9, −56.1, θ ≈ 12.1° ✓

−1

1

2

2

√3 2

√3 2

= ( )( )− ( ) ( )

√7.25

0° <θ < 90° −68.2° < θ − 68.2° < 21.8°

-1

√3 2

1.5


2 A

𝑦

− (−1)2 = √3

Trigonometric ratio cos(A + 60°) = cos A cos 60° − sin A sin 60°

1

=

𝑟=2

√22

sin 𝐴 =

R = √12 + 2.52 = √7.25 2.5 α = tan−1 ( ) ≈ 68.2°

cos(θ − 68.2°)

1

cos A = −√ = − =

A +B = 1.5 1 cos θ +2.5 sin θ = 1.5 cos θ + 2.5 sin θ = 1.5 [shown] ✓ A7(ii)  

B2(i) Coordinates A is obtuse ⇒ 2nd quadrant

= −1 ✓ S T

A α α C

α, 360 − α 56.1, 303.9

B2(ii) Recap

sin 𝐴 =

√3 , cos 𝐴 2

sin 2A = 2 sin A cos A=

=−

1

2 −1 √3 2( )( ) 2 2

=−

√3 2

✓

B2(iii) cos 2A = 2 cos 2 A − 1= 2 (1) − 1 = − 1 ✓ 4

2

B2(iv) sin 3A = sin 2A cos A + cos 2A sin A −1 √3 )( ) 2 2 √3 √3 − 4 4

= (− =

1

√3 2

+ (− ) ( ) 2

=0✓


sleightofmath.com

379


Rev Ex 13

B3(a) 4 sin 3x cos x − 4 cos 3x sin x + 1 = 0 4(sin 3x cos x − cos 3x sin x) = −1 4 sin(3x − x) = −1 sin(2x)

=−

B3(d) cos 2x − 3 cos x (2 cos 2 x − 1) − 3 cos x 2 cos 2 x − 3 cos x − 2 (2 cos x + 1)(cos x − 2)

1 4

cos x = −

α ≈ 14.5° ⇒ 3rd or 4th quadrant 0° < x < 180° 0° < 2x < 360°

α=

⇒ 2nd or 3rd quadrant

α T

α C

0≤x≤π S A α α T C x = π − α, π + α =

= 2 cos x = 2 cos x =0 =0 sin x = 1 0° < x < 360°

3(

1

𝑥

90° 180° 270° 360°

−1

−1

x = 90°, 270° ✓

x = 90° ✓

,

3

1 cos 2x

✓ = 8 sin 2x

) = 8 sin 2x

3

= 8 sin 2x cos 2x

3

= 8(

3

= 4(sin 4x)

sin 4x

=

1

1

0°
)

4

S

A

α T

α C

S

A

α T

α C

2π + α, 3π − α 7.13, 8.58 1.78, 2.14 ✓

B4(a) sin(x − y) =p (i) sin x cos y − cos x sin y = p q − cos x sin y = p q−p = cos x sin y cos x sin y =𝑞−𝑝 −(1)

√5


sin 𝑥 cos 𝑦 360 + α, 540 − α 386.6, 513.4 225°, 288.4° ✓

=𝑞

−(2)

sin(x + y) = sin(x) cos(y) + cos(x) sin(y) =q +q − p ∵ (2) & (3) = 2q −p ✓ B4(a) (ii)

tan x tan y

= = =


2

4x = α, π − α, ≈ 0.848,2.29, x ≈ 0.212, 0.573,

√5 sin(2x − 63.4°) = 1

2x − 63.4° ≈ α, 180 − α, ≈ 26.6,108.4, x ≈ 45°, 108.4°,

sin 4𝑥

3

0≤x≤π 0 ≤ 4x ≤ 4π

R = √12 + 22 = √5 2 α = tan−1 ( ) ≈ 63.4° =

3


𝑥

B3(c) sin 2x = 2 cos 2x + 1 sin 2x − 2 cos 2x = 1

sin(2x − 63.4°)

2π 4𝜋

B3(e) 3 sec 2x

𝑦

1

 

3

A

𝑦

90° 180° 270° 360°

or

2

S

2x = 180 + α, 360 − α ≈ 194.5, 355.5 x ≈ 97.2°, 172.8° ✓ B3(b) sin 2x 2 sin x cos x sin x cos x − cos x cos x (sin x − 1) cos x = 0 or 0° < x < 360°

π

1

=1 =1 =0 =0 cos x = 2 (no solution)

sleightofmath.com

sin x cos x sin y cos y

sin x cos y cos x sin y q q−p

✓

380


Rev Ex 13 B5(b) cos 2 x = 1+cos 2x ✓

B4 (a) sin 2x sin 2y = 2 sin x cos x (2 sin y cos y) (iii) = 4 sin x cos y (cos x sin y) = 4q (q − p)✓

𝑦

2

y = 4 cos 2 x = 4(

B4(b) A + B + C = 90° C = 90° − (A + B)

1+cos 2x 2

)

𝑦 = 2 cos 2𝑥 + 1

3

−1

1

−1

O

180° 𝑥

-1

= 2 + 2 cos 2x −1 = 2 cos 2x + 1 ✓ i.e. 𝑎 = 2, 𝑏 = 2, 𝑐 = 1

Show: tan A tan B + tan B tan C + tan C tan A = 1 Amplitude = |𝑎| = |2| = 2 LHS = tan A tan B + tan B tan C + tan C tan A = tan A tan B + tan B tan[90° − (A + B)]

Period

90° ] tan A −(A + B) = tan A tan B + tan B cot(A + B) + cot(A + B) tan A = tan A tan B + cot(A + B) [tan B + tan A] + tan [

= tan A tan B + = tan A tan B +

1 tan(A+B) 1

= tan A tan B + (

tan A+tan B

𝑏

=

360 2

= 180°

Workings Domain 0° ≤ x ≤ 180° Axis with y=1±2 Amplitude Shape +cos

(tan A + tan B)

1−tan A tan B

360

max. = 1 + (2) = 3 ✓ min. = 1 + (−2) = −1 ✓

(tan A + tan B)

tan A+tan B 1−tan A tan B

=

) (tan A + tan B)

180−0

Cycle

= tan A tan B +1 − tan A tan B =1 = RHS [shown] ✓

180

=1

B5(a) For 2 cos x − 3 sin x, R = √22 + 32 = √13 3

α = tan−1 ( ) ≈ 56.3° 2

B6(a) LHS = cos(x+y)

(2 cos x − 3 sin x)2 = [√13 cos(x + 56.3°)] = 13 cos 2 (x + 56.3°) ≤ 13 [shown] ✓

2

∵ cos 2 (x + 56.3°) ≤ 1

0° < x < 360° 56.3° < x + 56.3° < 416.3° x + 56.3° ≈ α, 180 − α, 360 + α, 540 − α = 0,180, 483.7 x = 123.7°,

−sin(x−y)

(cos x cos y−sin x sin y)+(cos x cos y+sin x sin y)

= (sin

Trigonometric equation (2 cos x − 3 sin x)2 = 13 13 cos 2 (x + 56.3°) = 13 cos(x + 56.3°) = ±1


+cos(x−y)

sin(x+y)

=

x cos y+cos x sin y)−(sin x cos y−cos x sin y)

2 cos x cos y 2 cos x sin y

= cot y = RHS [proven] ✓ 𝑦

B6(b) LHS

1

90° 180° 270° 360°

=

𝑥

1+sec 2θ tan 2θ

=

−1

= 180 + 𝛼, 360 − 𝛼,

=

1 cos 2θ sin 2θ cos 2θ

1+

(2 cos2 θ−1)+1 2 sin θ cos θ cos θ sin θ

=

=

cos 2θ+1 cos 2θ sin 2θ cos 2θ

=

cos 2θ+1 sin 2θ

2 cos2 θ 2 sin θcosθ

= cot θ = RHS [proven] ✓

360, 303.7° ✓

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381


Rev Ex 13

B6(c) LHS = sin2 2θ (cot 2 θ − tan2 θ) = sin2 2θ ( = sin2 2θ ( = sin2 2θ [ 2

cos2 θ

sin2 θ

−

sin2 θ cos2 θ 4 cos θ−sin4 θ

)

𝐴𝐵 >0 17 sin(𝜃 − 28.1°) > 0

)

sin2 θ cos2 θ (cos2 θ+sin2 θ)(cos2 θ−sin2 θ) (sin θ cos θ)2

= sin 2θ [

B7(iii) Given: 0° < θ < 90°

(1) cos 2θ 2

1 2

]

AB

AB = 17 sin(𝜃 − 28.1°) 118.1 θ O 28.1

]

( sin 2𝜃) cos 2θ

= sin2 2θ (1 4

sin2 2θ

= 4 cos 2θ = RHS [proven] ✓ B7(i) AB = OB −OA = 15 sin θ −8 cos θ 

R = √152 + 82

= 17 ✓

8

 α = tan−1 ( ) ≈ 28.1° ✓ 15 ∴ AB = 17 sin(θ − 28.1°) B7(ii) AB =5 17 sin(θ − 28.1°) = 5 sin(θ − 28.1°)

=

θ = 90

)

5 17


∴ 28.1° < 𝜃 < 90° B7(iv) shaded area = area of △ OPB 1

1

2

2

= (15 sin θ)(15 cos θ) − (8 sin θ)(8 cos θ) = = = =

225 2 161 2

sin θ cos θ sin θ cos θ

2 161 4

( sin 2𝜃) 2

sin 2θ ✓ ≤ sin 2θ

161 4

≤

161 4

⇒ max = 161 4

−32 sin θ cos θ

161 1

B7(v) −1 −

0° <θ < 90° [by inspection] −28.1° < θ − 28.1° < 61.9° S A θ − 28.1° ≈ α, 180 − α α α ≈ 17.1,162.9 T C θ ≈ 45.2° ✓

− area of △ OAQ

sin2θ =

sin 2θ

≤1

sin 2𝜃 ≤

161 4

161 4 161

𝑦 1

4

=1

90° 180° 270° 360°

𝑥

−1

0° < θ < 90° 0° < 2θ < 180° 2θ = α, 180 − α = 90 θ = 45° ✓


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382


Ex 14.1 3(a)

Ex 14.1 1(a)

d dx

(3x −7 ) = 3

d

(x −7 )

dx

d

55

5 d

( √x 2 ) =

dx 2

=

= d dx

(3x 2 d

d dx

√x3

3(c)

= −4

dx

d

d dx

= = = 2(d)

d dx

= = = 2(e)

d dx

=

dx

(1)

−0

d dx

(x 2 )

= −4(3x 2 ) +5(2x) = −12x 2 +10x ✓ 2(c)

d

(x) −

dx

+5

(

π2 x 2

3 π2 d

−

2x4 5

(x 2 )

3(d)

3 dx π2

(2x)

3 2π2 3

(

20 3 d

dx

(12)

−

πx9

11 8 (x )

d

100 x

2 d

d

3

(9x 2 − d dx

x2

a

b

dx 2

2

4(a) dx

(x 2 )

=

d

a

d

6

(

dx x3 d

d

d

6

✓

x3

1

− + 3) x

dx

(x −3 )

−

−4 )

d dx

+

x4

1

+

−2 )

d dx

(3)

+0

✓

x2

(3x + 2√x − 3) d dx

(x)

d

+2

1

(x 2 )

−

dx 1 1 +2 ( x −2 ) 2 1

+

√x

d dx

(3)

−0

✓

(8x 2 + 3x − √x) d dx

(x 2 ) +3

d dx

(x) −

+3(1)

= 16x

+3

d

(x −1 )

−(−x

18

(

2x2 +4x x

d dx

d

1

(x 2 )

dx 1 1 − ( x −2 ) 2 1

−

2√x

✓

)

(2x + 4) d

dx

(x)

= 2(1) =2✓

(x 3 ) 2 dx b 0 − (3x 2 ) 2 3b − x2 ✓ 2

(x −2 )

= 8(2x)

=

b d

d dx

+

=2

( ) −

✓

= 18x

dx d

=

x2

−3(−2x −3 )

=8

(x 9 )

11 dx π − (9x 8 ) 11 9 − πx 8 ✓ 11

( − x3 ) =

100

= 9(2x)

dx π d

(x −1 )

)

(x 2 ) −3

=3 3(f)

dx

−

= 3(1)

)

d

(x 2 ) +100

= 200x

=3

(x 4 )

5 dx 2 − (4x 3 ) 5 8 3

(3a) +b

)

+100(−x −2 )

dx

−

✓

= 100(2x)

=− 3(e)

(3a + bx 2 ) dx

d dx

2 x2

5

20 dx 3 (8x 7 ) 20 6 7 x 5

d

(100x 2 +

= 6(−3x

−0

=0 +b(2x) = 2bx ✓ 2(f)

d

=6

− x ✓

x

3x8

−

d

)

−

(x −1 )

−

=9

+4(1) +4 ✓

(x 3 )

d dx

=4

= 100

2

(x 5 )

(−4x 3 + 5x 2 − 12) d

(x) +2

+2(−x −2 )

dx

+ 4x − 1)

= 3(2x) = 6x

dx

✓

5

(x 2 ) +4

dx

x

d

= 4(1)

dx

=3

2(b)

( √x 2 )

2 dx 5 2 −3 ( x 5) 2 5 1

=

2(a)

3(b)

5

2 dx 5 d

2

(4x + )

=4

= 3(−7x −8 ) = −21x −8 ✓ 1(b)

d dx

+

d dx

(4)

+0

dx 2


sleightofmath.com

383


d dx

= =

(

Ex 14.1

x2 −6x+4

d

6(b)

)

x

4

dx d dx

(x − 6 + ) (x)

x d

−

=1

dx

(6)

+4

d dx

(

d dx

(x −1 )

4

−

= 1.5 t

✓

x2

6(c)

dx

=2

d

(x 2 )

dx

= 2(2x)

d dt

d

−

3

( )

dx

= =

(

( )

dx 2

−0

+

2x2

( +

5(a)

d dx

=

2 x2

✓

6(d)

dx 2

(

dx

d

dx

6(e) )

(3x1.5 d

dx

(x1.5 )

9

= √x 2 dx

=

(

2x

d dx

=3

(3x d

dx

(x)

= 3(1) =3 6(a)

d dt

d

+

1 2

(x )

−

dx 1 1 + ( x −2 ) 2 1

+

6x2 −√x+2

1

− x −2 )

+ x1.5

= 3(1.5x 0.5 )

d

d

−

dx

1 2

(x )

3 1 − (− x −2 ) 2 1

+

2√x

1

1

− x −2 −

2 1 d

−4(2.5t1.5 )

= 24t

−10t 2 ✓

2x√x

✓

1 − 2

(x )

+

d dx

6(f)

(x −1 )

−

1 x2

+

= 2(2t) = 4t

+1 +1 ✓

+t d


d dt

d dt

(9)

+0

d dt

−3

(t) −

d dt

(3)

−0

− 2t −1 ) −2

d dt

(t −1 )

−2(−t −2 ) +

2 t2

✓

y = 4x 2 − 6x + 1 dx

− 1) (t) −

+12(1) +12 ✓

=2

dy

d = (2t 2 dt d = 2 (t 2 )

(2t

= 2(1)

7(a)

(t + 1)(2t − 1)

d dt

=2

✓

+

dt

d (2t + 1)(t − 2) [ ] dt t d 2t 2 − 3t − 2 = ( ) dt t =

+(−x −2 )

4x√x

d (2t + 3)2 dt d = (4t 2 + 12t + 9) dt d d = 4 (t 2 ) +12 (t) = 4(2t) = 8t

+ x −1 )

2 dx 3 1 1 − (− x −2 ) 2 2 1

dt

dt

= 12(2t)

dt

)

+

dt

[4t 2 (3 − √t)]

3

✓

√x

=3

5(b)

(x −2 )

3x2 +x−1

d

(1)

dt

d

+2(−2x −3 ) 4

dt

d (12t 2 − 4t 2.5 ) dt d d = 12 (t 2 ) −4 (t 2.5 )

)

( ) +2

x3

−

dt d

=

dx 2 d 1

=−

(1 + √t)(1 − √t)

= −1 ✓

)

=0

−2 ✓

=0

x2 +4

2x2 d 1

−2(1)

dt

dt d

(t)

d (1 − t) dt d d = (1) − (t)

dx 2x 3 d (x −1 ) − 2 dx 3 − (−x −2 ) 2 3

= 4x 4(d)

0.5

d dt

=

dx 2 d 5

−

−2

= √t 2

)

d 5 3 = (2x 2 − − ) dx 2 2x d d 5 2 = (2x ) − ( )

− 2t)

3

4x3 −5x−3 2x

t(√t − 2) d = (t1.5 dt d = (t1.5 ) dt

+4(−x −2 )

−0

=1 4(c)

d

dt

= 8x − 6

Gradient at (2,5):

(1)

dy

−0

|

dx x=2

= 8(2) − 6 = 10 ✓

sleightofmath.com

384


Ex 14.1 8(b)

y = √x(2 − x) = 2√x −x√x = 2x dy

=

dx

=

1 2

−x

y = 2x 2 + 3x dy

= 4x + 3

dx

3 2

1 1 2 ( x −2 ) 2 1

Curve at y = 2: 2x 2 + 3x =2 2 2x + 3x − 2 = 0 (2x − 1)(x + 2) = 0

3 1

− x2 2 3

− √x 2

√x

1

x= dy

|

dx x=9

= =

1

3

− √9 2

√9 1

1

Gradients at x = , −2: 2

3

dy

2

dx x=1

− (3)

3

=−

or x = −2

2

Gradient at x = 9:

25 6

1

|

= 4( ) + 3 = 5✓ 2

2

✓

dy

= 4(−2) + 3 = −5 ✓

|

dx x=−2

7(c)

y= =

(x+1)(2x−3) x 2x2 −x−3

8(c)

x

= 2x −1 −

x

= x + 4x −1

= 2 −0 −3(−x =2

+

|

dx x=−1

=1−

3

Curve at y = 5:

=2+

x2 +4

3 (−1)2

dx

=5

x 2

x +4 = 5x 2 x − 5x + 4 = 0 (x − 1)(x − 4) = 0 x = 1 or x = 4

y = x 2 − 2x dy

4 x2

x2

=5✓ 8(a)

x

= x + 4(−x −2 )

dx

−2 )

Gradient at x = −1: dy

4

3

dy dx

x

=x+

= 2x −1 −3x −1 dy

x2 +4

y=

= 2x − 2 Gradients at x = 1,4:

Curve at y = −1: x 2 − 2x = −1 2 x − 2x + 1 = 0 (x − 1)2 =0 x=1

dy

dx

Gradient at x = 1: |

dx x=1

= 2(1) − 2

= 1 − (4)2 = ✓

4

3 4

= 4x − 5

Gradient at y − axis (x = 0): dy

=0✓


|

y = 2x 2 − 5x + 1 dy

dy

= 1 − (1)2 = −3 ✓

dx x=4

9(a)

4

|

dx x=1 dy

|

dx x=0

sleightofmath.com

= 4(0) − 5 = −5 ✓

385


y=

Ex 14.1

x−4

10(a) y = 3x 2 + 6x + 2

x

dy

4

=1 −

dx

x

= 1 −4x −1 dy

Gradient is 3: dy

= 0 −4(−x −2 )

dx

=

x

x

4

6x + 6 = 3

x2

x

dx x=4

y=

= (4)2

=−

1 2

1

2

2

1

1

2

4

1 4

10(b) y = ax 2 + b

=

x

1

= ax 2 +bx −1

✓

4

x+2

dy

x

dx

=1 +

2

⇒ (− , − ) ✓

4

|

1

2

=0

Gradient at x = 4:

9(c)

=−

y|x=−1 = 3 (− ) + 6 (− ) + 2

=4

dy

=3

dx

Curve crosses x − axis (y = 0): y =0 x−4

= 6x + 6

= a(2x) +b(−x −2 )

2

= 2ax

x

b

−

x2

= 1 +2x −1 Gradient is 2 at x = 1: dy

dy

= 0 +2(−x −2 )

dx

=−

2

=2

2a − b b

x2

Curve crosses y = x: x

|

dx x=1

=

=2 = 2a − 2

−(1)

Gradient is −1 at x = 4:

x+2

dy

x

2

|

dx x=4 b

x =x+2 2 x −x−2 =0 (x + 1)(x − 2) = 0 x = −1 or x = 2

8a −

16

= −1 = −1

−(2)

sub (1) into (2): 8a −

2a−2 16 1−a

= −1

Gradient at x = −1,2

8a +

dy

64a + 1 − a 63a

= −8 = −9

a

=− ✓

|

=

|

=

dx x=−1 dy dx x=2

2 − (−1)2 2 − (2)2

= −2 ✓ 1

=− ✓ 2

= −1

8

1

−(3)

7

sub (3) into (1): 1

16

7

7

b = 2 (− ) − 2 = −


sleightofmath.com

✓

386


Ex 14.1

Line: 4y − 3x = 7 4y = 3x + 7 y

3

7

4

4

= x+

13

y=

3

5

8

8

x+

4 3 2 x 8 2

7

3

4 3

8 9

4

8

1

=9 − −(2)

x2

Tangent ∥ x − axis: dy

=0

dx

= x2 +

− x−

x −1

= 9 +(−x −2 )

dx

sub (1) into (2): 3

1

= 9x +x

−(1)

dy

= x2 +

x

= 9x +

Curve: 3x 2 − 8y = −5 −8y = −3x 2 − 5 y

9x2 +1

5

9−

8

1

1

=0

x2

=9

x2 2

=0

x

3x − 6x − 9 = 0 x 2 − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1 or x = −3

= 1

1 9

1

(x + ) (x − ) = 0 3

3

1

x=−

or

3

x=

dy

y|x=1 = 6

1

1

= x 4

14 3

3

4

4

y=

9

4

4

= (3) = ✓

|

dx x=3

12

3

2 π 1

3

+

√x 1

2

dy dy

π√x

3

= x 2 +3x −2

= (−1) = − ✓

|

dx x=−1

⇒ ( ,6 ) ✓

3

Gradient at x = −1, −3: dy

3

⇒ (− , −6) ✓

3

dx

3

y|x=−1 = −6 3

Curve gradient:

1

1

π 1

1

3

= ( x −2 ) +3 (− x −2 )

dx

=

2 2 π

2

−

4√x

3 2x√x

2

y = x(x − 2) −3 2 = x(x − 4x + 4) −3 = x 3 − 4x 2 + 4x −3 dy dx

Gradient at x = 1: dy

|

dx x=1

=

= 3x 2 − 8x + 4 15(i)

Gradient is 7:

dx

=7

dx

3x 2 − 8x + 4 = 7 3x 2 − 8x − 3 = 0 (3x + 1)(x − 3) = 0 1

x=−

or

3

y|x=−1 = −4 3

22 27

1

22

3

27

⇒ (− , −4

)✓

π

−

3

4√(1) 2(1)√(1) π 3 4

− ✓ 2

y = x 3 + px + q dy

dy

=

= 3x 2 + p

Curve at (3,16): (16) = (3)3 + p(3) + q 16 = 27 + 3p + q q = −3p + 11

x=3 y|x=3 = 0

Gradient at (3,16) is 20:

⇒ (3,0)✓

dy

|

dx x=3 2

= 20

3(3) + p = 20 27 + p = 20 p = −7 ✓ q|p=−7 = 10 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

387


Ex 14.1

15(ii) Curve & its gradient with p = −7, q = 10: y = x 3 − 7x − 10 dy dx

19(i)

f(0) = 0 ∴x=0✓

= 3x 2 − 7

Gradient is 20: dy

= 20

dx

3x 2 − 7 = 20 2 3x − 27 =0 2 x −9 =0 (x + 3)(x − 3) = 0 x=3 or x = −3 (taken) y|x=−3 = (−3)3 − 7(−3) + 10 =4 ⇒ (−3,4) ✓ 16

f(x) = x[x 2 + kx + (2k − 3)]

19(ii) Observe quadratic factor [x 2 + kx + (2k − 3)] i.e. a = 1, b = k, c = 2k − 3 Discriminant For 2 non − real roots, b2 − 4ac <0 2 k − 4(1)(2k − 3) < 0 k 2 − 8k + 12 <0 (k − 2)(k − 6) <0 +

f(x) = √x

− 2

f(x + δx) = √x + δx

= lim

δx→0

= lim

19(iii) f(x) = x 3 + kx 2 + (2k − 3)x f ′ (x) = 3x 2 + 2kx + 2k − 3 i.e. a = 3, b = 2k, c = 2k − 3 Roots: α & β

δx

δx→0

6

2
f(x+δx)−f(x)

f ′ (x) = lim

+

√x+δx−√x √x+δx+√x δx √x+δx+√x x+δx−x

δx→0 δx(√x+δx+√x) δx

= lim

δx→0 δx(√x+δx+√x) 1

Sum of roots

= lim = = 17

y

δx→0 (√x+δx+√x) 1


= (−

= (x 3 + 1)(x 2 + 1)(x + 1)

=

= (x 5 + x 3 + x 2 + 1)(x + 1) +x 4 +x

5

+x 3 +x 3

=

+x +x

2

=

+1

= x 6 + x 5 + x 4 + 2x 3 + x 2 + x + 1 dy dx

c

b a

=− =

a

2k

3 2k−3 3

(α − β)2 = (α + β)2 −4αβ

✓

= x6

=

∵ √x + δx → √x as δx → 0

√x+√x 1 2√x

=α+β =−

3

)

4k2

−4 ( −

2k−3

3 8k−12

9 4k2 −24k+36

)

3

9 4 9 4

(k 2

− 6k + 6)

= (k − 3)2 ✓ 9

2 distinct roots ⇒ (α − β)2 ≠ 0

= 6x 5 + 5x 4 + 4x 3 + 6x 2 + 2x + 1 ✓

18 f(x) = (x − 1)(x + 1) (x 2 + 1) (x 4 + 1) = (x 2 − 1) (x 2 + 1) (x 4 + 1) (x 4 + 1) = (x 4 − 1) = (x 8 − 1) … = (x1024 − 1) = x 2048 − 1

2k 2

4 9

… (x1024 + 1) … (x1024 + 1) … (x1024 + 1) … (x1024 + 1)

k

(k − 3)2

≠0 ≠3✓

(x1024 + 1)

f ′ (x) = 2048x 2047 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

388

A math 360 sol (unofficial) 19(iv) |α − β|

≤

(α − β)2 4 9

≤

(k − 3)

2

≤

2

(k − 3) k 2 − 6k + 9 k 2 − 6k + 8 (k − 2)(k − 4) +

− 2

Ex 14.1

2

20

3 4

dy dx

9 4

= 5x 5

y = x2 ✓

9

y=

≤1 ≤1 ≤0 ≤0

21

+ 4

⇒2≤k≤4

2 5 2 x 2

Step 1: Step 2: Step 3: Step 4:

+1✓ x + x + ⋯ + x = x2 1 + 1 + ⋯ + 1 = 2x x = 2x [wrong] 1=2

The ‘number of x’ should not be disconnected from x and be excluded from being differentiated

Combine inequalities, 2 ≤ k ≤ 4 and k ≠ 3 and 2 < k < 6 and k ∈ ℤ

2

3

4

6

and k ∈ ℤ

⇒ k=4✓


sleightofmath.com

389


Ex 14.2 2(b)

Ex 14.2 1(a)

d dx

) =2

dx

(x + 2)

= 2(c)

d 3 [ ]= dx (3−4x)3

d dx

(2x − 1)

=

3

= 4(2x − 1) ⋅ 2 = 8(2x − 1)3 ✓ 1

( x + 2)

3

d dx

= 2(d)

5

dx 4

4

1

= 5 ( x + 2) ⋅ 4

4

1

= 5 ( x + 2) ⋅ 4

d

d 6 [ ] dx (2−x)2

[(3 − 4x)−3 ]

( x + 2)

dx 4

=

1 4

=

= ( x + 2) ✓ 1(d)

d dx

d

12 − (2−x)3 12 ✓ (2−x)3

dx

(1 − 4x)10

= 10(1 − 4x)

9

⋅

d dx

(√2x − 3) =

=

d dx

= 3(b)

(2 − 3x 2 )4

= 4(2 − 3x 2 )3 ⋅

d dx

d dx

1

d dx 1

[(2x − 3)2 ] 1

1

(2 − 3x 2 )

√2x−3

1

d dx 1

dx

(−x + x

1

2 )3

= 3(−x + x 2 )2

⋅

d dx

3(c)

(−x + x 2 )

d

(

4

) =4

dx 2x+7

d dx

=

=

√5−3x2

1

1

−4 (2x+7)2 8 − (2x+7)2

= d dx

(2x + 7)

=

⋅2 ✓

3(d)

1

√x2 +2x+2

⋅ (−6x) ✓

d dx

(x 2 + 2x + 2)

✓

d (√x 2 − x + 1) dx 1 d = [(x 2 − x + 1)2 ] dx 1

1 2

= = sleightofmath.com

(5 − 3x 2 )

⋅ (2x + 2)

2√x2 +2x+2 x+1

= (x 2 − x + 1)−2 ⋅


d dx

1 d [(x 2 + 2x + 2)2 ] dx

= (x 2 + 2x + 2)−2 ⋅

[(2x + 7)−1 ]

⋅

(√x 2 + 2x + 2)

2

= 4[−(2x + 7)−2 ] ⋅ =

d dx

= 3(−x + x 2 )2 ⋅ (−1 + 2x) = 3(−x + x 2 )2 (2x − 1) ✓ 2(a)

1

2√5−3x2 3x

=− d

[(5 − 3x 2 )2 ]

2

=

(2x − 3)

✓

= (5 − 3x 2 )−2

= 4(2 − 3x 2 )3 ⋅ (−6x) = −24x(2 − 3x 2 )3 ✓ 1(f)

d dx

⋅2

2√2x−3 1

(√5 − 3x 2 ) =

(2 − x)

⋅ (−1)

2

= 10(1 − 4x)9 ⋅ (−4) 9 = −40(1 − 4x) ✓ 1(e)

d dx

= (2x − 3)−2 ⋅

(1 − 4x)

(3 − 4x)

[(2 − x)−2 ]

4 4

3(a)

d dx

⋅ (−4)

= 6[−2(2 − x)−3 ] ⋅

1

4

5 1

d dx

(6x 2 + 5)

✓

9 − (3−4x)4 36 ✓ (3−4x)4

=6

d dx

⋅ 12x

+5)2 24x − (6x2 2 +5)

= 3[−3(3 − 4x)−4 ] ⋅

= 4(2x − 1) ⋅

d

[(6x 2 + 5)−1 ]

= − (6x2

(2x − 1)4 3

1(c)

d dx

2

d

= 5(x + 2)4 ⋅ 1 = 5(x + 2)4 ✓ d

2

(x + 2)5

= 5(x + 2) ⋅

dx

(

= 2[−(6x 2 + 5)−2 ] ⋅ 4

1(b)

d

dx 6x2 +5

1 2√x2 −x+1 2x−1 2√x2 −x+1

d dx

(x 2 − x + 1)

⋅ (2x − 1) ✓ 390


Ex 14.2

y = (3x − 1)4 dy

4(d)

= 4(3x − 1)

dx

3

⋅

d dx

1

y = (4x−5)3 = (4x − 5)−3

(3x − 1)

= 4(3x − 1)3 ⋅ 3 = 12(3x − 1)3 ✓

dy

= −3(4x − 5)−4 ⋅

dx

3

= − (4x−5)4 12

1

Curve at y = y = √5 − 2x = (5 − 2x) dy

1 2

1

1

= (5 − 2x)−2 ⋅

dx

2

1

=

⋅4

= 12(2)3 = 96 ✓

|

dx x=1

4(b)

(4x − 5)

= − (4x−5)4 ✓


d dx

2√5−2x 1

=−

√5−2x

d dx

(5 − 2x)

⋅ (−2) ✓

:

27 1

1 (4x−5)3

=

(4x − 5)3 (4x − 5)3 4x − 5 x

= 27 = 33 =3 =2

27


12

|

= − (4(2)−5)4 = −

dx x=2

1

Gradient at x = :

4

✓

27

2

dy

|

dx x=1

=−

2

4(c)

1 1 √5−2( ) 2

=−

1 √5−1

=−

1 2

✓

5(a)

2x−3

d

5

dx d

(2 − √x)

dx

5

= 6(2 − √x) ⋅ [0

= −(2x − 3)−2 ⋅

dx

5

= 6(2 − √x) ⋅ [

= (2x − 3)−1 dy

6

[(2 − √x) ]

= 6(2 − √x) ⋅

1

y=

d dx

1

= − (2x−3)2

d dx

5

= 6(2 − √x) ⋅ (−

(2x − 3)

=−

⋅2

(2) −

1

d

(x 2 )]

dx 1 −1 − x 2] 2 1 2√x

)

5

3 √x

(2 − √x) ✓

2

= − (2x−3)2 5(b) Curve at y = 1: 1 2x−3

2x − 3 x

d

1

dx [(1−1)3 ] x

d 1 −3 = [(1 − ) ] dx x

=1 =1 =2

1 −4

= −3 (1 − ) x

=−


|

dx x=2

2

2

= − (2(2)−3)2 = − (1)2 = −2 ✓

=− =− =−


sleightofmath.com

3 1 4 (1− ) x

3 1 4 (1− ) x

3 1 4 (1− ) x

⋅

d dx d

⋅[

1

(1 − )

dx

x

(1) −

d dx

(x −1 )]

⋅ [0 −(−x −2 )] ⋅

1 x2

3 1 4 x2 (1− ) x

391


d

1

[

Ex 14.2 6(a)

2]

y = (x + 2)2 √x + 2 1

dx 2(3−2)

= (x + 2)2 (x + 2)2

x

=

−2 1 d 3 [( − 2) ] 2 dx x 1

3

2

x

5

= (x + 2)2

−3

d

= [−2 ( − 2) ] ⋅ =− =−

1

dy

dx x d

dx

⋅ [3

3 3 ( −2) x

1 3 x

3

( − 2) dx

(x −1 ) −

3

⋅ [3(−x −2 ) −0]

3

⋅ (− 2 )

d dx

2 5

(2)]

1

3

3 x

= 5(d)

d dx

3 3 x

3

x2 ( −2)

2

= =

6(b)

d dx

−

dy

1 2

⋅

1

d

2

1 (2−3x2 )2

(2√x + 2)

dx

⋅ [2

2√2√x+2 1

1 2

d

(x ) +

dx

d dx

= 1.5√2 − 3x 2 (2)]

1 − 2

1

⋅ [2 ( x ) +0] 2

2√2√x+2 1

⋅

2√2√x+2 1

d dx

(2 − 3x 2 )

⋅ (−6x)

= −9x√2 − 3x 2 ✓ 7(i)

x 1+x

1

=

(1+x)−1 1+x 1

=1−

√x

[shown] ✓

1+x

✓ 7(ii)

d

(

x

dx 1+x d

=

1 3

[(x − ) ] x

1 2

d

x

dx

=

1

(x − )

d

x

dx

(x) −

d dx

(x −1 )]

=

x

1 2

1

x

x2

8(i) [(√x + 2x) ]

= 4(√x + 2x)

3 3

⋅

d dx d

1 2

(x ) +2

dx 1 3 1 2x) ⋅ [ x −2 2 3 1 2√x

(1) −

dx

) [(1 + x)−1 ]

− [−(1 + x)−2 ⋅ −[−(1 + x)

1 (1+x)2

−2

d dx

(1 + x)]

⋅ 1]

✓

d dx

2x−1 (x−1)2

≡

A

B

x−1

+ (x−1)2

Cover-up rule:

(√x + 2x)

⋅[

= 4(√x + 2x) (

1+x d

)✓

4

= 4(√x + 2x)

dx

1

(1 −

=0

= 3 (x − ) ⋅ [1 −(−x −2 )] = 3 (x − ) (1 +

dx d

)

=0

x

1 2

= 4(√x +

= 1.5(2 − 3x 2 )0.5 ⋅

dx

1 2

d

(2−3x2 )

[(2√x + 2) ]

= 3 (x − ) ⋅ [

dx

√2−3x2

3

= 3 (x − ) ⋅

5(f)

2

= (2 − 3x 2 )2

1 2

2√x(2√x+2)

5(e)

(2−3x2 )

y= =

= (2√x + 2) 2

=

3

= (x + 2)2 ✓

✓

1

=

(x + 2)

3 2

2 5

x

( −2)

d dx

= (x + 2) ⋅ 1

( −2)

=−

3

5

= (x + 2)2 ⋅

x = 1:

(x)]

+2(1)]

1 ( )2

Substitution:

+ 2) ✓

x = 0:

−1

= −A + B

1

=2✓

A 2x−1

∴ (x−1)2 =


⇒ B=1✓

=B

sleightofmath.com

2 x−1

1

+ (x−1)2

392


Ex 14.2

8(ii)

10

y = (1 − x)4

d 2x−1

dy

dx (x−1)2

dx

=

d

[

2

d dx

[(x − 1)−1 ]

= 2[−(x − 1)

−2 ]

⋅

+ d dx

d dx

[(x − 1)−2 ]

(x − 1) +2[−2(x − 1)

= 2[−(x − 1)−2 ] ⋅ 1

−3 ]

2

(1 − x)

Gradient is −4: ⋅

d dx

dy

(x − 1)

= −4

dx 3

−4(1 − x) = −4 1−x =1 x =0 y|x=0 = 1 ⇒ (0,1) ✓

+2[−2(x − 1)−3 ] ⋅ 1 2

= − (x−1)2

d dx

= 4(1 − x)3 ⋅ (−1) = −4(1 − x)3

1

+ (x−1)2 ]

dx x−1

=2

= 4(1 − x)3 ⋅

− (x−1)3

2

= − (x−1)3 [(x − 1) +1] 2

= − (x−1)3 (x)

11

3

y = √x 2 − 2x + 5

2x

1

= − (x−1)3 ✓ 9(a)

= (x 2 − 2x + 5)3

y = (x 2 − 2x − 4)3

dy

dy

= 3(x 2 − 2x − 4)2 ⋅

dx

d dx

3

At

y=

1 1

3

= − (1 + x)−2 ⋅

dx

2

=− =−

1

d dx

(1 + x)

12

3

2(1+x)2 1

dy

3 2(1+x)2

|

dx x=3

=−

3

=−

2(4)2

1

3

= (a − x)2 ⋅

dx

1

=0

= (a − x)2

2 3

d dx

(a − x)

1 2

= (a − x) ⋅ (−1) 2


= 0:

dx 2x−2

y = √(a − x)3

⋅1

3

dy

x =1 3 y|x=1 = √1 − 2 + 5 3 = √4 3 ⇒ (1, √4) ✓

√1+x

1

2

2

= (1 + x)−2 dy

2x−2

3(x2 −2x+5)3

= −12 ✓ 9(b)

(x 2 − 2x + 5)

3(x2 −2x+5)3

= (−12)(1 + 2 − 4)2

|

d dx

⋅ (2x − 2)

2 3(x2 −2x+5)3

=

Gradient at x = −1: dx x=−1

1

=

(x 2 − 2x − 4)

= 3(x 2 − 2x − 4)2 ⋅ (2x − 2) = (6x − 6)(x 2 − 2x − 4)2

dy

2

1

= (x 2 − 2x + 5)−3 ⋅

dx

1 16

3

= − √a − x 2

✓

Gradient of − dy

|

dx x=2 3

1

at x = 2:

3

=−

1 3 1

− √a − 2 = − 2 3


sleightofmath.com

√a − 2

=

a−2

=

a

=

2 9 4 81 166 81

✓

393


Ex 14.2

13 y = [(x 2 − 1)2 + 4]5

16(i)

y = (x 2 + 1)n dy

dy

= 5[(x 2 − 1)2 + 4]4 ⋅

dx

d dx

= 5[(x 2 − 1)2 + 4]4 ⋅ [

[(x 2 − 1)2

d dx

+4]

[(x 2 − 1)2 ]

= 5[(x 2 − 1)2 + 4]4 ⋅ [2(x 2 − 1) ⋅

dx

d dx

+

= n(x 2 + 1)n−1 (2x) = 2nx(x 2 + 1)n−1 [shown] ✓

d dx

(4)]

(x 2 − 1) +0]

16(ii) d2 y dx2

= 2nx [(n − 1)(x 2 + 1)n−2 ⋅

= 5[(x 2 − 1)2 + 4]4 ⋅ [2(x 2 − 1) ⋅ 2x]

d dx

(x 2 + 1)]

= 20x(x 2 − 1)[(x 2 − 1)2 + 4]4 ✓

+

14 y=

(4−√16−x2 )

2

−2

= 4nx 2 (n − 1)(x 2 + 1)n−2

dy

+2n(x 2 + 1)n−1

= 2n(x 2 + 1)n−2 [2x 2 (n − 1) + (x 2 + 1)]

dx −3

= −2(4 − √16 − x 2 ) =− =− =− =−

15

(2nx) ⋅ (x 2 + 1)n−1

= 2nx[(n − 1)(x 2 + 1)n−2 ⋅ 2x] +2n ⋅ (x 2 + 1)n−1

1

= (4 − √16 − x 2 )

=

d dx

2 (4−√16−x2 )

d dx d

(4 − √16 − x 2 )

⋅[

3

dx

2

(4)

−

d dx

⋅ [0 − (16 − x 2 )−2 ⋅

3

⋅ [0 − (16 − x 2 )−2 ⋅ (−2x)]

3

⋅

2

dx

2

√16−x2 (4−√16−x2 )

3

dy

−(1)

|

dx x=1 n)

n(2 1 3

1

= − (x 2 + 3)−2 ⋅ 2

=−

|

dx2 x=1

✓

√x2 +3

=−

d2 y

√16−x2

1

=−

= 2n(2n−1 ) = n(2n )

1

d dx

(x 2 + 3)

17

⋅ 2x

3 2(x2 +3)2

= 2n[(1)2 + 1]n−2 [2(1)2 (n − 1) +((1)2 + 1)] = 2n(2)n−2 [2(n − 1) + 2] = n(2)n−1 [2n] = n2 (2n )

1

dx

]]

= 2n(1)[(1)2 + 1]n−1

|

dx x=1

x

= (x 2 + 3)−2 dy

1 2

(16 − x 2 )]

1

1

(4−√16−x2 ) −2x

d

2)

3

2 (4−√16−x2 ) 2

dy

[(16 − x

1

1

(4−√16−x2 )

y=

⋅

:

d2 y

|

dx2 x=1 2 (2n )

:n : n [shown] ✓ f(x) = √1 + √x = (1 + √x)

1 2

x 3

(x2 +3)2 x 1

2

−(2)

x2 +3 √x2 +3

=

sub (1) into (2): dy

=−

dx

(x 2 + 3) (x 2 + 3)

dy dx dy dx

−

1

f ′ (x) = (1 + √x)

x x2 +3

= y =

= −xy

=

+ xy = 0 [shown] ✓


sleightofmath.com

1 2√1+√x 1 2√1+√x 1 2√1+√x 1 4√x+x√x

1 2

⋅

d dx d

(1 + √x)

⋅[

dx

(1) + 1

d dx

1

(x 2 )]

1

⋅ [0 + x −2 ] 2

⋅

1 2√x

[shown] ✓

394


Ex 14.2

x2 + y2 = 1 y2 = 1 − x2

19 1

y=

1 1 1 + + x2 +1 x2 +2 x2 +3

= ±√1 − x 2

y

=( dy

=±

dx

=± =±

1 2√1−x2 1 2√1−x

⋅

d dx

1 x2 +1

1

+

+

x2 +2

1

)

−1

x2 +3

(1 − x 2 ) dy

⋅ (−2x) 2

dx

= −(

x √1−x2

1 x2 +1

+

1 x2 +2

+

1

1

=−

1

1

1

2

( 2 + 2 + 2 ) x +1 x +2 x +3

1

Gradient at x = :

)

−2

x2 +3

⋅

d

|

dx x=1 2

=±

d

1 ( ) 2 2 √1−(1) 2

1

√4

2

√3

=± ×

(

=±

=±

1 2 3 √ 4

1 √3

=±

1 2 √3 √4

=−

1

✓ 1 1

1

1

2

( 2 + 2 + 2 ) x +1 x +2 x +3

⋅ +

1

(

+

=−

1

1

1

2

dx d

=

dy

1 2 1 1 1 ( 2 + 2 + 2 ) x +1 x +2 x +3

2x 2 1 1 1 ( 2 + 2 + 2 ) x +1 x +2 x +3

1

+

)

)

x2 +3

(x 2 + 2)−1

2 −1 [+ dx (x + 3) ]

[−(x 2 + 1)−2 ][2x] ⋅ [+[−(x 2 + 2)−2 ][2x]] +[−(x 2 + 2)−2 ][2x]

⋅{

[− (x2

2x

+1)2

2x

⋅ (−2x) [(x2 ⋅ [(x2

] + [− (x2

+ [− (x2

( 2 + 2 + 2 ) x +1 x +2 x +3

=−

+

x2 +2

2x

1

1

+

(x 2 + 1)−1

dx d

2 1 1 1 ( 2 + 2 + 2 ) x +1 x +2 x +3

=−

d

dx x2 +1 x2 +2 x2 +3 1 1 1

dx x2 +1

2

dy

⋅

1 +1)2

1 +1)2

+2)2

] }

]

+ (x2

1

+ (x2

+3)2

+2)2

1 +2)2

+ (x2

+ (x2

1 +3)2

1 +3)2

]

]

|

dx x=1

= = = =


2(1) 2 1 1 1 ( 2 + 2 + 2 ) (1) +1 (1) +2 (1) +3

2 1 1 1 2 ( + + ) 2 3 4

2 13 2 ( ) 12

1

⋅ [((1)2

1

+1)2

+ ((1)2

1

1

1

+2)2

1

+ ((1)2

+3)2

]

⋅ [(2)2 + (3)2 + (4)2 ] 1

1

1

4

9

16

⋅( + +

)

122 169

sleightofmath.com

395


Ex 14.3 2(a)

Ex 14.3 1(a)

d dx

d dx

d dx

[(2x − 1)3 ]

= x ⋅ 3(2x − 1)2 ⋅ 2 = 6x(2x − 1)

2

+

d dx

(x) ⋅ (2x − 1)3

+1 ⋅ (2x − 1)3 +(2x − 1)

3

= −2√x(1 − x)

+

=

= (2x − 1)2 (8x − 1) ✓ 1(b)

dx

= [(x − 1)(x + 2)2 ]

= (x − 1) ⋅

d dx

[(x + 2)2 ] +

d dx

2(b)

⋅ (x + 2)2

= (x − 1) ⋅ 2(x + 2) ⋅ 1

+1

= 2(x − 1)(x + 2)

+(x + 2)2

= (x + 2) [2(x − 1)

=

(x − 1) ⋅ (x + 2)2

= (x + 2) (2x − 2

dx

+x + 2)

=

= 3x(x + 2) ✓

= =

= (1 − 2x) ⋅

d dx

4]

[(3x + 2)

= 4]

+

d dx

2√x 1−x 2√x 1−x

(1 − 2x) ⋅ (3x + 2)

4

2(c)

d

d

d

1

= (x 2 + 1) ⋅

d dx

[(1 + x)2 ]

+

d dx

(x 2 + 1) ⋅ (1 + x)2

=

⋅ (1 + x)2

=

= (x 2 + 1) ⋅ 2(1 + x) ⋅ 1

+2x

= 2(x 2 + 1)(1 + x)

+2x(1 + x)2

= 2(1 + x) [(x 2 + 1)

+x(1 + x)]

= 2(1 + x) (x 2 + 1

+x + x 2 )

[x

1 1+3x √1+2x

+1

⋅ √1 + 2x

+(1 + 2x)]

(1 + 3x)

√1+2x

✓

(x 2 √1 − 2x 2 )

=−

=

⋅2

+√1 + 2x

√1+2x

−3x − 2)

[(x 2 + 1)(1 + x)2 ]

(√1 + 2x) + (x) ⋅ √1 + 2x dx

x

= x2 ⋅

dx

✓

√1+2x

−(3x + 2)]

1(d)

+(1 − x)]

1

= 2(3x + 2)3 [6(1 − 2x)

=

2√x

[−4x

2√1+2x

= x2 ⋅

= 2(3x + 2) (4 − 15x) ✓

(1−x)2

d

dx

−2(3x + 2)4

3

⋅ (1 − x)2

(x√1 + 2x)

= 12(1 − 2x)(3x + 2)3

= 2(3x + 2) (6 − 12x

1 2√x

+(1 − x)2 ]

2√x

+(−2) ⋅ (3x + 2)4

3

(√x) ⋅ (1 − x)2

[−4x(1 − x)

(1−x)(1−5x)

= (1 − 2x) ⋅ 4(3x + 2)3 ⋅ 3

dx

d dx

(1 − 5x)

2√x

=x⋅

= (x + 2) 3x

[(1 − 2x)(3x + 2)

1

=x⋅

1(c) d

d dx

+(x + 2)]

+ +

=

= (2x − 1) [6x +(2x − 1)]

[(1 − x)2 ]

dx

= √x ⋅ 2(1 − x) ⋅ (−1)

2

d

d

= √x ⋅

[x(2x − 1)3 ]

=x⋅

[√x(1 − x)2 ]

d dx

(√1 − 2x 2 ) 1

2√1−2x2

⋅ (−4x)

2x3 1 2x

√1−2x2 2x √1−2x2

d dx

+2x

(x 2 ) ⋅ √1 − 2x 2 ⋅ √1 − 2x 2

+2x√1 − 2x 2

√1−2x2

√1−2x2

+

[−2x 3

+2x(1 − 2x 2 )]

[−x 2

+(1 − 2x 2 )]

(1 − 3x 2 )

2x(1−3x2 ) √1−2x2

✓

= 2(1 + x) (2x 2 + x + 1) ✓


sleightofmath.com

396


Ex 14.3

2(d) d dx

3(a) 1)√3x 2

[(4x −

d

= (4x − 1) ⋅

dx

= = = = =

2√3x2 +1

+

⋅ 6x

1 √3x2 +1 1 √3x2 +1

dx

=

d

x

1

1

= (x + 2)2 (3 +

3

d

1 2√x+1

d

+

d dx 3

⋅1

2

dx

3 2

(x + 1) ⋅ √x + 1

+ x

3 x2 +1

1 2

=

⋅ √x + 1

1

[x 2 + 1

+3√x(x + 1)]

=

(x√x + 1

+3x√x + 3√x)

=

(4x√x + 3√x + 1)

2√x+1

4x√x+3√x+1 2√x+1

= =

✓

=

[(2√x + 3)√1 − 4x]

= (2√x + 3) ⋅ = (2√x + 3) ⋅

= = = =

d dx

1 −2 √1−4x

√1−4x

1 √x√1−4x 1 √x√1−4x 1−6√x−8x √x√1−4x

d dx

(2√x + 3)

⋅ (−4) + [2 (

−2(2√x+3)

√1−4x√x

=

(√1 − 4x) +

2√1−4x

2

2

x

x2

+ +

1 2√x

) + 0]

1 √x

⋅ √1 − 4x

=

⋅ √1 − 4x

=

⋅ √1 − 4x

d dx

(x + 2)]

1

2

x

x2

)

1

+(3x 2 − 2x) ⋅ √5 + 6x

2√5+6x

⋅6

x √5+6x x √5+6x x √5+6x x √5+6x

+x(3x − 2)√5 + 6x

[3x 2 (x − 1)

+x(3x − 2)(5 + 6x)]

[3x(x − 1)

+(3x − 2)(5 + 6x)]

(3x 2 − 3x

+(3x − 2)(6x + 5)]

(3x 2 − 3x

+18x 2 + 3x − 10)

(21x 2 − 10)

x(21x2 −10) √5+6x

4

dx

+(3x 2 − 2x)√5 + 6x

√5+6x 1

(x 3 − x 2 ) ⋅ √5 + 6x

+

3x2 (x−1)

√5+6x

d

(√5 + 6x)

