Cape Pure Maths Unit 1 Solutions 2011-2016.PDF

CAPE PURE MATHEMATICS UNIT 1 CAPE UNIT 1 2016 SOLUTIONS Question 1

a.

i.

Given 𝑓(𝑥) = 2𝑥 3 − 𝑥 2 + 𝑝𝑥 + 𝑞. If 𝑥 + 3 is a factor then 𝑓(−3) = 0. 𝑓(−3) = 2(−3)3 − (−3)2 + 𝑝(−3) + 𝑞 = 0 −54 − 9 − 3𝑝 + 𝑞 = 0 −3𝑝 + 𝑞 = 63

----------

(1)

Also 𝑓(−1) = 10 then 𝑓(−1) = 2(−1)3 − (−1)2 + 𝑝(−1) + 𝑞 = 10 −2 − 1 − 𝑝 + 𝑞 = 10 −𝑝 + 𝑞 = 13

------

(2)

Subtracting (1) from (2) we have 2𝑝 = −50, 𝑝 = −25 Substitute 𝑝 = −25 into (2) we have −(−25) + 𝑞 = 13, 𝑞 = −12 ii.

𝑓(𝑥) = 2𝑥 3 − 𝑥 2 − 25𝑥 − 12 𝑥 + 3 is a factor therefore using long division we have 2𝑥 2 − 7𝑥 − 4 𝑥+3

2𝑥 3 − 𝑥 2 − 25𝑥 − 12 2𝑥 3 + 6𝑥 2 −7𝑥 2 − 25𝑥 −7𝑥 2 − 21𝑥 −4𝑥 − 12 −4𝑥 − 12 0

𝑓(𝑥) = (𝑥 + 3)(2𝑥 2 − 7𝑥 − 4) = (𝑥 + 3)(𝑥 − 4)(2𝑥 + 1) 1

When 𝑓(𝑥) = 0, 𝑥 = −3, − 2 , 4 1

SOLUTIONS TO CAPE PURE MATHEMATICS UNIT 1 EXAM 2016

b.

When 𝑛 = 1, 61 − 1 = 5 which is divisible by 5 therefore the statement is true for 𝑛 = 1 Assume the statement is true for 𝑛 = 𝑘. Therefore 6𝑘 − 1 = 5𝑚 where 𝑚 𝜖 𝐍. When 𝑛 = 𝑘 + 1 we have 6𝑘+1 − 1 = 6(6𝑘 − 1) + 5 = 6(5𝑚) + 5 = 5(6𝑚 + 1) which is divisible by 5. Therefore the statement is true for 𝑛 = 𝑘 + 1 Since the statement is true for 𝑛 = 1, 𝑘 and 𝑘 + 1. It is true for all natural numbers n.

c. p T T F F ii.

q T F T F

𝐩→𝐪 T F T T

𝐩˅𝐪 T T T F

𝐩∧𝐪 T F F F

𝐩 → 𝐪 and (𝐩˅𝐪) → (𝐩 ∧ 𝐪) are logically equivalent because both have the same truth values in their output column.

Question 2

a.

(𝐩˅𝐪) → (𝐩 ∧ 𝐪) T F T T

log 2 (10 − 𝑥) + log 2 𝑥 = 4 log 2 (10 − 𝑥)𝑥 = 4 𝑥(10 − 𝑥) = 24 10𝑥 − 𝑥 2 = 16 𝑥 2 − 10𝑥 + 16 = 0 (𝑥 − 2)(𝑥 − 8) = 0 𝑥 = 2, 8

2

b.