√5+6x

✓

y = x√3 + x 2

[typo in question]

√1−4x dy

√x

dx

=x⋅ =x⋅

+1 − 4x)

= =

(1 − 6√x − 8x)

=

✓


(x + 2)3

)✓

3(x3 −x2 )

[−2√x(2√x + 3) +(1 − 4x)] (−4x − 6√x

1 x2

) ⋅ (x + 2)3

[(x 3 − x 2 )√5 + 6x]

= (x 3 − x 2 ) ⋅

3

1

d dx

= (x 3 − x 2 ) ⋅

3

2√x+1

1 x2

x2

[x 2 (x − 1)√5 + 6x]

+ √x√x + 1 2

2√x+1

2√x+1

x

1

3(b)

(√x + 1)

dx

1

(1 + ) ⋅ (x + 2)3

− −

x

= (x + 2)2 (3 + −

✓

1

3

d dx

+ (0 −

−

x

(24x 2 − 3x + 4)

+

−

x

[(x 2 + 1) √x + 1]

1

[(x + 2)3 ]

= (x + 2)2 [3 (1 + )

= (2√x + 3) ⋅

=

dx

1

2(f) dx

d

x

+12x 2 + 4)

= (x + 1) ⋅

=

1

= (1 + ) ⋅

(12x 2 − 3x

3 2

=

x

= 3 (1 + ) (x + 2)2

= (x + 1) ⋅

=

1

[(1 + ) (x + 2)3 ]

+4(3x 2 + 1)]

3 2

=

⋅ √3x 2 + 1

d dx

[3x(4x − 1)

√3x2 +1 d

=

= (1 + ) ⋅ 3(x + 2)2 ⋅ 1

24x2 −3x+4

2(e)

(4x − 1) ⋅ √3x 2 + 1

+4√3x 2 + 1

√3x2 +1 √3x2 +1

d dx

+4

3x(4x−1) 1

1

[ (x + 1)(x + 2)3 ]

dx x

√3x 2 + 1 1

= (4x − 1) ⋅

d

+ 1]

=

sleightofmath.com

d dx

(√3 + x 2 ) + 1

2√3+x2

⋅ 2x

x2 √3+x2 1 √3+x2 1 √3+x2 3+2x2 √3+x2

d dx

(x) ⋅ √3 + x 2

+1 ⋅ √3 + x 2 +√3 + x 2

[x 2

+(3 + x 2 )]

(2x 2 + 3) ✓

397


Ex 14.3 7

y = x√x − 1 dy

=x⋅

dx

=x⋅ = = = = =

d

(√x − 1) +

dx

1 2√x−1

⋅1

x

2√x−1 1 2√x−1 1 2√x−1

dy

(x) ⋅ √x − 1

+1

d

[(x − 4)4 ]

+

= √x ⋅ 4(x − 4)3 ⋅ 1

+

= 4√x(x − 4)3

+

= √x ⋅

dx

⋅ √x − 1

+√x − 1

2√x−1 1

d dx

y = √x(x − 4)4

[x

+2(x − 1)]

=

(x

+2x − 2)

=

(3x − 2)

=

3x−2 2√x−1

=

dx

1 2√x (x−4)3 2√x (x−4)3 2√x

d dx

(√x) ⋅ (x − 4)4

1 2√x

⋅ (x − 4)4

(x−4)4 2√x

[8x(x − 4)3

+(x − 4)4 ]

[8x

+(x − 4)]

(9x − 4)

(x−4)3 (9x−4) 2√x

✓

Gradient at (5,10): dy

|

dx x=5

=

3(5)−2 2√5−1

=

13 2(2)

=

13 4

dy

At

✓

dx

1 2√x

6

= 0:

(x − 4)3 (9x − 4) = 0 4

x = 4 or x = ✓

y = (x + 1)3 (x − 1)

9

dy dx 3

= (x + 1) ⋅

d dx

(x − 1)

+

d dx

[(x + 1)

3]

8 ⋅ (x − 1)

y=

x2 −1 x2 +1 (x2 +1)−2

= (x + 1)3 ⋅ 1

+3(x + 1)2 ⋅ 1 ⋅ (x − 1)

=

= (x + 1)3

+3(x + 1)2 (x − 1)

=1 −

= (x + 1)2 [(x + 1)

+3(x − 1)]

= 1 −2(x 2 + 1)−1

= (x + 1)2 (x + 1

+3x − 3) dy

= (x + 1)2 (4x − 2)

dx

= (x2 at

dy dx 4x

x

(x 2 + 1)

⋅ 2x ⋅ 2x

+1)2

=0 =0 (0)2 −1

y|x=0 = (0)2

sleightofmath.com

d dx

= 0:

(x2 +1)2

=8✓


2

+1)2 4x

Gradient at x = −1,1: dy | =0✓ |

= −2[−(x 2 + 1)−2 ] ⋅ = (x2

Curve cross x − axis: y =0 2 (x 2(x + 1) − 1) = 0 x = −1 or x = 1

dx x=1

2 x2 +1

= 2(x 2 + 1)−2

= 2(x + 1)2 (2x − 1)

dx x=−1 dy

x2 +1

+1

= −1 ✓

398


Ex 14.3 11

y = x√4 − x 2 dy

d

=x⋅

dx

dx

(√4 − x 2 ) 1

=x⋅ = = = = =

2√4−x −x2

√4−x2 1

+

⋅ (−2x) 2

d dx

y = (x − a)√x − b

(x) ⋅ √4 − x 2

dy

⋅ √4 − x 2

+1 +√4 −

= (x − a) ⋅

dx

= (x − a) ⋅

x2

=

[−x 2 +(4 − x 2 )]

√4−x2 1

=

(4 − 2x 2 )

√4−x2 1 √4−x2 2(2−x2 )

=

2(2 − x 2 )

=

x√4 − x 2

=x

x(√4 − x 2 − 1)

=0

or √4 − x 2 = 1 4 − x2 = 1 3 − x2 = 0 x2 =3 x = −√3 or

x=0

|

=

dx x=√3 dy

|

2(2)

=

dx x=−√3

10

dx

d dx

(x − a) ⋅ √x − b

+1

⋅ √x − b

+√x − b

2√x−b x−a +(2x−2b) 2√x−b 3x−a−2b 2√x−b

Curve intersects x-axis (y = 0) with x = b + 1: 0 = (b + 1 − a)√b + 1 − b 0 = (b − a + 1)√1 0=b−a+1 a=b+1 Gradient at A: dy

|

dx x=b+1

= =

x = √3

√4 2(2−3)

=

=

=2✓

√4−3 √4−3

2√(b+1)−b (3b+3)−a−2b 2√1 b+3−a 2 b+3−(b+1) 2

∵a=b+1

2 2 = 1 [shown] ✓

= − = −2 ✓

2(2−3)

3(b+1)−a−2b

=

2 1

= −2 ✓

Curve & its gradient y = 3x 2 − 7x + 2 dy

(1)

2√x−b x−a +2(x−b)

= dx x=0 dy

1 2√x−b

x−a

Gradient at x = 0, −√3, √3: |

(√x − b) +

√4−x2

Curve cross y = x: y =x

dy

d dx

= 6x − 7

mtan = 6x − 7 Line & its gradient 5x − y = 0 y = 5x mline = 5 Tangent ∥ Line: mtan = mline 6x − 7 = 5 x =2 y|x=2 = 3(2)2 − 7(2) + 2 =0 ⇒ (2,0) ✓


12 d dx

=

[f(x)g(x)h(x)] d dx

f(x)[g(x)h(x)]

= f(x) ⋅

d dx

[g(x)h(x)]

+

d dx

f(x)[g(x)h(x)]

= f(x) ⋅ [g(x)h′ (x) + g ′ (x)h(x)]

+f ′ (x)

= f(x)g(x)h′ (x) +f(x)g ′ (x)h(x)

+f ′ (x)g(x)h(x) ✓

13(i)

[g(x)h(x)]

f(a) = 0 By factor theorem, (x − a) is a factor ✓

13(ii) The gradient of f(x) at x = a is 0

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399


Ex 14.4 1(d)

Ex 14.4 1(a)

d

(

5x

d 3x 2 ( ) dx 1 − 4x

)

dx 2x+1

=

=

6x − 24x 2 + 12x 2 (1 − 4x)2

5 = (2x + 1)2 ✓

=

6x − 12x 2 (1 − 4x)2

d

(

1−x

dx 1−2x

(1 − 4x) ⋅ 6x − 3x 2 ⋅ (−4) = (1 − 4x)2

✓ 1(e)

)

1 (1 − 2x)2

= (

x2

d 2 d (x ) − x 2 ⋅ (x + 3) dx dx = (x + 3)2 (x + 3) ⋅ 2x − x2 ⋅ 1 = (x + 3)2 2x 2 + 6x − x2 = (x + 3)2 =

x(x + 6) (x + 3)2

✓

)

2x−1

2(x 2 − x − 1) (2x − 1)2

✓

)

dx x+3

(x + 3) ⋅

x2 +1

(2x − 1) ⋅

✓ d

(

d 2 d (x + 1) − (x 2 + 1) ⋅ (2x − 1) dx dx = (2x − 1)2 (2x − 1) ⋅ 2x − (x 2 + 1) ⋅ 2 = (2x − 1)2 2 4x − 2x − 2x 2 − 2 = (2x − 1)2 2x 2 − 2x − 2 = (2x − 1)2

d d (1 − x) − (1 − x) ⋅ (1 − 2x) dx dx = (1 − 2x)2 (1 − 2x) ⋅ (−1) − (1 − x) ⋅ (−2) = (1 − 2x)2 2x − 1 + 2 − 2x = 2 (1 − 2x) =

d dx

(1 − 2x) ⋅

1(c)

d d (3x 2 ) − 3x 2 ⋅ (1 − 4x) dx dx (1 − 4x)2

d d (5x) − 5x ⋅ (2x + 1) dx dx = (2x + 1)2 (2x + 1) ⋅ 5 − 5x ⋅ 2 = (2x + 1)2 10x + 5 − 10x = (2x + 1)2 (2x + 1) ⋅

1(b)

(1 − 4x) ⋅

1(f)

d 2x 3 ( ) dx 1 − x =

(1 − x) ⋅

d d (2x 3 ) − 2x 3 ⋅ (1 − x) dx dx (1 − x)2

=

(1 − x) ⋅ 6x 2 − 2x 3 ⋅ (−1) (1 − x)2

=

6x 2 − 6x 3 (1 − x)2

=

6x 2 − 4x 3 (1 − x)2

+ 2x 3

2x 2 (3 − 2x) = (1 − x)2 ✓


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400


Ex 14.4

=

1 2√x (1 + x)2

1 (1 + x 2√x = (1 + x)2

=

2

(√1 − x) √1 − x ⋅ 1

−x⋅

=

1−x +

√1 − x

=

=

1 2√1 − x 1 2√1 − x

1 ⋅ (−1) 2√1 − x

x 2√1 − x

1−x [2(1 − x) + x]

+ x)

− 2x + 1

d

(

x+1

dx √1−4x


− 2x]

− 2x)

=

)

d √1 − 4x ⋅ dx (x + 1)

− (x + 1) ⋅ 2

d − 4x dx √1

(√1 − 4x) √1 − 4x ⋅ (1)

− (x + 1) ⋅

= √1 − 4x =

1 − 4x

1 [(1 − 4x) − 4x = √1 1 − 4x 1 (1 − 4x − 4x √1 = 1 − 4x 1 (3 − 2x) − 4x √1 = 1 − 4x 3 − 2x = (1 − 4x)√1 − 4x ✓

sleightofmath.com

1 ⋅ (−4) 2√1 − 4x

1 − 4x 2(x + 1) + √1 − 4x

1−x

1 (2 − x) 2√1 − x = 1−x 2−x = 2√1 − x(1 − x) ✓

2x 2x √ +1

2(d)

1−x (2 − 2x

1 ⋅ (2) 2√2x + 1

− 2x ⋅

1 [2(2x + 1) 2x +1 √ = 2x + 1 1 (4x + 2 2x + 1 √ = 2x + 1 1 (2x + 2) 2x + 1 √ = 2x + 1 2x + 2 = (2x + 1)√2x + 1 ✓

d (√1 − x) dx

2

d 2x + 1 dx √

2x + 1

=

2√x(1 + x)2

−x⋅

− 2x ⋅

2√2x + 1

− 2x)

d x ( ) dx √1 − x d √1 − x ⋅ dx (x) =

d √2x + 1 ⋅ dx 2x

=

1−x

=

)

√2x + 1 ⋅ 2

− √x ⋅ 1

✓ 2(b)

2x

(√2x + 1)

1 (1 − x) 2√x = (1 + x)2 =

(

dx √2x+1

d d (1 + x) ⋅ (√x) − √x ⋅ (1 + x) dx dx = (1 + x)2 (1 + x) ⋅

d

2(c)

d √x ( ) dx 1 + x

+ 2(x + 1)]

+ 2x + 2)

401


Ex 14.4

2(e) d 3x 2 ( ) dx √2x 2 − 3 √2x 2 − 3 ⋅

=

3

y= dy

d (3x 2 ) dx

− (3x 2 ) ⋅ 2

d √2x 2 − 3 dx

=

dx

=

(√2x 2 − 3)

= 6x√2x 2 − 3 =

=

1 − 3x 2 ⋅ [ ⋅ (4x)] 2√2x 2 − 3 2x 2 − 3

√2x 2 − 3 ⋅ (6x)

−

=

6x 3 √2x 2 − 3

dy

6x

=

4

− 6x 3 ]

dy

− x2]

−(3x2 )⋅

d (1−4x2 ) dx

(1−4x2 )2 (1−4x2 )⋅(6x)

−3x2 ⋅(−8x)

(1−4x2 )2 6x−24x3 +24x3 2 2 (1−4x ) 6x (1−4x2 )2

= (1−4)2 =

2 3

✓

1−x

= =

(1−x)⋅

d (x−2) dx

−(x−2)⋅

d (1−x) dx

(1−x)2 −(x−2)⋅(−1) (1−x)2 +x−2 (1−x)2

(1−x)⋅1 1−x −1

= (1−x)2

(x 2 − 3) Curve cross x − axis: y =0

✓

x−2 1−x

2(f) d 5x ( ) dx √1 − x 2 √1 − x 2 ⋅

x

=0 =2

Gradient at x = 2:

d 5x dx

− (5x) ⋅ (√1 − x 2 )

2

dy

d √1 − x 2 dx

|

dx x=2

−1

= (1−2)2 = −1 ✓

1 − 5x ⋅ [ ⋅ (−2x)] 2√1 − x 2 1 − x2

√1 − x 2 ⋅ (5) = 5√1 − x 2 =

d (3x2 ) dx

x−2

=

dx

√2x 2 − 3(2x 2 − 3)

=

y=

(1−4x2 )⋅

6

|

dx x=1

2x 2 − 3

1 [6x(2x 2 − 3) 2−3 √2x = 2x 2 − 3 6x [(2x 2 − 3) 2−3 √2x = 2x 2 − 3 6x (x 2 − 3) 2−3 √2x = 2x 2 − 3

3x2 1−4x2

+

5x 2 √1 − x 2

1 − x2

1 [5(1 − x 2 ) 2 √1 − x = 1 − x2 5 [(1 − x 2 ) 2 √1 − x = 1 − x2 5 (1) 2 √1 − x = 1 − x2 5 = (1 − x 2 )√1 − x 2 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

+ 5x 2 ]

+ x2 ]

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402


y=

Ex 14.4

x+2

6

√3x+1

dy dx

y=(

5x+3 4 10x−6

)

dy dx

d √3x + 1 ⋅ dx (x + 2) = √3x + 1 ⋅ 1 = √3x + 1 =

d − (x + 2) ⋅ 3x + 1 dx √ 3x + 1 1 − (x + 2) ⋅ ⋅ (3) 2√3x + 1 3x + 1 3(x + 2) − 2√3x + 1 3x + 1

1 [2(3x + 1) 2√3x + 1 = 3x + 1 1 (6x + 2 2√3x + 1 = 3x + 1 1 (3x − 4) 2 3x + 1 = √ 3x + 1 3x − 4 = 2√3x + 1(3x + 1)

= 4( = 4( = 4( = 4( = 4(

− 3(x + 2)] =− − 3x − 6)

5x+3 3

d

10x−6

dx 10x−6

5x+3 3

(10x−6)⋅

) ⋅ ) ⋅

(

5x+3

10x−6

5x+3 3

) ⋅

d d (5x+3) −(5x+3)⋅ (10x−6) dx dx (10x−6)2

(10x−6)⋅5

10x−6

5x+3 3

) ⋅

)

−(5x+3)⋅10 (10x−6)2

50x−30

−50x−30 (10x−6)2

10x−6

5x+3 3

−60

) ⋅ (10x−6)2

10x−6

240(5x+3)3 (10x−6)5

Curve at y = 16: y = 16 (

5x+3 4

) = 16

10x−6 5x+3

10x−6

=2

5x+3

or

10x−6

= −2

5x + 3 = 20x − 12 15x = 15

5x + 3 = −20x + 12 25x =9

x

x

=1

=

9 25


|

dx x=1

=

3−4 2√4(4)

=−

1 2(2)4

=−

1 16

✓

Gradient at x = 1, dy

|

dx x=1 dy

|

dx x= 9

25


sleightofmath.com

= =

9

:

25 −240(5(1)+3)3 (10(1)−6)5 9 25

−240(5( )+3) 9 25

(10( )−6)

5

= −120 ✓ 3

=

1000 3

✓

403


y=√

Ex 14.4 8

1−x

x2 +3

y=√ dy

dy dx

1−x

⋅

2√ 2 x +3

= = = = = = =

At

1

=

dy

1 √1−x √x2 +3

2

√x2 +3 2√1−x √x2 +3 2√1−x √x2 +3 2√1−x √x2 +3 2√1−x 1 2√1−x

⋅

d

⋅ ⋅ ⋅

1−x

)

dx x2 +3 (x2 +3)⋅

d (1−x) dx

−(1−x)⋅

d 2 (x +3) dx

=

(x2 +3)2

⋅[ ⋅

(

=

dx

(x2 +3)⋅(−1)

−(1−x)⋅2x

(x2 +3)2

−x2 −3

x−a b−x 1 x−a 2√ b−x

1

2

d

⋅

⋅ √x−a

(

x−a

dx b−x

(b−x)⋅

)

d (x−a) dx

−(x−a)⋅

d (b−x) dx

(b−x)2

√b−x

]

=

√b−x 2√x−a

⋅

=

√b−x 2√x−a

⋅

(b−x)⋅(1)

−(x−a)⋅(−1) (b−x)2

−2x+2x2 (x2 +3)2

x2 −2x−3

b−x

+x−a (b−x)2

(x2 +3)2

=

(x−1)(x−3) (x2 +3)2

=

(x−1)(x−3)

b−a √b−x ⋅ 2√x−a (b−x)2 b−a 3

2√x−a(b−x)2

3 (x2 +3)2

(x−3)(x+1) 3

2√1−x(x2 +3)2

Gradient at x = dy

= 0:

|

dx x=a+b

dx (x−3)(x+1) 3

2√1−x(x2 +3)2

2√(

=0 2√

= =

=

3 a+b a+b 2 )−a(b−( )) 2 2

3 a+b 2a 2b a+b 2 − ( − ) 2 2 2 2

b−a 3 b−a b−a 2 2√ ( ) 2 2

b−a 2(

=

:

b−a

=

x = −1 or x = 3 ✓

2

b−a

=

2

a+b

b−a 2 ) 2

b−a (b−a)2 2 22

b−a (b−a)2 2

2

= (b − a) ⋅ (b−a)2 =


sleightofmath.com

2 b−a

[shown] ✓

404


Ex 14.4

1 3

x+h

y= √ dy dx

=

x−k

1 x+h

= (

3 x−k

= =

=

=

)

x+h 3 ( ) x−k 2 − 3

1

⋅ ⋅

2 x+h 3 3( ) x−k

1 1 2 x+h 3 3[( ) ] x−k

1 1 2 x+h 3 3[( ) ] x−k

1

⋅

⋅

d

−(1) (

x+h

dx x−k (x−k)⋅

d (x+h) dx

−(x+h)⋅

d (x−k) dx

dy

(x−k)2 (x−k)⋅1

k

=

dx

mtan =

−(x+h)⋅1

x−k

9(ii)

dy dx

−1

2√x

−1

Tangent & its gradient

−x−h

y=

(x−k)2

25 4

mtan = 0 ⋅ (x−k)2

−(2)

Using intersection, k√x − x =

25

−(1)

4

Using gradients, 1 3y2

⋅

−(k+h) (x−k)2

k 2√x k

− 1= 0 =1

2√x

3y 2 (x − k)2

− 1)

2√x k

(x−k)2

=

dx

k √x

= k√x − x

sub (1) into (2): dy

y = x(

)

−k−h

1 2 x+h 3 3[( ) ] x−k

10(i) Curve & its gradient

= −(k + h) [shown] ✓

k

= 2√x

−(2)

Tangent ∥ to y − axis:

sub (2) into (1):

dy

⇒∞

(2√x)√x − x =

⇒∞

2(x) − x

=

x

=

dx 1 1 2 x+h 3 3[( ) ] x−k

⋅

−k−h (x−k)2

⇒ x = −h or x = k (rej ∵ k > x > h) y|x=−h = 0 ⇒ (−h, 0)

k|x=25 = 2√ 4

25 4

25 4 25 4 25 4 5

= 2( ) = 5✓ 2

10(ii) Curve gradient with = k = 5: dy dx

=

5 2√x

−1

As x → ∞, ⇒

dy dx

5 2√x

→0

→ −1

∴ gradient of l is − 1 l passes through O and gradient is −1 ⇒ y = −x ✓


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405


Ex 14.4

1 + x + x 2 + ⋯ + x n−1 + x n

=

xn+1 −1 x−1

Diff wrt x, 0 + 1 + 2x + ⋯ + (n − 1)x n−2 + nx n−1 1 + 2x + 3x 2 + ⋯ + (n − 1)x n−2 + nx n−1

= = =

(x−1)

d (xn+1 −1) dx

−(xn+1 −1)

d (x−1) dx

(x−1)2 (x−1)[(n+1)xn ]

−(xn+1 −1)(1)

(x−1)2 (x−1)(n+1)xn

−xn+1 +1

(x−1)2

✓

11(ii) x = 3, n = 10: 1 + 2(3) + 3(3)2 + ⋯ + (10 − 1)(3)10−2 + 10(3)10−1 = 1 + 2(3) + 3(3)2 + ⋯ + 9(3)8 + 10(3)9

=

(3−1)(10+1)310

−310+1 +1

(3−1)2 2(11)310

−311 +1 4

= 280483 ✓ 12

y=

y

′

u2 v3

= = =

v3 ⋅

d (u2 ) dx

−u2 ⋅

d (v3 ) dx

(v3 )2 v3 ⋅(2u⋅u′ ) 2uu′ v3

−u2 ⋅(3v2 ⋅v′ ) v6 −3u2 v2 v′

v6


✓

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406


Rev Ex 14 A3

Rev Ex 14 A1(a)

d dx

=

(√1 − 3x 2 )

= (1 − 3x 2 ) 2

=

A1(b)

d dx

1 − 2

⋅

1 2√1−3x2 3x

=−

dx

[(1 − 3x 2 )2 ]

dx 1

√1−3x2

d dx

=x⋅

dx

=

=

[(1 + 2x)

4]

+

d dx

(x) ⋅ (1 + 2x)

+1 ⋅ (1 + 2x)

= 8x(1 + 2x)3

+(1 + 2x)4

4

= (1 + 2x)3 (8x +1 + 2x) = (1 + 2x)3 (10x + 1) ✓ d

(

1+2x √x

)= =

d

(

1

=− x 2

=− A2

+ 2√x)

dx √x 1 d (x −2 ) dx 1 −3 2

1

+2

(x 2 )

dx 1 1 +2 ( x −2 ) 2 1

+

2x√x

√x

A4

9

y = (2x−5)2 = 9(2x − 5)−2

✓

dy

y = (2x + 3)√5x − 9

= 9[−2(2x − 5)−3 ] ⋅

dx

= 9[−2(2x − 5)

dy

=

dx

= (2x + 3) ⋅ = (2x + 3) ⋅ = = = = =

d dx

(√5x − 9) + 1

2√5x−9

⋅5

5(2x+3) 1 1 2√5x−9 1 2√5x−9

−3 ]

d dx

(2x − 5)

⋅2

36 − (2x−5)3

(2x + 3) ⋅ √5x − 9

+2

Gradient at x = −1: ⋅ √5x − 9

dy

|

dx x=−1

36

= − (2(−1)−5)3 =

36 343

✓

+2√5x − 9

2√5x−9 2√5x−9

d dx

+ (x + 1)]

⇒ n=0✓

1

d

1 ⋅ (−2) 2√1 − 2x

1 − 2x

1 [(1 − 2x) − 2x √1 = 1 − 2x 1 (2 − x) − 2x √1 = 1 − 2x 2−x = √(1 − 2x)3

4

d − 2x dx √1

1 − 2x x+1 + √1 − 2x

√1 − 2x

= x ⋅ 4(1 + 2x) ⋅ 2

dx

− (x + 1) ⋅

√1 − 2x ⋅ 1

✓

− (x + 1) ⋅

1 − 2x

⋅ (−6x)

3

A1(c)

d √1 − 2x ⋅ dx (x + 1)

=

(1 − 3x 2 )

[x(1 + 2x)4 ] d

x+1 √1−2x

dy 1

d

y=

[5(2x + 3)

+4(5x − 9)]

(10x + 15

+20x − 36)

(30x − 21)

30x−21 2√5x−9

⇒ k = 30 ✓


sleightofmath.com

407


y=

Rev Ex 14

x2 −1

B1(b)

x2 +1

dx

dy dx =

=

=

=

(x 2 + 1) ⋅

d 2 d (x + 1) − (x 2 − 1) ⋅ (x 2 − 1) dx dx (x 2 + 1)2

(x 2 + 1) ⋅ 2x

=− =−

− 2x 3 + 2x

=−

(x 2 + 1)2

(x 2

=−

4x + 1)2

B2

4x

x

y|x=0 = (0)2

1

dx 9(8−7)2

5 33 5

−

(4√x − 1)

8 5

⋅ [4

8 5

(4√x − 1)

66 5√x

(4√x − 1)

−

d dx

⋅ [4 (

8 − 5

⋅ 8 5

d

(√x) − (1)] dx 1

2√x

)−0]

2 √x

✓

1

2

4

1 2

Curve

+1

y = 4x +

= −1

dx

]

8

9

x

−3

d

9

2 3

8 ( −7) x

]

2

( − 7)

4x +

dx x

2

d dx

(x −1 ) −

d dx

2

(7)]

3

16 3

9x2 ( −7)

x2

1

1

2 1

4

= x+ c

x 8

= c

x

4

= 14x +

32 x

−(1)

Equate gradients: 1

8

8 x

x+

8

c

⋅ (− 2 )

=4−

2 8

x

9( −7)

8 x

⋅ [8

⋅ [8(−x −2 ) −0]

3 8 9( −7) x

8 x2 8

Tangent intersects curve:

8

7 1

= 4 +8(−x −2 )

mtan = 4 −

= [−2 ( − 7) ] ⋅ = [−

x

=4 −

−2 1 d 8 [( − 7) ] 9 dx x 1

8

= 4x +8x −1

x

=

33

1

dy

=−

5

−

(4√x − 1)

= x+ c

⇒ (0, −1) ✓

=−

33

mtan =

=0 (0)2 −1

=

d

=0

(x2 +1)2

[

8

Tangent 4y = 2x + c y

=0

dx

−

3

(x 2 + 1)2

dy

d

3 d − [(4√x − 1) 5 ] dx

= 11 [− (4√x − 1) 5 ] ⋅ (4√x − 1) 5 dx

− (x 2 − 1) ⋅ 2x

2x 3 + 2x

3

[11(4√x − 1) 5 ]

= 11

tangent ∥ x − axis:

B1(a)

−

d

=

x2 x2

✓

=

8

8 x2

7 2 2 7 16

x

= ±√

7

−(2)

sub (2) into (1): c| 16 = 16√7 ✓ x=√

c|

7

x=−√

16 7

= −16√7 ✓

Note: Consider the discriminant approach © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

408


Rev Ex 14

Curve y = x 3 − 3x 2 − 9x + k dy dx

B5(i)

y= dy

2

= 3x − 6x − 9

ax2 x−b (x−b)⋅

=

dx

mtan = 3x 2 − 6x − 9

Using intersection: x 3 − 3x 2 − 9x + k = 0 k = −x 3 + 3x 2 + 9x

−ax2 ⋅

d (x−b) dx

(x−b)2 −ax2 ⋅1

(x−b)⋅2ax

=

Tangent x − axis ⇒ y = 0 mtan = 0

d (ax2 ) dx

(x−b)2

=

2ax2 −2abx (x−b)2

=

ax2 −2abx (x−b)2

=

ax(x−2b) (x−b)2

−ax2

1

Curve at ( , 6): 3

Using gradients: 3x 2 − 6x − 9 = 0 x 2 − 2x − 3 =0 (x − 3)(x + 1) = 0 x=3 or x = −1 k|x=3 = 27 ✓ k|x=−1 = −5 ✓

6

=

2 − 6b = a

1 2 3

a( )

1 −b 3

a 9

= 18 − 54b 1

tangent at ( , 6) is horizontal: 3

B4

dy

y = x√a − 3x 2 dy

=x⋅

dx

=x⋅ = = = =

d dx

(√a − 3x 2 ) 1

2√a−3x −3x2

√a−3x2 1 √a−3x2 1 √a−3x2 a−6x2

+

d dx

+√a − 3x 2

[−3x

2

+(a − 3x

2 )]

1

a|b=1 = 18 − 54 ( ) = 9 ✓

(a − 6x 2 )

6

B5(ii) Curve & its gradient with a = 9, b = 1: 6

√a−3x2

y= dy dx

9x2 x−

=

1 6

1 3 1 2 (x− ) 6

9x(x− )

For horizontal tangent:

Tangent at (1, b) ∥ to y = 2x + 1: mtan = mline

√a−3

6

6

b = √a − 3

|

1

or b = ✓

(rej ∵ a ≠ 0)

(b) = (1)√a − 3(1)2

dx x=1 a−6

=0

a=0

Curve at (1, b):

dy

=0

3 1 1 a( )( −2b) 3 3 2 1 ( −b) 3

⋅ √a − 3x 2

⋅ (−6x) +1 2

|

dx x=1

(x) ⋅ √a − 3x 2

dy dx 1 9x(x− ) 3 1 2 (x− ) 6

=2 =2

=0 =0

x = 0 or x =

a−6 = 2√a − 3 2 a − 12a + 36 = 4(a − 3) a2 − 12a + 36 = 4a − 12 a2 − 16a + 48 = 0 (a − 12)(a − 4) = 0 a = 12 or a = 4 (rej)

1 3

(taken)

y|x=0 = 0 ⇒ (0,0) ✓

b|a=12 = 1√12 − 3 = 3 ✓


sleightofmath.com

409


Ex 15.1 2(a)

Ex 15.1 1(a)

y = x 3 − 5x + 8 dy dx

y = x 3 + 3x 2 dy dx

= 3x 2 + 6x

Tangent Point:

= 3x 2 − 5

Normal Point:

y|x=1 = 13 − 5(1) + 8 =4

(−1,2) dy

Gradient:

⇒ (1,4) 2

|

= 3(−1) + 6(−1)

dx x=−1

−1

Gradient:

dy | dx x=1

= −3 Tangent: y − y1 =

dy

|

dx x=−1

Normal:

(x − x1 )

y − y1

y=x +

2

Gradient:

=2−

2

= −1 dy

|

dx x=1

7

2

2

y

= 10 8

= 10

x

2x 2 + 8 = 10x x2 + 4 = 5x 2 x − 5x + 4 = 0 (x − 4)(x − 1) = 0 x = 4 or x = 1 (rej ∵ x > 1) ⇒ (4,10)

= 1 − (1)2

Tangent: y − y1 =

2

1

x2

(1,3) |

1

x 8

Normal Point:

x2

dx x=1

[x − (1)]

8

y = 2x + dy

2

dy

2

= x+ ✓

y

2x + Tangent Point:

1

(x − x1 )

2

2

= 1 +2(−x −2 ) =1 −

=

−2

1

1

dx dx

−1

|

y−4 = x−

= x +2x dy

=

−1

= dy

y − (4) =

2(b)

x −1

−1 3(1)2 −5

dx x=1

y − (2)= (−3)[x − (−1)] y − 2 = −3 (x − x1 ) y − 2 = −3x − 3 y = −3x − 1 ✓ 1(b)

=

(x − x1 )

y − (3)= (−1)[x − (1)] y − 3 = −(x − 1) y − 3 = −x + 1 y = −x + 4 ✓

Gradient:

Normal:

−1 dy | dx x=4

=

y − y1

−1 2−

=−

8 (4)2

−1

= dy

|

2 3

(x − x1 )

dx x=4


sleightofmath.com

2

y − 10

=−

y − 10

=− x+

y

=− x+

3 2

(x − 4) 8

3 2

3 38

3

3

✓

410


y= dy dx

Ex 15.1

x2 +5

3(b)

x+1

= = = =

(x+1)⋅

d (x2 +5) dx

−(x2 +5)⋅

y = √1 − 2x dy

d (x+1) dx

dx

(x+1)2

=

−(x2 +5)⋅1

(x+1)⋅2x (x+1)2 2x2 +2x

=

1

2√1−2x 1

=−

−x2 −5

⋅

d

2√1−2x dx 1

(1 − 2x)

⋅ (−2)

2√1−2x

(x+1)2

Tangent & Normal Point: y =3 √1 − 2x = 3 1 − 2x = 9 −2x =8 x = −4 ⇒ (−4,3)

x2 +2x−5 (x+1)2

Tangent & Normal Point:

y|x=3 =

32 +5

=

3+1

7 2

7

⇒ (3, ) 2

Gradient:

dy

|

dx x=3

Tangent: y − y1

=

32 +2(3)−5 (4)2

=

7

dy

|

dx x=3 5

=

5

Gradient:

8

(x − x1 )

2

y−

8 5

8 13

8

8

= x−

2

y Normal:

15

= x+

y − y1

= − dy

1

dy

1 √1−2(−4)

✓

4

3 1

3 5

3

y − y1

3

= − dy

7

1

2

( )

y− y− y

7 2 7 2

=−

5 8

8 5 8

[x − (3)]

y − (3) = −

(x − 3)

y−3 y−3 y

=− x+

24

5 8

5 83

5

10

=− x+

1

(x − x1 )

|

dx x=−4

dx x=3

y−( ) =−

3

=− x+ ✓

y Normal:

1

[x − (−4)]

1

y−3 =− x−

=−

(x − x1 )

|

dx x=−4 1 3

(x − x1 )

|

=−

y − 3 = (− )

8

5

|

dx x=−4

Tangent: y − y1 =

y − ( ) = ( ) [x − (3)] 7

dy

1 −

1 3

[x − (−4)]

(x + 4) =3 = 3x + 12 = 3x + 15 ✓

✓ 4(i)

dy dx

=

2−x √4x−x2 +1

Tangent Point: Gradient:

P(1,2) dy

|

dx x=1

Tangent: y − y1

=

2−(1) √4−1+1

=

dy

=

|

dx x=1 1

1 2

(x − x1 )

y − (2) = ( ) [x − (1)] 2

4(ii)

Normal:

1

1

2 1

2 3

2

2

y−2

= x−

y

= x+ ✓

y − y1

= − dy

1 |

(x − x1 )

dx x=1

y − (2) = − y−2 y © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

1 1 2

[x − (1)]

= −2x + 2 = −2x + 4 ✓ 411


y= dy dx

Ex 15.1

3x+1

6

1−x

= = =

(1−x)⋅

d (3x+1) dx

(1−x)⋅3 3−3x

−(3x+1)⋅

dy

d (1−x) dx

dx

(1−x)2 −(3x+1)⋅(−1) (1−x)2 +3x+1 (1−x)2

= 6x − 2

Line & its gradient 4y + x = 2 4y = −x + 2

4

= (1−x)2 Normal Point:

Curve & its gradient y = 3x 2 − 2x + 5

y

mline = − At x-axis, y =0 3x+1 1−x

x

1 3

⇒ (− , 0)

Normal:

−1

y − y1

1

=

4 4 2 ( ) 3

y

=

= − dy

1 4(9) 16

=−

1 |

dx x=−1 3

y − (0) = − y

4

mtan ⋅ (− )

= −1

mtan

=4

4

3

dy | dx x=−1 3

2

1

1

1

Gradient:

1

4

Tangent Gradient: ∵ Tangent ⊥ line m1 ⋅ m2 = −1 mtan ⋅ mline = −1

=0 =−

1

=− x+

=−

4 9 4 9 4

Point:

9

(x − x1 ) 1

[x − (− )] 3

6x − 2 =4 6x =6 x =1 2 y|x=1 = 3(1) − 2(1) + 5 =6 ⇒ (1,6)

1

(x + )

=− x− 9

4

3

4 27

Tangent: y − y1 y − (6) y−6 y

✓

7

= mtan (x − x1 ) = (4)[x − (1)] = 4x − 4 = 4x + 2 ✓

1st curve & its gradient y = x 3 + 2x 2 − 4x + 5 dy dx

= 3x 2 + 4x − 4

2nd curve & its gradient 2

y = x 3 + 3x 2 − x − 1 3

dy dx

= 2x 2 + 6x − 1

tangents are ∥: m1 = m2 2 3x + 4x − 4 = 2x 2 + 6x − 1 x 2 − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = 3 or x = −1 ✓


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412


Ex 15.1

y = ax 5 dy

= 5ax

dx

9(b) 4

y

Curve at (1, b): b = a(1)5 =a dy dx dy

=

3 2

=

5a

=

a

=

b|a= 3 = 10

dy

2 3

dx

10 3 10


✓

1

l

2

2

− x + = x 2 − 2x + 3

=0 =

dy

2

= 2x 2 − 3x + 6

l

Origin (0,0)

Tangent: y − y1

Using gradients, |

dx x=0

(x − x1 )

−

y − (0) = (0)[x − (0)] y =0✓ 9(a)

3

= x2 − x + 3

2

|

2x−2

Using intersection,

✓

dx x=0

2

−1

mnorm =

dy

2

= 2x − 2

l

8(ii)

l

2 1

Curve y = x 2 − 2x + 3

3

2 3

1

=− x+

mnorm = −

at (1, b):

|

dx x=1

Normal 2y + x = l 2y = −x + l

1 2

2x − 2 2x x

=−

1 2x−2

=2 =4 =2

Note: Consider the discriminant approach Tangent y = 3x + k mtan = 3

l|x=2 = 2(2)2 − 3(2) + 6 =8✓ 10(i) y = px 3

Curve y = 2x 2 − 5x + 3 dy dx

y B(0,8) 2 𝑂 A

= 4x − 5

mtan = 4x − 5

y = px 3 dy

Using intersection, 3x + k = 2x 2 − 5x + 3 k = 2x 2 − 8x + 3

dx

= 3px 2

Gradient at A mtan =

Using gradients, 4x − 5 = 3 4x =8 x =2

dy

|

dx x=2

= 3p(2)2 = 12p ✓

10(ii) Point A At A, y = px 3 has coordinate of x = 2 y|x=2 = p(2)3 = 8p ⇒ A(2,8p)

k|x=2 = 2(2)2 − 8(2) + 3 = −5 ✓

Gradient of AB A(2,8p) B(0,8) mAB =


x

sleightofmath.com

8p−8 2−0

= 4p − 4 ✓

413


Ex 15.1

10(iii) Equate gradients: mtan = mAB 12p = 4p − 4 8p = −4 1

=− ✓

p 11(i)

12(ii) Tangent at P(2, −3): (−3) = (1 + 2a)(2) − a2 −3 = 2 + 4a − a2 a2 − 4a − 5 =0 (a − 5)(a + 1) = 0 a = 5 or a = −1 ✓

2

y = 2x 2 − kx + 3 dy dx

Tangent at a = 5: y = (1 + 2(5))x −(5)2

= 4x − k

mtan =

dy

|

dx x=1

=4−k ✓

Tangent at a = −1: y = (1 + 2(−1))x −(−1)2

11(ii) Point A y|x=1 = 2(1)2 − k(1) + 3 =5−k ⇒ A(1,5 − k)

13(i)

Gradient of AB A(1,5 − k) B(5,1) (5−k)−(1) (1)−(5)

=

y=

dy dx

4−k

4

=

dx

4−k

Gradient:

8 x3

4

y|x=a = ⇒ (a,

−4

a2

4 a2

Gradient: mtan =

+1

+ 1) dy

|

dx x=a

8

=−

a3

Tangent: y − y1

= mtan (x − x1 )

4

+ 1) = (− 3 ) [x − (a)]

y−(

a2

8

a

a2 y − 4 − a2 = −

= 1 + 2x

Tangent Point:

+1

= 4(−2x −3 ) +0

Point:

Curve & its gradient y = x + x2 dy

−2

Tangent

4k − 16 = 4 − k 5k = 20 k =4✓ 12(i)

+1

x2

=−

−4

Equate gradients: mtan = mAB 4−k

−1 ✓

= −x

= 4x

mAB =

−25 ✓

= 11x

Gradient at A

8 a 8

(x − a)

a2 y − 4 − a2 = − x + 8 2

2

y|x=a = a + a ⇒ (a, a + a2 ) dy

|

dx x=a

a 8

a y

= − x + a2 + 12

y

=−

a 8

a3

x+1+

12 a2

✓

= 1 + 2a

Tangent: y − y1

=

dy

|

dx x=a

(x − x1 )

y − (a + a2 ) = (1 + 2a)[x − (a)] y − (a + a2 ) = (1 + 2a)x −a − 2a2 y = (1 + 2a)x −a2 ✓


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414


Ex 15.1

13(ii) Gradient of PQ P(b, 0),Q(0, b) mPQ =

0−b b−0

14(ii) Normal Point: A(−1,6) or C Gradient: ∵ AC ⊥ BD

= −1

mAC =

Equate gradients: mtan = mPQ − −

8

=

a3 8

Normal:

a3 a

5 1

0−b

y − 6 = − (x + 1)

b−0

y

1

5 1

5 29

5

5

=− x+

Point C 29

5

5

)

meets −axis (y = 0). y =0 1

29

5

5

− x+ 1

Tangent at (b, 0): (0) = −(b) + 4 b =4✓ B(−2.2,0) D(0,11) F(−1,0)

1

At C, normal (y = − x +

12

= − (2)3 x + 1 + (2)2 = −x + 1 + 3 = −x + 4

14(i)

5 1

y−6 =− x−

=8 =2✓

8

5

y − y1 = mAC (x − x1 ) 1

Tangent with a = 2: y

−1

=

y − 6 = − [x − (−1)]

= −1

a3

−1 mBD

5

x

=0 =

29 5

x = 29 ⇒ C (29,0) ✓ 14(iii) Area △ ABC = 1 (BC)(AF) 2 1

y = x + 7x + 12

= [29 − (−2.2)][6]

dy

= 3(31.2) = 93.6 ✓

2

dx

2

= 2x + 7

mtan = 2x + 7 (11)−(0)

mBD = (0)−(−2.2) = 5 Equate gradients: mtan = mBD 2x + 7 = 5 2x = −2 x = −1 y|x=−1 = (1)2 + 7(1) + 12 = 6 ⇒ A(−1,6) ✓


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415


Ex 15.1

𝐂𝐮𝐫𝐯𝐞 & its gradient (y − 2)2 = x y−2 = ± √x y = 2 + √x or 2 − √x dy dx

1

=

2√x

or −

Line & its gradient x − 2y = 4 2y =x−4

=

= x−2 2

=

2

Point:

⇒

1

=

2

=

1

=

2√x

=

1

=

2

x =1 y|x=1 = 2 + √1 =3 ⇒ (1,3)

y−3

16

dy

=

y

1

(x − 1)

2 1

1

2 1

2 5

2

2

y−3

= x−

y

= x+ ✓

16

5−x3 2√ x

1 5−x3 2√ x

1 5−x3 2√ x

|

=

at (1,2):

5−x3

(1,2) lie in y = √

x


⋅

−(5−x3 )⋅

x⋅(−3x2 )

d (x) dx

−(5−x3 )⋅1 x2

⋅ ⋅

−3x3

−5+x3 x2

−x3

−5 x2

⋅ (−2x −

1

5 x2

)

5

5−(1)3 2√ (1)

1

⋅ [−2(1) − (1)2 ] ⋅ (−2 − 5)

2√4

4

d dx

(y 2 ) + dy dx

dy dx

dy

1

(1,2) does not lie in

d dx

+1 +y 2

(x) ⋅ y 2

=0

⋅ y2

=0 =0 = −3x 2 − y 2

dx

=

dx

5−(1)3

2 = −2 [inconsistent]

y = −√

)

d (5−x3 ) dx

dy

−3x2 −y2 2xy

At (1,2), dy

5−x3

x

Method 2 (Without subjecting y)

2xy

x

(2) = −√

2 = 2 [consistent]

5−x3

(x)2

3x 2 +x ⋅ 2y ⋅

at (1,2):

5−(1)3 (1)

x⋅

(

7

3x 2 +x ⋅

or y = −√

x

(2) = √

1

3x 2 +2xy

5−x3

⋅

d dx

x 3 +xy 2 =5 Differentiating with respect to x:

x

y=√

3

5−x 2√

=−

5−x3

=

1

=

Method 1 (Subjecting y) xy 2 = 5 − x 3 2

3

5−x 2√

dx x=1

= mtan (x − x1 )

Tangent: y − y1

1

⋅

x

Tangent Gradient: ∵ tangent ∥ line, mtan = mline mtan

1 5−x3 2√ x

x

1

mline =

2√x

=

dx

1

y

1

dy

dx

=

−3(1)2 −(2)2 2(1)(2) 7

=− ✓ 4

5−x3 x

sleightofmath.com

416


Ex 15.1

Curve & its gradient y = x3 + 5 dy dx

= 3x 2


y|x=1 = 6 ⇒ (1,6) dy

|

dx x=1

Tangent: y − y1

=3 =

dy

|

dx x=1

(x − x1 )

y − (6) = (3)[x − (1)] y−6 = 3x − 3 y = 3x + 3 ✓


17(ii) Point At point where curve meets tangent, x3 + 5 = 3x + 3 3 x − 3x + 2 =0 (x − 1)( ) (x − 1)(x 2 ) 2 (x − 1)(x −2) 2 (x − 1)(x + x − 2) = 0 (x − 1)(x + 2)(x − 1) = 0 x=1 or x = −2 (taken) y|x=−2 = 3(−2) + 3 = −3 ⇒ (−2, −3) ✓

sleightofmath.com

417


Ex 15.2 1(a)

Ex 15.2

y = 4x − 1

y= dy

=

dx dy dx

2x x+1

=4✓

=

>0 ⇒ y is increasing ✓

= =

1(b)

y = x 2 + 2x dy dx

1(c)

5

y= dy

3

dx

= 𝑥2 − 1

(x+1)2 −2x(1) (x+1)2 2x+2 −2x (x+1)2 2 (x+1)2 2 (x (x+1)(2)

x

=

3

= √x ✓ 2

=

> 0 for x > 0 ⇒ y is increasing ✓ 2(a)

(x2 +1)⋅

x2 +1

−2x2 (x2 +1)2

1−x2 +1)2

For decreasing function:

+ − + −1 1

f ′ (x) = −3x 2 ✓ < 0 for 𝑥 > 0 ⇒ f(x) is decreasing ✓

x < −1 or x > 1 [shown] ✓ 6(a)

y = x 2 + 2x − 4 dy

′ (x)

5

= 6x − 5 ✓ < 0 for x < 0 ∵ x 5 < 0 ⇒ f(x) is decreasing ✓ f

dx

y = x 3 + 3x dy dx

3(ii)

dy dx

= 2x + 2

For increasing function: dy dx

3(i)

<0

⇒ 1 − x2 < 0 ∵ (x 2 + 1)2 > 0 x2 − 1 >0 (x + 1)(x − 1) > 0

f(x) = −(x 3 + 1)

f(x) = x 6 − 5x

<0

dx 1−x2 (x2 +1)2

′ (x)

= −1 ✓ <0 ⇒ f(x) is decreasing ✓

2(c)

−x⋅2x

(x2 +1)2

dy

2(b)

d d (x) −x⋅ (x2 +1) dx dx (x2 +1)2

(x2 +1)⋅1

= (x2

f(x) = −9 − x f

d (x+1) dx

x2 +1

= dx

−(2x)⋅

+ 1) > 0 for all x values except x ≠ −1 ⇒ y is increasing for all real values of x [shown] ✓

y = √x 3 − 1

dy

d (2x) dx

>0 ∵

= 2x + 2 ✓

> 0 for x > 0 ⇒ y is increasing ✓

(x+1)⋅

>0

2x + 2 > 0 x > −1 ✓

= 3x 2 + 3 ✓ > 0 for all x values ∵ x 2 ≥ 0

⇒ y increases for all real values of x


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418


y = 2x 3 − 3x 2 + 6 dy dx

Ex 15.2 9(i)

dy

−2x3 +2x (x2 +1)2

4x +1)2

For increasing function: f ′ (x) > 0

y = 3x 2 + 4x − 3 = 6x + 4

4x (x2 +1)2

>0

4x x

> 0 ∵ (x 2 + 1)2 > 0 >0✓

9(ii)

For decreasing function: f ′ (x) < 0 ⇒x <0✓

10

T = −5x 2 + 90x + 65

<0

dx

2x3 +2x

= (x2

x < −1 or x > 1 ✓

dy

−(x2 −1)⋅2x (x2 +1)2

=

+ − + −1 1

For decreasing function:

d d (x2 −1) −(x2 −1)⋅ (x2 +1) dx dx (x2 +1)2

(x2 +1)⋅2x

=

6x 2 − 6 >0 x2 − 1 >0 (x + 1)(x − 1) > 0

dx

(x2 +1)⋅

f ′ (x) =

>0

dx

dy

x2 +1

= 6x 2 − 6x

For increasing function:

7(a)

x2 −1

f(x) =

6x + 4 < 0 2

<− ✓

x 7(b)

3

3

dT dx

2

y = 2x − 9x + 12x − 3 dy dx

= 6x 2 − 18x + 12

For cooling: dT

+

− 1

+

−10x + 90 < 0 10x > 90 x >9✓ 11(a)

𝑦

2

10 3 4(x + 1) > 0 x+1 >0 x > −1 ✓ 8(ii)

<0

dx

For decreasing function: 6x 2 − 18x + 12 < 0 x 2 − 3x + 2 <0 (x − 1)(x − 2) < 0

8(i)

= −10x + 90

𝑥

𝑂

11(b)

𝑦

𝑦 = (𝑥 + 1)3

−1 𝑂

11(i)

𝑥

y = f(x)

(a) y = x 3

For decreasing function: f ′ (x) < 0 ⇒ x < −1 ✓


𝑦 = 𝑥3

(b) y = (x + 1)3

sleightofmath.com

Interval Nature

Sign of

x>0 x<0 x > −1 x < −1

+ + + +

Increasing Increasing Increasing Increasing

dy dx

419


Ex 15.2

11(ii) No.