𝑥+3

Given the function 𝑓(𝑥) = 𝑥−1 , 𝑥 ≠ 1. If the function is one-to-one then, 𝑓(𝑎) ≠ 𝑓(𝑏), 𝑎 ≠ 𝑏, (𝑎, 𝑏) ≠ 1 𝑎+3 𝑏+3 ≠ 𝑎−1 𝑏−1 (𝑎 + 3)(𝑏 − 1) ≠ (𝑏 + 3)(𝑎 − 1) 𝑎𝑏 − 𝑎 + 3𝑏 − 3 ≠ 𝑎𝑏 − 𝑏 + 3𝑎 − 3 4𝑏 ≠ 4𝑎,

𝑎≠𝑏

Therefore a and b are distinct and hence a maps to f(a), and b maps to f(b) For any 𝑥 𝜖 𝐑, where 𝑥 ≠ 1 and 𝑓 −1 (𝑥) =

𝑓(𝑓

−1 (𝑥))

𝑥+3 𝑥−1

𝑥+3 +3 𝑥+3 = 𝑓( )=𝑥−1 𝑥+3 𝑥−1 𝑥−1−1

𝑥 + 3 + 3𝑥 − 3 =( ) 𝑥+3−𝑥+1 =

4𝑥 =𝑥 4

Therefore the function is a one-to-one and onto because for (𝑥, 𝑦) 𝜖 𝐑, where (𝑥, 𝑦) ≠ 1, 𝑥 = 𝑓 −1 (𝑦) <=> 𝑦 = 𝑓(𝑥)

c.

i.

Given the roots of the equation 2𝑥 3 − 5𝑥 2 + 4𝑥 + 6 = 0 are 𝛼, 𝛽, and 𝛾. 5 𝑥 3 − 𝑥 2 + 2𝑥 + 3 = 0 2 𝑥 3 − (𝛼 + 𝛽 + 𝛾)𝑥 2 + (𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)𝑥 − 𝛼𝛽𝛾 = 0 5 𝛼+𝛽+𝛾 = , 2

𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾 = 2,

3

and 𝛼𝛽𝛾 = −3


c.

ii.

An equation whose roots are

1 𝛼2

,

1 𝛽2

and

1 𝛾2

has

1 1 1 𝛽2 𝛾 2 + 𝛼 2 𝛾 2 + 𝛼 2 𝛽2 + + = 𝛼 2 𝛽2 𝛾 2 𝛼 2𝛽2𝛾 2 =

(𝛼𝛽 + 𝛽𝛾 + 𝛼𝛾)2 − 2𝛼𝛽𝛾(𝛼 + 𝛽 + 𝛾) (𝛼𝛽𝛾)2

5 (2)2 − 2(−3) ( ) 19 2 = = (−3)2 9 1 1 1 𝛾 2 + 𝛼 2 + 𝛽2 + + = 𝛼 2𝛽2 𝛽2𝛾 2 𝛼 2 𝛾 2 𝛼 2𝛽2𝛾 2 (𝛼 + 𝛽 + 𝛾)2 − 2(𝛼𝛽 + 𝛼𝛾 + 𝛽𝛾) (𝛼𝛽𝛾)2

=

5 2 (2) − 2(2)

= 1 𝛼 2𝛽2𝛾 2

𝑥3 −

(−3)2

=

1 (𝛼𝛽𝛾)2

=

1 1 = (−3)2 9

19 2 1 1 𝑥 − 𝑥− =0 9 4 9

36𝑥 3 − 76𝑥 2 − 9𝑥 − 4 = 0

4

=−

1 4

Question 3 a.

i.

Prove sec 2 𝜃 =

LHS

=

cosec 𝜃 cosec 𝜃−sin 𝜃

1 sin 𝜃 1 sin 𝜃 − sin 𝜃

1 = sin 𝜃2 1 − sin 𝜃 sin 𝜃 =

1 1 = 2 1 − sin 𝜃 cos 2 𝜃

= sec 2 𝜃 𝐏𝐫𝐨𝐯𝐞𝐧.

ii.

Given

cosec 𝜃 cosec 𝜃−sin 𝜃 4

=

4 3

3

sec 2 𝜃 = 3 , cos2 𝜃 = 4 cos 𝜃 = ±

√3 2

this gives an acute angle

𝜋 6

𝜋 5𝜋 7𝜋 11𝜋 Therefore 𝜃 = 6 , 6 , 6 , 6

b.

i.