12(ii) For decreasing function:

By Definition, if

dy dx

≥ 0 for all x, then the function is

(non-strictly) increasing. If function is increasing, necessarily

dx

for x >

√2x−1

= = = =

d √2x−1⋅dx(3x+4)

could be 0, not

𝑥<

−

7 3

1

7

2

3

1

dy

−(3𝑥+4)]

dx

2𝑥−1 (6𝑥−3 −3𝑥−4)

= = =

For increasing function:

=

>0 3

3

>0

=

=

2𝑥−1 (3𝑥−7)

3 (2x−1)2

dx 3x−7

∵ (2x − 1)2 > 0 for x >

(2x−1)2

3x − 7 > 0

2

x≠a

x−a

3x+4 √2x−1

2𝑥−1 3x−7

dy

1

13(i) y = x2 −2ax−2,

2x−1 [3(2𝑥−1)

and x >

⇒
1 (2) −(3x+4)⋅ 2√2x−1

3√2x−1

= √2𝑥−1

3

Combining inequalities:

d −(3x+4)⋅ √2x−1 dx 2

2

<0

2

2x−1

1

x>

<

1

7

1

√2x−1⋅3

1 √2𝑥−1

3

∵ (2x − 1)2 > 0 for x >

<0

3x − 7 x

(√2x−1)

= √2𝑥−1

=

dx

<0

dx 3x−7 3 (2x−1)2

>0

dx

3x+4

12(i) y = dy

dy

dy

dy

(x−a)⋅

d (x2 −2ax−2) dx

−(x2 −2ax−2)⋅

d (x−a) dx

(x−a)2 −(x2 −2ax−2)⋅1

(x−a)⋅(2x−2a)

(x−a)2 2x2 −4ax+2a2 −x2 +2ax+2 2 (x−a) (x2 −2ax)

+2a2 +2

(x−a)2 [(x−a)2 −a2 ]

+2a2 +2

(x−a)2 (x−a)2

+a2 +2 (x−a)2

a2 +2

1

= 1 + (x−a)2 ✓

2

13(ii) For increasing function:

7

dy

3

>0

dx a2 +2

Combine inequalities: 𝑥>

7 3

⇒x>

and x >

1 2

7 3


1 + (x−a)2 > 0 2+a2 (x−a)2

> −1

⇒ all ℝ values of a ∵ a2 + 2 > 0 and (x − a)2 > 0 for x ≠ a ✓

sleightofmath.com

420


Ex 15.3 4(i)

Ex 15.3 1(i)

h(t) = =

t(10−t)

r =3+

5 4

= t − t2

4(ii)

dr

−1

= 0 +2 [(1+t)2 ] ⋅ (1)

dt 5

5

2 5

4 5

2

2

−2

= (1+t)2

h′ (t) = − (2t) = − t

dr

|

5

2 3

2

42 1

h′ (2) = − (2)

= − cm s −1 ✓

−1

The balloon is deflating.

= km s 2

1(ii)

2

=−

dt t=3 5

1+(0)

= 5 cm ✓

4 2

2

r|t=0 = 3 +

4 10t−t2 5

2 1+t

5

8

✓ 5(i)

5

h′ (6) = − (6) 2

1

f(t) =

10 000 000

2

(6t 2 + 5)2

1

= − km s −1 ✓

11am: t = 60

2

2(i)

l = dl

t3 3

− 4t + 10

= t2 − 4

dt

dl

5(ii)

6(i)

v

s= + 8

ds

ds

v 40 1

|

= + 8

60 40

5

=1 h ✓ 8

= 2t 6(ii) |

dt t=2

= 2(2)

7(i)(a)

= 4cm s −1 ✓ 3(ii)

80

8

r =t +2

dr

v2

= +

dv

2

dt

10 000 000

1

dv v=60 dr

144(60)3 +120(60)

= 3.11 ≈ 3 people per min ✓

t − 4 = −4 t2 =0 t =0✓ 3(i)

[2(6t 2 + 5)][12t]

10 000 000

f ′ (60) =

= −4

dt 2

(6(60)2 + 5)2

10 000 000 144t3 +120t

=

For length is decreasing at 4 mm/s: dl

1

f ′ (t) =

=5

t −4 =5 2 t −9 =0 (t + 3)(t − 3) = 0 t = −3 (rej ∵ t ≥ 0) or t = 3 ✓ 2(ii)

10 000 000

≈ 46 ✓

For length is increasing at 5 mm/s: dt 2

1

f(60) =

C(x) =

=

t

The initial radius is 2 cm and is increasing at an increasing rate. © Daniel & Samuel A-math tuition 📞9133 9982

8

19 200 x

Annual gasoline costs if car gets 3km l−1 = C(3)

r r = t2 + 2 2 𝑂

5

The time taken to stop at v = 60km h−1 is 1 h ✓

sleightofmath.com

19 200 3

= $6400 ✓

421

A math 360 sol (unofficial) 7(i)(b) Annual gasoline costs if car gets 8km l−1 = C(8) =

Ex 15.3 7(ii)

19 200

Average rate of change in Ben’s annual gasoline costs =

8

= $2400 ✓

=

C(8)−C(3) 8 2400−6400 8−3

= −$800 per km/l ✓ 7(iii)

C ′ (x) = − C


sleightofmath.com

′ (5)

=−

19200 x2 19200 52

= −$768 per km/l ✓

422


Ex 15.4 1(c)

Ex 15.4 1(a)

dx dt

dx dt

= 2,

dy

|

dt x=1

=?

= 2,

dy dx dy dt

x 1

dy

x2

dt

dy

×

dx

= (4x −

1

x2 2

= 8x −

dx dt

) ×2

|

dt x=1

dx dt

dy

= 2,

|

dt x=2

dy

dx

×

dx

dt

At y = 10, x 3 + 2 = 10 x3 =8 x =2

= 8(1) −

2 12 −1

= 6 units s 1(b)

=

= 3x 2 × 2 = 6x 2

x2 dy

At x = 1,

=?

= 3x 2

dx

1

= 4x − =

|

dt y=10

y = x3 + 2 dy

y = 2x 2 +

dy

dy

✓

=?

= 6(2)2

|

dt x=2

= 24 units s −1 ✓ 1(d)

dx dt

= 2,

dy

|

dt y=2

=?

3

y = (2x−3)3 y=

= 3(2x − 3)−3

dy dy dx

= dy dt

dx

= 3[−3(2x − 3)−4 ] ⋅ 2

= = =

=

18 − (2x−3)4 dy 18 − (2x−3)4 36 − (2x−3)4

At x = 2,

= ×

dx

dy

|

dt x=2

x x+1 (x+1)(1) −x(1) = (x+1)2 x+1 −x (x+1)2 1 (x+1)2

dx dy

dt

=

dt

×2

dy

×

dx

dx dt

1

= (x+1)2 × 2 2

= (x+1)2 36

= − [2(2)−3]4

At y = 2,

= −36 units s −1 ✓

x x dy

|

dt x=−2


x x+1

sleightofmath.com

=2 = 2x + 2 = −2 2

= (−2+1)2 = 2 units s −1 ✓

423


dy dt

= 4,

dx

|

dt x=3

Ex 15.4 2(c)

=?

dt

y = x 3 − 2x 2 dy

= 4,

dy

= 3x − 4x =

dt

dy dx

=

dt

dt dx

dy

dt

4

dx

=

dt x=3

dt

= 4,

y= dy dx

=

= =

4 = dx dt

dt x=2

4

1

15

×

√2x+7

dx dt dx dt

= 4√2x + 7

dt

✓

At y = 3, √2x + 7 = 3 2x + 7 = 9 x =1

=?

1+x

=

dt

|

=

×

dx

3x2

=

dy

dx

4 3(3)2 −4(3)

⋅2

dy

4 =

3x2 −4x

At x = 3, | dy

=?

√2x+7

=

dt

dx

2(b)

2√2x+7 1

=

dx

×

4 = (3x 2 − 4x) × dx

|

dt y=3

1

=

dx

dy

dx

y = √2x + 7

2

dx

dy

=

dx

(1+x)⋅

d (3x2 ) dx

−3x2 ⋅

d (1+x) dx

|

dt x=1

= 4√2(1) + 7 = 12 unit s −1 ✓

(1+x)2

2(d)

−3x2 ⋅1

(1+x)⋅6x

dt

(1+x)2 6x+6x2

dy

dx

|

dt y=5

=?

−3x2 (1+x)2

y = x(x − 4) = x 2 − 4x

6x+3x2 (1+x)2 dy dx 6x+3x2 (1+x)2

× ×

dx

dy

dt dx

dx

dt

dy

4(1+x)2

dt

6x+3x2

4

dx

At x = 2, |

= 4,

dt x=2

=

4(1+2)2 6(2)+3(2)2

3 2

unit s −1 ✓

=

dy

×

dx

= (2x − 4) ×

dx

=

= 2x − 4

=

dt

=

dx dt dx dt

4 2x−4 2 x−2

At y = 5, x 2 − 4x =5 2 x − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1 (rej) dx

|

dt x=5


sleightofmath.com

=

2 5−2

=

2 3

units s −1 ✓

424


y= dy

2x−1

= = = 3(ii)

4(ii)

x+1

=

dx

Ex 15.4

dt dy

(x+1)⋅

d (2x−1) dx

−(2x−1)⋅

dx dy

d (x+1) dx

(x+1)2 (x+1)⋅2 −(2x−1)⋅1 (x+1)2 2x+2 −2x+1 (x+1)2 3 ✓ (x+1)2

y|x=0 =

−1 1

dy

|

dx x=0

=3

dt

dt dx dt

=3 =3

= −1 5(a) 3

=

⋅

dx

dx

(x + 1)2 =4 2 x + 2x + 1 =4 2 x + 2x − 3 =0 (x + 3)(x − 1) = 0 x = −3 or x = 1 ✓

⇒ A(0, −1) Gradient:

=3

dx 12 (x+1)2

Tangent Point:

dy

((0)+1)

2

=3

Let A ≡ area of circle & r ≡ radius dA dr = 2π, | =? dt dt r=6 A = πr 2 [area of circle]

Tangent: y − y1

=

dy

|

dx x=0

dA

(x − x1 )

dr

y − (−1) = 3 [x − (0)] y = 3x − 1 ✓ 3(iii)

dx dt

= 0.03,

dy dt

dA dt

=

dt

= =

dy

×

dx 3 (x+1)2 0.09 (x+1)2

dr

y= dy dx

dr

dr

×

dt dr

=

dt

1 r

dt

At r = 6,

× 0.03 5(b)

dr

|

dt r=6

=

1 6

cm s −1 ✓

Let A ≡ area of circle & r ≡ radius dA dr = 10π, | =? dt dt r=2

0.09

|

dt x=0

4(i)

dA

dx

At A(0, −1), dy

=

2π = 2πr ×

at A(0, −1) = ?

dt dy

= 2πr

= (0+1)2 = 0.09 units s −1 ✓

A = πr 2 [area of circle] dA

2x−10

dr

x+1

= = =

(x+1)⋅

d (2x−10) dx

(x+1)⋅2 2x+2

−(2x−10)⋅

dA

d (x+1) dx

dt

(x+1)2 −(2x−10)⋅1 (x+1)2 −2x+10 (x+1)2

= 2πr =

dA dr

×

10π = 2πr × dr dt

12

=

dt dr dt

5 r

At r = 2,

= (x+1)2

dr

dr

|

dt r=2

5

= = 2.5 m s −1 ✓ 2

2

> 0 [shown] ∵ (x + 1) > 0 for x ≠ −1 ✓


sleightofmath.com

425


Ex 15.4

dV dx = 3, = ? dt dt

7(b)

Let A be Surface Area, V be Volume & x be length of side dA

2

dt

V = x + 3x [given] dV dx dV dx

= 2x + 3 ×

dx

3 = (2x + 3) × dx

=

dt

6(ii)

dx

3 2x+3

|

dt x=3

=

|

dt x=6

=

A=x dA dx dA dt

3

=

1 3

cm s −1

3

=

1 5

dx

5

=

× ×

cm s −1 ✓

dV

=

=

3 A 2 ( ) 2 6

⋅

dV

dA

×

dA

=

1

A

4

6

1

√

1

1

A

4

6

= √

6

dt

20

A 6 A

At x = 1, 1 = √

dV

dt dx

dt A=6

8(i)

x

At A = 4, x 2 = 4 ⇒ x = 2

dt x=2

−(2)

= √ × 0.2

dx

dt

6

1

dV

[area of square]

dA

A

A 2 ( ) 6

= 2x =

⇒x=√

3

A 6

2(6)+3

= 2x

|

= x2

6

V = (√ ) =

3

dx

dx

A

sub (2) into (1):

dt

✓

2(3)+3

10 dt

−(1)

dt dx

Let A be area & x be side dA dx = 10, | =? dt dt A=4 2

=?

A = 6x 2 ⇒

dt

7(a)

|

dt x=1

dx

dA dx

dV

V = x3

dV

=

= 0.2,

|

=

1 20

6

⇒1=

A 6

⇒A=6

6

√ = 0.05 cm3 s −1 ✓ 6

Let y ≡ Length & x ≡ Breadth dx dy = 4, =? dt dt y = 2x [given] dy =2 dx

5

= = 2.5 cm s −1 ✓ 2

dy dt

=

dy dx

⋅

dx dt

= (2) ⋅ (4) = 8 cm/s


sleightofmath.com

426


Ex 15.4

Let A ≡ Area of rectangle dx dt

= 4,

dA

|

10(ii)

dt

=?

dt x=8

dx

= 2,

da

|

dt x=4

=?

1

a = x(x + 1) A = xy −(1) y = 2x −(2) sub (2) into (1): A = x(2x) = 2x 2

= da dx

da dA

=

dt

dA

×

dx

dt

dx

At x = 8,

dA

11(i)

dp

dA

|

dt p=3

1

3

2

2

dt

=

dt

× 3

2

2

=( + p =

2)

dp dt

dA dx dA

|

dt p=3

10(i)

dt 3

4

4

dt

=? −(1)

Let a be area of triangle 1

a = (2x)(x + 1) sin 150° 2

1 2

= 8x + 1 =

dA

×

dx

dx dt

= (8x + 1) × 1.2 = 9.6x + 1.2

= + (3)2 = 7 units 2 s −1 ✓

= x (x + 1)

dA

× 0.5

1 3 2 + p 4 4

1

= 1.2,

sub (2) into (1): A = x(4x + 1) = 4x 2 + x

At p = 3, dA

= 2(4) + 1 = 9 cm2 s −1

Q(2x, y) lies on y = 2x + 1: y = 2(2x) + 1 = 4x + 1 −(2)

=?

dA dp 1

dt

×2

A = xy

= + p2

dA

dx

Let A ≡ Area of parallelogram OPQR dx

= p + p3 ✓ = 0.5,

×

dx

dt x=4

(1 + p2 ) 2

da

da

= (p) 1

2

At x = 4, |

= 16(8) = 128 cm2 s −1 ✓

|

2 1

2

2

1

= 2x + 1

dt x=8

2 1

=

1

2

1

dA

2

= (x + )

A = (height) (sum of bases)

dt

1

1

dt

= 4x × 4 = 16x

dp

2

= (2x) +

= 4x

dx

9(ii)

1

+ x

=x+

dA

9(i)

2 1 2 x 2

C

At x = 1.5,

dA

|

dt x=1.5

= 9.6(1.5) + 1.2 = 15.6 units/s

x+1 150° A 2x B

1

= x (x + 1) [shown] ✓ 2


sleightofmath.com

427


Ex 15.4

11(ii) Let z be length of diagonal OQ dx dz = 1.2, | =? dt dt x=1.5

13

dx dt

By distance formula,

= 0.6,

−(1)

dy

=

dx

= sub (2) into (1): = √4x 2 + (4x + 1)2

dy

=

dt

= √4x 2 + (16x 2 + 8x + 1)

=

= √20x 2 + 8x + 1 dz dx

= =

dz dt

= = =

1 2√20x2 +8x+1 20x+4

dz

√20x2 +8x+1 24x+4.8

dt

dy

× 1.2

dt x=1.5

≈ 5.36 units per sec ✓

14

dP dt

dt

z dz

dI

dx

×

dP

×

dx −x

dx dt

× 0.6

√36−x2 −0.6x √36−x2

=

= −50,

−0.6(3.6) √36−3.62

= −0.45

dz

|

dt x=144

=?

𝑥 𝑧

270

= √x 2 + 72900 =

dz dt

= = =

|

dy

1 2√x2 +72900 x

⋅ 2x

√x2 +72900

dI dt

= 800I × 2 = 2400I At I = 3,

⋅ (−2x)

√36−x2

|

= dP

2√36−x2 −x

By Pythagoras’ Theorem, z 2 = x 2 + 2702

= 800I =

𝑥

Let z be distance between helicopter & lorry dx

P = I2 R = 400I 2 ∵ R = 400

dI

𝑦

∴ The rate at which the ladder is sliding down is 0.45m s −1

dI dP = 2, | =? dt dt I=3

dP

1

dt x=3.6

√20x2 +8x+1 dz

6

dx

× 20x+4

=?

At y = 4.8, x 2 + 4.82 = 62 x2 = 12.96 x = 3.6

√20x2 +8x+1

At x = 1.5, | 12

=

⋅ (40x + 8)

dx

|

dt y=4.8

= √36 − x 2

y

Q(2x, y) lies on y = 2x + 1, y1 = 2(2x) + 1 = 4x + 1 −(2)

z

dy

By Pythagoras’ Theorem, x 2 + y 2 = 62 y2 = 36 − x 2

z = √(0 − 2x)2 + (0 − y)2 = √4x 2 + y 2

Let x be distance from wall & y be distance from ground

dt I=3

= 2400(2) = 4800 W s −1 ✓


dz dx

× x

√x2 +72900 −50x

dt

× (−50)

√x2 +72900

At x = 144,

sleightofmath.com

dx

dz

|

dt x=144

= −23

9 17

m s −1 ✓

428


Ex 15.4

Let x ≡ top of man′ s shadow from the lamp y ≡ man from the lamp By similar triangles, 2

=

7

|

1

1

1

2

3

6

≈ 11.1 kmh−1

6

y 17(a)

x

d dr

= x✓

(πr 2 ) = 2πr

𝛿𝐴 𝑟 𝛿𝑟

7

dy dt dy dx dy dt 5 3 dx dt

16(i)

h after B reaches P

dt t=−1

5

y

3

dx

2

x

1

⇒ t = (− ) + = −

7

x−y

2x = 7x − 7y 7y = 5x

15(ii)

16(ii)

5 dx

= ,

3 dt

=

LHS =?

=

=

dx 5 7 7

× ×

= ms

dx dt dx dt −1

3

= lim ✓

δr→0

= 2πr = RHS

B (0,30) at t = 0

17(b)

P A (0,0) at t = 0

+ (3600t 2 + 3600t + 900)

=

1 2√9225t2 +3600t+900 9225t+1800

d

4

( πr 3 ) = 4πr 2

dr 3

𝑟 𝛿𝑟 = =

⋅ (18450t + 3600)

√9225t2 +3600t+900

d

4

( πr 3 )

dr 3 d dr

dt

=

9225t+1800

δV δr→0 δr

= lim

= lim

√9225t2 +3600t+900

−1

≈ 87.2 kmh

(4πr2 )δr δr

𝛿𝑟

δV ≈ (Surface Area) (breadth) (δr) ≈ (4πr 2 )

∵ δV = (4πr 2 )δr

dx | ≈ 87.2 kmh−1 dt t=1

= lim 4πr 2 δr→0

3


4𝜋𝑟 2 𝛿𝑉

(V)

δr⇒0

dx

𝛿𝑉

LHS

= √9225t 2 + 3600t + 900 =

δr

= lim 2πr

x = √[(−75t) − 0]2 + [0 − (30 + 60t)]2

dt

δA ≈ (length) (breadth) ≈ (2πr) (δr)

∵ δA = (2πr)δr

A(−75t, 0) B(0,30 + 60t)

= √5625t 2

𝛿𝑟

(2πr)δr

δr→0

t ≡ time after A reaches P x ≡ distance between the two motorists

dx

2𝜋𝑟 𝛿𝐴

=

7 dy

(πr 2 )

d (A) dr δA = lim δr⇒0 δr

5

=

d dr

= RHS

sleightofmath.com

429


Rev Ex 15 A2(ii) For ⊥ tangents: m1 ⋅ m2

Rev Ex 15 A1(i)

10

y= dy

x

( |

−x

=−

dx

dy

10 x2

dx x=−3 1

−1

10 3

1

−3=

3

⇒ (3, ) 3

−1 dy | dx x=3

=

=1

=

10

− 2 −1 3

9

A3(i)

19

dy

−1

y − y1

= dy

(x − x1 )

|

1

y−

=

3 1

y−

=

3

y

=

9 19 9 19 9 19

(x − 3) x− x−

Gradient:

27 19 62 57

2

−

x2

−1 =−

10

=

x2 2

7 2 7

5

dy

2

10 −2

dx

−2

y|x=2 =

10 2

−2

=3 ⇒ (2,3) ✓

8y = x 2 − kx + 17 1

1

17

8

8

8

= x 2 − kx +

y dy

1

1

4

8

= 9(1)2 − 5 = 4

|

dx x=1

=

dy

|

dx x=1

(x − x1 )

=

dy

|

dx x=1

9x 2 − 5 =4 2 x −1 =0 (x − 1)(x + 1) = 0 x=1 or x = −1 (taken) y|x=−1 = 3(−1)3 − 5(−1) + 4 =6 ⇒ (−1,6) ✓

x=2

= −3 ⇒ (−2, −3)✓ A2(i)

dy

A3(ii) tangent ∥ tangent at A: m1 = m2

2

x =4 x = −2 or y|x=−2 =

A(1,2)

y − (2) = (4)[x − (1)] y−2 = 4x − 4 y = 4x − 2 ✓

A1(ii) Tangent has gradient − : =−

= −64 = −64 =0 =0 =2✓

Tangent: y − y1

✓

7

10

8

= 9x 2 − 5

Tangent Point:

dx x=3

dx

= −1

y = 3x 3 − 5x + 4 dx

dy

)

(k + 6)(k − 10) k 2 − 4k − 60 k 2 − 4k + 4 (k − 2)2 k

1

Normal:

dx x=5 1

8

y|x=3 =

Gradient:

) ( |

[− (6 + k)] [ (10 − k)] = −1

Normal Point:

= −1

dy

= x− k

dx

Tangent gradient at x = −3, x = 5: dy

1

1

= (−3) − k

|

dx x=−3

4

=−

8 1

6

− k

8 1

8

= − (6 + k) ✓ 8

dy

|

dx x=5

1

1

4 5

8 1

= (5) − k =

4 1

− k 8

= (10 − k) ✓ 8


sleightofmath.com

430


Rev Ex 15

Curve

A5

y = ax + dy dx

b

=a−

b

dy

x2

dx

−1

mnorm = dy

|

=

dx x=2

−1 b a− 2 2

=

−1 b a− 4

=

−4 4a−b

> 0 [shown] ✓ for x > 2 ∵ 2x > 0, (x − 2) > 0, (x − 1)2 > 0 A6(i)

V= dp

b (2)

A6(ii)

dp

b = 14 − 4a

−(1)

dV

=

= 3,

dV dp 60

=−

=1

p2

dV

|

dt p=20

× ×

=?

dp dt dp dt

At p = 20, dV

= 2x + 6 x

=−

60 202

×3

= −0.45 units 3 s −1 ✓ A7(i)

Let A ≡ area of circular patch r ≡ radius Equate area of circle, πr 2 =

for a = 2, b = 6

−2x + 11

|

dt p=20

A4(ii) Point Q At Q, normal (y = −2x + 11) meets y = 2x +

−4x + 11 −

✓

p2

= −2

2 = 4a − b −(2) (1) sub into (2): 2 = 4a − (14 − 4a) 2 = 4a − 14 + 4a 2 = 8a − 14 16 = 8a a =2✓ b|a=2 = 14 − 4(2) =6✓

x

60

|

Equate gradients,

6

p

=−

dt

4a−b

60

dt p=20

= 7 − 2a

4a−b 2

2x2 −4x (x−1)2 2x(x−2) (x−1)2

=

Curve at P(2,7):

−4

(x−1)2 4x2 −4x −2x2 (x−1)2

=

dV

(7) = a(2) +

(x−1)(4x) −(2x2 )(1)

= =

Normal y + 2x = 11 y = −2x + 11 mnorm = −2

2

x−1

x

Gradient of normal at P(2,7)

b

2x2

y=

6

dA dt

(t)

πr 2 = (4)(16) = 64 64

r2

=

r

=√

π 64 π

x

=0

4x 2 − 11x + 6 = 0 (x − 2)(4x − 3) = 0 x = 2 (taken) or x =

3 4 3

6

4

3 4

y|x=3 = 2 ( ) + 4

=

19

2 3 19

⇒ Q( , ) 4


2

sleightofmath.com

431

A math 360 sol (unofficial) A7(ii)

dA dt

dr

= 4, |

dt t=16

Rev Ex 15 B1(i)

=?

y= dy

A = πr 2 dA dr

dx

x √x−2 d

=

dt

=

dA

×

dr

4 = 2πr × dr

=

dt

dr dt dr

=

dt

=

2 πr

= 64

At t = 16, r = √ ,

=

π

dr

|

dt t=16

A8(i)

dθ dt

2

=

π√

π ds

= ,

2 dt

= 64 π

2 π(

8 √π

)

=?

=

2 8√π

=

1

ds

(√x−2)

4√π

x−2

=

−

√x−2

dt

d (√x−2) dx

=

[2(x−2)

−x]

x−2 1 2√x−2 1 2√x−2

(2x−4

−x)

x−2 (x−4)

x−2 x−4 2(√x−2)

3

[proven] ✓

Point:

y|x=4 =

(4) √(4)−2

⋅

=

4 √2

= 2√2

⇒ (4,2√2) Gradient: − dy

dθ

x 2√x−2

B1(ii) Normal

=8 ds

1 ⋅1 2√x−2

x−2 1 2√x−2

1 |

dx x=4

ds

2

−x⋅

√x−2⋅1

cm s −1 ✓

s = rθ [arc length] = 8θ ∵ r = 8

dθ

−x⋅

= 2πr =

dA

√x−2⋅dx(x)

dθ

Normal:

=

−1

⇒ −∞

(4)−4 2(√(4)−2)

3

x=4✓

dt π

= (8) ⋅ ( ) 2

= 4π cm s −1 ✓ A8(ii)

dθ dt

π dA

= ,

2 dt

=?

1

A = r 2 θ[sector area] 2 1

= (8)2 θ

∵r=8

2

= 32θ dA dθ dA dt

= 32 =

dA dθ

⋅

dθ dt π

= (32) ⋅ ( ) 2

= 16π cm2 s −1 ✓


sleightofmath.com

432


Rev Ex 15

B2(a) 1st curve y = ax 2 + bx + 2 dy dx

B2(b) Tangent y = 2x − 16 mtan = 2

= 2ax + b

Curve y = ax 3 + bx

1

Tangent gradient at (1, ): 2

dy

= 2a + b

dy

1st curve at (1, ):

1

Gradient of tangent at x = 2

1

mtan =

mtan =

|

dx x=1

dx

2

2

= a(1)2 + b(1) + 2

= 3ax 2 + b

dy

|

dx x=2

= 3a(2)2 + b = 12a + b

3

a=− −b

At point of contact, y|x=2 = 2(2) − 16 = −12 ⇒ (2, −12)

2

2nd curve y = x 2 + 6x + 4 dy dx

= 2x + 6 (2, −12) lies on curve: −12 = a(2)3 + b(2) −12 = 8a + 2b −12 − 8a = 2b b = −6 − 4a

Normal gradient at (−2, −4): mnorm = − dy

1 |

=−

dx x=−2

1 2(−2)+6

=−

1 2

1

Tangent to 1st curve at (1, ) ∥ normal to 2nd curve 2

at (−2, −4): mtan

= mnorm

2a + b

=−

3

2 (− − b) + b

=−

−3 − 2b + b

=−

−3 − b

=−

2

=

b

=− 5

2

2

1 2 1 2 1 2 1 2

5

−b 3

Using gradients, 12a + b =2 12a + (−6 − 4a) = 2 8a − 6 =2 8a =8 a =1 b|a=1 = −6 − 4(1) = −10 ✓

2 5 2

a|b=−5 = − − (− ) = 1 ✓ 2


sleightofmath.com

433


Rev Ex 15

Curve y = (x + 1)2 dy

B4

Curve y=

= 2(x + 1)

dy

mtan = 2(x + 1)

dx

dx

1

3

4

4

= x+

mline =

P(a, b) dy

|

dx x=a

=− =

ab a2 dy

=−

b a

(x − x1 )

|

dx x=a b

y − (b) = (− ) [x − (a)] a

∵ Tangent ⊥ line, mtan ⋅ mline = −1 1

mtan ⋅ ( )

= −1

mtan

= −4

4

Tangent:

x2

Tangent: y − y1

4

Point:

ab

=−

Gradient:

1

Tangent Gradient:

x

Tangent Point:

Line 4y = x + 3 y

ab

dy

|

dx x=−3

(x − a)

y−b

=−

y−b

=− x+b

y

= − x + 2b

a b a b a

Point Q At Q, tangent meets x-axis (y = 0) y =0

2(x + 1) = −4 x+1 = −2 x = −3 (−3 y|x=−3 = + 1)2 =4 ⇒ (−3,4) y − y1 =

b

b

− x + 2b = 0 a b

− x

= −2b

a

x = 2a ⇒ Q(2a, 0)

(x − x1 )

[x − (−3)] y − (4)= (−4) (x + 3) y − 4 = −4 y − 4 = −4x − 12 y = −4x − 8 ✓

Point R At R, tangent meets y-axis (x = 0) b

y|x=0 = − (0) + 2b a

= 2b ⇒ R(0,2b) Distance PQ & RP P(a, b) Q(2a, 0)

R(0,2b)

PQ = √(a − 2a)2 + (b − 0)2 = √a2 + b 2 RP = √(0 − a)2 + (2b − b)2 = √a2 + b 2 ∴ PQ = RP [shown] ✓ B5(i)

v

v2

4

60

s= + ds

ds

1

v

4

30

= +

dv

|

dv v=45


sleightofmath.com

✓

1

(45)

4

30

= +

=

7 4

km2 h−1 ✓

434


Rev Ex 15

B5(ii) If one is travelling at 45km h−1 , for every 1km h−1 increase in speed, the stopping distance increases by 1.75 km ✓ B6(i)

dp dt dA dp

22

y=√ dy

x

22 2√ −x x

1

=

22 2√ −x x

(− (

22 x2

− 1)

x2

1 x

) =

= 2,

dt dy

=

dt

= =

dy

|

dt x=2

−22−x2 22 2x2 √ −x x

−22−x2

=

3

=

|

dt p=5

1

=? 1

= (2p) + (3p2 ) 2

=

−22−x2

=

22−x2 2x2 √ x

=

2 3 2 + p 2

dA

×

dp 3 (p + p2 ) 2 9 3p + p2 2

✓

2x2 √22−x2

2x2 √ √22−x2 dx

dA dt

−22−x2

−22−x2

=

dA

= 3,

=p

−x 1

=

dx

B6(ii)

B7(ii)

dA

At p = 5,

|

dt p=5

B8(i) ×

dx −22−x2 3

2x2 √22−x2

dt

×3

9

= 3(5) + (5)2 2

= 127.5 ✓

=?

dy

dp

dx

Let x ≡ man from lamp y ≡ shadow length

dt

dx dt

×2

dy

= 2,

dt

1.5

=? y

5

x

By similar triangles,

−22−x2

y

3

x2 √22−x2

1.5

=

x+y 5 3

3

2 3

2

5y = x + y At x = 2, dy

|

dt x=2

=

7 −22−22 3 22 √22−22

=

−26 2√2√18

=

−13 √36

=−

13 6

2

cm s −1

y = x

y

2 3

= x 7

✓ B7(i)

dy

y = x2 y|x=p = p2

dx dy dt

1

A = (RP)(PQ) 2 1

= [p − 2 1

= =

=

2 1 2 p 2

1

dy dx 3

⋅

dx dt

7

6

= ms −1 7

= (p + 1)p2 = (p2 + p3 )

7

= ( ) ⋅ (2)

(−1)](p2 )

2 1

3

B8(ii) z ≡ top of shadow from lamp z=x+y

+ p3 ✓ 2

Speed of the top of his shadow = = =

dz dt d dt dx dt

(x + y) +

=2 + =


sleightofmath.com

20 7

dy dt 6 7 −1

ms

✓

435


Ex 16.1 3(i)

Ex 16.1 1(a)

y = x 2 − 5x + 1 dy dx

=

At stationary point, =0

dx

=

2x − 5 = 0 x

=

2

2

5 2

5

2

2

= −5

=

1

1

2

4

4

(x 2 − 4) ⋅ 2

− (2x − 5) ⋅ 2 (x 2

−

4)2

2x 2 − 8

− 4x 2 + 10x (x 2 − 4)2

=

−2(x 2 − 5x + 4) (x 2 − 4)2

=

−2(x − 4)(x − 1) (x 2 − 4)2

y = 5 − 6x + x 2 dx

d d (2x − 5) − (2x − 5) ⋅ (x 2 − 4) dx dx (x 2 − 4)2

−2x 2 + 10x − 8 = (x 2 − 4)2

1

⇒ (2 , −5 ) ✓

dy

(x 2 − 4) ⋅

5

y|x=5 = ( ) − 5 ( ) + 1

1(b)

x2 −4

dy dx

= 2x − 5 ✓

dy

2x−5

y=

= −6 + 2x ✓

✓ At stationary point, dy

3(ii)

=0

dx

dy

−6 + 2x = 0 x =3 y|x=3 = −6 + 2(3) = −4 (3, ⇒ −4) ✓ 2(i)

y=

At stationary point, dx −2(x−4)(x−1) (x2 −4)2

dx

4(i)

y = 2x 3 − 9x 2 + 12x − 4

2x+1 dy

x−1

= = =

(x−1)⋅

d (2x+1) dx

−(2x+1)⋅

(x−1)2 −(2x+1)⋅(1) (x−1)2 −2x−1 (x−1)2

(x−1)⋅(2) 2x−2

4(ii)

At stationary point, dy

=0

dx

−3

3

= 6x 2 − 18x + 12 = 6(x 2 − 3x + 2) = 6(x − 1)(x − 2) [shown]✓

d (x−1) dx

6(x − 1)(x − 2) = 0 x = 1 or x = 2 y|x=1 = 1 y|x=2 = 0 ⇒ (1,1) ✓ ⇒ (2,0) ✓

= (x−1)2 ✓ 2(ii)

=0

x = 4 or x = 1 ✓

dx dy

=0

3

− (x−1)2 ≠ 0 ∵ − (x−1)2 < 0 for x ≠ 1 ⇒ no stationary pt ✓ 4(iii)

d2 y dx2 d2 y

= 12x − 18 |

= −6 < 0 ⇒ max pt (1,1) ✓

|

=6 >0

dx2 x=1 d2 y dx2 x=2


sleightofmath.com

⇒ min pt (2,0) ✓

436


Ex 16.1 6(b)

y = (x − 5)√7 + x

y = x 4 − 8x 2 + 2 dy

dy

d

= (x − 5) ⋅

dx

dx

= = = = 5(ii)

d

√7 + x + dx (x − 5) ⋅ √7 + x 1

= (x − 5) ⋅

dx

2√7+x

(1) +(1)

At stationary point,

⋅ √7 + x

dy

dx 3x+9 2√7+x

x

+√7 + x

2√7+x x−5

4x 3 − 16x =0 x 3 − 4x =0 2 x(x − 4) =0 x(x + 2)(x − 2) = 0 x=0 or x = −2 or x = 2 y|x=0 = 2 y|x=−2 = −14 y|x=2 = −14 ⇒ (0,2) ⇒ (−2, −14) ⇒ (2, −24)

+2(7+x) 2√7+x

x−5

+14+2x 2√7+x

3x+9 2√7+x

✓

d2 y

=0 =0

d2 y

= −3

dy

−3

−3+

−

0

+

sign

dx

= 3x 2 − 24x + 36

dy

=0 2

3x − 24x + 36 = 0 x 2 − 8x + 12 =0 (x − 2)(x − 6) = 0 x=2 or x = 6 y|x=2 = 32 y|x=6 = 0 ⇒ (2,32) ⇒ (6,0)

At stationary point, =0 2

3x − 12 =0 2 x −4 =0 (x + 2)(x − 2) = 0 x = −2 or x = 2 y|x=−2 = 16 y|x=2 = −16 ⇒ (−2,16) ✓ ⇒ (2, −16) ✓ d2 y dx2

d2 y dx2 d2 y

= 6x

2|

dx x=−2 d2 y

|

dx2 x=2

= 6x − 24 |

= −12 < 0

⇒ max. pt (2,32) ✓

|

= 12 > 0

⇒ min. pt (6,0) ✓

dx2 x=2 d2 y

d2 y

⇒ min pt (2, −14) ✓

= 32 > 0

dx

dx

⇒ min pt (−2, −14) ✓


= 3x 2 − 12

dy

⇒ max pt (0,2) ✓

= x(x − 6)2 = x(x 2 − 12x + 36) = x 3 − 12x 2 + 36x

y

dy

y = x 3 − 12x dx

= 32 > 0

|

dx

dy

|

dx2 x=2

⇒ min pt ✓ 6(a)

= −16 < 0

dx2 x=−2 d2 y

6(c) −3−

|

dx2 x=0 d2 y

Sign Test: x

= 12x 2 − 16

dx2

y|x=−3 = [(−3) − 5]√7 + (−3) = −16 ⇒ (−3, −16) ✓ 5(iii)

=0

dx

x−5


= 4x 3 − 16x

= −12 < 0

= 12 > 0


⇒ max pt (−2,16) ✓

dx2 x=6

⇒ min pt (2, −16) ✓

sleightofmath.com

437


y = 2x +

Ex 16.1

18

7(b)

x

y = x2 +

= 2x + 18x −1 dy

dy

= 2x + 16(−x −2 )

dx

= 2 − 18x −2 =2−

x

= x 2 + 16x −1

= 2 + 18(−x −2 )

dx

16

= 2x − 16x −2

18

= 2x −

x2

16 x2

At turning point,


dy

dy

=0

dx 18

2−

=0

x2

2

=

=0

dx

2x −

18

16

=0

x2

2x

x2

2

=

16 x2

3

x =9 x2 − 9 =0 (x + 3)(x − 3) = 0 x = −3 or x=3 y|x=−3 = −12 y|x=3 = 12 ⇒ (−3, −12) ✓ ⇒ (3,12) ✓

2x = 16 x3 − 8 =0 2 (x − 2)(x + 2x + 4) = 0 x=2 y|x=2 = 12 ⇒ (2,12) ✓

d2 y

d2 y

dx2

= −18(−2x −3 ) =

d2 y dx2 d2 y

| |

dx2

36

=2+

x3

x=−3

dx2 x=3

= 2 − 16(−2x −3 )

=

=

36 −27

36 26

<0

>0


⇒ max pt (−3, −12) ✓

d2 y

|

dx2 x=2

32 x3

= 6 > 0 ⇒ min pt (2,12) ✓

⇒ min pt (3,12) ✓

sleightofmath.com

438


Ex 16.1

4x2 +9

y=

7(d)

x

= 4x +

9

dy

x

dx

= 4x + 9x −1

x2

y=

x+1

= =

dy

= 4 + 9(−x −2 )

dx

=

= 4 − 9x −2 =4−

dx 9

4x 2

2

2

or

x=

y|x=−3 = −12

3

d2 y

2

3

3

2

2

dx2

⇒ (− , −12) ✓ ⇒ ( , 12) ✓

dx2

= −9(−2x =

d2 y

=

|

dx2 x=3 2

=0

(x+1)2 ⋅

d (x2 +2x) dx

−(x2 +2x)⋅

d (x+1)2 dx

(x+1)4 (x+1)2 ⋅(2x+2)

−(x2 +2x)⋅2(x+1) (x+1)4 −(x2 +2x)⋅(2)

(x+1)⋅(2x+2) (x+1)3 2x2 +4x+2

−2x2 −4x (x+1)3

2

=−

2

d2 y

=

−3 )

x3

|

= =

18

dx2 x=−3

=0

2

y|x=3 = 12

2

=0

x=0 or x = −2 y|x=0 = 0 y|x=−2 = −4 ⇒ (0,0) ⇒ (−2, −4)

3

(x + ) (x − ) = 0

d2 y

−x2

(x+1)2

dx x2 +2x (x+1)2 x(x+2) (x+1)2

=0

4 3

2

(x+1)2 2x2 +2x

dy

=9 9

x=−

(x+1)2 (x+1)(2x) −x2 (1)


=0

x2

3

d (x+1) dx

x2 +2x

=0

x −

−x2 ⋅

x2

dy

2

d (x2 ) dx

= (x+1)2

9


4−

(x+1)⋅

=

16 3

16 3

<0

>0

= (x+1)3

3

⇒ max pt (− , −12) ✓ 2

d2 y

3

⇒ min pt ( , 12) ✓ 2

|

dx2 x=0 d2 y

|

=2 >0

dx2 x=−2


sleightofmath.com

= −2 < 0

⇒ min pt (0,0) ✓ ⇒ max pt (−2, −4) ✓

439


Ex 16.1

x −1 2 x +1 x +0x +0 −(x 2 +x) −x +0 −(−x −1) 1

8(i)

y = 8x +

= 8x + x 2

dy

1

= 8 + (−2x −3 )

dx

2

= 8 − x −3

x2

y=

=8−

x+1

= (x − 1) +


=0

dx dx

1

= 1 + [−(x + 1)−2 ](1)

8−

= 1 − (x + 1)−2

8

=

x3

=

x

=

1

= 1 − (x+1)2 At stationary point, dy

=0

x3

1 x3 1 8 1 2

y|x=1 = 6

=0

dx

1 x3

1 x+1

= (x − 1) + (x + 1)−1 dy

1 2x2 1 −2

2

1

⇒ ( , 6) ✓

1

1 − (x+1)2 = 0

2

1 = (x + 1)2 1 = x 2 + 2x + 1 0 = x 2 + 2x x(x + 2) = 0 x=0 or x = −2 y|x=0 = 0 y|x=−2 = −4 ⇒ (0,0) ✓ ⇒ (−2, −4) ✓

8(ii)

d2 y

= −(−3x −4 )

dx2

= d2 y

3 x4

|

dx2 x=1

= 48 > 0

2

1

d2 y dx2

⇒ min pt ( , 6) ✓ 2

= 0 −[−2(x + 1)−3 ](1) 9(i)

2

= (x+1)3

y = x 3 − 6x 2 + 3 dy dx

d2 y

|

dx2 x=0 d2 y 2|

=2 >0

dx x=−2

= −2 < 0

⇒ min pt (0,0) ✓

9(ii)

= 3x 2 − 12x ✓


⇒ max pt (−2, −4) ✓

=0

dx 2

3x − 12x = 0 x 2 − 4x = 0 x(x − 4) = 0 x = 0 or x = 4 ✓ 9(iii)

dy

<0

dx

3x 2 − 12x < 0 x(x − 4) < 0 +

− 0

+ 4

⇒ 0 < x < 4 [shown]


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440


y=

Ex 16.1

2x−1

11(ii) At k = 12,

x2 +2

3

x = √− dy dx

=

(x 2 + 2)(2) (x 2

− (2x − 1)(2x) + 2)2

2x 2 + 4

dy

− 4x 2 + 2x

dx2 d2 y

(x 2 + 2)2

|

= 36 > 0

⇒ min pt (−1, −9) ✓ 12

y = ax +

b x2

= ax + bx −2

−2(x − 2)(x + 1) (x 2 + 2)2 ✓

dy

=

= a + b(−2x −3 )

dx

=a−

2b x3

Curve at (3,5):

10(ii) At stationary point, dx −2(x−2)(x+1) (x2 +2)2

= 36x 2

dx2 x=−1

−2(x 2 − x − 2) (x 2 + 2)2

dy

= −1

= 12x 3 + 12

dx d2 y

−2x 2 + 2x + 4 = (x 2 + 2)2 =

12

y = 3x 4 + 12x y|x=−1 = −9 ⇒ stationary pt (−1, −9)

d d (x 2 + 2) ⋅ (2x − 1) − (2x − 1) ⋅ (x 2 + 2) dx dx = (x 2 + 2)2 =

12

b

5 = 3a +

=0

b 9

=0

9

= 5 − 3a

b = 45 − 27a

−(1)

x = 2 or x = −1 ✓ 10(iii)

At stationary pt (3,5):

dy dx −2(x−2)(x+1) (x2 +2)2 (x−2)(x+1) (x2 +2)2

>0

dy

>0 <0

(x − 2)(x + 1) < 0

a− ∵ (x 2 + 2)2 > 0

+ − + −1 2

27

=0 =0 =0

−(2)

sub (1) into (2): a−

⇒ −1 < x < 2 [shown] ✓ 11(i)

|

dx x=3 2b a − (3)3 2b

a−

2(45−27a) 27 2(9)(5−3a) 27

=0 =0

2

y = 3x 4 + kx

a − (5 − 3a) = 0

dy

3a − 10 + 6a = 0 9a = 10

dx

3

3

= 12x + k

a

At stationary pt: dy

10 9

✓

b|a=10 = 15 ✓

=0

dx

= 9

3

12x + k = 0 x3 x

=− 3

= √−

k 12 k 12

⇒ only 1 stationary pt ✓


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441


1st curve & its gradient y = 2x 2 − 4x + 5 dy dx

Ex 16.1 14(ii) At a = 9, dy

= 4x − 4

dy

d2 y

=0

dx

15(i)

dy

= a − 2b(2x − 1)−2 2b

= a − (2x−1)2 (2,7) lies on curve, (7) = a(2) + 7

At turning pt (1,3):

dy

= 21 − 6a

|

a−

2b

=0

9

−(2)

sub (1) into (2): a− a−

2(21−6a) 9 2(7−2a) 3

=0 =0

3a − 14 + 4a = 0 7a = 14 a =2✓

=0

6(−3) + 2a(−3) = 0 54 − 6a =0 a =9✓ b|a=9 = −8 ✓


=0 2b

At stationary pt (−3,19): |

−(1)

a − (2(2)−1)2 = 0

= 6x 2 + 2ax

dx x=−3 2

b 3

= 7 − 2a

dx x=2

Curve at (−3,19): (19) = 2(−3)3 + a(−3)2 + b 19 = −54 + 9a + b b = 73 − 9a

dy

b 2(2)−1

At stationary pt (2,7):

y = 2x 3 + ax 2 + b dx

= 2a +

b 3 b

=0

3 + 2a + 1 = 0 2a = −4 a = −2 ✓ b|a=−2 = 3 ✓

dy

= a + b[−(2x − 1)−2 ⋅ (2)]

dx

= 3x + 2ax + 1

|

b 2x−1

= ax + b(2x − 1)−1

2nd curve at (1,3): 3=1+a+1+b b=1−a

14(i)

y = ax +

2

dx x=1

= −18 < 0

⇒ max pt (−3,19) ✓

2nd curve & its gradient y = x 3 + ax 2 + x + b

dy

|

dx2 x=−3

4x − 4 = 0 x =1 y|x=1 = 3 ⇒ turning pt (1,3)

dx

= 12x + 18

dx2

At turning point,

dy

= 6x 2 + 18x

dx d2 y

Put a = 2 into (1): b|a=2 = 9 ✓

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442


Ex 16.1 16(ii) Curve at (1,1): 1 = 12 (1 − k)2 1 = 1 − 2k + k 2 0 = k 2 − 2k k 2 − 2k = 0 k(k − 2) = 0 k=0 or k = 2 (rej ∵ k ≠ 0)

15(ii) At a = 2 & b = 9, y = 2x + dy dx

9 2x−1 18

= 2 − (2x−1)2


=0

dx 18

2 − (2x−1)2

=0

2

= (2x−1)2

18

∴ y = x 2 (x − 2)2 dy

(2x − 1)2 =9 2 4x − 4x + 1 = 9 4x 2 − 4x − 8 = 0 x2 − x − 2 =0 (x − 2)(x + 1) = 0 x=2 or x = −1 (taken) y|x=−1 = −5 ⇒ (−1, −5) ✓

dx


dx

=0

dx

2x(x − 2)(2x − 2) = 0 2x(x − 1)(x − 2) = 0 x=0 or x = 1 or y|x=0 = 0 y|x=1 = 1 ⇒ (0,0) ✓ ⇒ (1,1) ✓

15(iii) 1st derivative dy

16(iii) 1st derivative −2

dy

= 2x(x − 2)(2x − 2)

dx

= 2x(2x 2 − 6x + 4) = 4x 3 − 12x 2 + 8x

2nd derivative dx2

= −18[−2(2x − 1)−3 ] ⋅ = −18[−2(2x − 1) =

d2 y

|

d dx

(2x − 1)

−3 ]

. (2)

2nd derivative d2 y

72 (2x−1)3

dx2

8

dx2 x=2

d2 y

= >0 3

y=x dy dx

2 (x

2

− k)

=x ⋅

d dx

d2 y

[(x − k)

d2 y 2]

+

d dx

= x ⋅ 2(x − k) +2x = 2x(x − k)(2x − k) ✓

(x 2 )

=8 >0

|

= −4 < 0 ⇒ max pt (1,1) ✓

|

=8 >0

⋅ (x − k)

dx2 x=2

2

⋅ (x − k)2

⇒ min pt (0,0) ✓

|

dx2 x=1

2

2

= 12x 2 − 24x + 8

dx2 x=0

⇒ (2,7) is min pt ✓ 16(i)

x=2 y|x=2 = 0 ⇒ (2,0) ✓

18

= 2 − (2x−1)2 = 2 − 18(2x − 1)

d2 y

= 2x(x − 2)(2x − 2)

⇒ min pt (2,0) ✓

16(iv) 𝑦 𝑦 = 𝑥 2 (𝑥 − 2)2 (1,1) (0,0)


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(2,0)

𝑥 ✓

443


y= dy dx

1 20

=

Ex 16.1

7

18(ii) 1st derivative

x 5 − x 3 + 3x 2 6

1 4 x 4

−

7 2 x 2

dy dx

+ 6x


At stationary point, dy dx 1 4 x 4 4

dy

−

+ 6x

3x 2 − 12x + 9 = 0 x 2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 x=1 or x=3 y|x=1 = 0 y|x=3 = −4 ⇒ (1,0) ⇒ (3, −4)

=0

x − 14x + 24x = 0 x(x 3 − 14x + 24) = 0 x(x − 18(i)

Curve y = x 3 + ax 2 + bx + c

2nd derivative d2 y dx2

touch x − axis at x = 1 & cross x − axis at x = 4 y = k(x − 1)2 (x − 4) Compare x 3 : k = 1 y = (x 2 − 2x + 1)(x − 4) = x 3 −2x 2 +x = −4x 2 +8x −4 = x 3 − 6x 2 + 9x − 4 Compare x 2 : a = −6 ✓ Compare x1 : b = 9 ✓ Compare x 0 : c = −4 ✓

=0

dx

=0 7 2 x 2 2

= 3x 2 − 12x + 9

d2 y

= 6x − 12 |

= −6 < 0

⇒ max pt (1,0) ✓

|

=6 >0

⇒ min pt (3, −4) ✓

dx2 x=1 d2 y

dx2 x=3

18(iii) Tangent Recall

y = x 3 − 6x 2 + 9x − 4 dy dx

Point: Gradient: Tangent:

&

2

= 3x − 12x + 9

y|x=0 = −4 ⇒ (0,4) dy

|

dx x=0

y − y1

=9 =

dy

|

dx x=0

(x − x1 )

(x − 0) y − (−4) = 9 y = 9x − 4 ✓


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444


Ex 16.2 1(d)

Ex 16.2 1(a)

dy

dx

=

√1−2x

dy

=0

dx 1−3x

=4>0

=0 1

= ✓

x

3

⇒ min ✓

Sign test x

= x 2 + (2x − 1)2 = x 2 + (4x 2 − 4x + 1) = 5x 2 − 4x + 1

dy dx

dx

At stationary value, dy

y= dy

=0

dx

dx

10x − 4 = 0

dx2

= ✓ 5

−

(x2 +1)⋅

d d (x) −x⋅ (x2 +1) dx dx (x2 +1)2

(x2 +1)⋅1

−x⋅2x

(x2 +1)2 x2 +1

−2x2 (x2 +1)2

dy

= 4(x − 2) ⋅ (1)