𝑓(𝜃) = sin 𝜃 + cos 𝜃 𝑟 sin(𝜃 + 𝛼) = 𝑟 sin 𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 𝜃 Therefore 𝑟 cos 𝛼 = 1 and 𝑟 sin 𝛼 = 1 𝜋

Hence, tan 𝛼 = 1, 𝛼 = 4 , and 𝑟 = √2 𝜋

sin 𝜃 + cos 𝜃 = √2 sin (𝜃 + 4 )

ii.

The max value of 𝑓(𝜃) is √2 𝜋

This occurs at (𝜃 + 4 ) =

𝜋 2

Therefore the smallest non-negative value of 𝜃 is 5

𝜋 4


c.

Prove tan(𝐴 + 𝐵 + 𝐶) =

tan 𝐴 + tan 𝐵 + tan 𝐶 − tan 𝐴 tan 𝐵 tan 𝐶 1 − tan 𝐴 tan 𝐵 − tan 𝐴 tan 𝐶 − tan 𝐵 tan 𝐶

tan(𝐴 + 𝐵 + 𝐶) = tan(𝐴 + (𝐵 + 𝐶)) =

=

tan 𝐴 + tan(𝐵 + 𝐶) 1 − tan 𝐴 tan(𝐵 + 𝐶)

tan 𝐵 + tan 𝐶 tan 𝐴 + 1 − tan 𝐵 tan 𝐶 tan 𝐵 + tan 𝐶 1 − tan 𝐴 (1 − tan 𝐵 tan 𝐶 )

tan 𝐴 (1 − tan 𝐵 tan 𝐶) + tan 𝐵 + tan 𝐶 1 − tan 𝐵 tan 𝐶 = (1 − tan 𝐵 tan 𝐶) − tan 𝐴 (tan 𝐵 + tan 𝐶) 1 − tan 𝐵 tan 𝐶 =

tan 𝐴 + tan 𝐵 + tan 𝐶 − tan 𝐴 tan 𝐵 tan 𝐶 1 − tan 𝐴 tan 𝐵 − tan 𝐴 tan 𝐶 − tan 𝐵 tan 𝐶 Proven

Question 4

a.

i.

Given sin 𝜃 = 𝑥, sin2 𝜃 = 𝑥 2 1 − sin2 𝜃 = cos 2 𝜃 cos 𝜃 = √1 − 𝑥 2 tan 𝜃 = =

sin 𝜃 cos 𝜃 𝑥 √1 − 𝑥 2

𝐒𝐡𝐨𝐰𝐧

6

Given 𝑦 = tan 2𝑡 and 𝑥 = sin 𝑡

ii.

𝑦=

2 tan 𝑡 𝑥 , and tan 𝑡 = 2 1 − tan 𝑡 √1 − 𝑥 2

𝑥 2( ) 2 √1 − 𝑥 𝑦= 2 𝑥 1−( ) √1 − 𝑥 2 2𝑥 2 = √1 − 𝑥2 𝑥 1− 1 − 𝑥2 2𝑥 𝑥2 = √1 − 1 − 𝑥2 − 𝑥2 1 − 𝑥2 =

2𝑥(1 − 𝑥 2 ) √1 − 𝑥 2 (1 − 2𝑥 2 )

2𝑥√1 − 𝑥 2 𝑦= 1 − 2𝑥 2 b.

1 2 Given 𝐮 = (−3) and 𝐯 = (1) 2 5

i.

|𝐮| = √1 + 9 + 4 = √14 |𝐯| = √4 + 1 + 25 = √30 𝐮∙𝐯

cos 𝜃 = |𝐮||𝐯|

ii.

= c.

2 − 3 + 10 √14 × √30

= 0.439

At any time the point 𝑃(𝑥, 𝑦) is 2𝑎 from the origin and a from the x-axis. Therefore its distance from the y-axis is given by √4𝑎2 − 𝑎2 = 𝑎√3 using Pythagoras Theorem. Hence, 𝑥 = 𝑎√3 and 𝑦 = 𝑎. 𝑥 = 𝑦√3, 𝑦=

𝑥 √3 7


d.