4(x − 2)3 = 0 x =2✓

Sign test x dy

= 4[3(x − 2) = 12(x − 2) |

dx2 x=2

2]

=0

1−x =0 (x + 1)(x − 1) = 0 x = −1 or x = 1 ✓

=0

dx

=0

dx 1−x2 (x2 +1)2 2

3

dy

+1)2

At stationary value,


d2 y

0

1−x2

= 4(x − 2)3

dx2

=

= (x2

= 10 > 0

y = (x − 2) + 3

d2 y

3

x

=

4

dx

1+

3

x2 +1

=

⇒ min ✓

dy

1

3

sign +

2

x

1−

⇒ max ✓

= 10x − 4 1(e)

d2 y

⋅ √1 − 2x

√1−2x

d2 y

dy

(x) ⋅ √1 − 2x

+ √1 − 2x

√1−2x 1−3x


y

d dx

⋅ (−2) + 1

4x − 8 = 0 x =2✓

dx2

1(c)

2√1−2x x

+

=0

dx

1(b)

1

=−


√1 − 2x

dx

=x⋅

= 4x − 8

dy

d

=x⋅

dx

y = 2x 2 − 8x + 3 dy

y = x√1 − 2x

dx

⋅ (1)

sign

−1− −

−1

−1+

0

+

⇒ min at x = −1 ✓

2

dy

=0

dx

⇒ not max/min ✓


x

1−

1

1+

sign

+

0

−

⇒ max at x = 1 ✓

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445


Ex 16.2

2x + y = 10 y = 10 − 2x

4(iii)

dA

= 8 + 800(−x −2 )

dx

= 8 − 800x −2 =8−


dA

dA

dx

= 10 − 4x

At stationary value, dA

800

dx

=

dx

d2 A

d2 A

Fence = 36 2x + y = 36 y = 36 − 2x [shown] ✓

3(ii)

A = xy = x(36 − 2x) = 36x − 2x 2 [shown] ✓

3(iii)

dA

|

5(i)

dx2

By Pythagoras Theorem, Height = √(5x)2 − [

=0

4(ii)

2

]

= √25x 2 − 9x 2 = 4x Area of trapezium

= −4 < 0

1

A = (height)(sum of bases) 2 1

= (4x)[(6x + y) + y] 2

Volume = 400 [given] x(8)h = 400 h

(6x+y)−y 2

= √(5x)2 − (3x)2

⇒ max 4(i)

= 1.6 > 0

Wire length = 104 y + 5x + 5x + (6x + y) = 104 16x + 2y = 104 2y = 104 − 16x y = 52 − 8x ✓

36 − 4x = 0 x =9✓ d2 A

x3

⇒ min ✓

At stationary value, dx

1600

dx2 x=10

= 36 − 4x

dA

= −800(−2x −3 ) =

<0

⇒ max ✓

dx

x2

x = 100 x = −10 or x = 10 ✓ (rej ∵ x > 0)

dx2 2 = −4

800

2

10 − 4x = 0 x = 2.5 A|x=2.5 = 12.5 d2 A

=0

x2

8

=0

x2

=0

dx

8−

3(i)

800

A = xy = x(10 − 2x) = 10x − 2x 2

=

50 x

= 2x(6x + 2y) ∵ y = 52 − 8x, A = 2x[6x + 2(52 − 8x)] = 2x(104 − 10x) = 208x − 20x 2 [shown]✓

✓

A = 8x + 2xh + 2(8)h 50

50

x

x

= 8x + 2x ( ) + 2(8) ( ) = 8x +

800 x

+ 10 ✓ [shown]

= 8x + 800x −1 + 10


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446


dA dx

= 208 − 40x

Ex 16.2 7

y = 2x 3 − 12x 2 + x + 9 dy dx

= 6x 2 − 24x + 1


Normal gradient (n)

=0

dx

1

= − dy

208 − 40x = 0 x

= =

y|x=5.2 d2 A dx2

dx

208

=−

40 52

= −(6x − 24x + 1)−1

10

= 5.2 ✓ = 10.4 ✓

dn dx

= −40 < 0

1

⋅ (12x − 24)

−24x+1)2 12x−24

= (6x2

−24x+1)2

x 4 y = 32 y

=

32


x4

12x−24 (6x2 −24x+1)2

= x2 +

32 x4

[shown] ✓

2

= x + 32x −4

dz

= 2x + 32(−4x −5 )

dx

8(i)

= 2x − 128x −5 = 2x −

128

dx

2x −

128 x5

128 x5

d2 z

ℎ

400 πr2

=0 =0

= 2πr 2 + 2πr (

= 2x

= 2πr 2 +

800 r

400 πr2

)

[shown] ✓

= 2πr 2 + 800r −1

= 2 − 128(−5x −6 ) =2+

d2 z

=

𝑟

Area A = 2πr 2 + 2πrh

2x 6 = 128 6 x = 64 x = −2 or x=2 (rej ∵ x > 0) ✓

dx2

Volume V = 400 2 πr h = 400 h

x5

At stationary value, dz

=0

12x − 24 =0 x =2 y|x=2 = −21 ⇒ P(2, −21)✓

z = x2 + y

6(ii)

= −[−(6x 2 − 24x + 1)−2 ⋅ (12x − 24)] = (6x2

⇒ max A ✓ 6(i)

1 6x2 −24x+1 2

|

dx2 x=2

640 x6

= 12 > 0

⇒ min z ✓


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447


Ex 16.2

1st derivative dA dr

9(ii)

= 4πr + 800(−r

1st derivative dA

−2 )

dx

= 4πr − 800r −2 800 = 4πr − 2 r

= 4x − 108x −2

4πr −

800

dA

4x −

800

4x

r2

3

4πr

=

πr 3

= 200 3

=√

r

=0

dx

=0

r2

x2


=0

dr

108

= 4x −


= 4x + 108(−x −2 )

108 x2

=0 =

108 x2

x = 27 x =3✓ h|x=3 = 2 ✓

200 π

≈ 3.99cm ✓ 2nd derivative d2 A

2nd derivative d2 A dr2

dx2

= 4π − 800(−2r −3 ) = 4π +

=4+

1600 r3

d2 A

> 0 for r > 0 ⇒ min A 9(i)

Volume x(2x)h = 36 h

= =

ℎ 2𝑥 𝑥

2x2 18

By Pythagoras Theorem, 25x2

4

16

5

= x 4

Perimeter PQRST = 30 2x + 2y +2RS = 30 x+y + RS = 15 5

x + y + ( x) = 15 4

= 2x + 6x ( 2 )

9

y

x

x

= 12 > 0

3x 2

18

108

x3

RS = √x 2 + ( ) = √

x2

2

216

⇒ min 10

36

|

dx2 x=3

Area A = 2[x(2x) + xh + 2xh] = 2(2x 2 + 3xh) = 2x 2 + 6xh = 2x 2 +

= 4 − 108(−2x −3 )

= 15 − x 4

[shown] ✓ A = triangle STR

= 2x 2 + 108x −1

1

3x

2 3 2 x 4 3 2 x 4 3 2 x 4

4

= (2x) ( )

+2xy

=

+2xy

= =

= 30x −


sleightofmath.com

+ rectangle PQRT

9

+2x (15 − x) +30x − 15x2 4

4 9 2 x 2

[shown]

448


Ex 16.2

1st derivative dA dx

= 30 −

15

11(iv) At stationary value, dA

x

2

=0

dx

48x − At stationary value, dA

48x

=0

dx

30 −

15 2

=−

15 2

<0

4

9

4

Area A = xy = x(40 − x) = 40x − x 2

3 2

SN|x=4 = 9 ✓ By Pythagoras Theorem, 6x 2

AA′ = √(5x)2 − ( ) = 4x 2

2

A 5x

C

Volume = 4500 (△ area) h = 4500 1

x2

+ (15 − x)

= 15 − x

11(i)

6000

Perimeter 2x + 2y = 80 y = 40 − x

10(ii) SN = 3x +y =

=

12(a) Function x ≡ length y ≡ breadth

⇒ max 4 3x

=0

x = 125 x =5 A|x=5 = 1800 cm2 ✓

x= 0

2nd derivative dx2

x2

3

x =4 A|x=4 = 60 cm2

d2 A

6000

5x

A’ 6x

dA

B

h

=

x2

=0

dx

40 − 2x = 0 x = 20 A|x=20 = 400 m2 ✓

✓

11(ii) A = 2(triangle)

= 40 − 2x

dA

= 4500 375

dx


(6x)(4x)h = 4500

12x 2 h

1st derivative

+sides

1

2nd derivative

2 2

d2 A

= 2 [( ) (6x)(4x)] +h(6x + 5x + 5x) = 24x + 16xh = 24x 2 + 16x ( = 24x 2 +

6000 x

dx2

⇒ max

375 x2

= −2 < 0

)

[shown] ✓

= 24x 2 + 6000x −1 11(iii)

dA dx

= 48x + 6000(−x −2 ) = 48x −

6000 x2

✓


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449

A math 360 sol (unofficial) 12(b) Function x ≡ length y ≡ breadth

Ex 16.2 13

Function E2 R

W ∝ (r+R)2 E2 R (r+R)2 R kE 2 [(r+R)2 ]

W=k Area xy = 16 y

=

16

=

x

1st derivative dW

Perimeter P = 2x + 2y

dR 16

2

= kE [

= 2x + 2 ( ) x

= 2x +

32 x

r + R − 2R ] = kE 2 [ (r + R)3

1st derivative = 2x + 32(−x −2 )

dx

r−R ] = kE 2 [ (r + R)3

= 2 − 32x −2 =2−

32 x2


At stationary value, dP

dW

2−

32 x2

32

r−R

kE 2 [(r+R)3 ] = 0

=0

=r✓

R

=2

x2

=0

dR

=0

dx

d d R − R ⋅ (r + R)2 dR dx ] (r + R)4

(r + R)2 (1) − R ⋅ 2(r + R) ] = kE 2 [ (r + R)4

= 2x + 32x −1

dP

(r + R)2 ⋅

x2 = 16 x = −4 or x = 4 (rej ∵ x > 0)

Sign Test R

P|x=4 = 16 cm ✓

⇒ work done is greatest when R=r

dW dR

r− sign +

r 0

r+ −

2nd derivative d2 P dx2

= −32(−2x −3 ) =

64 x3

>0 ∵x>0

⇒ min P


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450


Ex 16.2

Perimeter

15 1

𝑦

x + 2y + ( ) πx = 6 x + 2y +

2 πx

𝑥

=6

2

2y

=6−

πx 2

−x

= 6 − x ( + 1)

y

= 3 − x ( + 1)

Area A = rectangle

+semicircle

x 𝑎

Rectangle area, A = xy

π

2y

x ≡ length y ≡ breadth A ≡ area of rectangle

y

−(1)

2

1

π

2

2

1

x 2

2 1

2 2

= xy

+ π( )

= xy

+ πx

−(1)

By Pythagoras Theorem, x 2 + y 2 = (2a)2 y 2 = 4a2 − x 2 y = √4a2 − x 2 or y = −√4a2 − x 2 (rej ∵ y > 0)

−(2)

8

−(2)

sub (2) into (1): sub (1) into (2): 1

π

1

2 1 π

2

8 1

A = x√4a2 − x 2

A = x [3 − x ( + 1)] + πx = 3x − ( + 1) x 2 2 π 1

= 3x − ( + ) x 4 1

2

+ πx

dx

2

8

1

= 3x + ( π − − ) x 4 1

=

2

2

= 3x + (− − ) x 2 = 3x − (

8 π+4 8

=

2

) x2

=

1st derivative dA

=3−(

dx

=3−(

π+4 8 π+4 4

dA

3−( (

4

)x

=

= 3(

π+4

= =

dx2

⋅ √4a2 − x 2

(−2x) +(1)

+√4a2 − x 2

−x2

+(4a2 −x2 ) √4a2 −x2

4a2 −2x2 √4a2 −x2

=0 =0

x = √2a or x = −√2a (rej ∵ x > 0)

12 π+4 12

)−(

π+4

36 π+4 18 π+4

−

y|x=√2a = √4a2 − (√2a)

✓ 12

) (

π+4

8

= −2 (

π+4 8

2

= √4a2 − 2a2 = √2a2 = √2a = x ∵ length = breadth, largest rectangle is square of side √2a

2 π+4

)

144 8(π+4)

Sign Test

[shown]

x dy

2nd derivative d2 A

(x) ⋅ √4a2 − x 2

=3

x

A|x= 12

d dx

⇒ 4a − 2x 2 = 0 −2x 2 = −4a2 x 2 = 2a2

)x = 0

π+4

1 2√4a2 −x2

−x2

√4a2 −x2 2

π+4 4

(√4a2 − x 2 )+

√4a2 −x2

dx 4a2 −2x2

)x

=0

dx

d dx


) 2x


=x⋅ =x⋅

+ πx 2

π

π

dA

8 1

2

2

8

2

dx

)<0

+

−

√2a

√2a

0

+

−

A is max when x = √2a i.e. when rectangle is a square

⇒ max ✓


sign

√2a

sleightofmath.com

451


Ex 16.2

y = 12 − x 2

17

A = 2xy = 2x(12 − x 2 ) = 24x − 2x 3 dA dx

= 24 − 6x

𝑦

𝑂

y 2 = 8x y = √8x or y = −√8x

2


dy

dA

dx

=0

dx

24 − 6x 2 = 0 x2 − 4 =0 x = 2 or x = −2 (NA) A|x=2 = 32 units 2 ✓ d2 A dx2 d2 A

|

= −24 < 0

= 2√2 (

1 2√x

)=

√2 √x

Normal Point:

(4,2)

Gradient:

m⊥ = − dy

= − √2

dx

√x

1

1

=−

√x √2

y − y1 = m⊥ (x − x1 )

Normal:

= −12x

dx2 x=2

𝑦 2 = 8𝑥 (4,2) 𝑥

y − 2= −

⇒ max

y=−

√x √2

√x √2

[x − 4]

(x − 4) + 2

Point where y 2 = 8x & normal [y = − √8x

√x √2

(x − 4) + 2] meet.

=−

√x √2

(x − 4) +2

√16x = −√x(x − 4) +2√2 4√x = −x√x + 4√x +2√2 x√x = 2√2 ⇒x=2 −(3) sub (3) into (1): y|x=2 = √8(2) = 4 ⇒ (2,4) Minimum distance from (2,4) to (4,2) = √(2 − 4)2 + (4 − 2)2 = √22 + 22 = 2√2 ✓


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452


Ex 16.2

Let P(a, b) be at point on y 2 = 8x, b2 = 8a a

18

b2

=

8

b2 ⇒ P ( , b) 8

Area is not the only factor of cost Most of the expense is incurred in joining the sides to the rims of the cans. Minimization of circumference must also be factored in.

b2

D = distance from P ( , b) to (4,2) 8

= √( = √( =√

b2 8 1 64

1

64

dD db

2

− 4) + (b − 2)2 b 4 − b 2 + 16) + (b 2 − 4b + 4)

b 4 − 4b + 20 1

=

1 2√ b4 −4b+20 64

1

=

1 2√ b4 −4b+20 64

⋅[

1 64

⋅(

1 16

(4b3 ) − 4] b3 − 4)

At min point: dD db 1 3 b 16 1 3 b 16 3

b b

a=

=0 −4 =0 =4 = 64 =4

42 8

=2

(2,4)


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453


Rev Ex 16 A3

Rev Ex 16 A1

q

y = (x − 2)√5 − x dy

= (x − 2) ⋅

dx

= (x − 2) ⋅ = = = =

2−x

d dx

√5 − x 1

2√5−x

Function p2 q = 9

(−1)

+

d dx

(x − 2) ⋅ √5 − x

+(1)

2−x

9 p2

z = 16p + 3q 9

= 16p + 3 ( 2 )

⋅ √5 − x

p

= 16p +

+√5 − x

2√5−x

=

27 p2

= 16p + 27p−2 +2(5−x)

2√5−x 2−x

1st derivative

+10−2x

dz

2√5−x

dp

12−3x

= 16 + 27(−2p−3 ) = 16 − 54p−3

2√5−x

= 16 −

54 p3

At stationary value, dy dx 12−3x 2√5−x

x

=0


=0

dp

=4

16 −

dz

y|x=4 = ((4) − 2)√5 − (4)

y = 4x +

x

2

= 4x + 27x

= 16

−1

=

p

= ✓

d2 z

27

dp2

x2

At turning point,

162 p4

>0∵p>0

⇒ min z

=0

dx

8x −

27 x2

8x 8x

3 2 2

( )

= −54(−3p−4 ) =

dy

9

=4✓ 2nd derivative

= 8x − 27x −2 = 8x −

8 3 2

q|p=3 =

= 8x + 27(−x −2 )

dx

27

p3

2

dy

=0

p3

p3

27

2

54

54

=2✓ A2(i)

=0

=0 =

3

A4(i)

27 x2

= 27

x

=

3 2

y|x=3 = 27

Wire length = 2400 4(length + breadth + height) = 2400 4(3x + x + h) = 2400 4(4x + h) = 2400 4x + h = 600 h = 600 − 4x ✓

2

3

⇒ ( , 27) ✓ 2

A2(ii)

d2 y dx2

= 8 − 27(−2x −3 ) =8+

d2 y

|

dx2 x=3

54 x3

= 24 > 0

2

A4(ii) Volume V = 3x(x)h = 3x 2 h = 3x 2 (600 − 4x) = 12x 2 (150 − x) ✓ = 1800x 2 − 12x 3

3

⇒ min pt ( , 27) ✓ 2


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454

A math 360 sol (unofficial) A4(iii) 1st derivative dV dx

Rev Ex 16 A6(i)

= 3600x − 36x

2

Volume = 100 2x(x)h = 100 h

h

50

=

2x

x2

x

At turning value, dV

Total surface area S = 2[2x(x) + 2x(h) + x(h)] = 2(2x 2 + 2xh + xh) = 2(2x 2 + 3xh) = 4x 2 + 6xh

=0

dx

3600x − 36x 2 = 0 x 2 − 100x =0 x(x − 100) =0 x = 0 or x = 100 ✓ (rej ∵ x > 0) 2nd derivative d2 V

= 3600 − 72x

dx2

A6(ii)

= 4x

2

= 4x

2

dS dx

d2 V dx2

|

x=100

50

= 4x 2

+6x ( 2 ) x

+

300 x

[shown] ✓

+300x −1

= 8x + 300(−x −2 ) = 8x − 300x −2

= −3600 < 0

= 8x −

⇒ max V

300 x2


A5(i)

dS

=0

dx

𝑦 𝑥 Wire length = 100 4x + 4y = 100 4y = 100 − 4x y = 25 − x✓

8x −

3

=

dx

3 75 2

x= √

2 75

d2 S

2

≈ 134 ✓

= 8 − 300(−2x −3 ) =8+

>0 ⇒ min ✓

= 2x −50 + 2x = 4x − 50

x2 75 3

dx2

= 2x +2(25 − x)(−1)

300

=√

x

A6(iii)

=0 =

x

A5(iii) 1st derivative dA

x2

8x

S| A5(ii) Total area A = x2 + y2 = x 2 + (25 − x)2 [shown] ✓

300

600 x3

∵x>0

At stationary value: dA

=0

dx

4x − 50 = 0 x

=

50 4

= 12.5 ✓ 2nd derivative d2 A dx2

=4>0

⇒ min


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455


Rev Ex 16

Perimeter

B1(i)

1

y = x2 +

x + x + (2πr) = 80 = 80 = 80 − πr

dy dx

1

x

= 40 − πr

= 2x −

Area A = semicircle + triangle 1

1

2 1

2 1

1

2

2

dy

+ (40 − πr)

2 1

1 1 2

2 πr2

2 2

2x −

A7(ii)

dA

=0

x3

=

16 x3

4

x = 16 x=2 or y|x=2 = 8 ⇒ (2,8)

+ (80 − πr)2 [shown] ✓ 8 1

= πr + [2(80 − πr)] ⋅ (−π)

dr

32

x

1

2

x3

=0

dx

2

= πr 2 + ( ) (80 − πr)2 =

32


= πr 2 + x 2 = πr

= 2x + 16(−2x −3 ) = 2x − 32x −3

2

2

x2

= 2x + 16x −2

2

2x + πr 2x

16

8 π

x = −2 y|x=−2 = 8 ⇒ (−2,8)

= πr − (80 − πr) 4

= πr −20π + = (π +

π2 4

π2 4

r

B1(ii)

dx2

(π +

4 π

d2 y

=0

(1 + ) r

= 20

4 π 4

(

4+π 4

)r

= 20

r

=

80 π+4

≈ 11.2 ✓ A7(iii)

d2 A dr2

=π+

π2 4

>0

⇒ min ✓

|

dx2 x=2 d2 y

) r − 20π = 0

(1 + ) r − 20

= 2x − 32(−3x −4 ) =2+

) r − 20π

At stationary value, π2

d2 y

|

96 x4

=8>0

dx2 x=−2

= −4 < 0

⇒ min pt (2,8) ✓ ⇒ max pt (−2,8) ✓

B2(i) y=

4 2−x

= 4(2 − x)−1

+

9 x−3

+9(x − 3)−1

dy dx

= 4 [−(2 − x)−2 ⋅

d dx

(2 − x)] +9 [−(x − 3)−2 ⋅

= 4[−(2 − x)−2 ⋅ (−1)] = 4(2 − x)−2 4

d dx

(x − 3)]

+9[−(x − 3)−2 ⋅ (1)] −9(x − 3)−2 9

− (x−3)2 ✓

= (2−x)2 d2 y dx2

= 4 [−2(2 − x)−3 ⋅

d dx

(2 − x)] −9 [−2(x − 3)−3 ⋅

= 4[−2(2 − x)−3 ⋅ (−1)] 8

= (2−x)3


sleightofmath.com

d dx

(x − 3)]

−9[−2(x − 3)−3 ⋅ (1)] 18

+ (x−3)3 ✓

456


Rev Ex 16

B2(ii) At stationary value, dy

B4(i)

AR =

=0

dx 4 (2−x)2 4 (x−2)2

9

− (x−3)2

=0 =

4(x − 3)2 4(x 2 − 6x + 9) 4x 2 − 24x + 36 5x 2 − 12x x(5x − 12) x=0

By Pythagoras’ Theorem,

=

9 (x−3)2

y|x=0 = −1

AS

− [ (10)]

S Q 13cm h

13cm P

2

2

B

C R 10cm

PQ BC x 10 6

12 − h = x 5

5

5

=

12

12

12

=

AR 12−h

6

h

5

⇒(

1

By similar triangles,

y|x=12 = −25

⇒ (0, −1)

−

√(13)2

A

BR2

= 12

= 9(x − 2)2 = 9(x 2 − 4x + 4) = 9x 2 − 36x + 36 =0 =0

or x =

√AB 2

= 12 − x 5

6

= (10 − x) ✓

h

, −25)

5

B4(ii) Area of △ PQR d2 y

1

|

dx2 x=0 d2 y dx2

B3

|

3

12 x= 5

=−

625 3

1

⇒ min pt (0, −1) ✓

= >0 <0

⇒ max pt (

12 5

A = xh 2 1

=

Function x ≡ length y ≡ breadth

dx

dA

2

dx

2

2

d2 A

1st derivative

dx2

1

= l − 2x

dA

B5(i)

BC 4−x

l − 2x= 0

4

y

1

2x

= l

x

= l 4

1

y|x=1l = l = x

6

=− <0 5

= =

4

A 3

AB y

𝑦

P

Y B

3 3

= (4 − x) ✓ 4

X

5 C

𝑥 4

3

= x ( ) (4 − x) 4

3

= x(4 − x) [shown] ✓ 4

3

= −2 < 0

= 3x − x 2 4

⇒ max A © Daniel & Samuel A-math tuition 📞9133 9982

XP

B5(ii) Area A = xy

⇒ square ✓ 2nd derivative dx2

=6

2 1

4

=0

5

By similar triangles, XC

=0

dx 1

5

⇒ max A

2

At maximum value,

6x

x =5✓ 2nd derivative

2

d2 A

6x

5

= x ( l − x) = lx − x

5

=0

6x 1

2

=6−

6−

Area A = xy 1

3x2

At maximum value,

1

= l−x

dx

(10 − x) [shown] ✓

5

B4(iii) 1st derivative dA

dA

2 5 3x

= 6x −

Perimeter 2x + 2y = l 2y = l − 2x y

6

= x [ (10 − x)]

, −25) ✓

sleightofmath.com

457


Rev Ex 16

B5(iii) 1st derivative dA dx

B6(ii)

3

dx

2

= (8 − x) ⋅

=3− x


= (8 − x) ⋅

=0

dx

dA

= (8 − x)

3

3− x=0 2

3 2

x

x

=2

1 2√28x−28

⋅ 28 +(−1)

14

⋅ √28x − 28

−√28x − 28

√28x−28

√28x−28 140−42x

=

3

d

√28x − 28 + dx (8 − x) ⋅ √28x − 28

112−14x−28x+28

=

=3

d dx

√28x−28

A|x=2 = 3(2) − (2)2 4

= 3 cm2 ✓ 2nd derivative

At stationary value:

d2 A

dx

dx2

dA

3

=− <0

140−42x

2

√28x−28

⇒ max A B6(i)

x

P (6 + 𝑥)

=0

(6 + 𝑥)

Q R (16 − 2𝑥) By Pythagoras Thm

10 3

x dy sign dx ⇒ max

− (P ′ Q)2

= √(6 + x)2

=

✓

Sign Test

P’

PP′ = √PQ2

=0

1

− [ (16 − 2x)]

2

2

10

10+

3

3

3

+

0

−

= √(x 2 + 12x + 36) − (64 − 16x + x 2 )

Volume V = 250π πr 2 h = 250π

= √28x − 28

h

= √(6 +

x)2

B7(i)

10−

− (8 − x)2

=

250 r2

Area A = 2πr 2 + 2πrh

Area of triangle 1

A = (QR)(PP ′ )

= 2πr 2 + 2πr (

2 1

= (16 − 2x)√28x − 28

2

= 2πr +

2

= (8 − x)√28(x − 1) [shown] ✓

500π r

𝑟 ℎ 250 r

2 )

[shown] ✓

2

= 2πr + 500πr −1

= (8 − x)√28x − 28


sleightofmath.com

458

A math 360 sol (unofficial) B7(ii) 1st derivative dA (a) = 4πr + 500π(−r −2 ) dr

Rev Ex 16 B7(ii) Cost 500π (b) (0.05) C = 2πr 2 (0.03) +

= 4πr − 500πr −2 = 4πr −

500π r2

+

= 0.06πr 2

+25πr −1


1st derivative

dA

dC

=0

dr

4πr −

500π r2

dr

r3 r

= 125 =5✓ 52

r2

dC

= 10 ✓

= 4π + d2 A

|

dr2 r=5

r2

=0

dr 25π

0.12πr − 0.12πr

= 4π − 500π(−2r −3 )

dr2

25π


2nd derivative d2 A

= 0.06πr 2 + 25πr −2 = 0.12πr −

500π

=

250

r

= 0.12πr − 25πr −2

=0

4πr

h|r=5 =

r 25π

= 0.06πr 2

= 25π 3

625

=√

h|

≈ 5.93 ✓ ≈ 7.11 ✓

r3

⇒ min

=0

r

1000π

= 4π + 8π > 0

r2

3

3 625 3

r= √

3

2nd derivative d2 C dr2

= 0.12 π − 25π(−2r −3 ) = 0.12π +

d2 C

|

dr2 r= 3√625

50π r3

≈ 1.13 > 0

3

⇒ min


sleightofmath.com

459


Ex 17.1 2(d) d 1 − 2 sin x ( ) dx cos x d cos x ⋅ (1 − 2 sin x) dx =

Ex 17.1 1(a)

d dx

(4 sin x − 3)

=4

d dx

(sin x) −

d dx

(3)

= 4 cos x ✓ 1(b)

d dx

=

1(c)

(x 2 − 5 cos x) d dx

d dx

= 2x

+5 sin x ✓

d

d

(x 2 ) +3

d dx

(tan x)

−2 cos 2 x

+ sin x − 2 sin2 x cos 2 x

3(a)

d dx

= 8x

+3 sec 2 x ✓

[(1 − cos x)3 ]

d dx

(sin x) +3

(cos x)

dx

3(b)

d

[(2 + 3 sin x)2 ]

+3(− sin x)

= 2 cos x

−3 sin x ✓

= 2(2 + 3 sin x) ⋅

d dx

(tan x) +

d dx

(x) ⋅ tan x

= x ⋅ sec x

+1

= x sec 2 x

+ tan x ✓

(x 2 2

=x ⋅

d dx

(cos x)

⋅ tan x

=

+

d dx

(x 2 )

⋅ cos x

+2x

⋅ cos x

3(d)

−x sin x ✓

dx

(sin x)

= (x + 1)2 cos x

d dx

2

d

= =

+

d dx

(√2 − tan x) 1

[(x + 1)2 ]⋅ sin x

=

⋅

d

2√2−tan x dx 1 2√2−tan x

=−

[(x + 1)2 sin x]

= (x + 1)2 ⋅

d dx

cos x)

= 2x cos x d

3(c)

=

= x 2 ⋅ (− sin x)

dx

(2 + 3 sin x)

= 6 cos x (2 + 3 sin x) ✓

2

dx

d dx

= 2(2 + 3 sin x) ⋅ 3 cos x (x tan x)

=x⋅

d

(1 − cos x)

= 3 sin x (1 − cos x)2 ✓

d

= 2 cos x

dx

dx

= 3(1 − cos x)2 ⋅ sin x

dx

d

d

= 3(1 − cos x)2 ⋅

(2 sin x + 3 cos x)

=2

2(c)

cos x ⋅ (−2 cos x)

2

+3 sec x

d

cos 2 x − (1 − 2 sin x) ⋅ (− sin x) 2 cos x

sin x − 2(sin2 x + cos 2 x) cos 2 x sin x − 2 = cos 2 x ✓

(4x 2 + 3 tan x) dx

d (cos x) dx

=

= 4(2x)

dx

2(b)

=

(cos x)

−5(− sin x)

=4

2(a)

(x 2 )−5

= 2x

dx

1(d)

=

− (1 − 2 sin x) ⋅

(2 − tan x)

⋅ (− sec 2 x)

sec2 x 2√2−tan x

✓

(√sin x + 2 cos x) 1 2√sin x+2 cos x 1 2√sin x+2 cos x cos x−2 sin x 2√sin x+2 cos x

d

⋅

dx

(sin x + 2 cos x)

⋅ (cos x

− 2 sin x)

✓

+2(x + 1) sin x ✓ 4(a)

d dx

(3 tan 2x) = 3

d dx

(tan 2x)

= 3 sec 2 2x ⋅

d dx

(2x)

= 3 sec 2 2x ⋅ (2) = 6 sec 2 2x ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

460


d dx

Ex 17.1

1

d

2

dx

[4 sin ( x)] = 4

1

5(d) d cos(x + π) [ ] dx sin 3x

[sin ( x)] 2

1

d

2 1

dx 2 1

= 4 cos ( x) ⋅

1

( x)

= 4 cos ( x) ⋅ ( ) 2 1

d d cos(x + π) − cos(x + π) ⋅ sin 3x dx dx = sin2 3x (sin 3x) ⋅ [− sin(x + π)] − cos(x + π) ⋅ (3 cos 3x) = sin2 3x − sin(3x) sin(x + π) − 3 cos(x + π) cos(3x) = 2 sin 3x sin(3x) sin(x + π) + 3 cos(x + π) cos(3x) =− sin2 3x ✓

2

(sin 3x) ⋅

= 2 cos ( x) ✓ 2

4(c)

d dx

π

[cos (2x − )] 3 π

= − sin (2x − ) ⋅ 3 π

d

π

(2x − )

dx

3

= − sin (2x − ) ⋅ (2) 3

π

= −2 sin (2x − ) ✓ 3

4(d)

d dx

=

[sin 3x + cos(4x − 5)] d dx

sin 3x

+

= cos(3x) ⋅

d dx

d dx

6

cos(4x − 5)

+[− sin(4x − 5)] ⋅

(3x)

5) = cos(3x) ⋅ (3) = 3 cos 3x

d dx

y = sin 2x dy

(4x −

d dx

d dx

+ d dx

d dx

+ cos x cos 3x −3 sin x sin 3x ✓

d

d dx

tan 5x

+

d dx

2

7

y = sin x + 2 cos x dy

d dx

= (1 + x 2 ) ⋅ sec 2 5x ⋅ 5 = 5(1 + x 2 ) sec 2 5x 5(c)

(5x)

+2x

dy

=0

dx

cos x − 2 sin x = 0 2 sin x = cos x

⋅ tan 5x

+2x tan 5x +2x tan 5x ✓

tan x

=

1 2

α ≈ 0.464 ⇒ 1st or 3rd quadrant

d x ( ) dx cos 2x d d (x) − x ⋅ (cos 2x) dx dx = cos 2 2x (cos 2x) ⋅ (1) − x ⋅ (−2 sin 2x) = 2 cos 2x cos 2x + 2x sin 2x = 2 cos 2x 1 + 2x tan x = cos 2x

= cos x − 2 sin x

At stationary point, (1 + x 2 ) ⋅

tan 5x = (1 + x 2 ) ⋅ sec 2 5x ⋅

3

=1✓

[(1 + x 2 ) tan 5x]

= (1 + x 2 ) ⋅

π

1

dx dx

(2x)

= 2( )

(sin x) ⋅ cos 3x

(3x) + cos x ⋅ cos 3x

= sin x ⋅ (− sin 3x) ⋅ 3 = cos x cos 3x 5(b)

= 2 cos

6

(cos 3x)

= sin x ⋅ (− sin 3x) ⋅

|

dx x=π

(sin x cos 3x)

= sin x ⋅

d dx

= (cos 2x) ⋅ (2) = 2 cos 2x ✓

+[− sin(4x − 5)] ⋅ (4) −4 sin(4x − 5) ✓ dy

5(a)

= (cos 2x) ⋅

dx

0 ≤ x ≤ 2π x = α, π + α ≈ 0.464, 3.61 ✓

(cos 2x) ⋅

8(a)

d dx

(2 sin3 x) = 2

d dx

S α T

A α C

(sin x)3

= 2 ⋅ 3(sin x)2 ⋅

d dx

(sin x)

2

= 2 ⋅ 3 sin x⋅ cos x = 6 sin2 x cos x ✓

= sec 2x (1 + 2x tan x) ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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461


d dx

d

(cos 2 3x) =

Ex 17.1 9(a)

(cos 3x)2

dx

= 2(cos 3x) ⋅

dt d dx

d dx

d

dy dx

dx

dy

[(tan 5x)2 ] d

= 4 ⋅ 2 tan 5x ⋅

dx

d

+

dx

dt

(3)

d dx

=

|

=?

dt x=π 6

= 1 − sin x dy

=

dx

×

dx

dt

= (1 − sin x) × 0.2 = 0.2 − 0.2 sin x

(tan 5x) + 0

= 4 ⋅ 2 tan 5x ⋅ (5 sec 2 5x) = 40 tan 5x sec 2 5x ✓ 8(d)

dy

y = x + cos x

[4 tan2 (5x) + 3]

=4

= 0.2,

cos 3x

= 2(cos 3x) ⋅ (−3 sin 3x) = −6 sin 3x cos 3x ✓ 8(c)

dx

π

dy

6

dt x=π

At x = ,

π

|

d

(x) +3

d

1 = 0.2 − 0.2 ( ) 2 = 0.1 unit s −1 ✓

[(sin 2x)5 ]

dx

+3 ⋅ 5(sin 2x)4 ⋅

=1

d dx

(sin 2x)

9(b)

dx dt

+3 ⋅ 5(sin 2x)4 ⋅ (2 cos 2x) +30 sin4 2x cos 2x ✓

=1 =1

6

6

[x + 3 sin5 (2x)] dx

= 0.2 − 0.2 sin ( )

π dy

=

,

|

=?

30 dt x=π 3

y = √1 + sin x 8(e) d dx

dy

[(1 + x) cos 7 2x]

= (1 + x) ⋅

d dx

(cos 7 2x)

= (1 + x) ⋅ 7 cos 6 2x ⋅

d dx

+ (cos 2x)

d dx

dx

(1 + x) ⋅ cos 7 2x

= dy dt

= 2( = 2( = 2( =

=

d cos 2x 2 ( ) dx 1 − x

= 2(

cos 2x 1−x cos 2x 1−x cos 2x 1−x cos 2x 1−x

) ⋅ ) ⋅

d dx

(

cos 2x 1−x

(1−x)⋅

d dx

(1 + sin x)

2√1+sin x dy

×

dx cos x

)

×

2√1+sin x

dx dt π 30

π cos x 60√1+sin x π

dy

3

dt x=π

At x = ,

|

≈ 0.0192 unit s −1 ✓

3

10(a) y = x − 2 sin x

d (cos 2x) dx

−(cos 2x)⋅

dy

d (1−x) dx

dx

= 1 − 2 cos x

(1−x)2

) ⋅

(1−x)⋅(−2 sin 2x) −(cos 2x)⋅(−1) (1−x)2


) ⋅

2(x−1) sin 2x +cos 2x (1−x)2

dx

4(x − 1) sin 2x cos 2x (1 − x)3

1 4(x − 1) ( sin 4x) 2 = (1 − x)3 =

= =

d cos 2 2x [ ] dx (1 − x)2 =

⋅

⋅ cos 2x 7

= (1 + x) ⋅ 7 cos 2x ⋅ (−2 sin 2x) + cos 2x = cos 6 2x [cos 2x − 14(1 + x) sin 2x] ✓ 8(f)

1 2√1+sin x cos x

7

+1

6

=

(1−x)3

=0

1 − 2 cos x = 0 −2 cos x = −1

+ 2 cos 2 2x

cos x α=

+ 2 cos 2 2x

+2 cos2 (2x)

(2x−2) sin 4x

dy

=

π

2

S

3


T

A α α C

0 ≤ x ≤ π: x = α, 2π − α

✓

π 5π

= , 3


1

sleightofmath.com

3

✓

462


Ex 17.1

10(b) y = x − tan x dy dx

y = x + 4 tan2 2x

12

= 1 − sec 2 x

dy

= 1 + 4(2 tan 2x)(sec 2 2x)(2)

dx

= 1 + 16 tan 2x sec 2 2x


=0

dx

1 − sec 2 x sec 2 x tan2 x + 1 tan x

=0 =1 =0 =0

dy

𝑦

dx

⇒ (0,0) ✓ ⇒ (π, π) ✓

α T

(1)

sin 3x 2 − cos 3x

y= dy dx

(2 − cos 3x) ⋅

=

3 2

α = 0.983 ⇒ 1st or 3rd quadrant S

1

13

−2 sin x + 3 cos x = 0 −2 sin x = −3 cos x =

)

2

=0

0 ≤ x ≤ π: x = α, π + α = 0.983 ✓

3

π cos 3

= 1 + 64√3 ✓


tan x

1

= 1 + 16(√3)

𝑥

= −2 sin x + 3 cos x

dx

2

π

2

−1

dy

3

1

y = 2 cos x + 3 sin x dy

π

3

= 1 + 16 tan ( ) (

y|x=0 = (0) − tan(0) = 0 y|x=π = (π) − tan(π) = π 11(i)

π

= 1 + 16 tan ( ) sec 2 ( )

6

90° 180° 270° 360°

0≤x≤π x = 0, π

|

dx x=π

=

(2 − cos 3x) ⋅ (3 cos 3x) − (sin 3x) ⋅ (3 sin 3x) (2 − cos 3x)2

=

6 cos 3x − 3 cos 2 3x − 3 sin2 3x (2 − cos 3x)2 − 3(sin2 3x + cos 2 3x) (2 − cos 3x)2

6 cos 3x

=

A α

d d (sin 3x) − (sin 3x) ⋅ (2 − cos 3x) dx dx (2 − cos 3x)2

6 cos 3x

=

−3 (2 −

C

cos 3x)2

Tangent ∥ to x − axis, 11(ii)

d2 y dx2 d2 y

dy

= −2 cos x − 3 sin x |

dx2 x=0.983

dx 6 cos 3x−3 (2−cos 3x)2

≈ −3.61 < 0

⇒ max pt ✓

=0 =0

6 cos 3x − 3 6 cos 3x

=0 =3

cos 3x

=

α=

π

∵ (2 − cos 3x)2 > 0

1 2

3


S

0
T

3x = α, 2π − α,

x © Daniel & Samuel A-math tuition 📞9133 9982

2π + α, 4π − α

π 5

7

3 3 π 5π

3 7π

= , π, = , 9

sleightofmath.com

9

,

A α α C

9

π

✓

463


Ex 17.1

Curve & its derivative y = 1 + cos x dy dx

14(ii) Normal: π

y = 2x − + 1 + 3

= − sin x

y|x=π = 1 + cos ( )

⇒ ( ,1 + 6

dy

√3 2

= √3 )= 2

dy dx

|

6

2

1

π

(− ) [x − ( )] 2

6 π

= −x + + 2 + √3

π

2 1

12

y

= − x+

1 2

= − dy

π 12

1 |

dx x=π 6

1

y − (1 +

√3 ) 2

=−

y − (1 +

√3 ) 2

= 2x −

𝜋

√3 4

6

1 2

+ 2 + √3

+1+

√3 2

✓

√3 2

x

= + 2 + √3

2

π 6

√3 2

π

2x

= −1−

x

= − −

(x − x1 )

3 π

1

6

2

=0 √3 2 √3 4

Area 1

= (base)

6

2 1

= 2x − + 1 +

(height)

π

π

1

2 6 1 5

6

2

= [( + 2 + √3) − ( − −

π

3

=0

π

3

[x − ( )]

3 π

√3 2

=− −1−

√3 2

✓

5 4 5

= (1 + 4

sleightofmath.com

5

= ( + √3) 2 2 4 = (1 +


√3 2

𝑥

− x

π

π

(− )

+1+

2x − + 1 +

6

y − y1

2

+1+

Normal at x-axis, y =0

6

π

2y

y

1

− x+

(x − x1 )

π x= 6

2y − 2 − √3 = −x +

Normal:

1

2

y − y1 y − (1 +

π

= − sin = −

dx x=π

12

Tangent at x − axis, y =0

√3 ) 2

|

1

− −

6

6

Tangent:

𝜋

6

π

π

2 √3 ) 2

𝑂

π

=1+

Gradient:

𝜋 6

6

1

y=− x+ ( ,1 +

Tangent & Normal Point:

Tangent:

𝑦

√3 2

√3 ) 2 √3 ) 2

√3 )] 4

(1 + (1 + (1 +

√3 ) 2 √3 ) 2 √3 ) 2

2

[shown] ✓

464


Ex 17.1

y = 2 sin 2x − 1 i.e. a = 2, b = 2, c = −1

15(iii) At point where tangent (y = 1) meets y = 2 sin 2x − 1 2 sin 2x − 1 = 1 sin 2x =1 π α= 2 ⇒ 1st or 2nd quadrant


=

2π b

2π

=

2

=π

𝑦 1𝑂 −1 −3

𝑦 = 2 sin 2𝑥 − 1

0 < x < 2π 0 < 2x < 4π

𝑥

2𝜋

✓

2x = α, π − α, π = ,

Workings Domain Axis with amplitude Shape Cycle

x

2 π

=

4

0 < x < 2π −1 ± 2

π

A

α T

α C

α + 2π, 3π − α π + 2π 2 5π

(taken),

4

⇒( 16(i)

+sin 2π−0

S

5π 4

, 1) ✓

f(x) = sin x + cos x f ′ (x) = cos x − sin x

=2 At turning points, f ′ (x) =0 cos x − sin x = 0 − sin x = − cos x tan x =1 π α= 4


15(ii) Curve & its gradient y = 2 sin 2x − 1 dy dx

0 ≤ x ≤ 2π

S

x = α, π + α

α T

= 2(cos 2x)(2) = 4 cos 2x

A α C

π 5π

= , Tangent Point:

4

π

y|x=π = 2 sin − 1

π

π

4

4

f ( ) = sin + cos

2

4

= 2(1) −1 =1

=

π 4

dy

4

√2 + 2

π

|

dx x=π

= 4 cos

4

π 2

⇒ ( , √2) ✓ 4

=0

5π

f ( ) = sin

Tangent: y − y1 = m(x − x1 )

4

π

y − 1 = (0) [x − ( )]

=

4

y

√2 2

π

= √2

⇒ ( , 1) Gradient:

4

=1✓

4 √2 − 2

+ cos −

5π 4

√2 2

= −√2 ⇒(


5π

sleightofmath.com

5π 4

, −√2) ✓

465


Ex 17.1

16(ii) For increasing function, f ′ (x) >0 cos x − sin x > 0 cos x > sin x 𝑦 5π π 4

⇒0≤x<

dS dθ

π

or

4

5π 4

At stationary value, dS

64 cos 2θ − 64 sin θ =0 cos 2θ − sin θ =0 2 1 − 2 sin θ − sin θ =0 2 sin2 θ + sin θ − 1 =0 (2 sin θ − 1)(sin θ + 1) = 0

< x ≤ 2π ✓

𝑂

π 4

⇒ 17(i)

π 4

5π 4

cos θ =

AF 8 BF 8

α=

⇒ AF = 8 sin θ ⇒ BF = 8 cos θ

1

6

6

S

A

α T

α C

2nd derivative d2 S dθ

A 8 cm 8 cm θ C B F

= 64(−2 sin θ) −64 cos θ = −128 sin 2θ −64 cos θ

d2 S

|

dθ2 θ=π

D

≈ −35.8 < 0

6

⇒ max +area of rectangle +(BE)(BC)

= (2 ⋅ BF)(AF)

+(BE)(2 ⋅ BF)

2 1

or sin θ = −1 (rej ∵ θ > 0)

2

= (BC)(AF) 2 1

2

θ = α, π − α π = ✓

E

Area S = area of triangle

π

1

⇒ 1st or 2nd quadrant π 0<θ<

✓

By trigonometry sin θ =

sin θ =

𝑦 = cos 𝑥 2𝜋 𝑥 𝑦 = sin 𝑥

4

=0

dθ

For decreasing function, f ′ (x) <0 cos x − sin x < 0 cos x < sin x 𝑦 5π

= 32(2 cos 2θ) −64 sin θ = 64 cos 2θ −64 sin θ

𝑦 = cos 𝑥 2𝜋 𝑥 𝑦 = sin 𝑥

4

𝑂

17(ii) 1st derivative

y = a tanm (x n ) = a[tan(x n )]m

18

dy

d

= a ⋅ m[tan(x n )]m−1

⋅

+64 cos θ

= a ⋅ m[tan(x n )]m−1

⋅ sec 2 (x n ) ⋅

= 64 ( sin 2θ)

+64 cos θ

= 32 sin 2θ

+64 cos θ ✓

= a ⋅ m[tan(x n )]m−1 = amn tanm−1 (x n )

⋅ sec 2 (x n ) ⋅ nx n−1 sec 2 (x n ) (x n−1 ) ✓

= (2 ⋅ 8 cos θ)(8 sin θ) +(4)(2 ⋅ 8 cos θ)

dx

2

= 64 sin θ cos θ 1 2

dx

[tan(x n )] d dx

(x n )

19(i) y = (k + x) cos x dy dx

= (k + x) ⋅

d dx

cos x

+

= (k + x) ⋅ (− sin x) = cos x d2 y dx2

= − sin x−[(k + x) ⋅

sleightofmath.com

(k + x) ⋅ cos x

+1 ⋅ cos x −(k + x) sin x d dx

(sin x) +

= − sin x−[(k + x) ⋅ cos x = − sin x−(k + x) cos x = −2 sin x −(k + x) cos x ✓ © Daniel & Samuel A-math tuition 📞9133 9982

d dx

d dx

(k + x) ⋅ sin x]

+1 − sin x

⋅ sin x]

466


d2 y dx2

Ex 17.1 22(i)

+y

= [−2 sin x − (k + x) cos x] + [(k + x) cos x] = −2 sin x [independent of k] ✓ 20(i)

d dx

d

(sec x)=

1

(

dx cos x cos x⋅

=

−1⋅

cos x⋅(0)

d cos x dx

−(− sin x)

= LHS [shown] ✓

sin x cos2 x

=(

1 cos x

)(

sin x cos x

tan x − tan3 x + 2 tan x 3 tan x − tan3 x = 1 − tan2 x − 2 tan2 x 1 − 3 tan2 x

=

cos2 x

=

22(ii)

d dx

)

(

3 tan x−tan3 x 1−3 tan2 x

)=

[

= 1 − (sec 2x tan 2x)(2) = 1 − 2 sec 2x tan 2x

dy

|

dx x=3

20(iii)

dx dt

= 2, =

dt

1−3 tan2 x

)]|

x=

=

dy

=3✓

=?

dy

×

dx

3 12

y = x + cos x

dx

dy

dt

dx

= 1 − sin x


≈ 3.21 ✓

sin x cos x (

1

1

1−sin2 x 2 cos2 x

1 − sin x = 0 sin x =1 π α=

)

1+sin x 1−sin x (1−sin x)+(1+sin x)

= sin x cos x (

)

2

)

d

=

[sin x cos x ( d dx

=− 1 1+sin x

𝑦 1

90° 180° 270° 360°

𝑥

−1

−2π ≤ x ≤ π x = −2π + α, −π − α, α, π − α

= 2 tan x ⇒ k=2✓

dx

=0

dx

+

π 3

cos2 π

dy

|

(3x)

3

≈ 1.61 ✓

= sin x cos x (

21(ii)

3 tan x−tan3 x

= (1 − 2 sec 2x tan 2x) × 2 At x = 3, dt x=3

21(i)

dx

(

=

|

d dx

= 3 sec 2 π

23(i) dy

d

= 1 − 2 sec 6 tan 6

dt x=3

(tan 3x)

= sec 2 3x ⋅ 3 = 3 sec 2 3x

20(ii) y = x − sec 2x

dx

d dx

= sec 2 3x ⋅

= sec x tan x ✓

dy

= tan 3x

2 tan x tan x + tan x + tan 2x 1 − tan2 x = = 1 − tan x tan 2x 1 − tan x ( 2 tan x ) 1 − tan2 x

cos2 x

=

1−3 tan2 x

RHS

)

d (1) dx

3 tan x−tan3 x

Show:

+

1 1−sin x

3π 2

,

π 2

✓

)]

2 tan x

= 2 sec 2 x ✓


sleightofmath.com

467


Ex 17.1

23(ii) Coordinates

= sin x (1 + cos x) = sin x + sin x cos x 1 = sin x + sin 2x 2

y

π

π

2

2

y|x=π = + cos ( ) 2

π = 2 π π ⇒( , )

1st derivative

2 2

dy

y|x=−3π = − 2

=− ⇒ (−

3π 2

,−

3π 2 3π

3π

+ cos (−

2

dx

)

at stationary pt:

2 3π 2

dy

)

cos x + cos 2x

=0

3x

x

2

2

2 cos ( ) cos (− ) = 0

= − cos x

dx2

=0

dx

2nd derivative d2 y

= cos x + cos 2x

3x

cos ( ) = 0

d2 y

π

|

dx2 x=π

2

0 ≤ x ≤ 2π:

= − cos ( ) = 0 2

2

x

cos ( ) = 0

or

2

0≤

π π

⇒ not max. or min. at ( , ) ✓ 2 2

3x 2

0 ≤ x ≤ 2π: x

≤ 3π:

0 ≤ < π: 2

𝑦

𝑦

1

d2 y

3

|

dx2 x=−3π

= − cos (− π)

2

not max. or min. at (−

⇒

=0

2

3π 2

,−

90° 180° 270° 360°

3π 2

)✓

3x

2

𝑦

𝑥

90° 180° 270° 360°

−1

2 3x

23(iv)