2𝑥 + 𝑦 + 3 = 0

------

𝑥2 + 𝑦2 = 9

(1)

-------

From (1), 𝑦 = −(2𝑥 + 3)

(2) -----

(3)

Substituting (3) into (2) we have 2

𝑥 2 + (−(2𝑥 + 3)) = 9 𝑥 2 + 4𝑥 2 + 12𝑥 + 9 = 9 5𝑥 2 + 12𝑥 = 0 𝑥(5𝑥 + 12) = 0 𝑥 = 0, −

12 5

When 𝑥 = 0, 𝑦 = −3 When 𝑥 = −

12 5

12

9

, 𝑦 = −2 (− 5 ) − 3 = 5

Therefore the points of intersection are (0, −3) and (−

12 9 5

, 5)

Question 5

a.

Given ∫(𝑥 + 1)1/3 𝑑𝑥, Using the substitution 𝑢 = 𝑥 + 1 we have 𝑑𝑢 = 1, 𝑑𝑥

𝑑𝑢 = 𝑑𝑥.

∫(𝑥 + 1)1/3 𝑑𝑥 = ∫(𝑢)1/3 𝑑𝑢 =

(𝑢)4/3 +𝐶 4 3

=

3 4/3 𝑢 +𝐶 4

=

3 (𝑥 + 1)4/3 + 𝐶 4 8

b.

0

V = 𝜋 ∫−1 𝑥 2 𝑑𝑦 𝑦 =3 − 1, 𝑥 = (𝑦 + 1)1/3 𝑥 2 = (𝑦 + 1)2/3 0

𝑉 = 𝜋 ∫(𝑦 + 1)2/3 𝑑𝑦 −1

3 0 = 𝜋 [ (𝑦 + 1)5/3 ] −1 5

c.

𝑎

=

3𝜋 cubic units 5

𝑎

Given ∫0 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑓(𝑎 − 𝑥) 𝑑𝑥 𝑎 > 0 1

1

𝑒𝑥 𝑒 1−𝑥 ∫ 𝑥 𝑑𝑥 = ∫ 𝑑𝑥 𝑒 + 𝑒 1−𝑥 𝑒 1−𝑥 + 𝑒 (1−(1−𝑥)) 0

0

1

𝑒 1−𝑥 = ∫ 1−𝑥 𝑑𝑥 𝑒 + 𝑒𝑥 0

Dividing both numerator and denominator by 𝑒 𝑥 we have 1

𝑒 1−2𝑥 = ∫ 1−2𝑥 𝑑𝑥 𝑒 +1 0

1 1 = [− ln(𝑒 1−2𝑥 + 1)] 0 2 1 = − [(ln(𝑒 −1 + 1)) − (ln(𝑒 + 1))] 2 1 𝑒+1 = − [ln | | − ln|𝑒 + 1|] 2 𝑒 1 = − [ln|𝑒 + 1| − ln 𝑒 − ln|𝑒 + 1|] 2 1 1 = − (−1) = 2 2

9


d.

i.

Bacteria grow exponentially at a rate of 2% per hour where 𝑦 = 𝑓(𝑡) is the number of bacteria present t hours later is given by the differential equation 𝑑𝑦 = 0.02𝑦 𝑑𝑡 Separating variables and integrating both sides we have ∫

𝑑𝑦 = ∫ 0.02 𝑑𝑡 𝑦

ln 𝑦 = 0.02𝑡 + 𝐶 𝑦 = 𝑒 0.02𝑡+𝐶 𝑦 = 𝑒 𝐶 𝑒 0.02𝑡 When 𝑡 = 0, 𝑦 = 1000, therefore 1000 = 𝑒 𝐶 , 𝑦 = 1000𝑒 0.02𝑡 ii.

When the bacteria population is double 𝑦 = 2000, 2000 = 1000𝑒 0.02𝑡 2 = 𝑒 0.02𝑡 ln 2 = 0.02𝑡, 𝑡=

ln 2 = 34.66 hrs 0.02

Question 6 a.

Given 𝑓(𝑥) = 2𝑥 3 + 5𝑥 2 − 𝑥 + 12 𝑓 ′ (𝑥) = 6𝑥 2 + 10𝑥 − 1 The gradient of the tangent at the point where 𝑥 = 3, is given by 𝑓 ′ (3) = 6(3)2 + 10(3) − 1 = 83 When 𝑥 = 3, 𝑓(3) = 2(3)3 + 5(3)2 − (3) + 12 = 112 Therefore the equation of the tangent at the point where 𝑥 = 3 is given by 𝑦 − 112 = 83(𝑥 − 3) 𝑦 = 83𝑥 − 137 10

2

b.

i.

Given 𝑓(𝑥) = {𝑥 + 2𝑥 + 3 𝑎𝑥 + 𝑏

𝑥≤0 𝑥>0

lim 𝑓(𝑥) = 02 + 2(0) + 3 = 3

𝑥→0−

lim 𝑓(𝑥) = 𝑎(0) + 𝑏 = 𝑏

𝑥→0+

ii.

For 𝑓(𝑥) to be continuous at 𝑥 = 0, lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) 𝑥→0

𝑥→0

Therefore 𝑏 = 3, and 𝑎 𝜖 𝐑.

iii.

Given 𝑓 ′ (0) = lim

𝑓(0+𝑡)−𝑓(0)

𝑡→0

𝑡

For 𝑥 ≤ 0 𝑎(0 + 𝑡) + 3 − (𝑎(0) + 3) 𝑡→0 𝑡

= lim

𝑎𝑡 + 3 − 3 𝑡→0 𝑡 𝑎𝑡 = lim = 𝑎 𝑡→0 𝑡 = lim

For 𝑥 > 0 (0 + 𝑡)2 + 2(0 + 𝑡) + 3 − (02 + 2(0) + 3) 𝑡→0 𝑡

𝑓 ′ (0) = lim

(𝑡)2 + 2(𝑡) + 3 − 3 = lim 𝑡→0 𝑡 𝑡 2 + 2𝑡 𝑡→0 𝑡

= lim

= lim 𝑡 + 2 = 2 𝑡→0

If the 𝑓(𝑥) is differentiable at 𝑥 = 0 then 𝑓′(0) = 2 Therefore 𝑎 = 2

11


c.

Given 𝑓(𝑥) = √𝑥 therefore 𝑓(𝑥 + ℎ) = √𝑥 + ℎ √𝑥 + ℎ − √𝑥 √𝑥 + ℎ + √𝑥 × ℎ→0 ℎ √𝑥 + ℎ + √𝑥

𝑓 ′ (𝑥) = lim

= lim

(𝑥 + ℎ) − 𝑥

ℎ→0 ℎ(√𝑥

= lim

+ ℎ + √𝑥) ℎ

ℎ→0 ℎ(√𝑥

= lim

ℎ→0 (√𝑥

=

+ ℎ + √𝑥) 1

+ ℎ + √𝑥)

1 2√2

12

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2015 EXAM

1.

a.

i.

The inverse ~𝐩 → ~𝐪 and the contrapositive ~𝐪 → ~𝐩

ii.

iii.

p

q

~𝐩

~𝐪

𝐩→𝐪

~𝐪 → ~𝐩

T

T

F

F

T

T

T

F

F

T

F

F

F

T

T

F

T

T

F

F

T

T

T

T

𝐩 → 𝐪 and ~𝐪 → ~𝐩 are logically equivalent because both final columns are the exactly same.

b.

Given 𝑓(𝑥) = 𝑥 3 + 𝑝𝑥 2 − 𝑥 + 𝑞 i.