1

𝑦 = 𝑥 + cos 𝑥

−1

= α, 2π − α, 2π + α, 4π − α π 3π

= ,

2 2 π

5π

,

3

𝑂 (−

3𝜋 2

,−

3𝜋 2

π

1

2π

3

2

3

y|x=π = sin ( ) + sin

𝑥

3

=

)

√3 2 3

=

π 2

3

𝜋 𝜋 2 2

x 2

x=π

2 5π

x = , π,

( , )

𝑥

+

1 √3 ( ) 2 2

= √3 4

✓

π 3

24

d dx

(sin x°) = = =

25(i) 25(ii)

𝑦

d dx π

[sin (

180 π 180

π 180 π

cos (

⇒ ( , √3) 3 4

x)]

180

1

x)

y|x=π = sin π + sin 2π 2

=0 ⇒ (π, 0)

cos(x°) ✓

𝜋 3 ( , √3) 3 4

𝑦 = |sin 𝑥 (1 + cos 𝑥)|

𝑂

3

=−

(𝜋, 0)

𝑥 5𝜋


3

3

, − √3) 4

1

10π

3

2

3

√3 2

1

+ (− 2

)

√3 ) 2

3 = − √3 4 5π 3 ⇒ ( , − √3) 3 4

𝑦 = sin 𝑥 (1 + cos 𝑥) (

5π

y|x=5π = sin ( ) + sin (

✓

sleightofmath.com

468


Ex 17.1

2nd derivative d2 y dx2 d2 y

|

0600 1800

π 3

≈ −2.60 < 0

⇒ max at ( , √3) 3 4

|

=0

⇒ not max/min (π, 0)

|

≈ 2.60 > 0

⇒ min at (

3

dx2 x=π d2 y

y = sin x

= − sin x − 2 sin 2x

dx2 x=π d2 y

26

dx2 x=5π 3


5π 3

3

, − √3) 4

sleightofmath.com

1200

0000

✓

Fastest when gradient is the steepest ⇒ 0900, 1500, 2100 & 0300

469


Ex 17.2 3(b)

Ex 17.2 1(a)

d dx

d

(ex + 1) =

dx

dx

1(b)

dx

d

(4ex − 6) = 4

dx d

=4

dx x

dy dx

(ex ) −

d dx

3(c) (6)

d dx

d

(xex ) = x ⋅

dx d

(x) ⋅ ex

3(d)

dx x

d dx

ex

( )=

= d dx

d x (e ) dx

−ex ⋅

x⋅ex

3

= e2x ✓

d (x) dx

dy dx

x2 ex (x−1)

4(a)

✓

x2

(e−2x ) = e−2x ⋅

d dx

(−2x)

= e−2x ⋅ (−2) = −2e−2x ✓ 2(b)

d dx

dy dx

d (e3−5x ) dx d 5e3−5x ⋅ (3 − dx 3−5x (−5)

d

d

(e3x + 7) =

(e3x ) +

dx 3x

=e

⋅

d dx

d dx

d dx

(ex + e−x ) =

d dx x

4(b)

(7)

=e

= ex = ex

dx

d (e−x ) dx d +e−x ⋅ (−x) dx −x (−1)

4(c)

+e ⋅ −e−x ✓

=

= = =

dx

= 4e4x−1 ✓


du dx

dy

⋅

du

du dx

y = √1 + ex Let u = 1 + ex y = √u ✓

dx

y = e3x . ex−1 = e4x−1 ✓ dy

⋅

= 5u4 ⋅ (1 + ex ) = 5(x + ex )4 ⋅ (1 + ex ) = 5(1 + ex )(x + ex )4 ✓

dy

3(a)

dy du

y = (x + ex )5 Let u = x + ex y = u5 ✓ dy

(3x) + 0

(ex ) +

=

5x)

= e3x ⋅ 3 = 3e3x ✓ 2(d)

2

= 2u ⋅ (−ex ) = 2(1 − ex ) ⋅ (−ex ) = 2ex (ex − 1) ✓

= 5e ⋅ = −25e3−5x ✓ dx

3 3

= e2x ✓

y = (1 − ex )2 Let u = 1 − ex y = u2 ✓

(5e3−5x ) = 5 =

2(c)

y = √e3x 1

x2 −ex ⋅1

x

=

2(a)

x⋅

= −2e2−2x ✓

= (e3x )2

= xe +1 ⋅ e = ex (x + 1) ✓ 1(d)

= ex+3 ✓

y = (e1−x )2 = e2−2x ✓ dy

(ex ) +

dx x

✓

(ex ) − 0

= 4e ✓ 1(c)

ex x+3

(1)

= ex + 0 = ex ✓ d

e2x+3

=e

d

(ex ) +

y=

=

sleightofmath.com

dy du 1 2√u 1 2√1+ex ex 2√1+ex

⋅

du dx

⋅ (ex ) ⋅ (ex ) ✓

470


Ex 17.2 6(a)

y = √x + ex Let u = x + ex y = √u ✓

y= dy

dx

= = = =

5(a)

dy

⋅

du 1

du dx

6(b)

✓

2√x+ex

= (x + 1) ⋅

d dx

(e2x )

= (x + 1) ⋅ e2x ⋅

d dx

(2x) +

=

d

+

= (x + 1) ⋅ e2x ⋅ 2 = e2x [2(x + 1) = e2x (2x + 3) ✓ 5(b)

dx

d dx

(x + 1) ⋅ e2x

dx

=e

−2x

⋅

+1 ⋅ e2x +1]

d dx

(sin x)

+

d

(e−2x )

dx −2x

+e

⋅

d dx

dy dx

6(c)

⋅ sin x

−2x

(ex

3 +2x

dx x3 +2x

=x⋅e

= x ⋅ ex = ex

3 +2x

⋅

)

d dx

+

d dx

(x 3 + 2x) +1

⋅ (3x 2 + 2)

+ex

3 +2x

[x(3x 2 + 2) 3 = ex +2x (3x 3 + 2x + 1) ✓

⋅ sin x =

(x) ⋅ ex ⋅ ex

3 +2x

=

3 +2x

dx

1−x

=e

⋅

d dx

d x (e ) dx

−ex ⋅

d (1+e2x ) dx

(1+e2x )2 (1+e2x )⋅ex

−ex ⋅2e2x

(1+e2x )2 ex (1+e2x

−2e2x )

(1+e2x )2 ex (1−e2x ) (1+e2x )2

(cos 2x)

6(d)

+1]

cos x

cos x ⋅

d 3x d e − e3x ⋅ cos x dx dx cos 2 x

cos x ⋅ 3e3x

− e3x ⋅ (− sin x) cos 2 x

e3x (3 cos x + sin x) 2 cos x

y=

x+e2x ex

dy dx +

d dx

(e1−x )

⋅

cos 2x = −e1−x ⋅ (−2 sin 2x) +(e1−x ) ⋅ cos 2x = −e1−x (2 sin 2x + cos 2x) ✓


e3x

✓

3 +2x

y = e1−x cos 2x dy

1+e2x

dx

= d

ex

(1+e2x )⋅

y=

3 +2x

=x⋅

✓

dy

(−2x) ⋅ sin x

=e cos x +e ⋅ (−2) −2x = e (cos x − 2 sin x) ✓

5(d)

= =

−2x

y = xex

1+e−x +xe−x (1+e−x )2

✓

= e−2x ⋅ cos x

5(c)

=

(x + 1) ⋅ e2x

y = e−2x sin x dy

−x⋅(−e−x )

(1+e−x )2

dx

y = (x + 1)e

dx

−x⋅

dy 2x

dy

y=

d x dx

(1+e−x )⋅1

=

⋅ (1 + ex )

2√x+ex 1+ex

d (1+e−x ) dx −x 2 (1+e )

(1+e−x )⋅

=

⋅ (1 + ex )

2√u 1

1+e−x

=

dx dy

x

sleightofmath.com

d d (x + e2x ) − (x + e2x ) ⋅ ex dx dx = e2x x (1 2x ) e ⋅ + 2e − (x + e2x ) ⋅ ex = e2x 2x (1 + 2e ) − (x + e2x ) = ex 2x 1−x+e = ex ✓ ex ⋅

471


Ex 17.2

y = 2ex + 1

9(ii)

dx

|

= p,

dt x=1 dy dx dy

= 2ex

dx x=1

dx2 d2 y

8(i)

=

= 2ex |

dx2 x=1

=

dt

= 2e ✓

|

d2 y

dy

dy

|

y = 3e2x

1 − e−2x 3 dx

|

10

1 3e2 p

3e2

×

dx dt dx dt

= p,

dt x=1

=−

at y = 3: 3e2x = 3 e2x = 1 2x = ln 1 =0 x =0✓

=?

×

dx

= (−

dt x=1

|

dt x=1

dy

At x = 1,

= 2e ✓

dy

) ⋅ (p)

✓

t

Q = −ECe−RC t

dQ

= −EC (e−RC ) ⋅

dt

d dt

t

(−

= −EC (e−RC ) ⋅ (− E

t RC

1 RC

)

)

t

= e−RC R

8(ii)

dy

dQ

= 3(2e2x )

dx

dt x=3

= 6e2x 11(a) dy

8(iii)

11(b)

6e2x = 3 1

e2x

=

2x

= ln ( )

2 1 2

1 2

d2 y

=

2

2x )(2)

= 6(e

dx2

R

(e2x−1 + x)3 d dx

(e2x−1 + x)

d 2 + e1−x dx tan 2x d d tan 2x ⋅ (2 + e1−x ) − (2 + e1−x ) ⋅ (tan 2x) dx dx = tan2 2x

1

= ln ( )

x

3

= 3(e2x−1 + x)2 ⋅ (2e2x−1 + 1) ✓

=3

dx

E

= e−RC ✓

= 3(e2x−1 + x)2 ⋅

Gradient is 3, dy

d dx

= 6e2 ✓

|

dx x=1

|

=

= 12e2x

tan 2x ⋅ (−e1−x ) −e1−x tan 2x

− (2 + e1−x ) ⋅ 2 sec 2 2x tan2 2x − 2 sec 2 2x (2 + e1−x ) tan2 2x

✓ d2 y

1 2

1 2

2[ ln( )]

|

dx2 x=1 ln(1) 2

= 12e

2

1

= 12eln2 1 = 12 ( ) 2 = 6✓ 9(i)

y= dy dx

e−2x 6 1

= (−2e−2x ) 6

1

= − e−2x ✓ 3


sleightofmath.com

472


Ex 17.2

11(c) d x cos x dx e1−x2 =

dx

2 d (x cos x) dx

−x cos x⋅ 2

e1−x ⋅(x⋅

d cos x dx

+

)

=

Gradient:

2

e1−x ⋅[x⋅(− sin x) +1 ⋅cos x]

−x cos x⋅e1−x ⋅(−2x)

2 2

dy

|

dx x=2

= 2e4−1 = 2e3

(e1−x ) 2

=

= 2e2x−1

Tangent & Normal Point: y|x=2 = e3 ⇒ (2, e3 )

2 d d x⋅cos x) −x cos x⋅e1−x ⋅ (1−x2 ) dx dx 2 1−x2

2

=

2 d (e1−x ) dx

e1−x

(e

=

y = e2x−1 dy

e1−x ⋅

2

=

13(i)

Tangent:

2

e1−x ⋅(cos x−x sin x)

+2x2 cos x⋅e1−x

2 2 (e1−x )

+2x2 cos x

(cos x−x sin x) 2 e1−x

(1+2x2 ) cos x−x sin x

12(i)

✓

2

e1−x

y=

x k (e2k

+e

−

Normal:

x 2k )

y = (e 2

(x − x1 )

y − (e3 ) y − e3 y − e3 y

= 2e x − (2)] = e3 (2x − 4) = 2e3 x −4e3 = 2e3 x −3e3 ✓

y − y1

= − dy

dx x=2 3[

1

(x − x1 )

|

1

1

1

|

=

dx x=2

y − (e3 ) = − (2e3 ) [x − (2)]

At k = : 2

dy

y − y1

x 1 2( ) 2

x

− 1 2( )

+e

2

y

)

=−

1 2e3

x+

1 e3

+ e3 ✓

1

= (ex + e−x ) = 12(ii)

dy dx

2 (ex +e−x ) 2 1

= (ex − e−x ) =

d2 y dx2

✓[verified]

2 ex −e−x 2

✓

1

= [ex −(−e−x )] 2 1

= (ex +e−x ) 2

=

ex +e−x 2

✓


sleightofmath.com

473


Ex 17.2

13(ii)

14(ii) 1st curve y = e2x−3

𝑦 = 𝑒 2𝑥−1

𝑦

(2, 𝑒 3 )

dy

= 2e2x−3

dx

𝑂

3

2 + 2e6

2

𝑥 dy

Normal: 1 𝑦 = − 3𝑥+

Tangent: 𝑦 = 2𝑒 3 𝑥 − 3𝑒 3

2𝑒

1 𝑒3

=

= 2e2(2)−3 = 2e ✓

+ 𝑒3

Point B At B, tangent (y = 2e3 x − 3e3 ) crosses x − axis (y = 0). y =0 3 (2x e − 3) = 0 x

|

dx x=2

2nd curve y = e3−x dy

= −e3−x

dx dy

3

|

dx x=2

2

= −e3−(2) = −e ✓

3

⇒ B ( , 0) 2

15(i) Point C

y = 5xe−x dy

At C, normal (y = −

1 2e3

x+

1 e3

= 5x ⋅

dx

3)

+e

−

=0 1 2e3

1 2e3

x+

1 e3

+ e3 = 0

d2 y

1

dx2

x

=

e3

3

+e

⇒ C(2 + 2e6 , 0) Area of △ ABC 1

= (base)

(height)

2 1

3

dy

2

dx

= [4 + 4e6 − 3] = 14(i)

4 e3 (1+4e6 ) 4

= 5e−x ⋅

d dx

⋅ e−x

(1 − x) +

d dx

(5e−x ) ⋅ (1 − x)

+(−5e−x ) ⋅ (1 − x) −(1 − x)] −1 + x)

=0 −x (1

5e − x) = 0 x =1✓ ⇒ x − coordinate of S is 1

e3

✓

At P, y = e2x−3 has x-coordinates of 2 y|x=2 = e2(2)−3 = e ⇒ P(2, e)

(5x) ⋅ e−x

15(ii) At stationary point,

= [(2 + 2e6 ) − ( )] (e3 ) 2 1

d dx

= 5e−x ⋅ (−1) = 5e−x [−1 = 5e−x (−1 = 5e−x (x − 2) ✓

= 2 + 2e6

x

(e−x ) +

= 5x ⋅ (−e−x ) +5 = e−x (−5x +5) = 5e−x (1 − x) ✓

crosses x − axis (y = 0). y

d dx

15(iii)

d2 y dx2 d2 y

at S, |

dx2 x=1

= 5e−1 (1 − 2) 5

P(2,2e) lies in y = ea−x e = ea−2 ⇒1 =a−2 a =3✓

=− ✓ e

16(i)

y = ex − 2x − 1 dy dx

= ex − 2

At stationary point, dy dx x

=0

e −2 =0 ex = 2 x = ln 2 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

474


Ex 17.2

16(ii) y|x=ln 2 = e(ln 2) − 2(ln 2) − 1 =2 − 2 ln 2 − 1 =1 − 2 ln 2 ≈ −0.386 ✓ d2 y

= ex

dx2 d2 y

18(iii) P|t=25 = 995 000 m(25) 650 000e = 995 000

|

dx2 x=ln 2

= eln 2

18(iv)

=2 >0 ⇒ min at x = ln 2

>0

dx x

y = e−x dy dx

25m

= ln (

m

=

m

=

= e−x

2 +2x

= e−x

2 +2x

⋅

d dx

(∵ m =

2 +2x

✓

18(vi)

2 +2x

−

ln 199−ln 130 (9) 25

]

1 sign +

1 0

dP

> 18 000

mP

> 18 000

P

>

mt

1 −

)

25

650 000e

+

t ≡ t years after beginning of 2006

>

18 000 m 18 000 m 9

e

>

mt

> ln (

t

>

t (∵ m =

325m

1 m

9

)

325m 9

ln (

325m

)

> 28.5 ln 199−ln 130 25

)

⇒ t = 29 ✓ 19(i)

P = 650 000 emt

m = ae−kt Initial of 100g m|t=0 = 100 ae0 = 100 a = 100 ✓

2006 ⇒ t = 0 P|t=0 = 650 000 ✓ dt

(30)

ln 199−ln 130

mt

=0 =1 = e−1+2 = e [shown] ✓

Max ✓

dP

[shown] ✓

≈ 13 000 ✓

(−x 2 + 2x)

=0

−2(x − 1)e−x x y|x=1

18(ii)

25

ln 199−ln 130 (8) 25 ]

⋅ (−2x + 2)

dx

18(i)

)

2 +2x

dy

dx

ln (

25 130 ln 199−ln 130

−650 000 [e

17(ii) At turning point,

dy

)

ln 199−ln 130

dt

x

1

199

130 199

25 P|t=30 = 650 000e ≈ 1 083 000 ✓

= 650 000 [e

= −2(x − 1)e−x

17(ii)

130

P|t=9 − P|t=8

e −2 >0 ex >2 x > ln 2 ✓ 17(i)

=

18(v) 2014 ⇒ t = 8 2015 ⇒ t = 9

16(iii) For increasing function, dy

199

e25m

= 650 000memt = m(650 000emt ) = mP

⇒

dP dt

∝ P [shown] ✓


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475


Ex 17.2

19(ii) m = 100e−kt

21(iv)

𝑛 𝑛 = 20𝑒 0.1𝑡

th

At 40 h, mass reduces to 90g m|t=40 = 90 −k(40) 100e = 90 −40k e = 0.9p −40k = ln(0.9p) k 19(iii)

=− (

m = 100e

1

ln

40

9 10

21

= ab(e−bx ) >0 ∵ a > 0, b > 0, e−bx > 0 ⇒ increasing for all x ∈ ℝ ✓

1

= m|t=0

100e [

1

e 40 [

1 40

ln

2 1

1 9 ln ]t 40 10

= (100)

9 10

=

ln ]t 9 10

]t

2 1

22

y = ae2x + be4x dy

2

= ln

f: x ↦ a(1 − e−bx )

f ′ (x) = −a(e−bx )(−b)

Decay to half: [

dx

1

= 2ae2x +4be4x

2 d2 y dx2

1 9 ( ln )t 40 10

m = 100e dm dt

=(

1 40

ln

5

9

2

10

= ln 20(i)

= a(2e2x ) +b(4e4x )

≈ 263h ✓

t 19(iv)

✓

f(x) = a − ae−bx

✓

1 9 ln )t 40 10

m

𝑥

𝑂

9 10

e

= 4ae2x (

) 100e

[

1 9 ln ]t 40 10

= 2a(2e2x ) +4b(4e4x )

1 9 ln )t 40 10

Show:

d2 y dx2

+16be4x =6

dy dx

− 8y

✓ RHS

n = 20e0.1t n|t=5 = 20e0.5 ≈ 33 ✓

= 6(2ae2x + 4be4x ) −8(ae2x + be4x ) = 12ae2x + 24be4x −8ae2x − 8be4x = 4a2x + 16be4x = LHS [shown] ✓

20(ii) n > 200 0.1t 20e > 200 0.1t e > 10 0.1t > ln 10 t > 10 ln 10 t > 23.026 ⇒ least t = 24 ✓ 20(iii)

dn

= 20(e0.1t )(0.1)

dt

= 2e0.1t dn

|

dt t=15

= 2e0.1(15) ≈ 9 snails /week ✓


sleightofmath.com

476


Ex 17.2

y = (2 + 3x)e−x

24

dy

ex − x = 0 ex = x 𝑦

dx d

= (2 + 3x) ⋅

dx

(e−x ) +

= (2 + 3x) ⋅ (−e−x ) +3 = e−x [−(2 + 3x) +3] = e−x (−2 − 3x +3) = e−x (1 − 3x)

d dx

(2 + 3x) ⋅ e−x

𝑦=𝑥

⋅ e−x

y1 = ex y1 = ex > 0 ∴ y1 > y2

y2 = x y2 = x <0

x = 0:

y1 = e0 = 1 ∴ y1 > y2

y2 = 0

x > 0:

y1 = ex

y2 = x

dy1

dy2

x < 0: dx2 d

d (e−x )⋅ (1 dx −x ) (1

(1 − 3x) +

dx

−x

𝑥

𝑂

d2 y

= e−x ⋅

𝑦 = 𝑒𝑥

= e ⋅ (−3) +(−e ⋅ −x = e [−3 −(1 − 3x)] −x = e (−3 −1 + 3x) = e−x (3x − 4)

− 3x)

− 3x)

dx

At stationary point, dy dx −x (1

e

⇒

=0 − 3x) = 0

x

=

dy dx

=e

dy1 dx

>

dx

dx

∵ y1 > y2 at x = 0

3

&

dx

>

dy2 dx

for x >

0, y1 > y2 for x > 0

1−

1

1+

3

3

3

0

−

sign +

dy1

=1

dy2

1

⇒ stationary pt exist Sign Test: x

x

∴ ex − x = 0 does not have a real solution ∵ y1 > y2 for all x

⇒ max 1

2nd derivative at x = : 3

d2 y

1 3

−( )

|

dx2 x=1

=e

1

[3 ( ) − 4] 3

3

1 −e−3

= <0

1

⇒ max at x =

3

1

Curve at x = : 3

y|

1 x= 3 1

1 1 1 −( ) −( ) = [2 + 3 ( )] e 3 = 3e 3 3 1

⇒ ( , 3e−3 ) ✓ 3

1

1

Insert ( , 3e−3 ) into y = 3e−x 3

−

1 3

1

1 3

−( )

(3e ) = 3e

[consistent]

1 − 3

⇒ ( , 3e ) lies on y = 3e−x 3


sleightofmath.com

477


Ex 17.2

Curve Eqn: y1 = ex Curve Grad:

dy1 dx

25

y = ekx

= ex dy dx

= kekx

Line Eqn: y2 = x Line Grad:

dy2 dx

d2 y

=1

dx2

= k(kekx ) = k 2 ekx

x ≤ 0:

y1 > 0 y2 ≤ 0 ⇒ y1 > y2

dn y dxn

26

x > 0:

y1 > 0 y2 > 0 ⇒ cannot confirm y1 and y2 don′ t intersect dy1

x > 0:

dx

= ex > 1 [exponential function is an

increasing function] dy2 dx

⇒

= k n ekx [by inspection] ✓

Exponential function: (constant)variable (variable)constant Power function ex is an exponential function with e as a constant & x as a constant It is wrongly treated as a power function & power rule is wrongly applied.

=1

dy1 dx

>

dy2 dx

⇒ y1 > y2 x ∈ ℝ:

y1 > y2 y1 − y2 > 0 ex − x > 0 ex − x ≠ 0 ✓


sleightofmath.com

478


Ex 17.3 2(d)

Ex 17.3 1(a)

d dx

= = 1(b)

d dx

d dx 1

(ln x) +2

=1− (

d dx 1

=

(ln x)

+

x

dx

3(a)

) =

(1+ln x)⋅

d (x) dx

d (1+ln x) dx 2 (1+ln x)

−x⋅

−x⋅

1 x

(1+ln x)2 1+ln x −1 (1+ln x)2 ln x ✓ (1+ln x)2

d

d dx

ln(5x + 1) =

5 5x+1

[ln(4x − 3)2 ] =

=2

d dx

= d dx

ln(8 − x 3 ) = = =

8 4x−3 1

8−x3 1 8−x3 3x2 x3 −8


⋅

=

dy

|

dx x=3

4

3(b)

⋅ (0

⋅ (−2)

⋅ ln(5 − 2x)

+ ln(5 − 2x) + ln(5 − 2x) ✓

1 2x−5 2

⋅ (2)

2x−5

=

2 2(3)−5

y = ln(3 − 2x)

dy

=

dx

= dy

)

|

dx x=1

1 3−2x 2

⋅ (−2)

2x−3

=

2 2(1)−3

= −2 ✓

✓ d

(5 − 2x) +1

(x) ⋅ ln(5 − 2x)

Curve crosses x − axis, y =0 ln(3 − 2x) = 0 3 − 2x = e0 =1 −2x = −2 x =1✓ (1,0) ⇒

4x−3

dx

d dx

=2✓

[ln(4x − 3)]

=2⋅(

d dx

+

y = ln(2x − 5)

=

[2 ln(4x − 3)]

dx

⋅

2x−5

dx

✓

d

5−2x 1

5−2x 2x

dy

d sin x ( ) dx ln x d d ln x ⋅ (sin x) − sin x ⋅ (ln x) dx dx = (ln x)2 1 ln x ⋅ cos x − sin x ⋅ x = (ln x)2 sin x cos x ln x − x = (ln x)2 ✓

dx

[ln(5 − 2x)]

Curve crosses x − axis, y =0 ln(2x − 5) = 0 2x − 5 = e0 =1 x =3 ⇒ (3,0) ✓

⋅ ln x

(1+ln x)⋅1

d

dx 1

(x − 1) ⋅ ln x

+ ln x ✓

=

2(c)

d

+1

x

=

2(b)

=x⋅

x

=

2(a)

(x)

1

dx 1+ln x

1(d)

dx

=x⋅

[(x − 1) ln x]

= (x − 1) ⋅

d

d

+2 ✓

x

x ln(5 − 2x)

=x⋅

(ln x + 2x)

= (x − 1) ⋅

1(c)

d dx

(8 − x 3 ) − 3x 2 )

✓

sleightofmath.com

479


y = 3 ln(x − 3)

Ex 17.3 4(ii)


dy

= 3(

dx

=

1

=0

dx ln x−1 = (ln x)2

Curve crosses x − axis, y =0 3 ln(x − 3) = 0 ln(x − 3) = 0 x−3 = e0 x−3 =1 x =4✓ (4,0) ⇒

0

ln x = 1 x =e y|x=e =

e =e ln e

⇒ (e, e) Sign Test x 3−

) ⋅ (1)

x−3

3

dy

x−3

dx

sign

−

e

e+

0

+

⇒ min dy

|

=

dx x=4

3 4−3

5

=3✓ 3(d)

y = ln(3x − 2)2 = 2 ln(3x − 2)

dy dx

= 2(

dx

|

y= dy dx

d [ln(x+1)] dx

−ln(x+1)⋅

d (x+1) dx

(x+1)2 1 (x+1)⋅ x+1

−ln(x+1)⋅1 (x+1)2

1

−ln(x+1) (x+1)2

>0

dx 1−ln(x+1) (x+1)2

>0

1 − ln(x + 1) ln(x + 1) x+1 x

) ⋅ (3)

6 3−2

>0 ∵ x > −1 <1
Combine inequalities, −1 < x < e − 1 ✓

=6✓ 4(i)

(x+1)⋅

dy

3x−2

=

=

For increasing function,

3x−2

dx x=1

for x > −1

x+1

=

6

= dy

1

ln(x+1)

=

Curve crosses x − axis, y =0 2 ln(3x − 2) = 0 ln(3x − 2) = 0 3x − 2 = e0 3x − 2 =1 x = 1✓ dy

y=

x ln x

= = =

6

d d ln x⋅ (x) −x⋅ (ln x) dx dx (ln x)2

ln x⋅1

dC dt

1 x

−x⋅( )

1

) (1)

t+1 1

t+1

> 0 for 0 ≤ t ≤ 10

⇒ increasing function ✓

✓


= 0.1 + ( = 0.1 +

(ln x)2 ln x −1 (ln x)2

C = 11.9 + 0.1t + ln(t + 1)

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480


Ex 17.3

7(a) d dx

=

(3x ln √x 2 d

dx

− 2x − 3)

=

3

[ x ln(x 2 − 2x − 3)]

dx 2 3

d

2

dx

= x⋅

2

[ln(x − 2x − 3)]

3

= x⋅ 2

1

d

⋅

x2 −2x−3 dx

3

1

2

x2 −2x−3

= x⋅ =

d

7(d)

+

d

3

2

( x) ⋅ ln(x − 2x − 3)

dx 2

(x 2 − 2x − 3)

+

3 2

= ⋅ ln(x 2 − 2x − 3) =

3

+ ln(x 2 − 2x − 3)

⋅ (2x − 2)

2

3x(x−1)

=

3

2

+ ln(x − 2x − 3) ✓

x2 −2x−3

7(b)

=

2

=

Method 1 d dx

[ln(sin x cos x)]

d 1 ln ( sin 2x) dx 2 d 1 = (ln + ln sin 2x) =

2 1

d

2

dx

(ln ) +

dx

=0

+

=0

+

1 sin 2x 1 sin 2x

7(e)

d dx

=

(ln sin 2x) d

(sin 2x)

=

⋅ (2 cos 2x)

=

⋅

dx

= 2 cot 2x ✓

dx

=

[ln(sin x cos x)]

d = (ln sin x + ln cos x) dx d d = (ln sin x) + (ln cos x) = =

dx 1

sin x 1 sin x

⋅

d dx

(sin x)

⋅ cos x

+ +

dx 1

⋅

cos x 1

cos x

d dx

(cos x)

⋅ (− sin x)

− tan x ✓

= cot x

=

d dx

[ln (3 + e

d dx

= =

1 − x 2

)]

1 d − x 2 ) = 1 ⋅ dx (3 + e − x 2 3+e

= = =

1

− x 3+e 2

1 1 − x 3+e 2

⋅[

d dx

(3) +

⋅ [0 +e

1 − x 3+e 2

d

d dx d

1 √x2 +1−x

⋅

1 2

(e

dx

2

⋅ [0 +e−2x ⋅ (− )]

− x e 2 1 − x 2(3+e 2 )

⋅

d

2√x2 +1 dx

⋅( ⋅(

1 2√x2 +1

d dx

(x)]

(x 2 + 1) −1]

⋅ 2x

−1)

x

−1)

√x2 +1

x − √x 2 + 1 ⋅( ) √x 2 + 1 − x √x 2 + 1 1 √x2 +1

ln [ d dx d dx d dx

✓

e2 (3x−2) 4−x

]

[ln e2 + ln(3x − 2)

− ln(4 − x)]

[2 + ln(3x − 2)

− ln(4 − x)]

(2) + +

d dx 1

3x−2

1

[ln(3x − 2)] ⋅

d dx

(3x − 2)

⋅ (3)

3x−2

− − −

3

+

3x−2

dx d dx

sin x

d dx

[ln(4 − x)]

1 4−x 1 4−x 1 4−x

⋅

d dx

(4 − x)

⋅ (−1) ✓

]

[3 ln(x + 2)

+ ln(cos 3x)

− ln(sin x)]

[3 ln(x + 2)]

+

d dx

[ln(x + 2)] 1

⋅

d

x+2 dx

=

1 x+2

+

(x + 2) +

⋅ (1)

+

3

d dx

[ln(cos 3x)]

1 cos 3x 1 cos 3x

⋅

d dx

−

(cos 3x) −

⋅ (− sin 3x) ⋅

8(a)

1 cos 3x

d dx

[ln(sin x)]

1 sin x

d dx

d

⋅

dx

(sin x)

(3x) 1 sin x

⋅ cos x

⋅ (− sin 3x) ⋅ 3 − cot x

−3 tan 3x

x+2

− cot x ✓

y = ln(2x − 1) dy

✓


1

⋅[

−

− ln(sin x)]

)]

(− x)]

2

(√x 2 + 1)

dx

−

1

1

d

⋅[

+ ln(cos 3x)

=3⋅ d

(√x 2 + 1 − x)

1

(x+2)3 cos 3x

1

=−

1 √x2 +1−x

d dx

[ln(x + 2)3

=3⋅

− x

dx

1 − x 2

1

1

ln [

=3

1

1

1 √x2 +1−x

⋅

7(f)

= 7(c)

1 √x2 +1−x

=0

Method 2 d

1 √x2 +1−x

=−

=

dx d

[ln(√x 2 + 1 − x)]

dx

=

2 2x−1

≠0

⇒ no stationary pts

sleightofmath.com

481


Ex 17.3

y = ln(3x − k) dy

=

dx

= =

1

d

⋅

3x−5 1

dx

10

(3x − k)

y = ln(2 + e−x )

⋅3

3x−5 3

dx dy = 3, | =? dt dt x=1

dy

3x−5

=

dx

=

Curve cross x − axis (y = 0) at x = 2, 6−k = e0 6−k =1 k =5

=−

(2,0)

Gradient:

− dy

1

=−

|

dx x=2

Normal:

=

dt 1 3 ( ) 3(2)−5

y − y1 = − dy

1 |

=−

3

y − (0)= −

[x − (2)]

3 1

2

3

3

11(i)

y = ln(x 2 − 4) dy

= 9(ii)

dx

1

=(

dx

|

dt x=3 dy

=

dt

=

dy

|

dt x=3

dy

= 2x ⋅

dx

|

dt x=3

× ×

x2 −4 dx

32 −4

3 2e+1

✓

d dx

(ln x) +

1

d dx

(2x) ⋅ ln x

+2

x

⋅ ln x

=? 11(ii) At stationary point, dy

dx

=0

dx

dt dx

2 + 2 ln x = 0 ln x = −1

dt

dt x=3 2(3)

=−

+2 ln x ✓

1

= ✓

x |

|

dt x=1

=2

dx 2x

=

2ex +1

✓ dy

dt

×3

2ex +1 3

= 2x ⋅

dy

At x = 3,

1

) (2x)

= p,

dx

×

y = 2x ln x

x2 −4 2x

x2 −4

1 +1

2ex

At x = 1,

dy

9(i)

2+e−x

(x − x1 )

=− x+ ✓

y

(e−x )

⋅ (−e−x )

dx

=−

dx x=2

1

d dx

dy

=−

1

⋅

2+e−x e−x

=−

dy

Normal Point:

1 2+e−x 1

e

= p, ×p

11(iii)

d2 y

=

dx2

2 x

6

d2 y

5

dx2 x=1

= p

|

=

e

2 1 e

( )

= 2e ✓ 12(i)

y= dy dx

ln x x2

= = = =


sleightofmath.com

x2 ⋅

d d (ln x) −ln x⋅ (x2 ) dx dx x4

x2 ⋅

1 x

−ln x⋅2x x4

x

−2x ln x x4

1

−2 ln x x3

✓

482


Ex 17.3

12(ii) At stationary point, dy

13

x

=0

dx 1−2 ln x

= λx 2 (ln 1 − ln x) = λx 2 (0 − ln x) = −λx 2 ln x 1st derivative

=0

x3

1

V = λx 2 ln ( )

1 − 2 ln x = 0 −2 ln x = −1

dV

1

ln x

=

x

= √e ✓

dx

2

d

= −λx 2 ⋅

dx 1

(ln x) +

= −λx 2 ⋅ ( )

12(iii) d2 y dx 2

d

(λx 2 ) ⋅ ln x

dx

+ (−2λx) ⋅ ln x

x

= −λx −2λx ln x = −λx(1 + 2 ln x)

d d (1 − 2 ln x) − (1 − 2 ln x) ⋅ (x 3 ) dx dx = x6 2 x 3 ⋅ (− ) − (1 − 2 ln x) ⋅ 3x 2 x = x6 x3 ⋅

−2x 2

=

−2

=

dV

−λx(1 + 2 ln x) = 0 x=0

or

1

ln x = −

(rej ∵ 0 < x < 1) x

2

1

=

√e

✓

Sign Test 1−

x dy

x4

dx

6 ln x − 5 x4

=

=0

dx

− (1 − 2 ln x) ⋅ 3x 2 x6 − (1 − 2 ln x) ⋅ 3 4 x − 3 + 6 ln x

−2

=

At turning point,

√e

sign

+

1+

1 √e

√e

0

−

⇒ max

✓

2nd derivative d2 V

d2 y

| 2

dx x=√e

= =

dx2

6 ln √e−5 (√e) 3−5

4

= −λx ⋅

e2

=−

d dx

(1 + 2 ln x) +

2

2

2 ✓

(−λx) ⋅ (1 + 2 ln x) ⋅ (1 + 2 ln x)

= −λx ⋅ ( )

+(−λ)

= −2λ

−λ(1 + 2 ln x)

x

e

d dx

= −λ(2 + 1 + 2 ln x) = −λ(3 + 2 ln x) d2 V

1

|

dx2 x= 1

= −λ (3 + 2 (− )) 2

√e

= −2λ <0 ∵λ>0 ⇒ max ✓ 14

y = 2x ln y = ln 2x ln y = x ln 2 diff wrt x: 1 dy

( ) = ln 2

y dx dy dx


sleightofmath.com

= (ln 2)y [shown] ✓

483


Ex 17.3

15 y = (2 + x 2 )3 (1 − x 3 )4 −(1) ln y = ln[(2 + x 2 )3 (1 − x 3 )4 ] = ln(2 + x 2 )3 + ln(1 − x 3 )4 = 3 ln(2 + x 2 ) +4 ln(1 − x 3 ) differentiating wrt x, 1 dy y

⋅

dx

=3⋅ =3⋅ =

dy dx

1 2+x2 1 2+x2

⋅

d dx

(2 + x 2 )

⋅ 2x

+4 ⋅

6x 2+x2 6x

=(

2+x2

+4 ⋅

− −

12x2 1−x3

)⋅y

1 1−x3 1

1−x3 12x2

⋅

d dx

(1 − x 3 )

⋅ (−3x 2 )

1−x3

−(2)

sub (1) into (2): dy dx

=(

6x 2+x2

= 6x (

− 1

2+x2

12x2 1−x3

−

⋅ (2 + x 2 )3 (1 − x 3 )4

)

2x 1−x3

)

⋅ (2 + x 2 )(1 − x 3 )4

=6x[(1 − x 3 ) − 2x(2 + x 2 )] ⋅ (2 + x 2 )(1 − x 3 )3 = 6x[(1 − x 3 ) − 4x − 2x 3 ]

⋅ (2 + x 2 )(1 − x 3 )3

= 6x(1 − 4x − 3x 3 )

(2 + x2 )2 (1 − x3 )3 ✓


sleightofmath.com

484


y=

Ex 17.3

1 5 x 10x ln x

(x+ ) (5+x)ex 1 5 x 10x ln x

(x+ ) (5+x)ex

ln y = ln [

]

1 5

+ ln(5 + x) + ln ex − ln 10x − ln(ln x)

= ln (x + ) x

1

= 5 ln (x + )

+ ln(5 + x) +x −x ln 10

x

− ln(ln x)

Diff wrt x, 1 dy

( ) = 5(

y dx

1 dy

( ) =

y dx dy dx

1

) (1 −

1 x 1 5(1− 2 ) x 1 (x+ ) x

x+

) + +

1

= y[

1 x2

5(1− 2 ) x 1 x

(x+ )

+

1 5+x

1 5+x 1 5+x

− ln 10 −

1 x ln x

+1 − ln 10

−

+1 − ln 10

−

1

1

( )

ln x x 1 x ln x

]✓

16(b) y = (cos5 x)(x2 +5)(ln x) x+1 ln y = ln [

e (cos5 x)(x2 +5)(ln x) ex+1

] + ln(x 2 + 5) + ln(ln x)

= 5 ln cos x diff wrt x, 1 dy

1

y dx

cos x

( ) = 5(

) (− sin x) + (

= −5 tan x dy dx

= y (−5 tan x +


+ 2x x2 +5

1

) (2x) + (

x2 +5 2x

x2 +5 1

+

x ln x

+

1 ln x 1

1

)( )

x ln x

x

−(x + 1) −1 −1

− 1) ✓

sleightofmath.com

485


Ex 17.3

y = ax − x ln x

18(ii) y = x ln x dy

Curve cross x − axis: y =0 ax − x ln x = 0 x(a − ln x) = 0 x=0 or ln x = a (rej ∵ x > 0) x = ea ✓ 17(ii)

dy dx

= a − [x ⋅ = a − (x ⋅

d dx 1

(ln x) +

d dx

=x⋅

d dx 1

(ln x) +

d dx

(x) ⋅ ln x

+1 ⋅ ln x

x

=1

+ ln x


=0

dx

1 + ln x = 0 (x) ⋅ ln x]

x

+1 ⋅ ln x)

x

=x⋅

dx

=

1 e

1

1

e 1

e

y|x=1 = − ln e

= a − (1 + ln x) = a − 1 − ln x

=−

e

1 1

⇒ M( , )✓ e e


=0

dx

d2 y

a − 1 − ln x = 0 ln x =a−1 x = ea−1 ✓

dx2 d2 y

=

1 x

|

dx2 x=1

y|x=ea−1 = a(ea−1 ) −(ea−1 ) ln(ea−1 ) = a(ea−1 ) −(ea−1 )(a − 1) = ea−1 [a −(a − 1)] = ea−1 (a −a + 1) = ea−1 ⇒ (ea−1 , ea−1 ) d2 y dx2 d2 y dx2

=− |

e

𝑦 𝑂

1 1 e

( )

=e >0 ⇒ min without reflection ⇒ max at M 19

1

x x is (variable)variable x n is (variable)constant ⇒ Tom is wrong ax is (constant)variable ⇒ Doris is wrong

x

x=ea−1

=−

1 ea−1

<0

Dave’s method can be derived as follows let y = x^x ln both sides, ln y = ln x x ln y = x ln x Differentiating wrt x,

∵ ea−1 > 0

⇒ max. ✓ 18(i)

=

𝑦 = |𝑥 ln 𝑥|

1 dy

𝑥 ✓

= x ( )+1(ln x)

( )

=1

y dx dy dx d dx


sleightofmath.com

1

( )

y dx 1 dy

(x x )

x

+ ln x

= y(1 + ln x) = x x (1 + ln x)

486


Clearly,

Ex 17.3

ln 3x

= ln 3 + ln x ≠ ln x The derivative might be the same because the derivative of the constant is ln 3 is 0.

21

No. d d ln|ax| = a ⋅ ln|x| dx dx For a = −1, LHS =

d dx

(ln 3x)

= =

d dx d dx

(ln 3 + ln x) (ln 3) +

=0 = d dx

(ln x)

=

+

d dx 1

d dx

ln|−x|

RHS = −1 ⋅ (ln x)

d dx

=

d dx

ln|x| = −

ln|x|

=

1 x

1 x

LHS ≠ RHS

x

1 x 1 x


sleightofmath.com

487


Rev Ex 17 A2(b)

Rev Ex 17 A1(a)

d dx

d dx

(x) −2

= 3(1)

d

(

d dx

(cos x)

A2(c)

x

d

d dx

(tan 3x)

+

= 2x ⋅ (3 sec 2 3x)

d dx

d dx

A2(d)

A2(e)

=3

d d dx

⋅ tan 3x

A3(i)

d dx

+

d dx

(3x 2 ) ⋅ ln x

+6x

x

⋅ ln x

+6x ln x ✓

=

ex [(2x + 1) (2x + 1)2

=

ex (2x−1) (2x+1)2

d

ln (

= d

(sin 8x)

d dx

d

✓ ) =

d dx

d

[ln(1 − 2x) − ln(1 + x)]

dx −2

2x−1

−

1 x+1

✓

(cot x) = − csc 2 x

sin2 x + cos 2 x sin2 x 1 =− 2 sin x

(x 2 + 2)

=−

= − csc 2 x [shown] ✓

[(x + 1)e−x ] dx

1+x

− 2]

d cos x ( ) dx sin x d d sin x ⋅ (cos x) − cos x ⋅ (sin x) dx dx = sin2 x sin x (− sin x) − cos x (cos x) = sin2 x

[sin(x 2 + 2)] = cos(x 2 + 2) ⋅

= (x + 1) ⋅

1−2x

Show:

(sin 8x)4 dx

d x d (e ) − ex ⋅ (2x + 1) dx dx (2x + 1)2

(2x + 1) ⋅ ex − ex ⋅ 2 (2x + 1)2

LHS

= cos(x 2 + 2) ⋅ 2x = 2x cos(x 2 + 2) A2(a)

(ln x)

)

(2x + 1) ⋅

(sin4 8x)

= 96 sin3 8x cos 8x ✓ d

ex

=

= 3 ⋅ 4(sin 8x)3 ⋅ (8 cos 8x)

dx

(

+2 tan 3x ✓

= 3 ⋅ 4(sin 8x)3 ⋅

A1(e)

+(2e2x ) ⋅ sin x +2 sin x) ✓

=

dx

(3 sin 8x) dx

d

=

4

=3

(e2x ) ⋅ sin x

dx 2x+1

(2x) ⋅ tan 3x

+2

= 6x sec 2 3x A1(d)

d dx 1

d dx

)

(2x tan 3x)

= 2x ⋅

(sin x) +

= 3x

d d (sin 2x) − sin 2x ⋅ (x) dx dx = x2 x ⋅ (2 cos 2x) − sin 2x ⋅ 1 = x2 2x cos 2x − sin 2x = 2 x ✓

dx

(3x 2 ln x) = 3x 2 ⋅ = 3x 2 ⋅

x⋅

A1(c)

d dx

+2 sin x ✓

sin 2x

d dx

= e2x ⋅ cos x = e2x (cos x

−2(− sin x)

=3 dx

(e2x sin x) = e2x ⋅

(3x − 2 cos x)

=3

A1(b)

d dx

(e−x )

= (x + 1) ⋅ (−e−x ) = e−x [−(x + 1)

+1]

= e−x (−x − 1

+1)

+

d dx

+1

(x + 1) ⋅ e−x ⋅ e−x

= −xe−x ✓


sleightofmath.com

488


Rev Ex 17 dy

A3(ii) t = 20 cot θ dt

2

= −20 csc θ

dθ dθ

=

dt

= −2(2 cos x) ⋅

dx

1

20

At turning points,

sin θ

dy

=0

dx dt

|

π θ= 4

=− =− =− =−

A4(i)

1 20 1

4 sin x cos x − sin x = 0 sin x (4 cos x − 1) = 0

π

sin2 ( ) 4

√2 20 2 1 2

2

sin x = 0

( )

40

rad s

or

0 < x < 2π:

( )

20 4 1

𝑦 −1

✓

1

90° 180° 270° 360°

y = 1 − 2 cos 2 x + cos x

𝑥

−1

x = 0, π ✓ Curve meets x − axis, y =0 1 − 2 cos 2 x + cos x =0 2 2 cos x − cos x − 1 =0 (2 cos x + 1)(cos x − 1) = 0 cos x = − α=

π

1 2

or

Sign Test x π− dy sign −

cos x = 1

1 4 α ≈ 1.32 ⇒ 1st or 4th quad. 0 < x < 2π: S A α α T C x = α, 2π − α ≈ 1.32,4.97 ✓ cos x =

π 0

π+ +

1.32 0

1.32+ −

4.97 0

4.97+ −

dx

0 < x < 2π:

3

⇒ min at x = π

𝑦

⇒ 2nd or 3rd quad. 1 0 < x < 2π: 𝑥 90° 180° 270° 360° S A −1 α α x = α, 2π − α T C = 0,2π (no x = π − α, π + α solution) 2π 4π = , ✓ 3

(cos x) − sin x

= −2(2 cos x) ⋅ (− sin x) − sin x = 4 sin x cos x − sin x

−20 csc2 θ 1 2

=− dθ

d dx

x dy dx

1.32− sign +

⇒ max at x = 1.32 x dy

3

dx

4.97− sign +

⇒ max at x = 4.97 A5(i)

y = xex−3

A5(ii)

dy

=x⋅

dx

d

(ex−3 ) +

dx x−3

d dx

(x) ⋅ ex−3

=x⋅e +1 x−3 = (x + 1)e ✓ dy

|

dx x=3

⋅ ex−3

= 4e0 =4✓


sleightofmath.com

489

A math 360 sol (unofficial) A6(i) A6(iii)

Rev Ex 17 A8(ii) n = 40: 18e0.2t = 40

𝑦 𝑦 = ln 𝑥

2

20 9

20

t = 5 ln ( )

2

9

✓ A6(ii) x 2 ex ln(x 2 ex ) ln x 2 + ln ex 2 ln x + x 2 ln x

= 12 = ln 12 = ln 12 = ln 12 = −x + ln 12

ln x ⇒a=−

1

1

2

2

9

0.2t = ln ( )

𝑥 1 1 𝑦 = − 𝑥 + ln 12

𝑂

20

e0.2t =

≈4✓ B1(a)

d dx

=

(sin 4x − 3 cos 2x) d

(sin 4x)

dx

= cos 4x ⋅

d dx

= − x + ln 12 ✓

= cos 4x ⋅ 4

= ax +b

= 4 cos 4x

1

−3

d dx

(cos 2x)

(4x) −3(− sin 2x) ⋅

d dx

(2x)

−3(− sin 2x) ⋅ 2 +6 sin 2x ✓

2

1

⇒ b = ln 12 2

B1(b)

≈ 1.24

d

1

(

dx 1+cos 4x

=

Draw 1

1

2

2

d

)

[(1 + cos 4x)−1 ]

dx

= −(1 + cos 4x)−2 ⋅

y = − x + ln 12 ✓

1

A7(i)

cos θ (

1

+

= − (1+cos

)

1+sin θ 1−sin θ (1−sin θ)+(1+sin θ)

= cos θ [

= cos θ ( =

1

1−sin2 θ 2 cos2 θ

)

B1(c)

d dx

2 cos θ

✓

(x cos 2 3x) d

=x⋅

= 2 sec θ [shown] ✓

4x)2

dx

(cos 2 3x)

= x ⋅ 2 cos 3x ⋅

d dx

=

[cos θ ( d dx

=2

1 1+sin θ

+

1 1−sin θ

)]

(2 sec θ) d

(

1

dx cos θ

=2⋅ =2⋅ =2⋅ = 2(

B1(d)

=

d cos θ⋅ (1) dx

d −1⋅ (cos θ) dx cos2 θ

−1⋅(− sin θ) cos2 θ

sin θ cos2 θ 1

sin θ

cos θ

cos θ

)(

)

B1(e)

+1

(x) ⋅ cos 2 3x ⋅ cos 2 3x

(x 2 − 3 tan2 4x) d dx

(x 2 )−3

d dx

(tan2 4x) d

(tan 4x)

−3 ⋅ 2 tan 4x ⋅

= 2x = 2x

−3 ⋅ 2 tan 4x ⋅ (4 sec 2 4x) −24 tan 4x sec 2 4x ✓

d

dx

[2 cos(1 − x 2 )]

=2

n = 18e0.2t

d dx

[cos(1 − x 2 )]

= 2[− sin(1 − x 2 )] ⋅

d dx

(1 − x 2 )

= 2[− sin(1 − x 2 )] ⋅ (−2x)

n|t=10 = 18e2 ≈ 133 ✓ © Daniel & Samuel A-math tuition 📞9133 9982

(cos 3x)

d dx

= 2x

dx

= 2 sec θ tan θ ✓ A8(i)

d dx

)

cos θ⋅0

d dx

+

= x ⋅ 2 cos 3x ⋅ (−3 sin 3x) + cos 2 3x = x ⋅ 2 cos 3x ⋅ (−3 sin 3x) + cos 2 3x = cos 2 3x −6x sin 6x ✓

k=2✓ A7(ii)

(1 + cos 4x)

⋅ (−4 sin 4x)

4x)2

4 sin 4x

= (1+cos

]

d dx

= 4x sin(1 − x 2 ) ✓ sleightofmath.com

490

A math 360 sol (unofficial) B2(a)

d

x2

(

dx ln 2x

Rev Ex 17 B3(i)

)

d 2 d (x ) − x 2 ⋅ (ln 2x) dx dx = (ln 2x)2 d ln 2x ⋅ 2x − x 2 ⋅ (ln 2 + ln x) dx = (ln 2x)2 1 ln 2x ⋅ 2x − x2 ⋅ x = (ln 2x)2 ln 2x ⋅

=

2x ln 2x (ln 2x)2

−x

y = sin 3x + cos 3 x dy

dy

dx

2(x2 −1)

6

1

√3 2

2

= 3(0) −3 ( ) ( )

✓

2

3 3

=− ( ) 8

d

d dx

[ln(2x + 1)] +

dx 2

=e

⋅

1

= = = d dx

=

dt

= 0.3,

d

dx −4

dt

= 0.3,

=?