If (𝑥 − 5) is a factor then 𝑓(5) = 53 + 𝑝(52 ) − (5) + 𝑞 = 0 125 + 25𝑝 − 5 + 𝑞 = 0 25𝑝 + 𝑞 = −120 … … …. (1) When divided by (𝑥 − 1) the remainder is 24 therefore 𝑓(1) = 13 + 𝑝(12 ) − (1) + 𝑞 = 24 1 + 𝑝 − 1 + 𝑞 = 24 𝑝 + 𝑞 = 24 … … …. (2) Subtract (2) from (1) we have 24𝑝 = −144, 𝑝 = −6 −6 + 𝑞 = 24, 𝑞 = 30

13


ii.

𝑓(𝑥) = 𝑥 3 − 6𝑥 2 − 𝑥 + 30 𝑥2 − 𝑥 − 6 𝑥−5

𝑥 3 − 6𝑥 2 − 𝑥 + 30 𝑥 3 − 5𝑥 2 −𝑥 2 − 𝑥 −𝑥 2 + 5𝑥 −6𝑥 + 30 −6𝑥 + 30 0

𝑥 3 − 6𝑥 2 − 𝑥 + 30 = (𝑥 − 5)(𝑥 2 − 𝑥 − 6) = (𝑥 − 5)(𝑥 − 3)(𝑥 + 2) c.

Given 𝑆(𝑛) = 5 + 52 + 53 + 54 + ⋯ + 5𝑛 (𝐿. 𝐻. 𝑆) and 4𝑆(𝑛) = 5𝑛+1 − 5 (𝑅. 𝐻. 𝑆) When 𝑛 = 1, 𝐿. 𝐻. 𝑆 = 4𝑆(1) = 4 × 5 = 20, and R. H. S = 52 − 5 = 20 Therefore result is true for 𝑛 = 1. Assume result is true for 𝑛 = 𝑘, therefore 4𝑆(𝑘) = 4(5 + 52 + 53 + 54 + ⋯ + 5𝑘 ) = 5𝑘+1 − 5 When 𝑛 = 𝑘 + 1, we have R.H.S = 4𝑆(𝑘 + 1) = 5𝑘+2 − 5 L.H.S = 4(5 + 52 + 53 + 54 + ⋯ + 5𝑘 + 5𝑘+1 ) = 4(𝑆(𝑘) + 5𝑘+1 ) = 4(𝑆(𝑘)) + 4(5𝑘+1 ) = 5𝑘+1 − 5 + 4(5𝑘+1 ) = 5 × 5𝑘+1 − 5 = 5𝑘+2 − 5 R.H.S = L.H.S therefore result is true for 𝑛 = 𝑘 + 1. Since the result is true for 𝑛 = 1, 𝑘 and 𝑘 + 1, it is true for all positive integer n. 14

2.

a.

i.

A function is one-to-one if each element in the domain maps to one and only one image in the co-domain and each element in the range is the image of only one element in the domain. Therefore given that 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 are one-to-one functions, (𝑔 ° 𝑓) is a one-to-one function because the co-domain of f is used as the domain for g and this makes (𝑔 ° 𝑓) = 𝑔: 𝐵 → 𝐶 a one-to-one function. 𝑓(𝐴)

𝑎1 𝑎2 b 𝑎3

ii.

𝑔(𝐵)

𝑏1 𝑏2 𝑏3

𝑔°𝑓(𝐶)

𝑐1 𝑐2 𝑐3

A function is onto if each element in the co-domain is mapped unto at least one element in the domain. Therefore given that 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶 are onto functions, (𝑔 ° 𝑓) is a onto function because the co-domain of f is used as the domain for g, and this makes (𝑔 ° 𝑓) = 𝑔: 𝐵 → 𝐶 an onto function.

b.

i.