(e2x ) ⋅ ln(5 − 4x) ln(5 − 4x)

=

|

dt x=π

=?

dy

×

dx

dx dt

π

At x = , 6

dy

9

|

dt x=π 6

+ ln(5 − 4x)] ✓

dy

6

1

d

dx 1 1 + ( e2x ) ⋅ 2 1

5−4x 4

|

dt x=π

⋅ ln(2x + 1) dx

[ln(5 − 4x)] +

dy

6

dt

4x−5

=

dx

dy

= e2x [

=

B3(ii)

[e ln(5 − 4x)]

1 x 2

=

⋅ ln(2x + 1)

1 x 2

1

d

(x 2 − 1)

+2x

2x+1

= e2x ⋅

dx

d dx

+2x ln(2x + 1) ✓

2x+1

B2(e)

π

6

=− ✓

= (x 2 − 1) ⋅

B2(d)

π

2

6

[(x 2 − 1) ln(2x + 1)]

B2(c)

π

= 3 cos −3 sin cos 2

|

dx x=π

(cos x)

2 4 9

= (x 2 − 1) ⋅

=

d dx

+3 cos 2 x ⋅ (− sin x) −3 sin x cos 2 x

= 3 cos 3x = 3 cos 3x

B2(b) d

+3 cos 2 x ⋅

= 3 cos 3x

dx

= (− ) ⋅ (0.3) 8

≈ −0.338 ✓

2

[ln(x + √x 2 + 1)] 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 x+√x2 +1 1 √x2 +1

⋅

d

(x +√x 2 + 1)

dx

⋅ [1

+

⋅ (1

+

⋅ (1

+

⋅(

1 2√x2 +1 1 2√x2 +1 x √x2 +1

√x2 +1 √x2 +1

⋅

d dx

(x 2 + 1)]

⋅ 2x)

)

+x

)

✓

[x 3 ln(cos 2 x)] d dx

[2x 3 ln(cos x)]

= 2x 3 ⋅ = 2x 3 ⋅ = 2x 3 ⋅

d

[ln(cos x )]

dx 1

cos x 1

⋅

(cos x)

(− sin x)

cos x 3 sin x

= −2x (

d dx

cos x

)

= −2x 3 tan x


+

d dx

(2x 3 ) ⋅ ln(cos x)

+2(3x 2 ) ⋅ ln(cos x) +6x 2 ⋅ ln(cos x) +6x 2 ln(cos x) +6x 2 ln(cos x)✓

sleightofmath.com

491


Rev Ex 17

y = sin x cos 3 x

Sign Test x

dy dx

= sin x ⋅

d dx

(cos 3 x) d

2

= sin x ⋅ 3 cos x ⋅

dx

+

d dx

dy

(sin x) ⋅ cos 3 x

(cos x) + cos x

= sin x ⋅ 3 cos 2 x ⋅ (− sin x)

+ cos 4 x

= −3(1 − cos 2 x) cos 2 x

+ cos 4 x

= −3 cos 2 x + 3 cos 4 x

+ cos 4 x

2

2

0

−

sign −

π+

⇒ (0,0) is not max or min

⋅ cos x x dy

⇒( , 6

dy

dx 5π

=0

⇒(

4 cos 4 x − 3 cos 2 x =0 2 (4 2 cos x cos x − 3) = 0

π

π+

6

6

6

0

−

16

5π

5π+

) is max 5π−

x dy

π−

sign +

dx π 3√3

= 4 cos 4 x −3 cos 2 x At turning point, dx

π

2

3

+ cos 4 x

= −3 sin2 x cos 2 x

dx

π−

6

6

6

sign +

0

−

6

,−

3√3 16

) is min

√3

cos x = 0 or cos x = ± 2 π 0 < x < π: α= 𝑦 6 ⇒ 1st, 2nd, 3rd, 4th quadrant 1 𝑥 0 < x < π: 90° 180° 270° 360° x = α, π − α, π + α, 2π − α −1 π 5π π x= , 6 6 x= 2 π π y|x=π = sin cos 3 2

2

2

(0)3

= (1) =0 π

⇒ ( , 0) 2

π 3

π

y|x=π = sin (cos ) 6

6

6

1 √3 2 2 1 3√3

3

= ( ) = ( =

2 3

16 π 3√3

⇒( , 6

16

8

)

√3

)✓

y|x=5π = sin

5π

6

6

1

= (− 2

=− ⇒(

5π 6

,−

3

16 3√3 16

(cos

√3 ) 2

5π 3 6

)

3

√3

)✓


sleightofmath.com

492


Rev Ex 17

B5(a) y = 2e4x + 8e−4x 1st derivative dy

4x )

dx

B5(b) y = x ln x − 2x dy

= 2(−4e

+ 8(−4e

= 8e4x

−32e−4x

dx

−4x )

=0

dx 4x

−4x

8e − 32e e4x − 4e−4x e4x e8x 8x

= ln 22 8 2

= ln 2 8 1

B6

= ln 2 4

1 4

−4( ln 2)

+ 8e

= 2e

−2

=1 = ln x − 1

+ ln x

−2

=0

Curve y = (2x + c) ln x dy

4

ln 2

dx

− ln 2

+ 8e

= 2(2)

+8eln2

=4

+8 ( )

= (2x + c) ⋅ = (2x + c) ⋅

1

=2+

1

c

d dx 1

(ln x) +

x

⇒ ( ln 2 , 8) ✓

1

=−

|

dx x=1

2nd derivative

dx2

= 8(4e4x ) − 32(−4e−4x )

Line 2y = 5 − 3x

= 32e4x

y

|

+128e−4x 1 4

4( ln 2)

1 x= ln 2 4

⋅ ln x

+2 ln x

x

mnorm = − dy

4

d2 y

(2x + c) ⋅ ln x

Normal gradient at A(1,0)

1

dx2

+2

d dx

2

=8

d2 y

(x) ⋅ ln x −2

ln x − 1 = 0 ln x =1 x =e y|x=e = (e) ln(e) −2(e) = e −2e = −e ⇒ (e, −e) ✓

8 1

1 4

d dx

+(1) ⋅ ln x

dx

1

4( ln 2)

+

=x⋅( )

dy

= ln 4

y|x=1 ln 2 = 2e

(ln x)


=0 =0 = 4e−4x =4 = ln 4

x

d dx 1 x

At turning point, dy

=x⋅

= 32e

+128 e

3

=− x+ 2

mline = −

1 4

−4( ln 2)

1 c 2+ +2 ln 1 1

2

−

+128 ( )

>0 1

⇒ min at ( ln 2 , 8) 4

B7(i)

1 2+c

1 2+c

2

2

= 32(2)

=−

3

1

+128eln2

1 2+c+0

5

Normal at A(1,0) ∥ line: mnorm = mline

1

= 32eln 2

=−

=−

3 2

2

2+c

=

c

=− ✓

3 4 3

t

m = 24e−8 7

m|t=7 = 24e−8 ≈ 10.0g ✓


sleightofmath.com

493


B7(ii) m

= m|t=0 t

2 1

24e−8 = (24)

Rev Ex 17 B8(ii) x|t=0 = 26 − 30e0 = −4 ✓

2

B8(iii) sub x = 25, 26 − 30e−0.2t = 25

= 12 −

t

= ln

8

t

1 2

= −8 ln

1 2

≈ 5.55h ✓ B7(iii)

dm

1

t − 8

=

−0.2t

= ln

t

= −5 ln

= 24 (− e )

dt

8

t − 8

= −3e |

dt t=8

B8(iv)

dx

1 30

= 6e−0.2t

3

dx

e

|

dt t=5

= 6e−1 ≈ 2.21° per minute ✓

x = 26 − 30e−0.2t

t →∞ −0.2t e →0 x → 26 ✓

x|t=9 = 26 − 30e−1.8 ≈ 21.0 ✓


1 30

= −30(−0.2e−0.2t )

= −3e−1 = − g h−1 ✓

B8(i)

30

≈ 17.0 ✓ dt

dm

1

e−0.2t

sleightofmath.com

494


Ex 18.1 4(a)

Ex 18.1

∫(6x + 3) dx x2

= 6 ( ) +3(x) +c 1(i)

d dx

1(ii)

2

(x 2 + 5x) = 2x + 5 ✓

∫(2x + 5) dx= x 2 + 5 + c ✓

= 3x 2 4(b)

+3x +c ✓

∫(3 − √x) dx 1

2(i)

d

(

x

dx 1+2x

)=

= ∫ (3 −x 2 ) dx

d d (1+2x)⋅ x −x⋅ (1+2x) dx dx (1+2x)2

(1+2x)⋅1

=

3

x2

= 3x − ( 3 ) +c

−x⋅2

2

2

(1+2x)2 1+2x −2x (1+2x)2

=

1

= 3x − x√x +c ✓ 3 4(c)

= (1+2x)2 [shown] ✓ 2(ii)

2

x3

1

= 2(

1+2x 2x

= 3(a)

= ∫(3x 2 + 6x) dx 3

4(d)

x4

3(b) 3(c)

∫(x − 1)(x + 2) dx

=

= 2 ( ) +c =

+c ✓

4(e)

∫(−5) dx = −5x + c ✓

x3 3

1

2

=

= x +c ✓ 3(d)

1

∫ (− x2 ) dx= − ∫ x −2 dx = −( =

3(e)

1

1

1 x−2 2

=− 3(f)

∫

2 √x

−1

−2 1

dx = 2 ∫ x

+2x√x +c ✓

2

2

∫(2x − √x) dx

3

= ∫ (4x 2 − 4x 2 + x) dx 5

x3

x2

x2

= 4 ( ) −4 ( 5 ) + ( ) +c 3

) +c

4

= x3 3

+c ✓ 5(a) dx

∫

2x2 +3 x2

8

x2

5

2

− x 2 √x +

+c ✓

dx = ∫(2 + 3x −2 ) dx

= 2x

= 2 ( ) +c

2

2

= 2x

1 x2 1 2

= 4√x

2

x2

= ∫(4x 2 − 4x√x + x) dx

) +c

dx

4x2 1 − 2

4(f)

+c ✓

x

∫ 2x3 dx = 2 ∫ x = (

x−1

1

−3

x2

= ( ) +3 ( 3 ) +c

3 2

3

3

x2

+c

2

−2x +c ✓

2

∫ √x(√x + 3) dx

3 3 2

x2

= ∫ (x + 3x 2 ) dx

∫ √x dx = ∫ x dx =

+

= ∫(x + 3√x) dx

1 2

x2

+3x 2 +c ✓

= ∫(x 2 + x − 2) dx

∫ 2x 3 dx = 2 ∫ x 3 dx 4 1 4 x 2

2

= x3

) +c

+c ✓

1+2x

x2

= 3 ( ) +6 ( ) +c

∫ (1+2x)2 dx = 2 ∫ (1+2x)2 dx x

∫ 3x(x + 2) dx

+3 ( −

3 x

x−1 −1

) +c +c ✓

+c ✓


sleightofmath.com

495


∫

x2 +1 2x2

dx = ∫ (

1

Ex 18.1

1

2 1 x−1

2 1

2 −1 1

2

2x

= x + ( = x − 5(c)

∫

x+1 √x

8(a)

1

+ x −2 ) dx

2

+c ✓

x2

y = 2 ( ) +3 ( 2

= x2

1

2 3

3

1

x2

x2

2 3 2

2

= =

6(b)

(3x+1)5 (3)(5) (3x+1)5 15

+c

x

x

+c ✓

8(b)

dy

⇒

(1−x)4

4

) +c

3

∫(1 − x)3 dx = (−1)(4) +c =−

−1

∴ y = x2 − + 1 ✓

+c

dx

(1−x)4

−

3

x−1

Curve at (−1,5): 5 = (−1)2 + 3 + c c=1

+c ✓

+2√x

∫(3x + 1)4 dx

3 x2

= 2x + 3x −2

) +c

dx = ∫ (√x + x −2 ) dx

= x

= 2x +

dx

= ( 3 ) + ( 1 ) +c

6(a)

dy

∝ (x 2 − 1) dy dx

= k(x 2 − 1) = kx 2 − k

+c ✓ x3

6(c)

∫(2x + 5)−3 dx =

(2x+5)−2

2

=− dy dx

3

+c

(2)(−2) 1 (2x+5)−2

= [

7

y = k ( ) −kx +c

−2 1

4(2x+5)2

1

= kx 3 3

] +c dy

+c ✓

dx dy

x3 3

=x

|

=9

k(2) − k = 9 3k =9 k =3 c|k=3 = 1

x2

y = 3( ) − ( ) + c 3

= 9 when x = 2,

dx x=2 2

= 3x 2 − x

2 1 2 − x 2

−kx +c

+c

y = 3 when x = 2, y|x=2 =3

Curve at (2,4), 1

(4) = (2)3 − (2)2 + c

1

2

3 2

4 =8−2+c c = −2

3

k(2)3 − k(2) + c = 3 k+c

=3 2

c

=3− k 3

1

∴ y = x2 − x2 − 2 ✓ 2

1

∴ y = (3)x 3 −3x +1 3

= x 3 − 3x + 1 y|x=3 = (3)3 − 3(3) + 1 = 19 ✓


sleightofmath.com

496


dV dt

Ex 18.1 12(i)

= 6(2t − 1)2 + 1

V = 6 [ (2)(3) ] + t + c

=x⋅

= (2t − 1)3 + t + c V|t=1 (2 − 1)3 + 1 + c 1+1+c c

=4 =4 =4 =2

=

(given)

= = =

3

∴ V = (2t − 1) + t + 2 ✓ dx dt

(x√x + 1)

=x⋅

(2t−1)3

10(a)

d dx

d dx

d

√x + 1 + dx (x) ⋅ √x + 1 1

2√x+1

⋅1

+1

x

⋅ √x + 1

+√x + 1

2√x+1 x

+2(x+1) 2√x+1

x

+2x+2 2√x+1

3x+2

[shown] ✓

2√x+2

12(ii) ∫ 3x+2 dx = 2 ∫ 3x+2 dx 2√x+1

= 3t 2 + 2

√x+1

= 2x√x + 1 + c ✓ t3

x = 3 ( ) + 2t + c

13(a) ∫ √6x − 1 dx

3

=t

3

+ 2t + c

1 2

= ∫(6x − 1) dx Initial radius is 1: x|t=0 =1 03 + 0 + c = 1 c =1

3

= =

x = t 3 + 2t + 1 ✓ 10(b)

dA dt

2

− t2

d

√6x + 5 1

+c ✓

=− = 11 = 11 = 11 = −3

(2x−7)−1 (2)(−1) (2x−7)−1 −2

+c +c

= −(2x − 7)−1 +c

13(c) ∫

1 2x−7

1 √3−2x

+c ✓ 1

dx = ∫(3 − 2x)−2 dx 1

=

(3−2x)2 1 2

(−2)( )

+c

= −√3 − 2x +c ✓

A = 2t 3 − t 2 + t − 3 ✓ dx

9

=2⋅

+t +c

Area is 11 when t = 2: A|t=2 2(2)3 − (2)2 + (2) + c 16 − 4 + 2 + c c

11(i)

(6x−1)√6x−1

=2⋅

t2

3

+c

= 2 ∫(2x − 7)−2 dx

A= 6( ) − 2( ) + t + c = 2t 3

3 2

(6)( )

13(b) ∫ 2 dx (2x−7)2

= 6t 2 − 2t + 1 t3

(6x−1)2

d (6x + 5) 2√6x + 5 dx 1 = ⋅6 =

=

⋅

2√6x+5 3 √6x+5

✓

11(ii) ∫ 1 dx = 1 ∫ 3 dx 3 √6x+5 √6x+5 1

= √6x + 5 + c ✓ 3 © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

497


dy

Ex 18.1 15

= x 2 (x − k)

dx

dx

= x 3 − kx 2 x4

dy

⇒

∝ x(2x 2 − 3) dy dx

= kx(2x 2 − 3) = 2kx 3 − 3kx

x3

y = ( ) − k( ) +c =

4 1 4 x 4

−

3 k 3 x 3

1

= kx k

4

3 8

x2

4

2 3k 2 x 2

4

2

(2, −2) lies on curve, 1

x4

y = 2k ( ) − 3k ( ) + c

+c

−

(−2) = (2)4 − (2)3 + c

(1,3) lies on curve,

−2

=4

3 = k(1)4 −

c

= k−6

− k 3

+c

8

−(1)

3

1

3k

2 1

2

(1)2 +c

3

3= k

− k

2

+c

+c

2

c =k+3

−(1)

(4,2) lies on curve, 1

k

4

3 64

(2) = (4)4 − (4)3 +c


2

= 64

−

9 = k(2)4 −

0

= 62

−

3 64 3

k

+c

k

+c

0

= 62

−

0

= 56

−

56 3

k

3 56 3

k

3k

2

2

(2)2 + c

9 = 8k − 6k + c 9 = 2k + c

−(2)

sub (1) into (2): 64

1

−(2)

sub (1) into (2): 9 = 2k + (k + 3) 6 = 3k k =2

8

+ ( k − 6) 3

k

k= 56 c|k=2 = (2) + 3 = 5

=3

1

8

3

∴ y = (2)x 4 − (2)x 2 + 5

c|k=3 = (3) − 6 = 2

2

3

2

4

2

= x − 3x + 5✓ ∴y=

1 4 x 4

3

−x +2✓

16

dy dx

= x(2 − 3x)

= 2x − 3x 2 x2

x3

y = 2( ) − 3( ) + c 2

= x2

3

− x3

+c

Curve at (1,2), (2) = (1)2 − (1)3 + c c=2 ∴ y = x2 − x3 + 2 Curve at (−2, p), p = (−2)2 − (−2)3 + 2 = 14 ✓


sleightofmath.com

498


d2 y dx2 dy dx

Ex 18.1 18(ii) With c1 = 1 & c2 = 5,

= 6x − 16

d2 y

x2

= 6 ( ) − 16x + a

dx2 dy

= 3x − 16x + a

dx

2 2

= 6x − 4

= 3x 2 − 4x + 1

y = x 3 − 2x 2 + x + 6 x3

x2

3

2

y = 3 ( ) − 16 ( ) + ax + b

At max value,

= x 3 − 8x 2 + ax + b

dy

=0

dx

3x 2 − 4x + 1 = 0 (3x − 1)(x − 1) = 0

(0,3) lies on curve, 3 = 03 − 8(0)2 + a(0) + b 3=b

x=

1 3

or x = 1(min)

y|x=1 = 5

(−1,0) lies on curve, 0 = (−1)3 − 8(−1)2 − a + 3 0 = −1 − 8 − a + 3 a = −6

3

d2 y

|

dx2 x=1

4 27

✓

= −2 < 0

3

1

⇒ max at x =

3

∴ y = x 3 − 8x 2 − 6x + 3 ✓ 19(i) 18(i)

d2 y dx2 dy dx

d

(

d

2x

dx √x+1

= 6x − 4

)= =

x2

= 6 ( ) − 4x + c1 2 2

=

= 3x − 4x + c1 x3

x2

3

2

=

y = 3 ( ) − 4 ( ) + c1 x + c2 =

= x 3 − 2x 2 + c1 x + c2

dy

|

dx x=1 2

√x+1⋅(2)

−2x⋅

1 2√x+1

x+1 2√x+1

−

x √x+1

x+1 2(x+1) −x (x+1)√x+1 2x+2

−x

3 (x+1)2

x+2

=

At min (1,5),

d

√x+1⋅dx2x −2x⋅dx√x+1 x+1

3

[shown] ✓

(x+1)2

=0

3(1) − 4(1) + c1 = 0 c1 =1 (1,5) lies on curve, (5) = (1)3 − 2(1)2 + c1 (1) + c2 5 = −1 + c1 + c2 5 = −1 + (1) + c2 ∵ c1 = 1 c2 = 5

19(ii) ∫

x+2 3 4(x+1)2

1

x+2

1

(x+1)2 2x

dx = ∫ 4

3

= ( =

dx

) +c

4 √x+1 x 2√x+1

+c ✓

20(a) ∫ x2 +2x dx x2 +2x+1 =∫

∴ y = x 3 − 2x 2 + x + 5 ✓

(x2 +2x+1) −1 x2 +2x+1

= ∫ (1 −

dx

1

) dx

x2 +2x+1 1

= ∫ (1 − (x+1)2 ) dx = ∫[1 − (x + 1)−2 ] dx (x+1)−1

= x − (1)(−1) +c = x+


sleightofmath.com

1 x+1

+c ✓

499


Ex 18.1

20(b) ∫ 3x2 −4x dx 9x2 −12x+4 =∫

22

1 (9x2 −12x+4) 3 9x2 −12x+4

1

= ∫[

1

= ∫[

dx

dx

y=

] dx (3x−2)2

4

3

3

=

= ∫ [ − (3x − 2)−2 ] dx = =

1

4

(3x−1)−1

3

(3)(−1)

− ⋅

3 1

+

3

4

dy

+c

4 x3 4 3

) +c

2 3

3

3

4

4 3

= a2 (2x

9

2

3

−

3 4 x4 12

+ c1 x + c2

= −1 + c1 = −1 =−

1

1

2 5

12

4 = − 4 = 4 = c2 =

4 3

+ x ) +c ✓ 4

− ( ) + c1 x + c2

5 3


= 3a2 ( x 2 + x ) +c 3 2

+ c1

1 x4

2 x2

c1

1

+

+ c1

3 1 3 x 3

|

1−

= 3a2 ∫ (x 2 + x 3 ) dx = 3a2 (

x2

dx x=1 13

20(c) ∫ 3a2 (√x + 3√x) dx 1

x3

tangent at (1,4) is y = 5 − x,

+c ✓

9(3x−1)

3 x2 3 2

=x−

] dx

4 3

1

= 1 − x2

=x−

9x2 −12x+4

−

3

dx2 dy

4 3

4 3

−

3

−

d2 y

∴y=

12 5 12 21

+ c1 (1) + c2

+ c1 + c2 5

+ (− ) + c2 3

∵ c1 = −

5 3

4 x2 2

−

x4 12

5

21

3

4

− +

✓

21(i) 5

d 1

− (2t + 1)2 ]

dt 5

3

3 2

1 5

= ⋅ (2t + 1) ⋅ 5 2

d dt

3 2

d dt

(2t + 1)

1

1 3

= ⋅ (2t + 1)2 ⋅ 2

− ⋅ (2t + 1)2 ⋅ 2

5 2

= (2t + 1)

1

1 3

(2t + 1) − ⋅ (2t + 1)2 ⋅

3

1 5

3

1

[ (2t + 1)2

3 2

3 2

1

−(2t + 1)2

= (2t + 1)√2t + 1

−√2t + 1

= √2t + 1[(2t + 1)

−1]

= 2t√2t + 1 ✓ 21(ii) ∫ t(√2t + 1) dt 1 = ∫ 2t√2t + 1 dt 2 5

1 1

= [ (2t + 1)2 =

2 5 1 10

(2t + 1)

5 2

1

3

3 1

3

− (2t + 1)2 ] +c − (2t + 1)2 6


+c ✓

sleightofmath.com

500


dy dx

Ex 18.1 24(i)

= kx + c

d2 y dx2

x2

dy

2

dx

y = k ( ) + cx + a

x2

= λ ( ) − 18x + a

k

= x 2 + cx + a At stationary pt (2, −3): |

2k + c c

18x + a

λ x3

x2

y = ( ) − 18 ( ) + ax + b 2 3 λ = x3 − 6

=0

dx x=2

2 λ 2 x − 2

=

2

dy

= λx − 18

=0 = −2k

−(1)

2

2

9x + ax + b

Origin (0,0) lies on curve, λ

0 = (0)3 − 9(0)2 + a(0) + b

(2, −3) lies on curve,

6

0=b

k

(−3) = (2)2 + c(2) + a 2

−3 −3 −3

= 2k + 2c + a = 2k 2(−2k) + a = −2k + a

(1,16) lies on curve, λ

16 = − 9 + a + b

−(2)

6 λ

16 = − 9 + a

∴b=0

6 λ


25 = + a

k

(1) = (0)2 + c(0) + a

−(1)

6

2

a

=1

Gradient at (1,16) is 9, dy

Put a = 1 into (2): −3 = −4k + 1 2k =4 k =2

|

=9

− 18 + a

=9

dx x=1 λ 2

a

= 27 −

λ 2

−(2)

sub (2) into (1): λ

Put k = 2 into (1): c = −2(2) = −4

λ

25 = + (27 − ) 6

2

1

25 = 27 − λ 3

∴y =

2 2 x 2 2

1

+ (−4)x + 1

3

λ

= x − 4x + 1 ✓

dx

6

a = 27 − = 24

= 2x − 4

Tangent Point: Gradient: Tangent:

= 6✓

24(ii) Put λ = 6 into (2):

23(ii) With k = 2, c = −4: dy

λ =2

2

∴ y == x 3 − 9x 2 + 24x ✓ (0,1) dy

|

= −4

y − y1

=

dx x=0

dy

|

dx x=0

(x − x1 )

y − (1)= −4(x − 0) y = −4x + 1 ✓


sleightofmath.com

501


Ex 18.1

24(iii) At a = 24, b = 0, λ = 6, y = x 3 − 9x 2 + 24x dy

26 (a) For f(x) = 1 g(x) = 2

= 3x 2 − 18x + 24

dx d2 y dx2

∫ f(x)g(x) dx = ∫(1)(2) dx = 2x + c

= 6x − 18

At turning pts, dy

[∫ f(x) dx][∫ g(x) dx] = [∫(1) dx][∫(2) dx] = x(2x) + c = 2x 2 + c ∴ ∫ f(x)g(x) dx ≠ [∫ f(x) dx][∫ g(x) dx]

=0

dx

3x 2 − 18x + 24 = 0 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x=2 or x = 4 y|x=2 = 20 y|x=4 = 16 ⇒ (2,20) ✓ ⇒ (4,16) ✓

26 (b) ∫ f(x) dx = ∫ 1 dx g(x) 2 1

= x+c 2

f(x) dx

d2 y dx2 d2 y

| |

x=2

dx2 x=4

= −6 < 0

⇒ max at (2,20)

=6>0

⇒ min at (4,16)

∫ g(x) dx [wrong notation ?] ∫ f(x)dx ∫ g(x)dx

=

Both are correct The arbitrary constant can factor in the difference in constant values. ∫(x + 1) dx ∫(x + 1) dx

= = = =

x2 2

2 2

x+a 2x+b 2

f(x) g(x)

dx ≠ ∫

f(x) dx g(x) dx

+ c1

x2 +2x+1 x2

2 dx

1

∴∫

(1)(2)

1 dx

= +d

+x+c

(x+1)2

=∫

+ c1 1

+ x + + c1 2

1 ⇒ c = + c1 2


sleightofmath.com

502


Ex 18.2 1(e)

Ex 18.2 1(a)

5

9 1

∫4

√x

9

1

5

∫2 3x dx = 3 ∫3 x dx

x2

=[1] 2

5

x2

1

9

dx = ∫4 x −2 dx

4

= 3[ ]

1 9

2 3

= 2 [x 2 ]

3

= [x 2 ]53

4

2

=

9

= 2[√x]4

3 2 (5 − 32 ) 2

= 24 ✓ 1(b)

9 1 ∫1 x 2 dx

9

3

= 2[√9

− √4]

= 2(3

− 2)

=2✓

x2

=[3] 2

1

1(f)

3 9

2

4 3

4

∫1 x√x dx = ∫1 x 2 dx

= [x 2 ] 3

x2

=[5]

2 9 = [x√x]1 3 2 = (9√9 − 1√1) 3 2 = [9(3) − 1] 3

2

2 5 2

=

5

2

8

2(a)

1 x3

= [2] 3

4

2(b)

0

∫−1(3x 2 − 2x + 5) dx 0

x3

+5) dx

x2

= [3 ( )−2 ( ) +5x]0−1

3

]

3

3

2

+5x]0−1

= [x

1 1 3

= [(0)3 −(0)2 +5(0)] −[(−1)3 − (−1)2 + 5(−1)]

1 1

=− (

3 3

−x

2

1

3 x 2

1

−(4 + 4)

= ∫−1(3x 2 −2x

3

− [x −1 ]32 3

18

− 4x]1−1

= −8 ✓

=− [ ]

=

−1

= (4 − 4)

3 −1 2

=

2

= [4(1)2 − 4(1)] −[4(−1)2 − 4(−1)]

dx = ∫2 x −2 dx 3x2 3 = [

1

= [4x 2

2

− 13 )

2 3 = [(23 )3 − 1] 4 3 = (22 − 1) 4 1 =2 ✓

1 x−1

x2

= [8 ( ) − 4x]

4

1

1

∫−1(8x − 4) dx

1

2

3

3 1

5 2

= (83

∫2

62

= 12 ✓

3 2 8 = [x 3 ] 4 1

1(d)

− 1)

5

3

2

5

− 12 ]

= (25

1

8 1 −1 ∫1 2 x 3 dx

1 5 2

= [4

= 17 ✓ 1(c)

4

5

1

=7✓

1

− ) 2

✓


sleightofmath.com

503


Ex 18.2

4

4(a)

∫1 (6x − 3√x) dx 1

4

= ∫1 (6x

− 3x 2 ) dx 4

3

x2

x2 2

3 2

= ∫ (1 + x −2 ) dx

1

1

4

= [3x 2

− 2x ]

= [3x 2

− 2x√x]1

1 4

= [x

− ]

= 32

=3

2

4(b)

2

x4

2

2

0

1

2

4

0

= [ x4 − x2 ]

= [( 1

4

4

2

3

−

2

(2)2 2

t2

2

2 2

1

− t2]

1

−

(2)2 ]

1

− [− (1) − (1)2 ] +2

2 1 2

4(c)

= ∫1 (x 2 − x − 2) dx −

− 2t) dt

= −2 ✓

2

3

− 2t) dt

− 2 ( )]

1

= −4

∫1 (x + 1)(x − 2) dx

=[

−1

)

t 1 [− (2) 1

=

=0✓

(2)3

t−1

= [−

−0

x2

dt

1

1

=0

=[

1

− (1)]

−0

4 3

t2 2 1 ∫1 (t2 2 ∫ (t −2

=

= [ (2)4 − (2)2 ] − [ (0)4 − (0)2 ]

x3

x 1 1 ] − [(1) (4)

3

2 1−2t3

∫1 =

x2

− 2 ( )]

4

−1 1 1 4

4

∫0 x(x 2 − 2) dx

=[

4

]

=3 ✓

= ∫0 (x 3 − 2x) dx

3(b)

+

= [(4) −

−1

x−1

= [x

= [3(1)2 − 2(4)√4] −[3(1)2 − 2(1)√1]

= 31 ✓ 3(a)

x +1 dx x2 1 4 1 = ∫ (1 + 2 ) dx x 1 4

= [6 ( ) − 3 ( 3 )] 2

4 2

∫

− 2x]

√x 1

4

− 2(2)] − [

3

2x − 1

(1)3 3

− (−

−

(1)2 2

− 2(1)]

1

3

1

x2

x2

= [2 ( 3 )

13 6

dx

= ∫1 (2x 2 − x −2 )

1

1

∫ 1

2

= (−3 )

4

)

2

4

1

= −1 ✓

=[ x 3

6

3 2

4

= [ x√x 3 4

dx 4

− ( 1 )] 2 1 2

1 4

− 2x ]

1

− 2√x]

4

1

= [ (4)√(4) − 2√(4)] 3

=

20 3

4

− [ (1)√(1) − 2√(1)] 3

+

2 3

1

=7 ✓ 3


sleightofmath.com

504


9

∫ 1

1 3−2x2

x2

x−1 −1

= [−

3

−1.5 )

dx

1 − x 2 1 − 2

9

− 2x

−

3

= [− (9) +

)] 1

4 9

4 √(9)

− [− (1) +

4 √(1)

6(a)

]

2 −1 1 1 −1 2 x ] 2 1 1 2

− 2x −

]

2x 1 1

1

2(2)

2(1)

] − [2(1)4 − 2(1) −

3

+

4 1

]

1 2

0

∫−1 x(x − a)(x + a) dx 0

= ∫ x(x 2 − a2 ) dx = ∫ (x 3

−2⋅

3 2

1 2

=[ =[

]

1 2

1

x4

x4 (0)4

1 4

=0

1

a2

4

− 4x ]

2 3

2

+

3

=

2

− [ (1)√(1) − 4√(1)] 3 10

6(b)

3

2

0

2

−1

0 a2 x2

−

4

=[

x2

− a2 ( )]

4

4

= [ (4)√(4) − 4√(4)]

− a2 x) dx

−1

= [ x√x − 4√x] 3

−

2

a2 (0)2 2

]

−1

] −[

(−1)4 4

1

a2

4

2

− +

−

a2 (−1)2 2

]

1

− ✓

2

4

4

∫0 √x(a − √x) dx

= ✓

4

3

5(b)

4

)]

0

2

= −2

1 x−1

4

3

]

2

x4

−1

3 2

3

) dx

= 28 ✓

]

4 2 ∫1 (√x − ) dx √x 4 1 1 = ∫ (x 2 − 2x −2 ) dx 1 3 1 4 x2 x2

2

2x2 1 −2 2

= [2x 4

√x 1

−1

=[ x

+ x

− 2x −

=0✓

=[

−2

= [2x 4

= 27

1

=1

5(a)

) dx

+

= [2(2)4 − 2(2) −

1 9

+ 4x 2 ] +

x

2 ∫1 (8x 3

1

−2

= [8 ( ) − 2x + (

) − 2(

[−3x −1

2

∫1 (8x 3 =

dx

9 ∫1 (3x −2

= [3 ( =

5(c)

3 − 2√x dx x2

=∫ =

Ex 18.2

= ∫ (a√x − x) dx

2

0

4

∫1 (x 2 − x2 ) dx = ∫ (x 2

− 4x −2 ) dx

3

1

x3 =[ 3 x3 =[ 3

x2

= [a ( 3 ) −

2

−4(

x −1 )] −1 1

2

2

2

3

+ 4x −1 ] 1

2

1

− [ (1)3 + (1)]

=4

−4

3

3 1

=

3

1

=

= ✓ 3


x2 2

4

]

x2

0 4

]

2 0 x2

4

]

2 0

2

(4)2

3

2

= [ a(4)√(4) −

4

= [ (2)3 + (2)] 3 2

−

= [ ax√x − 3

1 4 = [ x3 + ] 3 x1 4

3

= [ ax 2

2

1

1

4

= ∫0 (ax 2 − x) dx

2

sleightofmath.com

16 3 16 3

a−8

2

(0)2

3

2

] − [ a(0)√(0) −

]

−0

a−8✓

505


2 2t2 +a

∫1

Ex 18.2 8

dt

t2 2

= ∫ (2 + at −2 ) dt

= [2t −

t−1

2

)]

=

−1 1 −1 ]2 at 1 a2 a

a

= [2(2) − (2)] − [2(1) − (1)] =4− a

=[

2

x2

a

2

1

=

− 3x

2)

1 12 1 12

x2

x3

1

2

3

2

[2x 2 ]1a

= [3x 2 −2(1)

2

= [3(1) −

2a2 −2

=2

2a2 −2

= −2

2a

9(i)

− x 3 ]12 2

(1)3 ]

(2x − 3)3

(2(1) − 3)3 +

a

=0

a

=0✓

dy 2

−[3(2) −

(2)3 ]

dx

+2

6

=x⋅

d dx

=x⋅[

−4

= = =

k

5

1 12

7

+

1

]

4(2x−3) 0 7

]

4(2(1)−3)

(2(0) − 3)3 +

7

]

4(2(0)−3)

5 6

y = x√1 + 2x 2

=0

2

−

1 7 (2x−3)−1 [ ]] 2 (2)(−1) 0

=1✓

dx

= [6 ( ) − 3 ( )]

2

7(b)

1 ∫2 (6x

[4 ( )]

2(a)

2

= −1

a ∫1 4x dx

2

7

2

−[

= +2✓ 7(a)

1 1

−2 + a

2

dx

1 (2x−3)3 [ [ (2)(3) ] 2

=[

= [2t − ] t 1 a

2(2x−3)2

= ∫0 [ (2x − 3)2 − (2x − 3)−2 ] dx

1

= [2t + a (

1 (2x−3)4 −7

∫0

√1 + 2x 2 1

2√1+2x2

(4x)]

2x2

+

d dx

+1

(x) ⋅ √1 + 2x 2 ⋅ √1 + 2x 2

+√1 + 2x 2

√1+2x2 2x2

+(1+2x2 ) √1+2x2

1+4x2 √1+2x2

[shown] ✓

∫−2 3(2 + x)2 dx

= 64

k 3 ∫−2(2

= 64

= [x√1 + 2x 2 ]0

= 64

= (2)√1 + 2(2)2 −(0)√1 + 2(0)2

+ x)2 dx

k

3 ∫−2(x 2 + 4x [ [ [

x3

3 (k)3 3

−[ 1 3 1 3

x2

k

2

−2

+ 4 ( ) + 4x]

3 x3

+ 4) dx

+ 2x

2

k

+ 4x]

−2

+ 2(k)2 + 4(k)] (−2)3 3

=

2

=

=

+ 2(−2) + 4(−2)]

k 3 + 2k 2 + 4k +

8

k 3 + 2k 2 + 4k −

56

3 3

k 3 + 6k 2 + 12k − 56 (k − 2)( + + ⬚) 2 (k − 2)(k + + ⬚) 2 (k − 2)(k + + 28) 2 (k − 2)(k + 8k + 28) ⇒ k=2✓ © Daniel & Samuel A-math tuition 📞9133 9982

=

9(ii)

2 1+4x2

∫0

√1+2x2

dx 2

=6

64 3

−0

=6✓

64 3

10(i)

64

Prove: (x − 4)(x 2 + 4x + 16) = x 3 − 64 LHS = x 3 +4x 2 +16x −4x 2 −16x −64 = x 2 − 64 [proven]✓

3 64 3

=0 =0

=0

sleightofmath.com

506


Ex 18.2

10(ii) ∫3 x3 −64 dx −1 x−4 =

12(ii) C| = 24 000 [10 + 4 (354 )] x=3 5

3 (x−4)(x2 +4x+16) dx ∫−1 x−4 3

= ∫−1(x 2 + 4x =[ =[ =[

x3

≈ $316 000 ✓

+ 16) dx

5

x2

3

2

−1

+ 4 ( ) + 16x]

3 x3

3

+ 2x 2

3 (3)3 3

12(iii) C| = 24 000 [10 + 4 (554 )] x=5

+ 16x]

≈ $384 000 ✓ 13

dV dt

−1

= 20 000 (t − 6)

= 20 000t − 120 000

2

+ 2(3) + 16(3)] −[

= 75

(−1)3 3

+14

Net change in Value

+ 2(−1)2 + 16(−1)]

4

= ∫0 [20 000t − 120 000] dt

1 3

= 89 ✓ 3

11(i)

2 ∫0 x dx

x2

t2

4

2

0

= [20 000 ( ) − 120 000t]

1

= [160 000 − 480 000 − (0)]

2

=[ ]

= −$320 000

2 0

1

= [x 2 ]20

Total loss = $320 000 ✓

2 1

= (22 −02 )

14(i)

2 1

3

8

Given: ∫0 f(x) dx = ∫3 f(x) dx = 12

= (4) 2

8

11(ii)

2

∫ |x| dx −2 2

2

= ∫−2|x| dx

+ ∫0 |x| dx

0

= [− ] 2

] − [−

(−2)2 2

= [x]83 =8−3 = 29 ✓

] +2

=2 =4✓

+2

C = 8000 (30 +

8

+2(12) +24

14(iii) ∫8 2f(x) dx 3 =

12(i)

+12

8

+2

2 −2

=[

= 12 = 24 ✓

= ∫3 1 dx +2 ∫3 f(x) dx

+ ∫0 x dx

0

−(0)2

8

14(ii) ∫8[1 + 2f(x)] dx 3

2

= ∫−2(−x) dx x2

3

∫0 f(x) dx = ∫0 f(x) dx + ∫3 f(x) dx

=2✓

x 1 3 ∫0 t 4 dt)

8 2 ∫3 f(x) dx

= 2(12) = 25 ✓

x 1

+1 +1 +1

= 24 000 (10 + ∫0 t 4 dt) x

5

t4

= 24 000 [10 + [ 5 ] ] 4

4

0 5 4

x

= 24 000 [10 + [t ] ] 5

0

4 5

= 24 000 [10 + x 4 ] 5

4

C|x=1 = 24 000 [10 + (1)] 5

= $259 200 ✓


sleightofmath.com

507


Ex 18.2

8

∫3 [f(x) − mx] dx 8

8

∫3 f(x) dx −m ∫3 x dx x2

=0

8

−m [ ]

12

=0

2 3 m 2 8 − [x ]3 2 1 2

12

=0

2 1

12

− m(55)

=0

12

−

=0

55 2

2 55 2

m

m

−1

= −2 ∫0 x 2 dx

Given:

= −2m ✓ 16(ii) ∫−1(−x 2 ) dx = − ∫−1 x 2 dx 0 0 (c) = −m ✓

= −12

m 15(i)

0

= 2 ∫−1 x 2 dx

− m[(8) − (3)2 ] = 0

12

−

1 16(ii) ∫1 x 2 dx = ∫0 x 2 dx + ∫0 x 2 dx −1 −1 (b) 0 0 = ∫−1 x 2 dx + ∫−1 x 2 dx (equal area under curve)

=0

= 4 ∫−3 f(x) dx 4 ∫−1 f(x) dx

−1

24 55

17

✓

b

∫a f(x) dx ≥ 0 Not True

=a In theory

=b

b

4

∫ f(x) dx ≡ algebraic area ≠ geometric area

4

a

∫−3 f(x) dx = ∫−3 f(x) dx − ∫−1 f(x) dx =a−b✓

For example

15(ii) ∫−3 2f(x) dx = 2 ∫−3 f(x) dx −1 −1 𝑦

−1

= −2 ∫−3 f(x) dx

𝑦 = 𝑓(𝑥)

= −2(a − b) = 2b − 2a ✓

𝑃 𝑂

16(i)

𝑎

Q

𝑏

𝑥

𝑦 𝑦=𝑥

Algebraic area = P − Q Geometric area = P + Q

2

𝑥 Clearly f(x) can be negative between a and b 16(ii) ∫−1 x 2 dx = ∫0 x 2 dx (equal area under curve) 0 −1 (a) 1 ∫0 x 2 dx

= =

0 ∫−1 x 2 dx −1 − ∫0 x 2 dx

18

∵ symmetry in y − axis

In order to compute b

∫ f(x) dx a

f(x) must be continuous [no gaps] for a ≤ x ≤ b x ≠ 0 [gap]

= −m ✓

1

1 dx cannot be computed 2 −1 x

∴∫


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508


Ex 18.3 2(e)

Ex 18.3 1(a)

1(b)

1(c)

2(a)

π

∫06 (3 cos x − 2) dx π

∫(sin x + 2) dx = (− cos x) + 2x + c = − cos x + 2x + c ✓

= [3(sin x) − 2x]06

∫(1 − 3 cos x) dx = x − 3(sin x) + c = x − 3 sin x + c ✓

= [3 sin ( ) − 2 ( )] −[3 sin(0) − 2(0)]

π

− 2x]06

= [3 sin x π

π

6

6

1

π

= [3 ( )

− ]

2

3

π

2

3

∫(5 sec x + 3) dx = 5 tan x + 3x + c ✓

1

3(a)

∫ cos 2x dx = 2 sin 2x +c ✓

3(b)

∫ sin 3x dx = − 3 cos 3x +c ✓

3(c)

∫ 2 cos 4x dx = 2 ∫ cos 4x dx

π

π 6

∫0 cos x dx = [sin x]06 π 6

1

=

−0

2 1

1

1

= 2 ( sin 4x) +c

= ✓

4

2

π 4

1

= sin 4x +c ✓ 2

π 4

2

∫0 sec x dx = [tan x]0

3(d)

π

= tan ( ) − tan(0)

1

1

∫ −3 sin 2 x dx = −3 ∫ sin 2 x dx

4

=1 = 1✓

1

1

= −3 (− 1 cos x) +c

−0

2

2

1

= 6 cos x 2

2(c)

− 0)

= − ✓

2

= sin ( ) − sin(0)

2(b)

−(0

3

+c ✓

π

π

∫02 sin x dx = [− cos x]02

3(e)

∫ 4 sec 2 5x dx = 4 ∫ sec 2 5x dx 1

π

= 4 ( tan 5x) +c

= −[cos x]02

5

4

= tan 5x +c ✓ 5

π

= −[cos ( ) − cos(0)] 2

= −(0

−1)

4(a)

=1✓

x

∫ 2 cos (2 + 1) dx x

= 2 ∫ cos ( + 1) dx 2

2(d)

π

1

∫02 (1 − 2 sin x) dx = [x − = [x + π

π 2(− cos x)]02 π 2 cos x]02 π

=( =

π 2

2

x

−(0

+c ✓

= 4 sin ( + 1) 2

4(b)

2

+ 0)

2

2

= [( ) + 2 cos ( )] −[(0) + 2 cos(0)] 2 π

x

= 2 [ 1 sin ( + 1)] +c

+ 2)

∫ 3 sin(2 − x) dx = 3 ∫ sin(2 − x) dx = 3 [−

−2 ✓

1 −1

cos(2 − x)] +c +c ✓

= 3 cos(2 − x) 4(c)

π

∫ 4 sec 2 (8x − 4 ) dx π

= 4 ∫ sec 2 (8x − ) dx 4 π

1

= 4 [ tan (8x − )] +c 8

4

1

π

2

4

= tan (8x − ) © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

+c ✓ 509


Ex 18.3

∫(sec 2 x − 4 sin x) dx = tan x − 4(− cos x) +c = tan x + 4 cos x +c ✓

9(a)

π 3 π 12

π

∫ cos (2x + 3 ) dx 1

π

2

3

π 3

= [ sin (2x + )] π 5(b)

5(c)

∫(3 cos x − 2 sin x) dx = 3 sin x − 2(− cos x) = 3 sin x + 2 cos x ∫(4 cos x = 4 sin x

+c +c ✓

12

1

π

2

3

= [sin (2x + )]

π 3 π 12

1

π

π

π

π

2

3

3

12

3

= [sin (2 ( ) + ) − sin (2 ( ) + )]

+ 3 sec 2 x) dx + 3 tan x +c ✓

1

π

= (sin π

− sin )

2

6

π 4

π 4

= [0

∫0 (1 + tan2 x) dx = ∫0 sec 2 x dx

1

π

=− ✓ 2

π

= tan − tan 0 4

9(b)

π

∫π 2 sin(π − x) dx 2

= 1 −0

π

= 2 ∫π sin(π − x) dx 2

=1✓

= 2 [−

π

∫04 [sec 2 x + 1] dx = [tan x + π

π

4

4

=1

+

π

−1

2

π 4

π

= 2[cos(0)

− cos ( )]

= 2(1

−0)

2

=2✓

∫[− sin(2x + 1)] dx

9(c)

= − ∫ sin(2x + 1) dx

1

π

=[

1

= − [− cos(2x + 1)] +c

−

2

2

1

+c ✓

= cos(2x + 1) 2

π

1

=[ π

=[

∫ cos (2x + 4 ) dx = 2 sin (2x + 4 ) + c ✓

π

∫0 [x − cos ( 3 − 6 x)] dx x2

8(c)

2

x)]ππ 2 π

= + 1 [shown] ✓

8(b)

π

cos(π − x)]π

= 2[cos(π − (π)) − cos (π − ( ))]

−0

4

1

= 2[cos(π −

π x]04

= [tan + ] −[tan 0 + 0]

8(a)

−1)

2

= [tan x]04

7

2

1

x2

1 π − 6

π

π

3

6

6

π

π

π

3

6

6

π

π

π

3

6

0 1

+ sin ( − x)]

2 (1)2 2

0

+ sin ( − (1))] −[

∫ 6 sin(3x + 2) dx = 6 ∫ sin(3x + 2) dx

=[

1

= 6 [− cos(3x + 2)] +c

=[

3

= −2 cos(3x + 2)

1

sin ( − x)]

+c ✓

1 1

π 6

6 1

+ ( )]

2

=(

6

π

+ sin ( )]

2

π 2

1

3

+ )

2

π

1

1

2

π

(0)2 2

6

π

π

π

3

6

6

π

π

3

+ sin ( − (0))]

− [0

+ sin ( )]

−[0

+ ( )]

−

6 √3 2

π

3√3 π

= + (3 − 3√3) ✓


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510


Ex 18.3

π

1

dx

= [− cos 5t] 5

d

12(a)

∫0 sin 5t dt π

sin(x 2 + 2x)

= cos(x 2 + 2x) ⋅

0

d dx

(x 2 + 2x)

= cos(x 2 + 2x) ⋅ (2x + 2)

1 = − [cos 5t]π0 5

= (2x + 2) cos(x 2 + 2x) ✓

1

= − (cos 5π − cos 0) 5

12(b) ∫(x + 1) cos(x 2 + 2x) dx

1

= − (−1

−1)

5

1

= ∫(2x + 2) cos(x 2 + 2x) dx 2

2

1

= ✓

= sin(x 2 + 2x) + c ✓

5

10(i)

d dx

2

(x cos x)

=x⋅

d dx

(cos x) +

d dx

(x) ⋅ cos x

= x ⋅ (− sin x) +1 cos x = cos x −x sin x ✓

13 Both are correct The arbitrary constant can factor in the differences in constant value ∫ sin x cos x dx 1

10(ii) d (x cos x) = cos x − x sin x dx d x sin x = cos x − x cos x ✓ dx

d

∫ x sin x dx = ∫ (cos x − dx x cos x) dx = sin x 11

− x cos x +c ✓

∫(sin x − cos x)2 dx = ∫(sin2 x −2 sin x cos x + cos 2 x) dx = ∫(sin2 x + cos 2 x −2 sin x cos x) dx = ∫(1 − sin 2x) dx 1

=x

− (− cos 2x)

=x

+ cos 2x

=x

+ (1 − 2 sin2 x) +c1

=x

+ − sin2 x

+c1

=x

− sin2 x

+ + c1

=x

− sin2 x

+c ✓

2

1 2 1 2 1 2


= ∫ sin 2x dx 2

1

cos 2x

2

2

= (−

+ a)

1

1

4 1

2

= − cos 2x + a 1

1

2

4 cos2 x

1

= − [1 − 2 sin2 x] + a or = − [2 cos 2 x − 1] + a = =

4 sin2 x 2 sin2 x 2

1

1

4

2

− + a

=−

+ c1

=−

2 cos2 x 2

2

1

1

4

2

+ + a + c2

+c1 +c1

1 2

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511


Ex 18.4 2(c)

Ex 18.4 1(a) 1(b)

2x

∫e

dx =

3x

∫ 2e

1 2x e 2

1(c)

dx

1 2 ( e3x ) 3 2 3x e +c 3

0

= −2[e

= −2(

1

1

2(d)

−1)

e

2

1

∫1 e1−x dx = [−1 e1−x ]

2

1

2 1 x 2

= 2 (2e ) +c 1

= 4e2x

=

= −[e−1 −e0 ]

∫ 4e2x+1 dx = 4 ∫ e2x+1 dx 1 4 ( e2x+1 ) 2 2x+1

1

−1

= −[

+c

1−x )

e

1 e

2(e) +c

ln 5

∫0

5 (ex + 1) dx = [ex + x]ln 0

= [e(ln 5) + (ln 5)] −[e(0) + (0)] = 5 + ln 5 −1 = 4 + ln 5 ✓

= 3(−e1−x ) +c = −3e1−x +c ✓ 2

2(f)

∫0 ex dx = [ex ]20

−1]

e

=1− ✓

+c ✓

∫ 3e1−x dx = 3 ∫ e1−x dx 1

−[e1−x ]12

= −[e1−(2) −e1−(1) ]

+c ✓

= 2e

0

= e −e = e2 −1 ✓ 2(b)

1

e

∫ 2e dx = 2 ∫ e dx

2

]

1

= 2 ( 1 e2x ) +c

2(a)

(0)

= 2(1 − )✓

1 x 2

= 3(

1

−e−2

= −2(e−1 −1)

2

1(f)

2

1 − (2) 2

∫ e−2x dx= −2 e−2x +c

=

0

1 2 [−2e−2x ] 0

✓

1

1(e)

2

1

1

1 x 2

− x 1e 2 ]

= −2 [e−2x ]

+c

= − e−2x +c ✓ 1(d)

2

1

1

−

=

dx= 2 ∫ e

=

=[

+c✓ 3x

=

2 −1x ∫0 e 2 dx

ln 2

∫0

ln 2 3x

5e3x dx = 5 ∫0

e

1

ln 2

3

0

= 5 [ e3x ]

1 1 1 ∫0 e2x dx = [2 e2x ] 0

=

dx

5 3x ln 2 [e ]0 3 5

1 2x 1 [e ]0 2 1 = [e2 −e0 ]

= [e3(ln 2) −e3(0) ]

= [e2 −1] ✓

= [8

=

3 5

3

= [eln 2 3

2 1

5

2

3

−1] −1]

2

= 11 ✓ 3


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512


dy

Ex 18.4 5(c)

= e2x

dx

y=

1 2x e 2

1 2

∫0

2−x

dx = 2 ∫

= 2[

+c

−1

ln|2 − x|]

1 2x e 2

= 2 ln 2 ✓

2

5(d)

+2−

1 2 e 2

1

∫0

✓

1

3 2+3x

dx = 3 ∫0

1 2+3x

dx 1

1

= 3 [ ln|2 + 3x|] 3

4(a)

0

= −2(0 − ln 2)

e2

=2−

∴y=

1

1

= −2(ln 1 − ln 2)

+c =2

c

dx

= −2[ln|2 − x|]10

y = 2 when x = 1, y|x=1 = 2 1 2 e 2

1 2−x

2

1

∫ x dx = 2 ∫ x dx

= [ln|2 +

= 2 ln|x| + c ✓

= ln 5 − ln 2 5

1

= ln ✓

4(b)

∫ x+1 dx = ln|x + 1| + c ✓

4(c)

∫ 2x−1 dx = 2 ln |2x − 1| + c ✓

4(d)

∫ 2x+1 dx = 3 ∫ 2x+1 dx

1

2

1

3

0

3x|]10

6(a)

dy dx

1

=

1 2x+1

1

y = ln|2x + 1| + c 2

1

= 3 ⋅ ln|2x + 1| +c 2

3

= ln|2x + 1| 2

4(e)

2

1

1

∫ 1−2x dx = 2 ∫ 1−2x dx =2⋅

1 −2

2

1

1

∫ 4−3x dx = −3 ln|4 − 3x|

+c ✓ +c

1

= − ln|4 − 3x| +c ✓ 5(a)

=

=

1 2 1

1

2

2

∴ y = ln|2x + 1| + ✓ 6(b)

dy dx

3

44 ∫1 x dx

ln|1| + c = 0.5

c

ln|1 − 2x| +c

= − ln|1 − 2x| 4(f)

y = 0 when x = 0, y|x=0 = 0.5

+c ✓

=

1 3−2x 1

y = − ln|3 − 2x| + c 2

41 4 ∫1 dx x

= 4[ln|x|]14

y = 2 when x = 1, y|x=1 =2

= 4(ln 4 − ln 1)

− ln|1| + c = 2

= 4 ln 22

c

1 2

= 8 ln 2 ✓

=2 1

∴ y = 2 − ln|3 − 2x| ✓ 5(b)

1 2 ∫0 2x+1 dx

=

2

1 1 2 ∫0 dx 2x+1 1

1

= 2 [ ln|2x + 1|] 2

= [ln|2x +

0

1|]10

= ln 3 − ln 1 = ln 3 ✓


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513


2

1

Ex 18.4

2

2

9(i)

x − x ∫0 (e3 + e 3 ) dx 1

4

4

= ∫ (e3x + 2 + e−3x ) dx 4

1 3

3 4

3

3

4

4 (0) 3

−[ e 4

3 4

= ( e3 3

−

4

7(b)

∫0

e−x

B 3−2x

⇒A=1

=B

1

]

9(ii)

3 −4(0) e 3 ] 4

⇒ B = −2

1 1−x

−

2 3−2x

✓

1

∫ (1−x)(3−2x) dx = ∫(

3 −4 e 3) 4 3

1 1−x

−

2 3−2x

) dx

= − ln|1 − x| + ln|3 − 2x| + c

4

= ln |

4 − 3

10

1 ex +e2x

)

∴ (1−x)(3−2x) =

0

= (e − e ) + 2 ✓ 4

+

=A

1 1 (− )( 2

2

− )

+0

4 3

3

+ 2(0) −

+2

4

−(

4 − (1) 3

+ 2(1) − e

A 1−x

1

4

3

=[ e 4

4

1 ( )(1)

3

x= :

0

− e−3x ]

4

4 (1) 3

e−3x ]

4 − 3

3

= [ e3x + 2x

x = 1:

1

4

1

=

Cover-up rule:

0

= [ 4 e3x + 2x +

1 (1−x)(3−2x)

dy dx

3−2x 1−x

|+c✓

1

= (x−1)(x−2) ≡

A x−1

+

B x−2

dx Cover-up rule:

1

= ∫ (e2x + e3x ) dx 0 1

= [ e2x

1

1

3

0

+ e3x ]

2 1

1

1

1

2

3

2

3

1 3 e ) 3

1

1 0 e ) 3

x = 1:

1 ( )(−1)

x = 2:

1 (1)( )

=A

=B

⇒ A = −1 ⇒B=1

= [ e2(1) + e3(1) ] − [ e2(0) + e3(0) ] =

8(i)

1 ( e2 2

+

1

1

5

2

3

6

−(

+

2

∴

dx

=−

1 x−1

+

1 x−2

= e2 + e3 − ✓

y = − ln|x − 1| + ln|x − 2| + c

2x−1 (x+1)(x+2)

y = 2 at x = 3, y|x=3 =2 − ln 2 + ln 1 + c = 2 c = 2 + ln 2

=

A x+1

+

B x+2

Cover-up rule: x = −1: x = −2:

−3 =A ( )(1) −5 =B (−1)( )

2x−1

∴ (x+1)(x+2) = − 8(ii)

dy

2

∫1

3 x+1

⇒ A = −3 y = − ln|x − 1| + ln|x − 2| + 2 + ln 2

⇒B=5

= ln | +

5 x+2

2x−4 x−1

|+2✓

✓

2x−1 dx (x+1)(x+2) 2

3 5 ) dx + x+1 x+2 1 = [−3 ln|x + 1| + 5 ln|x + 2|]12 = [−3 ln 3 + 5 ln 4] −[−3 ln 2 + 5 ln 3] 2 = −3 ln 3 + 5 ln 2 +3 ln 2 − 5 ln 3 = 13 ln 2 − 8 ln 3 ✓ = ∫ (−


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514


1 (x−2)(x−1)2

A

=

x−2

+

Ex 18.4 B

1 12(ii) ∫5 dx 3 (x+3)(x−2)2

C

x−1

+ (x−1)2

5 1

= ∫3

Cover-up rule: 1 ( )(1)2

x = 2:

=A

=

1

[

+

25 x+3

1 5 1 ∫ [ 25 3 x+3

+

1

5

x−2 1 x−2

+ (x−2)2 ] dx + 5(x − 2)−2 ] dx

⇒A=1 = 1 (1)( )2

x = 1:

=C

⇒ C = −1

=

Substitution:

=

1 (−2)(−1)2

x = 0:

B ∴

B

+

−2

C

+ (−1)2

−1

=

= −1

1 (x−2)(x−1)2

= =

11(ii)

A

=

1 x−2 1 x−2

+ −

−1

+

x−1 1

−

x−1

=

−1 (x−1)2 1 (x−1)2

=

✓

=

5 1 ∫3 (x−2)(x−1)2 dx

13(i)

5

1 1 1 ] dx =∫ [ − − x − 1 (x − 1)2 3 x−2 =

5 1 ∫3 [x−2

−

1 x−1

− (x − 2)

−2 ]

= [ln|x − 2| − ln|x − 1| − [

1

2

4

1 (x+3)(x−2)2

1

1

4

2

2

(x−1)−1

=

x+3

+

x−2

1 25 1 25

−1

1 ( )(−5)2

x = 2:

1 (5)( )2

1

=

5

6

3

3

1

8

10

3

3

[ln − ln 6 + 4

10

9

3

[ln ( ) +

10

3

3

2

10

3

3

[2 ln +

3

]

]✓

y = ecos x = ecos x ⋅

d dx

(cos x)

= ∫ sin x ecos x dx

+

C (x−2)2

=A

3

1 25

x+3 1

[

+

+

= −(ecos x )

+(− cos x) +c

= −ecos x

− cos x

3

2 √e3x

∫0

ex−1

2 ex

dx = ∫0

ex−1

−2

−

1 25

x−2

1


+

−

C 4

+ 1

x−2

+c ✓

dx

2

⇒C=

B

25 x+3

+ ∫ sin x dx

= ∫0 e dx

⇒A=

=C

A

]

]

2 2

[ln ( ) +

1

14(b)

25

1

∫ (e3

1

=

5

1 1 3

x−1

⇒B=−

− 3 cos 2x) dx

1

1 2

3

= 3e3x−1 − sin 2x 2

1

= 2e ✓

e3x−1 −3 ( sin 2x) +c 1

=

∴ (x+3)(x−2)2 =

8

13(ii) ∫ sin x (ecos x + 1) dx

Substitution: 3(4)

]

x−2 3

= e[x]20

x = −3:

1

5

|−

x−2

3

5

[[ln | | − ] − [ln | | − 5]]

Cover-up rule:

x = 0:

x+3

]]

5

]]

14(a) B

1 25

[ln |

−1

5

= − ∫ − sin x ecos x dx + ∫ sin x dx

1

A

1 25

dx

= ln + ✓ 12(i)

1 25

[ln|x + 3| + ln|x − 2| + 5 [

(x−2)−1

= ecos x ⋅ (− sin x) = − sin x ecos x ✓

= [ln | | + ] − [ln | | + ] 4

1 25

dx

x−2 1 |+ ] = [ln | x−1 x−1 3

3

25

dy

5

3

1

+c ✓

25

1 5

(x−2)2 5

+ (x−2)2 ] ✓

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515


dy dx

=

Ex 18.4 16(ii) ∫ sec x dx = ln(sec x + tan x) + c ✓

1

2x2 +1

− √e−2x

√x 3

1

17

1

= 2x 2 +x −2 −e−4x 5

1

x2

x2

2

2 1 2

x 1 − x 4 1 − 4 1 − x 4

y = 2( 5 ) + ( 1 ) −( 4 5

= x2 5

+2x

1

e

+c ∫

4 5

5

1

=

(0)2 + 2(0)2 + 4e−4(0) + c = 2

4e0 + c c ∴y=

4 2 x √x 5

=2 = −2 + 2√x + 4e

1 4

− x

x2 +a x+a

18

x2 2

+a −a2 ) a + a2

a+a2 x+a

) dx

−ax +(a + a2 ) ln|x + a| + c ✓

Recognize that the derivative of ln(−x) is also d

−2✓

+a

dx

= ∫ (x − a +

=2

1

+a

−a +0x +ax) −ax −(−ax

) +c

+4e

y = 2 when x = 0, y|x=0

x x2 −(x 2

dx

ln(−x) =

1 x 1

⋅

d

−x dx 1

(−x)

= − ⋅ (−1) 16(i)

d dx

= = = = = = = =

x

ln(sec x + tan x) 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x 1 sec x+tan x

⋅

d dx

⋅[

=

dx

[(cos x)−1 ]

⋅ [−(cos x)−2 ⋅

d dx

1

possibility that the integral of can be either ln x + x

+ sec 2 x]

c or ln(−x) + c

(cos x) + sec 2 x]

19

⋅ [−(cos x)−2 ⋅ (− sin x) + sec 2 x] ⋅( ⋅(

sin x

+ sec 2 x)

cos2 x 1 cos x

⋅

x

The modulus sign is included to account for the

(sec x + tan x)

d

1

sin x cos x

+ sec 2 x)

⋅ (sec x tan x + sec 2 x)

d dx

2

(e−x ) = −2xe−x

∫ eax+b dx = but ∫ ef(x) ≠

eax+b a ef(x) f′ (x)

2

+c

+c

The student factored out the variable (−2x) outside the integral which makes it wrong.

sec x(tan x+sec x) sec x+tan x

= sec x ✓


sleightofmath.com

516


Rev Ex 18 A3(i)

Rev Ex 18

=

x3

−3 ( ) + c

4 x4

3

− x3

4

+c✓

=

A1(b) ∫ 2x2 −√x dx x

= 1

= ∫ (2x − x −2 ) dx x2

2

1 2

= 2( ) − = x2

+c

A2(a) ∫(2 + ex )2 dx = ∫(e2x + 4ex + 4) dx 1

= e2x +4ex + 4x + c ✓ 2

A2(b) ∫ e3x −2 dx = ∫(e2x − 2e−x ) dx ex

A2(c)

− sin2 x

1 2x 1 e −2 ( e−x ) 2 −1 1 2x e +2e−x +c 2

+c A3(ii)

✓

π 1 2 π 1−cos x 4

∫

dx

∫0 (cos x − 3 sin x) dx

= − ∫π2 − 4

π

=

π 4

4

π 1+cos 2 π sin 2

π

= [sin + 3 cos ] −[sin 0 + 3 cos 0] √2 2

= − [(

4

√2 2

−[0 + 3]

+ 3 ( )]

= − [(

= 2√2 − 3 ✓ π

dx

1+cos x 2 ] −[ sin x π

= [sin x + 3 cos x]0

=[

1 1−cos x π

= [sin x − 3(− cos x)]04

A2(d)

1 1 − cos x

π

4

π

∫π3 sin (2x − 3 ) dx

1+0 1

)

1

π

2

3

= [− cos (2x − )]

π 3 π 4

π 1 π 3 − [cos (2x − )]π 2 3 4 1 π π

−(

= − [1

−(

= − (1

−

= − (− = π

π

4

3

= − [cos (2 ( ) − ) − cos (2 ( ) − )] 2 1

= − (cos 2 1 1

=− (

2 2

3

π 3

3 π

2 √2

×

2 √2

π 4

1+cos

) −(

4

=

− cos x

[shown] ✓

π 4

π

− cos x − cos 2 x sin2 x

− sin2 x − cos 2 x sin2 x

=−

=

)

sin2 x + cos 2 x + cos x sin2 x 1 + cos x =− sin2 x 1 + cos x =− 1 − cos 2 x 1 + cos x =− (1 + cos x)(1 − cos x)

− 2√x + c ✓

=

sin x

=−

1

x2

1+cos x

sin x ⋅

= ∫(x 3 − 3x 2 ) dx x4

(

d d (1 + cos x) − (1 + cos x) ⋅ (sin x) dx dx = sin2 x sin x ⋅ (− sin x) − (1 + cos x) ⋅ cos x = sin2 x

A1(a) ∫ x 2 (x − 3) dx

=

d dx

π sin 4 √2 2 √2 2

1+

2+√2 √2

2 √2

)]

)]

)]

− 1)

)

√2 √2

= √2 ✓

− cos ) 6

√3 − ) 2

1

= (√3 − 1) ✓ 4


sleightofmath.com

517


1 (x−3)(2x+1)

=

A x−3

+

Rev Ex 18 A6(a) ∫a(3 − 2x) dx 1

B 2x+1

1 ( )(7)

=A

1

1

2

(− )( )

1

1 7

x=− :

7 2

∴ (x−3)(2x+1) = =

7

1

=

⇒A=

=B

x−3 1

7(x−3)

7

⇒B=−

−

+

1

2 7

2 7

2x+1

−

2 7(2x+1)

✓

k 6 ∫2 (1 2

−

[x −

] dx

7(x−3) 7(2x+1) 7 1 2 ) dx − ∫ ( 7 4 x−3 2x+1

(k − k3

1 = [ln|x − 3| − ln|2x + 1|]74 7 1 = [(ln 4 − ln 15) − (ln 1 − ln 9)] 7 1 4 = (ln + ln 9) 7 1

15 12

7

5

A5(a)

d dx

3 3

[ln(cos x)] =

1 cos x

⋅

d dx

(cos x)

1 ⋅ (− sin x) cos x = − tan x [shown] ✓ =

π 3

1

2

2

= −100

− x 2 ) dx

= −100

k x3

]

=−

3 2 k3 3

8

) − (2 − )

−k−

3

52

=−

100 6 100 6

=0

3

k − 3k − 52 =0 2 (k − 4)(k + + ⬚) 2 (k − 4)(k + + 13) 2 (k − 4)(k + 4k + 13) =0 (k − 4)[(k + 2)2 − 4 + 13] = 0 (k − 4)[(k + 2)2 + 9] =0 k =4✓

✓

A7(i)

dy dx

= 3x 2 + k

At turning pt (−2,6):

π 3

dy

∫0 tan x dx = − ∫0 − tan x dx =

1

A6(b) ∫k 6(1 − x 2 ) dx 2

1

= ln

2

x2

= [4 ( )]

[3x − x 2 ]1a = 2[x 2 ]12 (3a − a2 ) − (3 − 1) = 2(1 − 4) 3a − a2 − 2 = −6 2 3a − a + 4 =0 2 a − 3a − 4 =0 (a + 1)(a − 4) =0 a = −1 or a = 4 ✓

1 A4(ii) ∫7 dx 4 (x−3)(2x+1)

= ∫4 [

a

[3x − 2 ( )]


x2

1

= ∫2 4x dx

|

dx x=−2 2

π −[ln(cos x)]03

=0

3(−2) + k = 0 k = −12 ✓

π = − [ln (cos ) − ln(cos 0)] 3 1 = − [ln − ln 1] 2

= ln 2 ✓ A5(b)

d dx

[ln(ex + 1)] =

1

= ln 2 ex

∫0

ex +1

⋅

d

(ex + 1)

ex +1 dx 1 = x ⋅ ex e +1 ex ex +1

[shown] ✓

2 dx = [ln(ex + 1)]ln 0

= ln(eln 2 + 1) − ln(e0 + 1) = ln(2 + 1) − ln(1 + 1) = ln 3 − ln 2 3

= ln ✓ 2


sleightofmath.com

518


dy dx

Rev Ex 18 B1(c) ∫1 (4x+1)4 −7 dx 0 2(4x+1)2

= 3x 2 − 12

1 1 2

y = 3 ( ) − 12x + a 3

=[

Curve at (−2,6): 6 = (−2)3 − 12(−2) + a a = −10

=[ =

3

∴ y = x − 12x − 10

4

(4x+1)3

2

(4)(3)

(4x+1)3 (4(1)+1)3

15

5 4

−[ =

632

−

15

= 41

9 10

3

1

2

2

0

1 2

6

π 1−sin2 x

= ∫π

3

6

1−sin x

dx

π (1+sin x)(1−sin x) 1−sin x

6

dx

π

= ∫ (1 + sin x) dx π 6

4

x2 2

+

2

= [x − cos x]ππ

5 2

(1)3

6

1

1

2 8

6

π 6

=

1

− (1)2 + (1)2 ] 5

π

6

= [(π) − (−1)] − ( −

2] 5

π

= (π − cos π) − ( − cos )

4 1 2 x ] 2 1

= [ (4) − (4) + (4) 3

12

B2(c) ∫ππ cos2 x dx 1−sin x

= ∫π

8 5

8

11

= − ln 8 ✓

2

5 x2 5 2

3

]

= −3 ln 2

= [4 ( ) − 4 ( ) + ( )]

4

7

8(4(0)+1)

✓

= 3 ln

4

5

+

= 3[ln 2 − ln 4]

= ∫1 (4x 2 − 4x√x + x) dx

− x

24

B2(b) ∫3 3 dx = [3 ln|x − 5|]13 1 x−5

B1(b) ∫4(2x − √x)2 dx 1

=

8(4(1)+1)

(4(0)+1)3

= 2 ln 5 ✓

=8 ✓

4 [ x3 3

−[

= 2(ln 5 − ln 1)

=

3

]

= 2[ln|2x + 1|]20

3 4 1 = [(2x + 1)2 ] 3 0

x3

]

= 4 [ ln|2x + 1|]

3 4

3 1 3 [92 − 12 ] 3 1 = (27 − 1) 3 26 = 3

7

8(4x+1) 0 7

B2(a) ∫2 4 dx = 4 ∫2 1 dx 0 2x+1 0 2x+1

∫0 √2x + 1 dx = ∫0 (2x + 1) dx (2x + 1)2 =[ ] 3 (2) ( ) 2 0

1

−

60 7

1

]

(4)(−1) 0

2

323

1 2

4

− ⋅

+

24

(4x+1)−1

7

+

24

=4

At point where curve meets y − axis (x = 0), y|x=0 = (10)3 − 12(10) − 10 = −10 ⇒ (0, −10) ✓

2

1

=[ ⋅

= x 3 − 12x + a

B1(a)

7

= ∫0 [ (4x + 1)2 − (4x + 1)−2 ] dx

x3

2

5π 6

+1+

√3 2

√3 ) 2

✓

7 30

✓


sleightofmath.com

519

A math 360 sol (unofficial) B2(d)

Rev Ex 18

π 2 π 4

B4(a)

∫ 2 csc 2 x dx

8x+13 (1+2x)(2+x)2

=

A

B

+

1+2x

C

2+x

+ (2+x)2

π

= 2[− cot x]π2

[not in syllabus]

Cover-up rule:

4

B3(i)

1

9 ( )(1.5)2

π π = 2 [(− cot ) − (− cot )] 2 4 π π = 2 (cot − cot ) 4 2 = 2(1 − 0) =2✓

x=− :

y = x 2 √2x − 1

Substitution:

dy dx

= x2 ⋅

d dx

= = = = =

2√2x−1

⋅2

x2

x = 0:

+2x ⋅ √2x − 1

x2

13 (1)(4)

=

B

= −2

8x+13

∴ (1+2x)(2+x)2 = =

+2x(2x−1) √2x−1

x2

2

∫1

+4x2 −2x √2x−1

2

√2x−1 √2x−1

4

1+2x

B

C

2

4

+ +

1

1+2x 4

4

−

1+2x

2

= ∫1 [4 ⋅

[shown] ✓

A

+ −

−2 2+x 2 2+x

1

+ (2+x)2 1

+ (2+x)2 ✓

8x+13 dx (1+2x)(2+x)2

= ∫1 [

5x2 −2x x(5x−2)

=C

⇒C=1

+2x√2x − 1

√2x−1

−3 (−3)( )2

x = −2:

d

1

=A

⇒A=4

√2x − 1 + dx x 2 ⋅ √2x − 1

2

=x ⋅

2

1

2

−2⋅

1+2x

1

+ (2+x)2 ] dx

2+x 1

+ (2 + x)−2 ] dx

2+x

1

= [4 ⋅ ln|1 + 2x| − 2 ⋅ ln|2 + x| + 2

B3(ii) ∫5 x(5x−2) = [x 2 √2x − 1]5 1 1

= [2 ln|1 + 2x| − 2 ln|2 + x| − = [2 ln | = [2 ln |

1+2x 2+x

|−

1+2(2) 2+(2)

5

1

4

4

6

1

4

12

= 2 ln +

sleightofmath.com

1

]

1

]

2

2+x 1

2

2+x 1

|−

= [2 ln − ]


2

] (1)(−1)

1

√2x−1

= (5)2 √2(5) − 1 −(1)2 √2(1) − 1 = 75 −1 = 74 ✓

(2+x)−1

1 2+(2)

]

− [2 ln |

1+2(1) 2+(1)

|−

1 2+(1)

]

1

− [2 ln 1 − ] 3

✓

520


Rev Ex 18

B4(b) Partial fractions 1

B5(ii)

1

x2 +3x+2

A

= (x+2)(x+1) =

x+2

+

B

π

∫04 cos 3 2x dx π 4

x+1

= ∫ [(cos 2x)(cos 2 2x)] dx Cover-up rule:

0

1 ( )(−1)

x = −2:

π 4

=A

= ∫ [(cos 2x)(1 − sin2 2x)] dx

⇒ A = −1

0 π

1 (1)( )

x = −1:

=B

⇒B=1 1

∴

x2 +3x+2

=−

1 x+2

∫ x2 +3x+2 dx =

1

+

x+1

0

x+1

= ln |

π 4

dx

−

x+2

x+1

+c ✓

|

x+2

dx

1 6

1

− 0)

2

= 3(sin 2x) ⋅ cos 2x ⋅

dx

B6(a)

d dx

[ln(ln x)] =

∫2

1 x ln x

(2x)

= ln (

cos 2x

= ln (

⋅

ln x 1

d dx 1

(ln x)

x

x ln x

ln 4

)

ln 2 ln 22

)

ln 2 2 ln 2

)

ln 2

= ln 2 ✓

π

π

d

1 4 = ∫ 6 sin2 2x cos 2x dx 6 0 π 1 = [sin3 2x]04 6 1 π = (sin3 − sin3 0)

dx

(x n ln x)

= xn ⋅ = xn ⋅ =x

2

− 0)

⋅

= [ln(ln x)]42 = ln (

∫04 sin2 2x cos 2x dx

1 (1 6 1 = (1) 6 1 = ✓

1 ln x 1

= ln(ln 4) − ln(ln 2)

= 3(sin 2x) ⋅ cos 2x ⋅ 2

=

6

3

2

= 6 sin2 2x

1

= ✓

4

d

−

1

(sin 2x)

2

6

2

= d

1

2

3]

= 3(sin 2x)2 ⋅

6

π

= (1

dx

1

1

=

[(sin 2x)

6

−

2

(sin3 2x) d

−

= [sin 2x]0

= ln|x + 1| − ln|x + 2| +c

=

2

1

✓

1 ∫ (x+2)(x+1) dx 1 1

=∫

dx

π 4

= (sin − sin 0) −

1

d

1

= [ sin 2x]

Integral

B5(i)

π

= ∫04 cos 2x dx − ∫04 sin2 2x cos 2x dx

d dx 1

x n−1

(ln x) + +nx

d

dx n−1

+n x

(x n ) ⋅ ln x

⋅ ln x

n−1

ln x ✓

B6(b) n = 1:

6


sleightofmath.com

521

A math 360 sol (unofficial) d dx

Rev Ex 18 B7(a) ∫ 1 dN N

(x ln x) = 1 + ln x

⇒ ln x

=

d dx

ln|N| + a = kt ln|N| = kt − a N = ekt−a = bekt

(x ln x) − 1

∫ ln x dx = ∫[

d dx

(x ln x) − 1] dx

B7(b) N|t=0 = A be0 = A b=A ∴ N = ekt [shown] ✓

= x ln x −x + c ✓ n = 2: d dx

(x 2 ln x) = x + 2x ln x

⇒ 2x ln x =

d

dx 1 d

x ln x = (

= kt

x 2 ln x − x

2 dx

x 2 ln x − x)

∫ x ln x dx 1 d

=∫ [

2 dx 1 d

(x 2 ln x) − x] dx

= ∫ [ (x 2 ln x) − x] dx 2 dx 1

x2

2

2

= (x 2 ln x − ) + c ✓ n = 3: d dx

x 3 ln x = x 2 + 3x 2 ln x

⇒ 3x 2 ln x =

d

dx 1 d

x 2 ln x

= [

x 3 ln x − x 2

3 dx

(x 3 ln x) − x 2 ]

∫ x 2 ln x dx 1 d

=∫ [

3 dx 1 d

(x 3 ln x) − x 2 ] dx

= ∫ [ (x 3 ln x) − x 2 ] dx 3 dx 1

= (x 3 ln x

x3

−

3

3

)+c✓

n = m + 1: d dx

(x m+1 ln x)

= x m + (m + 1)x m ln x

⇒ (m + 1)x m ln x = x m ln x

=

d dx 1

(x m+1 ln x) − x m [

d

m+1 dx

(x m+1 ln x) − x m ]

∫ x m ln x dx =∫ = =

1

[

d

m+1 dx 1 d

m+1 1 m+1

(x m+1 ln x) − x m ] dx

∫ [dx (x m+1 ln x) − x m ] dx (x m+1 ln x −


xm+1 m+1

)+c✓

sleightofmath.com

522


Ex 19.1 5

Ex 19.1 1

4 ∫0 x(4 4

= − ∫ sin(x − π) dx

𝑦

− x) dx

0 x2

x3

4

3

0

= [4 ( ) − ( )] 2

x3

= [2x 2 −

𝑥 𝑂

4

]

3 2

6 − [2(0)2 −

(0)3 3

= ∫ [(0) − (1 − ex )] dx

]

𝑦 = cos 2𝑥

𝑂 π 1 = [sin 2x]04 2 1 π = (sin − sin 0) 2 1

𝑥

𝜋

7

4

Shaded area =

2

= (1

−0)

2 1

4 ∫1 [(y

= [[

𝑦 2

− 3) + 2] dy

(y−3)3 3

4

4

] + 2y] 1

2

= unit ✓ Shaded area

((0)−3)

= [[

𝑦

1

= ∫ e2x dx =

0 1 2x 1 [ e ] 2 0

=8

𝑦 = 𝑒 2𝑥

1 2x 1 [e ]0 2 1 = (e2 − e0 ) =

𝑂

−[ =5

+ (3)]

𝑂

3

3

2

+ (0)] +

3

𝑥 = 𝑦(2 − 𝑦)

3

𝑂

2

y y = [2 ( ) − ( )] 2 3 0

𝑥

2 1 = [y 2 − y 3 ] 3 0

0

3

((0)−1)

2

0

𝑦 = (𝑥 − 1)2 + 1

3

3

𝑦

= ∫ [2y − y 2 ] dy 2

0

((3)−1)

3

0 2

= ∫ [(x − 1)2 + 1] dx

=[

2

= ∫ [y(2 − y) − (0)] dy

𝑦

3

+ x]

] + 2(0)]

2

Shaded area

3

+

3

3

3

Shaded area

2

=[

1

((0)−3)

8

= (e2 − 1) unit 2 ✓

(x−1)3

] + 2(0)] − [[

= 9 unit ✓

2 1

4

3

2

𝑥

1

3

𝑥 = (𝑦 − 3)2 + 2 𝑥

1

2

3

𝑦 = 1 − 𝑒𝑥

= −[x − ex ]20 = −[(2 − e2 ) − (0 − e0 )] = −[2 − e2 + 1] = (e2 − 3) unit 2 ✓

0

0

𝑥

= − ∫ (1 − ex ) dx

𝑦

= ∫ cos 2x dx

2

𝑂

2 0

π 4

π 4

2

0

−0

Shaded area

1

𝑦

2

3

= [ sin 2x]

𝑥

𝑦 = sin(𝑥 − 𝜋)

Shaded area

= 10 unit 2 ✓ 2

𝜋

4

3 0 (4)3 [2(4)2 − ] 3 2

= 10

𝑂

0

= −[− cos(x − π)]π0 = [cos(x − π)]π0 = cos 0 − cos(−π) =1 −(−1) 2 = 2 unit ✓

𝑦 = 𝑥(4 − 𝑥)

= ∫ (4x − x 2 ) dx

=

𝑦

π

Shaded area =

Shaded area

3

𝑥

=1 1 3

1

1

3

3

= [(2)2 − (2)3 ] − [(0)2 − (0)3 ] 1 3 1

−0

= 1 unit 2 ✓ 3

2

= 6 unit ✓


sleightofmath.com

523


Ex 19.1 π

11(a) Area

At A, y = sin (2x + ) cuts x-axis (y = 0): 3

=0

=

sin (2x + ) = 0

=

y π 3

α π 2x +

= ✓

9(ii)

=[

3

Shaded Area π

π

= ∫03 sin (2x + ) dx 1

π

π 3

2

3

0

1

π

π 3

2 1

3

= [− cos (2x + )] = − [cos (2x + )]

1

𝐴

4

3

𝑂

+ (5)] − [ 1

4

√

Shaded area

𝑦

a 0 a

1

√

1

3 2

𝑦=𝑒

3 2

𝑂

𝑎

𝑥

1

3

√

x2

2

x

−

) dx

1

3 x2

2

2

3

√

x2

+ ∫ 3 (2 − 2

) dx ) dx

3 2 x √3 2

+4 +

1

3

2

2

2

3

3 2

3

−2 (2√ + 2

3 √

3 2

)

= 10 − 2 (2√ + 3√ )

2

ea

2

√

+ ∫ 3 (2 −

+ [2x + ]

3 √ 2

=2+3

unit ✓

1

3

2

2

2

3

= 10 − 4√ − 6√

10(ii) Shaded area = 0.5

e a

3 𝑥2

𝑥

3

= [2x + ]

= e−(0) −e−(a)

ea a

2

3 1

a 1 = [ e−x ] −1 0 = −[e−x ]a0 = [e−x ]0a

ea

𝑦 =2−

= ∫ 3 (2 −

−𝑥

0

1

+ (2)]

−6

4 1

= ∫ e−x dx

1−

4

∫1 − (2 − x2 ) dx

= ∫ e−x − (0) dx

1

(2)4

3

1 1 = − (−1 − ) 2 2 3 2 = unit ✓

=1

2

𝑦

𝑥

𝜋

𝑂

= − (cos π − cos )

10(i)

(5)4

3

π

2

4

4

𝑦 = sin (2𝑥 + ) 11(b)

0

+ x]

= 155 unit 2 ✓

𝜋

3

𝑥

𝑂 2 5

5

x4

= 161

𝑦

𝑦 = 𝑥3 + 1

+ 1] − (0) dx

x + 1 dx

=[

π

x

𝑦

2

=0 =π

3

5 ∫2 [x 3 4 ∫ 3

= 0.5 =

1

Note: Deal with the upper limits first then the lower limits.

2

=2 = ln 2 ≈ 0.693 ✓


sleightofmath.com

524


Ex 19.1 14(i)

𝑦 𝑦 = 𝑥2 + 1

y = x2 − x − 6 = (x − 3)(x + 2) 𝑦

A 𝑂

B

𝑥

1

4

Line PQ Point:

2−0

3

2 3 2

8

3

3

=− x+

=

=[

3

0

3

4

+ [−

+[

3 4

+

3 1

=[ 8

+ ] − (0)dx

=[

3

2 x2

8

3 2

3

4

=

+ [− [ ] + x]

= [( + 1) − (0)]

=

−2

4 2 + ∫1 [− x 3

1

=

3

+area of B

+ 1] − (0) dx

+ x]

−2

2) dx = − ∫−2(x 2 − x − 6) dx

1

x3

𝑦 = 𝑥2 − 𝑥 − 6 𝑂 𝑥 3

= ∫3 (x 2 − x − 6) dx

Shaded area = area of A 1 ∫0 [x 2

𝑦

3

= − ∫−2(x − 3)(x +

y − 2 = − (x − 1) y

✓

= ∫−2[(0) − (x − 3)(x + 2)] dx

2

y − y1 = m(x − x1 )

PQ:

𝑥

3

=−

1−4

3

14(ii) Area

P(1,2), Q(4,0)

Gradient: mPQ =

𝑂

−2

x2 3

(−

3

−

(−2)3 3

1

x2 2

−

−2

− 6x] (−2)2 2

3

− 6(−2)] − [

22

+

3 5

(3)3 3

−

(3)2 2

− 6(3)]

27 2

2

= 20 unit ✓ 6

4

8

x3

+ x] 3

16

+

3

1

32

)

1

3 8

3

3

− (− + )

15(i)

Curve: y = x2

−(1)

Line: x + y= 6 y =6−x

−(2)

]

9 3

2

= 4 unit ✓ 3

13

Shaded area

𝑦 𝑦 = (𝑥 + 1)(𝑥 − 2)

2

= ∫ 0 − (x + 1)(x − 2) dx

𝑂

−1

𝑥

2

−1

2

= − ∫ (x + 1)(x − 2) dx −1 2

= − ∫ (x 2 − x − 2) dx −1 −1

sub (1) into (2): x2 = 6 − x x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x = −3 or (rej ∵ x > 0)

x=2 y|x=2 = (2)2 =4 ⇒ A(2,4) ✓

= ∫ (x 2 − x − 2) dx 2

=[ =[ =

x3

−

3

(−1)3 3

x2 2

−

−1

− 2x] (−1)2 2

2

− 2(−1)] − [

7

+

6 1

(2)3 3

−

(2)2 2

− 2(2)]

10 3

2

= 4 unit ✓ 2


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525


Ex 19.1

15(ii) Shaded region

16(ii) Method 2 (wrt x-axis) Line x=4−y y=4−x

𝑦

2

= ∫ [(6 − x) − x 2 ] dx

𝑦 = 𝑥2

0 2

6

𝐴

= ∫ [6 − x − x 2 ] dx 2

x2 x3 = [6x − − ] 2 3 0

=7

(2)2

−

2

2

𝑂

(2)3 3

] − [6(0) −

1

(0)2 2

𝑥

6 −

(0)3 3

Lower curve x = 4y − y 2 (y − 2)2 − 4 = −x (y − 2)2 =4−x y−2 = ±√4 − x y = 2 ± √4 − x ⇒y = 2 − √4 − x

]

−0

3 1

= 7 unit 2 ✓ 3

16(i)

x = 4y − y 2

−(1)

x=4−y

−(2)

𝐵

𝑥

𝑦 = 2 − √2 − 𝑥

Shaded Area 3

= ∫0 [(4 − x) − (2 − √4 − x)] dx 3

= ∫0 (2 − x + √4 − x) dx = [2x − = [2x −

16(ii) Method 1 (wrt y-axis)

x2 2 x2 2

𝑥 =4−𝑦

3

+

3 2

=

2 5

]

(−1)( )

0 3

2

3

− (4 − x)2 ] 3

9

2

2 2

3

3 2

0 3

2

− [0 − (4)2 ] 3

2

3

+ [ (22 )2 ]

=[ − ] 4

3

3

(4−x)2

= [6 − − (1) ]

𝑦

3

3 2 3 + 2 3

6 1

= 6 unit 2 ✓

𝑄

6

1 𝑃

17(i)

𝑥

2𝑂

d dx

1

=

+Area of Q

d

(√3x + 7) = ⋅ (3x + 7) 2√3x+7 dx =

Shaded Area = Area of P

𝑂 3

Line x=4−y y=4−x

sub (1) into (2): 4y − y 2 =4−y 2 y − 5y + 4 =0 (y − 1)(y − 4)= 0 y=1 or y = 4 x|y=1 = 4 − (1) x|y=4 = 4 − (4) =3 =0 ⇒ B(3,1) ✓ ⇒ A(0,4) ✓

𝑥 = 4𝑦 − 𝑦

𝐴

𝑦 =6−𝑥

0

= [6(2) −

𝑦 𝑦 =4−𝑥

1 2√3x+7 3 2√3x+7

⋅3 ✓

1

= ∫0 [(4y − y 2 ) − (0)] dy 4

+ ∫1 [(4 − y) − (0)] dy 1

= ∫0 (4y − y 2 ) dy y2

y3

1

3

0

= [4 ( ) − ( )] 2

= [2y 2 −

y3

1

]

3 0

1

= [(2 − ) − (0)] 3

=

5

4

+ ∫1 (4 − y) dy + [4y −

1

4

]

2 1 1

4

2

1

+ [4y − y 2 ]

1

+ [(16 − 8) − (4 − )] 2

+

3

y2

9 2

2

= 6 unit ✓ 6


sleightofmath.com

526


Ex 19.1

17(ii) Shaded area = = =

18(ii) Method 2 (without rectangle area) Horizontal line y|x=a = 1 + a2

𝑦

3 1 dx ∫−1 √3x+7 2 3 3 dx ∫ 3 −1 2√3x+7 3 2 [√3x + 7]−1 3

𝑦= −1

𝑂

1 √3𝑥+7

𝑥

3

a

B = ∫0 [(1 + a2 ) − (1 + x 2 )] dx a

= ∫ (a2 − x 2 ) dx

2

= (√3(3) + 7 −√3(−1) + 7) 3 2

= (√16 = (4 = 18(i)

3

= [a2 x −

−√4)

3 2 3 4

0

= (a3 −

−2)

a3

]

= (a +

3

(0)3 3

]

𝑥=𝑎 𝑎

= a(1 + a2 ) − (a + = a + a3

𝑥

−a −

a3 3

19

1 3 a 3 3

3

−a

=0

𝑦 𝑦=

2 k

𝑂

a3

Area of P

3

2 3 a 3

k 10

∫2 ( x2 ) dx k

∫2 10x −2 dx 2

= a3 3

10 [

=0

x−1

𝑥2

[ ]

a = 0 or a = −√3 or a = √3 ✓ (both rej ∵ a > 0)

2

k k

k

sleightofmath.com

−

= area of Q 5 10

= ∫k ( 2 ) dx x

5

= ∫k (10x −2 ) dx = 10 [

x−1

5

]

−1 k

1 5

=[ ]

x 2

1

𝑥

5

k

]

−1 2

1 k

a − 3a =0 2 a(a − 3) =0 a(a + √3)(a − √3) = 0


10

𝑃

)

Region A = Region B a3

−a

a = 0 or a = −√3 or a = √3 ✓ (both rej ∵ a > 0)

18(ii) Method 1 (with rectangle area) Horizontal line y|x=a = 1 + a2 Region B = Rectangle − Region A

a+

3

a − 3a =0 a(a − 3) =0 a(a + √3)(a − √3) = 0

) unit 2 ✓

=

2

= a3

3

1 3 a 3 3

𝐴

1 𝑂

=B a3

a+

−0

3 a3

) − (0)

A 𝑦 = 1 + 𝑥2 𝐵

− [(0) + =a+

𝑦

]

3

3

3

a x3

= [(a) +

3 0

a3

= a3

Region A a = ∫0 (1 + x 2 ) dx 3 0 (a)3

a

]

2

unit 2 ✓

= [x +

x3

x k

1 2

1

1

5

k

= − = =

7 10 20 7

✓

527


y = x 2 − ax − b

Ex 19.1 20(iii) Line PR Points:

P(−1,0) & R(0, −3)

(1, −4) lies on curve, −4 = (1)2 − a(1) − b −4 = 1 − a − b b =5−a

Gradient:

mPR = (−1)−(0) = −3

PR:

y − y1 = m(x − x1 ) y − 0 = −3[x − (−1)] y = −3x − 3

At min (1, −4),

Line RQ Points:

R(0, −3) & Q(3,0)

Gradient:

mRQ =

RQ:

y − y1 = m(x − x1 ) y − 0 = 1(x − 3) y =x−3

dy dx

dy

= 2x − a

|

dx x=1

=0

2(1) − a = 0 a =2✓ b|a=2 = 5 − (2) = 3 ✓ 20(ii) Curve with a = 2, b = 3: y = x 2 − 2x − 3 Points P & Q At P & Q, curve cuts x − axis (y = 0): y =0 2 x − 2x − 3 =0 (x + 1)(x − 3) = 0 x = −1 or x=3 ⇒ P(−1,0) ✓ ⇒ Q(3,0) ✓

(0)−(−3)

(0)−(−3) (3)−(0)

=1

𝑦 −1 𝑃

3

𝑂

𝑄

𝑥

𝑅 𝑦 = 𝑥 2 − 𝑎𝑥 − 𝑏 Shaded Area = left region

+right region

0

Point R At R, curve cuts y − axis y|x=0 = (0)2 − 2(0) − 3 = −3 ⇒ R(0, −3) ✓

= ∫−1[(−3x − 3) − (x 2 − 2x − 3)] dx 3

+ ∫0 [(x − 3) − (x 2 − 2x − 3)] dx 0

3

= ∫−1[−x 2 − x] dx + ∫0 [−x 2 + 3x] dx = [−

x3 3

−

x2

0

]

2 −1 1

1

3

2

= [(0) − ( − )]

+ [−

x3 3

+

+ [(−9 +

3x2 2

3

]

0

27 2

) − (0)]

2

= 4 unit 2 ✓ 3

21(i)


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P(2,0) lies on y = kx − x 2 : 0 = 2k − 22 k=2✓

528


Ex 19.1 22(a) ∫2π sin x dx = 0 0

21(ii) Point Q Q(3, h) lies on y = 2x − x 2 : h = 2(3) − 32 = −3 ⇒ Q(3, −3) Line OQ Points:

𝐴 𝑂

(−3)−(0) (3)−(0)

𝐵

𝑥

Algebraic area = A − B = 0 ∵ (symmetric) 22(b)

= −1

a

∫ cos x dx = 0 0

y − y1 = m(x − x1 ) y − (−3) = −1(x − 3) y = −x

OQ:

𝑦

𝑦

O(0,0) & Q(3, −3)

Gradient: mOQ =

[typo in book]

𝜋 2𝜋

𝑦 = 2𝑥 − 𝑥 2 3

By inspection a = kπ, k ∈ ℤ+

𝑥

𝑂

𝑄(3, ℎ) 3

Area = ∫0 [(2x − x 2 ) − (−x)] 3

= ∫ [3x − x 2 ] dx 0

=[ =[

3x2 2

−

3(3)2

=4

2 1 2 1

x3

3

]

3 0 (3)3

−

3

]− [

3(0)2 2

−

(0)3 3

]

−0

= 4 unit 2 ✓ 2


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529


Ex 19.2 3

Ex 19.2 1

Points P & Q At P & Q, y = x 2 − 8x + 20 meets y = 8 x 2 − 8x + 20 = 8 𝑦 x 2 − 8x + 12 = 0 𝑄 𝑃 (x − 2)(x − 6) = x = 2 or x = 6

Point P x=0 Point Q 1

At Q, y = cos x meets y = : 2

cos x =

𝑦0= 8

x

=

1 2 π 3

𝑦=7

2

𝑦 = 𝑥 − 8𝑥 + 20 2

𝑂

Shaded Area

Shaded Area

𝑥

6

1 = ∫ [(cos x) − ] dx 2 0

6

= ∫ [(8) − (x 2 − 8x + 20)] dx 2 6

= ∫ [8 − x 2 + 8x − 20] dx 2 6

=∫

(−x 2

+ 8x − 12) dx

= [−

=(

6

3

x + 4x 2 − 12x] 3 2 (6)3 3

− [−

4

+ 4(6)2 − 12(6)] (2)3

+ 4(2)2 − 12(2)]

3

2

= 10 unit 2 ✓ 3

2

π 3

2

0

π

π

3

6

Shaded Area 7

= ∫1 [(8x − x 2 ) − 7]

𝑃 𝑂 1

𝑄 7

√3 2

π

− ) unit 2 ✓ 6

Points P & Q At P & Q, y = x 2 − 12x + 42 meets y = x + 2 x 2 − 12x + 42 = x + 2 𝑦 x 2 − 13x + 40 = 0 𝑦 = 𝑥 2 − 12𝑥 + 42 𝑄 (x − 5)(x − 8) = 0 𝑦 =𝑥+2 x = 5 or x = 8 𝑂

Shaded Area (x + 2) 8 ] dx = ∫5 [ 2 −(x − 12x + 42)

= [−

3

𝑦=7 𝑥

= [− = [−

x3 3

+

512 3

13 2 x 2

3

98 3

= −74

7

1

+

13 2

2

(64) − 40(8)]

+79

3

5

125 3

+

13 2

(25) − 40(5)]

1 6

2

= 4 unit ✓

1

2

2

+ 4(7) − 7(7)] − [−

=

𝑥

8

− 40x]

− [−

+ 4x − 7x]

(7)3

8

8

7

2

5

= ∫5 [−x 2 + 13x − 40] dx

= ∫1 [−x 2 + 8x − 7] dx = [−

𝑄

𝑃

Points P & Q At P & Q, y = 8x − x 2 meets y = 7 8x − x 2 =7 2 x − 8x + 7 = 0 𝑦 (x − 1)(x − 7) = 0 𝑦 = 8𝑥 − 𝑥 2 x = 1 or x = 7

x3

𝑂

= (sin − ) − (sin 0 − 0)

2

= [−

1

= [sin x − x]

𝑦 = 8𝑥 − 𝑥 2 1 𝑦 7 𝑥

𝑃

π 3

+

(1)3 3

+ 4(1)2 − 7(1)]

10 3

= 36 unit 2 ✓


sleightofmath.com

530


Ex 19.2

Points P & Q At P & Q, y = x + 3 meets y = −x 2 + 8x − 7 x+3 = −x 2 + 8x − 7 x 2 − 7x + 10 = 0 (x − 2)(x − 5) = 0 x = 2 or x = 5 𝑦 Shaded Area 𝑦 =𝑥+3 5 (−x 2 + 8x − 7) ] dx =∫ [ 𝑄 −(x + 3) 2 𝑃 𝑥 5 5 𝑂 2 2 = ∫ [−x + 7x − 10] dx 𝑦 = −𝑥 2 + 8𝑥 − 7 2 =

x3

7 [− + x 2 3 2 125 7

= [−

= −4

3 1

− 10x]

7

Area of region A 1

= ∫0 [(ex ) − 1] dx

𝑦 = 𝑒𝑥 𝑥=1 1

= =

1 ∫0 [(1) 1 ∫ 2

− (1 − x

A B

Area of region B 2 )]

𝑂 dx

𝑦=1

𝑥 1 𝑦 = 1 − 𝑥2

x dx

0 x3

1

5

=[ ]

2

1 = [13 − 03 ] 3 1 = unit 2 ✓

3 0

8

7

3

2

+ (25) − 50] − [− + (4) − 20] 2

+8

6

𝑦

= [ex − x]10 = [e1 − 1] −[e0 − 0] = e − 1 −1 = (e − 2) unit 2 ✓

2

3

3

1

= 4 unit 2 ✓ 2

6

8(i)

𝑦 𝑦 = 2 sin 𝑥 𝑦=𝑥 𝑥

𝑂

By inspection, y = 2 sin x and y = x π π intersect at (0,0) and ( , ) 2 2 𝑦 𝑦 = 2 sin 𝑥 Area of region π

= ∫02 [(2 sin x) − x] dx = [−2 cos x −

=−

]

𝑂

2

2

2 π2 8

] − [−2 cos 0 −

] −[−2(1)

02 2

]

− 0]

+2

8

= (2 −

π 2 2

( )

−

π2

𝑦=𝑥 𝑥 𝜋

2 0

π

= [−2 cos − = [0

π x2 2

π2 8

) unit 2 ✓


sleightofmath.com

Points P & Q At P & Q, y = 7 + 6x − x 2 meets x-axis (y = 0). 7 + 6x − x 2 = 0 x 2 − 6x − 7 = 0 (x − 7)(x + 1) = 0 x = 7 or x = −1 ⇒ P(7,0) ✓ ⇒ Q(−1,0) ✓ Point R At R, y = 7 + 6x − x 2 meets y − axis (x = 0): y|x=0 = 7 ⇒ R(0,7) ✓ Point S At S, y = 7 + 6x − x 2 meets y = 7. y =7 2 7 + 6x − x = 7 6x − x 2 =0 2 x − 6x =0 x(x − 6) =0 x=0 or x=6 (taken at R) ⇒ S(6,7) ✓

531


Ex 19.2 9(i)

𝑦 𝑦 = 7 + 6𝑥 − 𝑥 2

𝑅

𝑆

𝑄 𝑂

𝑃

(−2,0)

y = x 2 − 7x + 15

−(1)

y=7−x

−(2)

sub (1) into (2): 7−x = x 2 − 7x + 15 x 2 − 6x + 8 =0 (x − 2)(x − 4) = 0 x = 2 or x = 4

𝑥

Left region = = =

𝑦

−1 ∫−2 0 − (7 + 6x − x 2 ) dx −1 − ∫−2 (7 + 6x − x 2 ) dx −2 ∫−1 (7 + 6x − x 2 ) dx −2 x3 2

= [7x + 3x −

− [7(−1) + 3(−1) − +3

3 1

(−2)3

3 (−1)3 3

𝑂

]

1 2 2

= ∫1 [(x 2 − 7x + 15) − (7 − x)] dx

3

2

= ∫1 (x 2 − 6x + 8) dx

2

= 4 unit ✓ 3

=[ Middle region 0

=[

= ∫ (7 + 6x − x 2 ) dx

x3

− 3x 2 + 8x]

3

= [7x + 3x −

x3

− 3(2)2 + 8(2)]

3

−[

0

]

3 −1 (0)3 2

= [7(0) + 3(0) −

3

− [7(−1) + 3(−1) − +3

=6 ]

(−1)3 3

2

2

(1)3

−5

3 1

3

− 3(1)2 + 8(1)]

1 3

2

= 1 unit ✓ 3

] 9(ii)

3

2

1

(2)3

−1

=0

𝑥

4

Region A

]

2

2

𝑦 =7−𝑥

𝐵

3 −1

= [7(−2) + 3(−2) −

=

𝐴

]

2

2

𝑦 = 𝑥 2 − 7𝑥 + 15

Region B 4

= ∫2 [(7 − x) − (x 2 − 7x + 15)] dx

2

= 3 unit 2 ✓ 3

4

= ∫ (−x 2 + 6x − 8) dx Right region =

6 ∫0 (7 6

2 4

x3 x2 = [− + 6 ( ) − 8x] 3 2 2

+ 6x − x 2 ) − 7 dx

= ∫ (6x − x 2 ) dx

= [−

0

= =

[3x 2

−

x3

6

]

3 0 (6)3 [3(6)2 − ] 3 (0)3 2

− [3(0) −

3

= [−

x3 3

+ 3x 2 − 8x]

(4)3 3

= −5

+ 3(4) − 8(4)]

3

+6

(2)3 3

+ 3(2)2 − 8(2)]

2 3

4

= 36 −0 = 36 unit 2 ✓


1

2

2

− [− ]

4

= ✓ 3

sleightofmath.com

532


Ex 19.2

y = x 2 + 4x − 5 dy dx

10(ii) Method 1 (complement)

= 2x + 4

𝑦

𝑦 = 𝑥 2 + 4𝑥 − 5

Gradient of tangent is 10, dy

𝑃

= 10

dx

2x + 4 = 10 x =3 y|x=3 = 32 + 4(3) − 5 = 16 ✓

𝑅 1

𝑂

Area of PQR = area of PR𝑇 =

10(ii) Point P P(3,16)

3 ∫1 (x 2

=[

x3

=( Point R At R, y = x 2 + 4x − 5 cuts x − axis (y = 0): x 2 + 4x − 5 = 0 (x + 5)(x − 1) = 0 x = −5 or x = 1 (rej ∵ x > 0) ⇒ R(1,0) Line PQ Points: Gradient: PQ:

−area of ⊿PQT

+ 4x − 5) − (0) dx

+ 2x 2 − 5x]

5

1

− (1.6)(16) 2

1

+ 18 − 15) − ( + 2 − 5)

15

7

2

−12.8

3

unit 2 ✓

𝑦

B A 𝑂

1

3

7

𝑥

5

Area of shaded region PQR = region A +region B 7

= ∫15(x 2 + 4x − 5) − 0 dx 3

+ ∫7 [(x 2 + 4x − 5) − (10x − 14)] dx 5 7 5

= ∫1 [x 2 + 4x − 5] dx 7 5

= ∫1 [x 2 + 4x − 5] dx

7 5

=[

7

⇒ Q ( , 0) 5

x3 3

+ 2x 2 − 5x]

3

sleightofmath.com

15

5

3

+ ∫7 [(x − 3)2 ] dx 5

+[

(x−3)3 3

3

]7 5

7 2 5

1

3 13

3

+ ∫7 [x 2 − 6x + 9] dx

+ 2 ( ) − 7]

− [ + 2 − 5] =1

7 5

1

7 3 ( ) 5

=[


3

1

3 13

1

− (3 − ) (16)

Method 2 (break)

P(3,16) or Q mPQ = 10 y − y1 = mPQ (x − x1 ) y − 16 = 10(x − 3) 𝑦 − 16 = 10𝑥 − 30 y = 10x − 14

=

3 27

=1

Point Q At Q, PQ (𝑦 = 10𝑥 − 14) cuts x − axis (y = 0): y =0 10x − 14 = 0 10x = 14 x

𝑥

𝑄 3

+0

− (−

512

)

375

unit 2 ✓

533


Ex 19.2

Point A At A, y = 7 + 6x − x 2 cuts y − axis (x = 0): y|x=0 = 7 ⇒ A(0,7)

12(ii) Area of region A k 1

= ∫1 =

=[

A(0,7) lies on y = mx + c, (7) = m(0) + c c =7✓

𝐴

=1− =

− m)x − x x3

2

3 0

[(6 − m) ( ) − [(6 − m) ( (6 −

(6−m)2 2

− (mx + 7)] dx = 20 = 20

6−m

x2

]

)−

1 m)3 [ ] 6 3

(6 − m) 6−m m

= 20 (6−m)3 3

] − [0]

= 20 =

k

k

𝑥

✓

−

=

−

=

= 20

dx

𝐴 1

k

1

=

2]

k−1

= (k − 1)(k + 1)

𝑥

+ 6x − x

𝐵

𝑄(1,1) 𝑂

𝑃(𝑘, 𝑘)

12(iii) By complement Area of region B = area of trapezium −area of region A

=

2)

1 𝑥2

k 1

𝑦 + 6𝑥 − 𝑥 2

Area of shaded region 6−m ∫0 [(7 6−m ∫0 [(6

−1 1 1 k

= − ( − 1)

𝑦 = 𝑚𝑥 + 𝑐

6−𝑚

𝑂

𝑦=

]

x 1 1

𝑦 𝐵

𝑦

= [− ]

11(ii) Point B At B, y = mx + 7 meets y = 7 + 6x − x 2 mx + 7 = 7 + 6x − x 2 x 2 + mx − 6x = 0 x 2 + (m − 6)x = 0 x[x − (m − 6)] = 0 x=0 or x = −(m − 6) (taken at A) =6−m✓ 11(iii)

dx

x2 k −2 ∫1 x dx k x−1

5 6 5

2 k2 −1 2

k−1 k k−1 k

k3 −k−2k+2 2k k3 −3k+2 2k (k−1)(k2 +k−2) 2k

13

[deduced] ✓

𝑦

6 5

𝑦 = √1 − 𝑥 2

6 5

𝑂

6

𝑥 𝑦 =1−𝑥

5 6

125

1

∫0 [√1 − x 2 − (1 − x)] dx

6

1

= 125 =5 =1✓

= ( area of circle) − (area of triangle) 4

1

= [π(1)2 ] 4 π

1

4

2

1

− (1)(1) 2

= − ✓ 12(i)

y=x✓ 14

∞ 1

∫1

x2

dx represents the convergent value of area as

→ ∞✓


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534


Rev Ex 19 A3

Rev Ex 19 A1(i)

Area of region A

𝑦

p 8

= ∫1

x

2 dx

1 p

𝑦=

= 8 [− ]

x 1 1 1

𝐴

= −8 ( − ) p 8

1

𝑂

= (8 − ) unit 2 ✓

−(1)

x=3 −(2) sub (2) into (1):

8 𝑥2

𝐵

1 P

Coordinates y 2 = 3x 𝑦 = ±√3𝑥

𝑦 = ±√3(3) = ±3

𝑥

6

p

Shaded region A1(ii) A

=B

8− 8− 8− 8− 8−

8

=

p 8 8 8

= −8 ( − )

p 8

6

8

3

p

=− +

p

= =

7

d

1

9

9

+6

✓ A4

(sin x) +

Area of region A =

d

(x) ⋅ sin x

− sin x

= x ⋅ cos x +1 ⋅ sin x = x cos x + sin x = x cos x [shown] ✓

− sin x − sin x

dx

𝑥

= 12 unit 2 ✓

(x sin x + cos x)

=x⋅

𝑥=3

1

=6

28 3 12

𝑂 −3

= [3(3) − (3)3 ] − [3(−3) − (−3)3 ]

p

4

𝑦 2 = 3𝑥

3

3 1 = [3y − y 3 ] 9 −3

x p 1 1

p

dx

3 1 = ∫ [3 − y 2 ] dy 3 −3

= −8 [ ]

p

𝑦

1 3

x p 1 6

p

d

6 8 ∫p (x2 ) dx 1 6

= 8 [− ]

p

16

A2(a)

3

= ∫−3 [(3) − ( y 2 )] dy

dx

1 ∫0 x(x

− 1)(x − 2) dx

𝑦 𝑦 = 𝑥(𝑥 − 1)(𝑥 − 2)

1 2

= ∫ x(x − 3x + 2) dx

𝐴 𝑂 1

𝐵

2

𝑥

0 1

= ∫0 (x 3 − 3x 2 + 2x) dx =[

A2 At A, y = x cos x cut x − axis (y = 0): (b)(i) x cos x = 0 x = 0 or cos x = 0 π (rej ∵ x > 0) x =

x4 4

1

− x3 + x2 ]

0

1

= ( − 1 + 1) − (0) 4

=

2

1 4

π

⇒ A ( , 0) ✓ 2

A2 Shaded Area π (b)(ii) = ∫02 x cos x − (0) dx

Area of region B

π

= ∫02 x cos x dx = [x sin x + π

π

2

2

𝑂

π cos x]02 π

2

= − ∫1 x(x − 1)(x − 2) dx

𝑦 𝑦 = 𝑥 cos 𝑥 𝐴 𝑥 𝜋 2

=

2

2

x4 4

1

− x3 + x2 ]

2

4

−1

≈ 0.57 ✓


=[

1 = ( − 1 + 1) − (4 − 8 + 4) 4 1 =

= ( sin + cos ) −(0 + 1) π

1

= ∫2 (x 3 − 3x 2 + 2x) dx

∴A=B✓

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535


Rev Ex 19

Point P At P, y = x 2 − 2x + 2 meets y − axis (x = 0): y|x=0 = 2 ⇒ P(0,2) ✓

A6(ii)

𝑦 = 2𝑥 𝐴 𝑂

Point Q At Q, y = x 2 − 2x + 2 has x-coordinate of 3 y|x=3 = 32 − 2(3) + 2 =5 ⇒ Q(3,5) ✓ A5(ii) Area of A =

3 ∫0 x 2

=[ =[

x3 3 3

−[

4

3

− x + 2x] −

Area of shaded region = left region +right region

𝑦 𝑦 = 𝑥 2 − 2𝑥 + 2

− 2x + 2 dx

(3)2

(0)3 3

+ 2(3)]

−

(0)2

𝑃 𝑂

𝐵 𝐴 (3,0)

+ ∫4 6x − x 2 dx

= [x 2 ]40

+ [3x 2 − x 3 ]

𝑥

B1(i)

2

4

+[108 − 72]− [48 −

= 16

+36

−48 +

64

3 64

]

3

2

Curve & its gradient y = x 2 − 4x + 5 dx

= 2x − 4

Normal PQ Point: y|x=3 = 32 − 4(3) + 5 = 2 ⇒ (3,2)

−6

Gradient:

1

= 4 unit 2 ✓

− 𝑑𝑦

1 |

𝑑𝑥 𝑥=3

2

PQ: Point A At A, y = 6x − x 2 meets y = 2x 6x − x 2 = 2x 4x − x 2 = 0 x 2 − 4x = 0 x(x − 4) = 0 x=0 or x = 4 y|x=0 = 2(0) y|x=4 = 2(4) =0 =8 ⇒ O(0,0) ⇒ A(4,8) ✓

3

=4 −0

dy

Area of B = area of trapezium −area of A 1

6

3

= 6 −0 = 6 unit 2 ✓

A6(i)

2

1

= 25 unit ✓

+ 2(0)]

= (3)(2 + 5)

6

= ∫0 2x dx

1

0

𝑥

4 6

2

2

(3)3

𝑦

y − y1

=−

1 2(3)−4

= − dy

1 |

=−

1 2

(x − x1 )

dx 𝑥=3

1

y − (2) = (− ) [x − (3)] 2

4 − 2y 2y + x

=x−3 = 7 [shown] ✓

B1(ii) Point Q At Q, PQ (2y + x = 7) cuts x − axis (y = 0): 2(0) + x = 7 x =7 ⇒ Q(7,0) ✓

Point B At B, y = 6 − x 2 meets x − axis (y = 0): 6x − x 2 = 0 x(x − 6) = 0 x=0 or x = 6 ⇒ O(0,0) ⇒ B(6,0) ✓


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536

A math 360 sol (unofficial) B1(iii)

Rev Ex 19 B2(b) Coordinates y = x(x − 3) −(1)

𝑦 𝑦 = 𝑥 2 − 4𝑥 + 5 𝑃 𝑄 𝑥 3 7

𝑂

Area of shaded region = left region 3

= ∫0 (x 2 − 4x + 5) dx =[

x3 3

− 2x 2 + 5x]

y = −2 −(2) (1) (2): sub into x(x − 3) = −2 2 x − 3x + 2 = 0 (x − 1)(x − 2) = 0 x = 1 or x = 2

+right region 7 7

1

2

2

+ ∫3 ( − x) dx

3

7

1

7

2 49

4 49

3

+ [ x − x2 ]

0

= [9 − 18 + 15] − [0]

+[

=6 = 10 unit 2 ✓

+4

2

−

4

]−[

𝑦

21

2

𝑦 = −2

x

3

2

= ∫ [(−2) − x(x − 3)] dx

4

1 2 ∫1 (−2 − 3x − x 2 ) dx 1 ∫2 (x 2 + 3x + 2) dx 1 x3 3 [ − x 2 + 2x] 3 2 2 1 3 8

=

=

= [ − + 2] − [ − 6 + 4]

𝑥

5

1 𝑂

2

9

− ]

𝑦 = 𝑥(5 − 𝑥)

𝑂

𝑦 = 𝑥(𝑥 − 3)

Area

= B2(a)

y

3 1

=

6

2

3

2

unit ✓

✓ B3(i)

Area of left region

𝑦

2

= ∫0 [x(5 − x)] dx = =

d dx

[(x − 1)ex ]

= (x − 1) ⋅

2 ∫0 (5x − x 2 ) dx 2 5 1 [ x2 − x3 ] 2 3 0 8

𝑂

2

5

d dx x

(ex )

= (x − 1) ⋅ e = ex [(x − 1) +1] = xex [shown] ✓

𝑥

+

d dx

(x − 1) ⋅ ex ⋅ ex

+1

= [10 − ] − (0) 3

=

B3(ii) Area of shaded region A

22

0

3

= ∫−2[(0) − (xex )] dx

Area of right region 5

=

5

= =

= ∫2 [x(5 − x)] − (0) dx = ∫2 [5x − x 2 ] dx 1

5

3 125

2

5

= [ x2 − x3 ] =[ =

2 125

2 27

−

3

𝑦

−2 ∫0 xex dx [(x − 1)ex ]−2 0 (−3)e−2

1 −2 A

−[−1(e0 )]

B 0.5

𝑥

3

8

=1

3

≈ 0.594 unit 2 ✓

] − [10 − ]

−

𝑂

𝑦 = 𝑥𝑒 𝑥

e2

2

Area of shaded region B Ratio =

22 3

=44

0.5

:

= ∫0 [(1) − (xex )] dx

27

= [x − (x − 1)ex ]0.5 0

2

: 81 [shown] ✓

1

1

= [0.5 − (− ) e2 ] −[0 − (−1)e0 ] 2

1

1

= ( + √e) 2 2

−1 2

≈ 0.324 unit ✓


sleightofmath.com

537


Rev Ex 19

π 2 π 6

π

B5(ii) y-coordinate of P

∫ sin x dx = [− cos x]π2

4 π

y|x=π = 1 + ( ) = 2

6

=

π −[cos x]π2 6

Region B = area of trapezium −region A

π π = − (cos − cos ) 2 6

B4(ii)

√3 2

1 π

= ( ) (1 + 2) 2 4 3π

√3 ) 2

= − (0 − =

π 4

4

=(

unit 2 ✓

B6(i)

𝑦

dx

𝑦 = sin 𝑥

A B

= ex

Tangent Point:

1 2

Gradient: 𝑂

𝜋

𝜋

6

2

𝜋

𝑥

sin x =

⇒x=

2

sin x = 1

⇒x=

6 π

= = = = B5(i)

6 π 2 π 6 π 2 π 6

π 12 π 12 π 12 5 12

√3 2

π

+ ∫ 1 dx

− ∫π2 sin x dx

+[x]

√3 − 2

6

π

π

2

6

+ − π−

dy

(x − x1 )

|

dx x=2 2 (x

−

B6(ii) Area enclosed by curve & x − axis & lines x = 0, x = 2

√3 2

π 4

= [ex ]20 = e2 − e0 = e2 − 1 ✓

𝑦 𝑦 = sec 𝑥 𝑦 =1+

𝑃

𝐵

𝑥

2

Shaded region = (Bigger area) − △ ABC area

𝐵 𝐴 𝜋

𝑥

sleightofmath.com

1

= e2 − 1

− [2 − 1][e2 ]

= e2 − 1

− e2

=

4


𝑂

4

1 𝑂

𝐶

𝜋

π

= [tan x]04 π = tan − tan 0 4 = 1 unit 2 ✓

𝐴

𝑦 = 𝑒𝑥

2

2

= ∫ (sec 2 x) dx

𝑦

= ∫0 [(ex ) − 0] dx

[shown] ✓

Region A

0

=

Point B At B, tangent cuts x − axis (y = 0): e2 x − e2 = 0 x =1 ⇒ B(1,0) ✓

+ [∫π2 1 − sin x dx]

2 6

= e2

|

dx x=2

2

π

1 π

dy

y − e2 = e − 2) 2 y = e x − 2e2 + e2 y = e2 x − e2

π

Area of the shaded region = rectangle A +region B = (1 − )

y|x=2 = e2 ⇒ A(2, e2 )

Tangent: y − y1

Coordinates 1

− 1) unit 2 ✓

Curve & its gradient y = ex dy

1

8

−1

1 ( e2 2

2 1 2

2

− 1) unit [shown] ✓

538


Ex 20.1 1(i)

Displacement s = 2t 2 + t s|t=2 = 2(2)2 + 2 = 10 ✓

1(ii) 1

𝑠 = 2𝑡 2 + 𝑡 𝑂 𝑡

Velocity v =

ds dt

= 14t − 10

4(ii)

v|t=3 = 14(3) − 10 = 32 ms −1 ✓

5(i)

Displacement s = 7t 2 − 2t 3

2

Distance travelled in 1st 2s = s|t=2 −s|t=0 = 10 −0 = 10m ✓

Velocity v=

Velocity 1 v = 5t − t 2 2 v|t=0 = 0ms −1 ✓

2(ii)

Displacement s = 7t 2 − 10t

𝑠

−

2(i)

Ex 20.1

ds dt

= 14t − 6t 2 ✓ 5(ii)

Acceleration a =

At rest, v =0

dv dt

= 14 − 12t ✓

1

5t − t 2 = 0 2

t 2 − 10t = 0 t(t − 10) = 0 t = 0 or t = 10 ✓ 3(i)

5(iii)

a|t=2 = 14 − 12(2) = −10ms −2 ✓

6(i)

Acceleration a=t−9

Acceleration a = t − 5t 2

Velocity v = ∫ a dt = ∫ t − 9 dt

Accelerating at 0.05m/s 2 , a = 0.05 2 t − 5t = 0.05 2 20t − 100t =1 2 100t − 20t + 1 = 0 (10t − 1)2 =0 t = 0.1s ✓

=

Decelerating at 6 m/s 2 , a = −6 2 t − 5t = −6 5t 2 − t − 6 =0 (5t − 6)(t + 1) = 0 t = 1.2 or t = −1 (rej ∵ t > 0) ✓

2

− 9t + c

Initial velocity of −20 m/s, v|t=0 = −20 (0)2 2

3(ii)

t2

− 9(0) + c= −20

c

= −20 1

∴ v = t 2 − 9t − 20 ✓ 2

6(ii)

At rest, v 1 2 t 2 2

− 9t − 20

=0 =0

t − 18t − 40 = 0 (t − 20)(t + 2) = 0 t = 20s ✓ or t = −2 © Daniel & Samuel A-math tuition 📞9133 9982

sleightofmath.com

(rej ∵ t > 0)

539


Velocity v=

Ex 20.1 9(i)

1 8e−2

Velocity

Displacement s = ∫ v dt = ∫ 8e

−

1 2

Displacement s = 12t − t 3

v =

ds dt

= 12 − 3t 2

dt

1 − 2

= 8e t + c

At rest, v =0 2 12 − 3t =0 2 t −4 =0 (t + 2)(t − 2) = 0 t = −2 or t = 2 (rej ∵ t > 0)

t ≡ time after leaving O: s|t=0 =0 1

8e−2 (0) + c = 0 c =0 1

∴ s = 8e−2 t ✓ 7(ii)

a=

s|t=5 = 8e (5) =

8(i)

Acceleration

1 − 2

40 √e

Acceleration at rest, a|t=2 = −6(2) = −12ms −2 ✓ 9(ii)

Velocity v =

ds

Recall

s = 12t − t 3 v = 12 − 3t 2 a = −6

dt

= 3t 2 − 24t + 36

Displacement Next at O, s =0 3 12t − t = 0 t 3 − 12t = 0 t(t 2 − 12) = 0 t = 0 or t = −√12 or t = √12 (rej both ∵ t > 0)

v|t=1 = 3 − 24 + 26 = 15 ms −1 ✓ 8(ii)

dt

= −6t

m✓

Displacement s = t(t − 6)2 = t(t 2 − 12t + 36) = t 3 − 12t 2 + 36t

dv

At rest, v =0 2 3t − 24t + 36 = 0 t 2 − 8t + 12 =0 (t − 2)(t − 6) = 0 t = 2 or t = 6 ✓

Velocity Velocity at t = √12: v|t=√12 = 12 − 3(√12)

8(iii)

= 12 − 36 = −24ms −1

Acceleration a =

2

dv dt

= 6t − 24 a|t=3 = 6(3) − 24 = −6ms −2 ✓


sleightofmath.com

540


Recall

s = 12t − t 3 v = 12 − 3t 2 a = −6

Ex 20.1 10(ii) s|t=2 = 18(2)2 − 24 = 56 Distance travelled in 3rd sec = s|t=3 − s|t=2 = 81 − 56 = 25m ✓

Distance 𝑦 12

𝑣 = 12 − 3𝑡 2 𝑥 −2 𝑂 2

10(iii) a = dv dt

Distance travelled in first 3 seconds

= 36 − 12t 2

3

= ∫ |v| dt = = =

0 2 ∫0 |v| dt 2 ∫0 v dt 2 ∫0 v dt [s]20

When a = −12, a = −12 36 − 12t 2 = −12 2 12t − 48 =0 2 t −4 =0 (t + 2)(t − 2) = 0 t = −2 or t = 2 (rej ∵ t > 0)

3 + ∫2 |v| dt 3 + ∫2 −v dt 2 + ∫3 v dt +[s]23

= = (s|t=2 − s|t=0 ) +(s|t=2 − s|t=3 ) = 2s|t=2 −s|t=0 −s|t=3 = 2(16) −0 −27 = 23m ✓ 10

Velocity Velocity when a = −12: v|t=2 = 36(2) − 4(2)3 = 72 − 32 = 40ms −1 ✓

Displacement s = 18t 2 − t 4 Velocity v =

ds

11(i)

dt

= 36t − 4t 3 At rest, v =0 36t − 4t 3 =0 3 t − 9t =0 2 t(t − 9) =0 t(t + 3)(t − 3)= 0 t = 0 or t = −3 or t = 3 (rej both ∵ t > 0) Displacement s|t=3 = 18(3)2 − 34 = 162 − 81 = 81 ✓


Height h = 24t − 3t 2 Velocity v =

dh dt

= 24 − 6t v|t=3 = 24 − 6(3) = 6ms −1 ✓ 11(ii) At max height, v =0 24 − 6t = 0 t =4✓ Height h|t=4 = 24(4) − 3(4)2 = 96 − 48 = 48m

sleightofmath.com

541


Ex 20.1

11(iii) At h = 0, 24t − 3t 2 = 0 t 2 − 8t =0 t(t − 8) = 0 t = 0 or t = 8

13(i)

At rest: v =0 2t 2 − 3t − 2 =0 (2t + 1)(t − 2) = 0

Time of flight = 8 − 0 = 8s ✓ 12(i)

Velocity v = 2t 2 − 3t − 2

t=−

Displacement s = ∫ v dt 3t2

3

2

− 2t + c

dv dt

At t = 0, s = 3: s|t=0

= 3e0.4t (0.4) = 1.2e0.4t

2 3

2

e

2

(0)2

=3

− 2(0) + c = 3 =3

2

3

3

2

s = t 3 − t 2 − 2t + 3 Displacement at rest:

+ 6t + c

0.4 15 0.4t

−

3

Displacement with c = 3:

12(ii) Displacement s = ∫ v dt 3e0.4t

(0)3

c

Initial acceleration a|t=0 = 1.2e0 = 1.2ms −2 ✓

=

2

= t3 −

Acceleration

=

or t = 2 ✓

2

(rej ∵ t > 0)

Velocity v = 3(e0.4t + 2) = 3e0.4t + 6

a=

1

2

3

s|t=2 = (8) − (4) − 2(2) + 3

+ 6t + c

=

3 16 3

2

−6−4+3 5

=− m✓

t ≡ time after leaving O: s|t=0 =0 15 0 e 2

+ 6(0) + c

=0

c

=−

15 2

Displacement with c = − s=

15 0.4t e 2

3

+ 6t −

15 2

:

15 2

Displacement when t = 1: s|t=1 = =

15 0.4 e 2 15 0.4 e 2

+6−

15 2

3

− ✓ 2

12(iii) Velocity v = 3e0.4t + 6 > 0 ∵ e0.4t > 0 ∴ The particle will not return to O


sleightofmath.com

542


Ex 20.1

13(ii) Displacement when t = 3: s|t=3 2

14(i)

3

= (27) − (9) − 2(3) + 3 3

2

= 18 −

27 2

−6+3

27 2 1 =1 m✓

Velocity v = ∫ a dt = −t 2 + 6t + c

= 15 − 2

Initial Velocity of 7 m/s: v|t=0 =7 2 −(0) + 6(0) + c = 7 c =7

𝑣 𝑣 = 2𝑡 2 − 3𝑡 − 2 𝑂 −

Acceleration a = 2(3 − t) = 6 − 2t = −2t + 6

1

2

2

t Velocity with c = 7: v = 6t − t 2 + 7

Total distance travlled in first 3s At v = 7, v =7 2 −t + 6t + 7 = 7 −t 2 + 6t =0 t(t − 6) =0 t = 0 (rej) or t = 6 ✓

3

= ∫ |v| dt = = =

0 2 3 ∫0 |v| dt + ∫2 |v| dt 2 3 ∫0 −v dt + ∫2 v dt 0 3 ∫2 v dt + ∫2 v dt [s]02 +[s]32

= = s|t=0 − s|t=2 +s|t=3 − s|t=2 = s|t=0 −2s|t=2 +s|t=3 5

3

3

2

14(ii) Displacement s = ∫ v dt 1

= − t 3 + 3t 2 + 7t + c2

= (3) −2 (− ) + ( )

3

5

=7 ✓

t ≡ time after passing O: s|t=0 =0

6

1

3(0)2 − (0)3 + 7t + c2 = 0 3

c2

=0

Displacement with c2 = 0: 1

s = 3t 2 − t 3 + 7t 3

Position when v = 7: 1

s|t=6 = 3(36) − (216) + 42 3

= 78m ✓


sleightofmath.com

543


Acceleration a = 8 − 4t = −4t + 8 Velocity v = ∫ a dt = −2t 2 + 8t + c

Ex 20.1 15(iii) Acceleration At max speed: a =0 8 − 4t = 0 t =2 da dt

<0 ∴ max

Starts from rest: v|t=0 =0 2 −2(0) + 8(0) + c = 0 c =0

Velocity Max speed: v|t=2 = 8(2) − 2(2)2 = 16 − 8 = 8ms −1 ✓

Velocity with c = 0: v = −2t 2 + 8t At rest, v =0 2 −2t + 8t = 0 t 2 − 4t =0 t(t − 4) =0 t = 0 (NA) or t = 4 ✓

16(i)

Initial Velocity of 30 m/s: v|t=0 = 30 2 13(0) − 3(0) + c = 30 c = 30

2

= − t 3 + 4t 2 + c2 3

t ≡ time after leaving O: s|t=0 =0 2

Velocity with c = 30: v = −3t 2 + 13t + 30

− (0)3 + 4(0)2 + c2 = 0 3

=0

Velocity after 3s: v|t=3 = −27 + 39 + 30 = 42 ms −1 ✓

Displacement with c2 = 0: 2

s = − t 3 + 4t 2 3

Distance OA: 2

s|t=4 = 4(16) − (64) 3

1

= 21 m ✓ 3

Acceleration a = 13 − 6t = −6t + 13 Velocity v = ∫ a dt = −3t 2 + 13t + c

15(ii) Displacement s = ∫ v dt

c2

= −4

16(ii) At max distance: v =0 13t − 3t 2 + 30 = 0 3t 2 − 13t − 30 = 0 (3t + 5)(t − 6) = 0 t=−

5 3

or

t=6 ✓

(rej ∵ t > 0)


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A math 360 sol (unofficial) 16(iii) Displacement s = ∫ v dt = −t 3 +

13 2 t 2

Ex 20.1 17(iii) Distance Total distance travelled in first 2s 2

+ 30t + c2

= ∫ |v| dt =

t ≡ time after passing O: s|t=0 = 0 c2 = 0

=

= = s|t=1 − s|t=0 +s|t=1 − s|t=2 = 2s|t=1 − s|t=0 − s|t=2 = 2(47) − (40) − 42 = 12

Displacement with c2 = 0: s = −t 3 +

13 2 t 2

0 1 2 ∫0 v dt + ∫1 −v dt 1 1 ∫0 v dt + ∫2 v dt [s]10 +[s]12

+ 30t

Max Distance: s|t=6 =

13 2

average speed in first 2s 12 = 2 = 6ms −1 ✓

(36) − 216 + 30(6)

= 198m ✓ 17(i)

Displacement s = t 3 − 9t 2 + 15t + 40

s|t=2 = 8 − 36 + 30 + 40 = 42

Velocity v=

ds

s|t=6 = 216 − 324 + 90 + 40 = 22

dt

= 3t 2 − 18t + 15 At zero velocity, v =0 2 3t − 18t + 15 = 0 t 2 − 6t + 5 =0 (t − 1)(t − 5) = 0 t = 1 or t = 5

17(iv) 𝑣 𝑣 = 3𝑡 2 − 18𝑡 + 15 𝑂 1

Displacement s|t=1 = 1 − 9 + 15 + 40 = 47m ✓

Total distance travelled in first 6s 6

= ∫0 |v| dt 1

= ∫0 |v| dt

s|t=5 = 125 − 225 + 75 + 40 = 15m ✓

1

= ∫0 v dt 1

17(ii) Acceleration a=

dv dt

= 6t − 18 magnitude of 9: |a| = 9 a = −9 or a =9 6t − 18 = −9 6t − 18 = 9 t = 1.5 s t = 4.5s


𝑡

5

sleightofmath.com

5

+ ∫1 |v| dt 5

+ ∫1 −v dt 1

6

+ ∫5 |v| dt 6

+ ∫5 v dt 6

= ∫0 v dt

+ ∫5 v dt

+ ∫5 v dt

= [s]10

+[s]15

+[s]65

= 2s|t=1

+s|t=6 −s|t=0

−2s|t=5

= 2(47) = 46 ✓

+22

−2(15)

−(40)

545


Ex 20.1

Acceleration a = 6t − 8

18(iii) Acceleration At min velocity, a =0 6t − 8 = 0

Velocity v = ∫ a dt = 3t 2 − 8t + c

t

4 2

2 3

= − ms

✓

Velocity v = −3t 2 + 8t + 5 Acceleration a=

dv dt

= −6t + 8 ✓

or t = 2

19(ii) s = ∫ v dt t3

t2

3

2

= −3 ( ) + 8 ( ) + 5t + c

Displacement with c2 = 0: s = t 3 − 4t 2 + 4t Distance 2

= ∫03|v| dt 2

= ∫03 v dt

= −t 3 + 4t 2 + 5t + c ∵ t ≡ time after passing O, s|t=0 = 0 c =0 ∴ s = −t 3 + 4t 2 + 5t 19(iii) Velocity At speed of 2ms −1 , v = ±2 v =2 2 −3t + 8t + 5 = 2 3t 2 − 8t − 3 =0 (3t + 1)(t − 3) = 0 1

t = − or t = 3 ✓

= s|2 − s|t=0

3

3

2 3

2 2

2

3

3

3

(rej ∵ t > 0)

= [( ) − 4 ( ) + 4 ( )] − 0 27

3

−1

3

19(i)


32

3 4

3

Displacement s = ∫ v dt = t 3 − 4t 2 + 4t + c2

=

4

v|t=4 = 3 ( ) − 8 ( ) + 4

At rest: v =0 3t 2 − 8t + 4 = 0 (3t − 2)(t − 2) = 0 t=

3

Velocity Min velocity,

Initial velocity of 4m/s: v|t=0 =4 3(0)2 − 8(0) + c1 = 4 c1 =4 Velocity with c = 4: v = 3t 2 − 8t + 4

4

=

or v = −2 −3t 2 + 8t + 5 = −2 −3t 2 + 8t + 7 = 0 3t 2 − 8t − 7 = 0 t = = =

m✓

= 18(ii) Return to starting pt: s =0 3 2 t − 4t + 4t = 0 t(t 2 − 4t + 4) = 0 t(t − 2)2 =0 t = 0 (NA) or t = 2 ✓


= t=

8±√64−4(3)(−7) 2(3) 8±√148 6 8±√4×37 6 8±2√37 6 4±√37 3 4−√37 3

or t =

4+√37 3

✓

(rej ∵ t > 0)

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A math 360 sol (unofficial) 19(iv) Displacement s = ∫ v dt = −t 3 + 4t 2 + 5t + c2

Ex 20.1 20(ii) Displacement s = ∫ v dt 1

= − t 3 + 9t + c 3


t ≡ time from start: s|t=0 = 0 c=0

Velocity with c2 = 0: s = −t 3 + 4t 2 + 5t

Displacement with c = 0: 1

s = − t 3 + 9t 3

At displacement of 20 m: s = 20 −t 3 + 4t 2 + 5t = 20 3 2 t − 4t − 5t + 20 =0 2 (t − 4)(t − 5) =0 (t − 4)(t − √5)(t + √5) = 0

At starting pt again: s = 0: 1

− t 3 + 9t = 0 3

t 3 − 27t = 0 t(t 2 − 27) = 0 t(t + √27)(t − √27) = 0 t(t + 3√3)(t − 3√3) = 0 t = 0 or t = −3√3 or t = 3√3 (rej both ∵ t > 0) ✓

t = 4 or t = √5 or t = −√5 (rej ∵ t > 0) 20(i)

Velocity v = 9 − t2 = −t 2 + 9 At rest, v =0 2 −t + 9 =0 2 t −9 =0 (t + 3)(t − 3) = 0 t = −3 or t = 3 ✓ (rej ∵ t > 0)

20(iii) 18m from starting pt, s = 18 or

s

1 − t3 3 1 − t3 3 3

1 − t3 3 1 − t3 3 3

+ 9t

= 18

+ 9t − 18

=0

t − 27t + 54 =0 2 (t + 6)(t − 6t + 9) = 0 (t + 6)(t − 3)2 =0 t = −6 or t = 3 ✓ (rej ∵ t > 0) 21


sleightofmath.com

= −18 + 9t

= −18

+ 9t − 18

=0

t − 27t − 54 =0 2 (t − 6)(t + 6t + 9) = 0 (t − 6)(t + 3)2 =0 t = 6 ✓ or t = −3 (rej ∵ t > 0)

max. speed ⇒ a = 0ms −2 ✓

547


Rev Ex 20 A1(iii) Displacement s|t=3 = 27(3) − 33 = 54 s|t=4 = 27(4) − 43 = 44

Rev Ex 20 A1(i)

Displacement s = 27t − t 3 = −t 3 + 27t

Distance travelled in fourth sec = |s|t=4 − s|t=3 | = |44 − 54| = 10cm ✓

Velocity v=

ds dt

= −3t 2 + 27 At rest: v 27 − 3t 2 t2 − 9 (t + 3)(t − 3) t = −3 or (rej ∵ t > 0)

A1(iv)

𝑣 = 27 − 3𝑡 2 =0 =0 =0 =0 t=3

−3

𝑂

3

t

Distance travelled in the first 4s 4

= ∫0 |v| dt 3

4

3

4

3

3

= ∫0 |v| dt + ∫3 |v| dt

Acceleration a=

𝑣

= ∫0 v dt + ∫3 −v dt

dv dt

= ∫0 v dt + ∫4 v dt

= −6t

= [s]30 Acceleration at rest: a|t=3 = −6(3) = −18ms −2 ✓ A1(ii) Displacement next at O: s −t 3 + 27t t 3 − 27t t(t 2 − 27) t(t + √27)(t − √27)

+[s]34

= 2s|t=3 −s|t=0

−s|t=4

= 2(54) −(0) = 64m ✓

−(44)

=0 =0 =0 =0 =0

t = 0 or t = −√27 or t = √27 (rej ∵ t > 0) Velocity Velocity when next at O, v|t=√27 = 27 − 3(√27)

2

= −54 cms −1 ✓


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548


Rev Ex 20

Velocity

A3(i)

v = 12 sin

t

Displacement 3t

s=

2

t2 +9

Velocity Speed of 8 m/s: |v| = 8 v = 8 or 12 sin sin

t

v

=8

2

t

sin

3

α ≈ 0.729 t 2

= −8

12 sin

2

=

2

ds

v=

t 2

t 2

t 2

≈ 0.729 t ≈ 1.46 ✓

(t2 +9)2 (t2 +9)⋅3

=

2 3

3t2 +27 −6t2 (t2 +9)2 27−3t2

= (t2

=π−α

+9)2 −3(t2 −9) (t2 +9)2

=

≈ 2.41 t ≈ 4.82 (NA)

v|t=1 = Acceleration a=

=

dv dt

6 25

−3(12 −9) (12 +9)2

ms −1 ✓

−3(e2 −9) A3(ii) v| t=3 = (32 2 = 0✓

t 1

= 12 cos ( ) 2 2

= 6 cos

−3t(2t)

(t2 +9)2

=

α ≈ 0.729

=α

d dt

=

= −8 =−

dt (t2 +9)⋅ (3t) −3t⋅(t2 +9)

+9)

t

∴ Particle comes to instantaneous rest when t = 3 ✓

2

a|t=1.46 = 6 cos (

1.46 2

≈ 4.47 ms

)

−2

✓

A3(iii) For positive velocity, v >0 −3(t2 −9) (t2 +9)2 2

A2(ii) Displacement s = ∫ v dt

>0

= −24 cos + c

−3(t − 9) > 0 t2 − 9 <0 (t + 3)(t − 3) < 0

t ≡ after leaving O: s|t=0 =0

+ − + −3 3

t

2

0

−24 cos + c1 = 0

⇒ −3 < t < 3

2

−24 + c c

∵ (t 2 + 9)2 > 0

=0 = 24

−3 < t < 3 and t ≥ 0 ∴0≤ t<3✓

Displacement with c = 24: t

s = −24 cos + 24 ✓

A4(i)

2

Displacement s = 12 sin(2t) − 6 Velocity v=

ds dt

= 12(cos 2t)(2) = 24 cos 2t ✓ Acceleration v=

dv dt

= 24(− sin 2t)(2) = −48 sin 2t ✓ © Daniel & Samuel A-math tuition 📞9133 9982

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549


Rev Ex 20

A4(ii) Displacement At fixed point: s =0 12 sin 2t − 6 = 0 sin 2t α=

=

π

A5(i)

1

Acceleration a=2−t = −t + 2 Velocity v = ∫ a dt

2

1

6

= − t 2 + 2t + c 2

2t = α π 2t = t

=

6 π

12

Initial Velocity of 6 m/s: v|t=0 = 6 c=6

s✓

A4(iii) −1 ≤ sin(2t) ≤ 1 −12 ≤ 12 sin(2t) ≤ 12 −18 ≤ 12 sin(2t) − 6 ≤ 6

Velocity with c = 6: 1

v = − t 2 + 2t + 6 2

Velocity when t = 5:

Max distance = |−18| = 18 m

1

v|t=5 = 10 − (25) + 6 2

= 3.5 ms −1 ✓

A4(iv) When s = 3: s =3 12 sin(2t) − 6 = 3 sin 2t

=

3

−(1)

4

A5(ii) Velocity At turning point: v =0 1

− t 2 + 2t + 6= 0 2

Speed when s = 3, |v| = |24 cos 2t| = |24√1 − sin2 2t| Sub (1) into (2):

t 2 − 4t − 12 =0 (t − 6)(t + 2) = 0 t = 6 or t = −2 (rej ∵ t > 0)

−(2)

3 2

|v| = |24√1 − ( ) |

Displacement s = ∫ v dt

4

7

= |24√ |

1

= − t 3 + t 2 + 6t + c2

16

= |24

6

√7 | 4

∵ t ≡ time after passing O, s|t=0 = 0 ⇒ c2 = 0

= 6√7 ms −1 ✓

1

∴ s = − t 3 + t 2 + 6t 6


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550


𝑣 𝑣=

Rev Ex 20

1 − 𝑡2 2

−2 𝑂

6

+ 2𝑡 + 6

A6(ii) Displacement s = ∫ v dt = −2e−t + 3t 2 − 2t + c

𝑡

Start from rest: s|t=0 = 0 −2e0 + c = 0 c =2

Distance travelled 6

= ∫0 |v| dt 6

= ∫0 v dt

Displacement with c = 2: s = −2e−t + 3t 2 − 2t + 2

= s|t=6 −s|t=0 = [36] −[0] = 36m ✓ A6(i)

Acceleration a = 2(3 − e−t ) = 6 − 2e−t = −2e−t + 6

s|t=1 = −2e−(1) + 3(1)2 − 2(1) + 2 2 =− +3−2+2 e ≈ 2.26m ✓ B1(i)

Velocity 144

v = (2t+3)2 − 4k

Velocity v = ∫ a dt = 2e−t + 6t + c

Initial velocity of 12 m/s: v|t=0 = 12 144 (2(0)+3)2

Start from rest: v|t=0 = 0 2 + c= 0 c = −2 Velocity with c = −2: v = 2e−t + 6t − 2

− 4k = 12

16 − 4k −4k k

= 12 = −4 =1✓

B1(ii) Velocity with k = 1: 144

v = (2t+3)2 − 4

Velocity when t = 2: v|t=2 = 2e−(2) + 6(2) − 2 ≈ 10.3 ms −1 ✓

At rest, v 144 (2t+3)2 144 (2t+3)2 (2t+3)2

=0

−4

=0 =4 =

144

1 4

(2t + 3)2 = 36 2 4t + 12t + 9 = 36 2 4t + 12t − 27 = 0 (2t − 3)(2t + 9)= 0 t=

3 2

or

t=−

9 2

(rej ∵ t > 0)


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551


Rev Ex 20

B1(iii) Velocity v = 144(2t + 3)−2 − 4

B2(ii) At rest, v =0 4e−t −

Acceleration a=

5

=0 2

dv

4e−t

=

dt

−t

e

=

−t

= ln

t

= − ln

= 144[−2(2t + 3)−3 ](2) 576 − (2t+3)3

=

2

5 1 10 1 10 1 10

≈ 2.30 ✓

1

Acceleration at t = : 2

a|t=1 = −

B2(iii) Acceleration

576 43

2

a=

= −9ms −2 ✓

=−

(2t+3)−1 (2)(−1)

72 2t+3

a|t=2.30 = −4e−(2.30) = −0.4 ms −2 ✓

] − 4t + c B3(i)

− 4t + c

∵ t ≡ time after leaving O, s|t=0 =0 −

72 3

v=

72

− 4t + 24

Acceleration

1

Displacement at t = :

a=

2

∴ s|t=1 = − 2

=−

72 1 2( )+3 2

72 4

1

− 4 ( ) + 24 2

− 2 + 24

Displacement 2

s = 4 − 4e−t − t 5

Velocity v=

ds dt

= 4e−t −

2 5

Initial velocity: v|t=0 = 4 −

dv dt

= 42 − 12t a|t=1 = 42 − 12(1) = 30ms −2 ✓

= 4m ✓ B2(i)

dt

v|t=1 = 42(1) − 6(1)2 = 36 ms −1 ✓

Displacement with c = 24: 2t+3

ds

= 42t − 6t 2

= 24

s=−

Displacement s = 21t 2 − 2t 3 Velocity

+c =0

c

dt

= −4e−t

Displacement s = ∫ v dt = 144 [

dv

B3(ii) Velocity For change in motion, v =0 2 42t − 6t = 0 t 2 − 7t =0 t(t − 7) =0 t=0 or t = 7 ✓ (rej ∵ t > 0)

2 5

= 3.6ms −1 ✓


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552


Rev Ex 20

B3(iii) s|t=7 = 21(49) − 2(343) = 343 s|t=0 = 0 s|t=10 = 2100 − 2000 = 100

B4(ii) At rest, v =0 2 t − 8t + 7 =0 (t − 1)(t − 7) = 0 t = 1 or t = 7 𝑣 𝑂1

Distance 𝑣

𝑡

7

Point A:

𝑣 = −6𝑡 2 + 42

1

s|t=1 = − 4 + 7 3

7

𝑂

=3

𝑡

1 3

Point B: Total distance travelled for 1st 10s =

10 ∫0 |v| dt

=

7 ∫0 |v| dt 7

= ∫0 v dt 7

s|t=7 =

3

=− 10 + ∫7 |v| dt

− 4(49) + 49

98 3

Distance between A & B = s|t=1 − s|t=7 = 36m ✓

10

+ ∫7 −v dt 7

= ∫0 v dt

+ ∫10 v dt

= [s]70

7 +[s]10

B4(iii) Distance Distance travelled in 1st 9s 9

= 2s|t=7 − s|t=0 −s|t=10 = 2(343) − 0 = 586 ✓ B4(i)

343

= ∫ |v| dt

−100 = =

Velocity v = t 2 − 8t + 7 Displacement

=

0 1 7 9 ∫0 |v| dt + ∫1 |v| dt + ∫7 |v| dt 1 7 9 ∫0 v dt + ∫1 −v dt + ∫7 v dt 1 1 9 ∫0 v dt + ∫7 v dt + ∫7 v dt [s]10 +36 +[s]97

= = s|t=1 − s|t=0 +36

s = ∫ v dt

=

1

= t 3 − 4t 2 + 7t + c

10 3

− 0 +36

+s|t=9 − s|t=7 98

+ [−18 − (− )] 3

= 54 ✓

3

∵ t ≡ time after passing O, s|t=0 = 0 c=0 Displacement with c = 0: 1

s = t 3 − 4t 2 + 7t ✓ 3


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553


Rev Ex 20

B4(iv) Displacement s=

1 3 t 3

B5(i)

2

− 4t + 7t

Velocity v = t 2 − 8t + 7 Acceleration a=

Velocity v = at 2 + b Initial Velocity of 3 m/s: v|t=0 = 3 ⇒b =3

dv dt

= 2t − 8

Velocity with b = 3: v = at 2 + 3

At zero acceleration: a =0 2t − 8 = 0 2t =8 t =4✓

Displacement s = ∫ v dt

Point O: s|t=0 = 0 Point C:

t ≡ time after passing O: s|t=0 = 0 ⇒c =0

1

= at 3 + 3t + c 3

1

s|t=4 = (4)3 − 4(4)2 + 7(4) 3

= −14

2

Displacement with c = 0:

3

Point B: s|t=7 = −

1

s = at 3 + 3t

98

3

3

Back at O after 3s: s|t=3 = 0 9a + 9 = 0 9a = −9 a = −1

OC = |s|t=4 | 2

= |−14 | 3

= 14

2 3

Velocity Velocity with a = −1: v = −t 2 + 3

BC = |s|t=4 − s|t=7 | = |(−

44 3

98

) − (− )| 3

= 18

Speed = |v|t=3 | = 6 ms −1 ✓

OC < BC ∴ OC is nearer to O ✓


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554


Rev Ex 20

B5(ii) At turning point, v =0 2 −t + 3 =0 (t + √3)(t − √3) = 0 t = −√3 or (rej ∵ t > 0)

B6(i)

t = √3

Acceleration 18 − 3t for 0 ≤ t ≤ 4 a={ t + 2 for t > 4 Velocity 0 ≤ t ≤ 4: V = ∫ a dt 3

= 18t − t 2 + c 2

Displacement Initial velocity of v m/s: V|t=0 = v c =v

1

s = − t 3 + 3t 3

3

1

s|t=√3 = − (√3) + 3√3 3

Velocity with c = v:

= 2 √3

3

V = 18t − t 2 + v

𝑣

−√3

𝑂

2

𝑣 = −𝑡 2 + 3 𝑡 √3

Velocity of 50 m/s when t = 4: V|t=4 = 50 3

18(4) − (4)2 + v = 50 2

Total Distance travelled between t = 0 and t = 3

48 + v v

3

= ∫0 |v| dt √3

= ∫0 |v| dt =

√3 ∫0 v dt √3

3

+ ∫√3|v| dt

B6(ii) 0 ≤ t ≤ 4: 3

V = 18t − t 2 + 2

3 + ∫√3 −v dt

2

√3

= ∫0 v dt

+ ∫3 v dt

= [s]√3 0

+[s]√3 3

= 2s|t=√3 −s|t=0

−s|t=3

= 2(2√3) −0

−0

= 50 =2✓

t > 4: 1 V = t 2 + 2t + c2 2 V|t=4 = 50: 1 2

= 4√3 ✓

(4)2 + 2(4) + c2 = 50

c2

= 34

t > 4: 1

V = t 2 + 2t + 34 2

Velocity at t = 5 = V|t=5 1 = (5)2 + 2(5) + 34 2 = 56.5 ms −1 ✓


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555


Rev Ex 20

B6(iii) Check for turning point At turning pt, V = 0 0 ≤ t ≤ 4: V 18t −

3 2 t 2

=0

+2 =0

3t 2 − 36t − 4 = 0 36±√(−36)2 −4(3)(−4)

t =

2(3)

36 ± 8 √21 6

=

4

= 6 ± √21 3 4 = 6 − √21 < 0 (rej ∵ t > 0) 3 4 = 6 + √21 > 4 (rej ∵ t ≤ 4) 3 ∴ no turning pt t > 4: V 1 2 t 2 2

=0

+ 2t + 34

=0

t + 4t + 68 =0 [(t + 2)2 − (2)2 ] + 68 = 0 (t + 2)2 + 64 =0 ∴ no turning pt Distance 𝑉

𝑉=

1 2 𝑡 + 2𝑡 + 34 2

3 𝑉 = − 𝑡 2 + 18𝑡 + 2 2 𝑡 4

𝑂

Total distance travelled 5

= ∫0 |V| dt 4

5

= ∫0 |V| dt

+ ∫4 |V| dt

4

5

= ∫0 V dt 4

+ ∫4 V dt 5 1

3

= ∫0 (18t − t 2 + 2) dt + ∫4 ( t 2 + 2t + 34) dt 2

=

[9t 2

−

1 3 t 2

+ 2t]

2

4 0

= [144 − 32 + 8] − 0

1 + [ t3 + t2 6 125

+[

6

+ 34t]

5 4

+ 25 + 170]

−[

64 6

+ 16 + 136]

1

= 173 m ✓ 6


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556

a-math-360-sol-29-sept-2014.pdf

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