4

4

3 − (9)𝑥 − (81)𝑥 = 0 3−

4 4 − 2𝑥 = 0 𝑥 9 9

3(92𝑥 ) − 4(9𝑥 ) − 4 = 0 multiplying both sides by 92𝑥 (3(9𝑥 ) + 2)(9𝑥 − 2) = 0 (3(9𝑥 ) + 2) = 0, 2 9𝑥 = − , 3

not possible

(9𝑥 − 2) = 0 9𝑥 = 2 𝑥=

log 2 = 0.315 log 9

15

factorising the equation


ii.

|5𝑥 − 6| = 𝑥 + 5 when 5𝑥 − 6 > 0 we have 5𝑥 − 6 = 𝑥 + 5 4𝑥 = 11 𝑥=

11 4

When 5𝑥 − 6 < 0 we have −(5𝑥 − 6) = 𝑥 + 5 −5𝑥 + 6 = 𝑥 + 5 1 = 6𝑥 1

𝑥=6 c.

Given 𝑁 = 300 + 5𝑡 i.

When 𝑡 = 0, 𝑁 = 300 + 1 = 301

ii.

When 𝑁 = 3(301) we have 903 = 300 + 5𝑡 5𝑡 = 603 𝑡=

3.

a.

i.

log 603 = 3.98 hours log 5

cos 3𝑥 = cos(2𝑥 + 𝑥) = cos 2𝑥 cos 𝑥 − sin 2𝑥 sin 𝑥 = (2 cos2 𝑥 − 1) cos 𝑥 − 2 sin 𝑥 cos 𝑥 sin 𝑥 = 2 cos3 𝑥 − cos 𝑥 − 2 sin2 𝑥 cos 𝑥 = 2 cos3 𝑥 − cos 𝑥 − 2(1 − cos2 𝑥) cos 𝑥 = 2 cos3 𝑥 − cos 𝑥 − 2 cos 𝑥 + 2 cos3 𝑥 = 4 cos3 𝑥 − 3 cos 𝑥

16

ii.

cos 6𝑥 − cos 2𝑥 = 0 cos 6𝑥 = 4 cos3 2𝑥 − 3 cos 2𝑥 4 cos 3 2𝑥 − 3 cos 2𝑥 − cos 2𝑥 = 0 4 cos 3 2𝑥 − 4 cos 2𝑥 = 0 4 cos 2𝑥 (cos2 2𝑥 − 1) = 0 4 cos 2𝑥 = 0 2𝑥 = 𝑥=

𝜋 3𝜋 5𝜋 7𝜋 , , , 2 2 2 2

𝜋 3𝜋 5𝜋 7𝜋 , , , 4 4 4 4

cos 2 2𝑥 − 1 = 0 cos 2 2𝑥 = 1 cos 2𝑥 = ±1 2𝑥 = 0, 𝜋 2𝜋, 3𝜋 4𝜋 𝜋 3𝜋 𝑥 = 0, , 𝜋, , 2𝜋 2 2 b.

i.

𝑓(2𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃 𝑟 sin(2𝜃 + 𝛼) = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃 𝑟 cos 𝛼 = 3 and 𝑟 sin 𝛼 = 4 4

tan 𝛼 = 3 , and 𝑟 = √32 + 42 4

𝛼 = tan−1 (3) = 0.927 and 𝑟 = 5 3 sin 2𝜃 + 4 cos 2𝜃 = 5 sin(2𝜃 + 0.927) ii.

Maximum value of occurs when 𝑓(𝜃) = 5 1 1 = 7 − 𝑓(𝜃) 7 − 5 =

1 2

Minimum value of occurs when 𝑓(𝜃) = −5 1 1 1 = = 7 − 𝑓(𝜃) 7 + 5 12 17


4.

a.

i.

Given 𝐶1 : 𝑥 = √10 cos 𝜃 − 3; 𝐶2 : 𝑥 = 4 cos 𝜃 + 3; From 𝐶1 : cos 𝜃 =

𝑦 = √10 sin 𝜃 + 2

𝑦 = 4 sin 𝜃 + 2 𝑥+3

;

√10

sin 𝜃 =

𝑦−2 √10

cos2 𝜃 + sin2 𝜃 = 1 𝑥+3 2 𝑦−2 2 ( ) +( ) =1 √10 √10 (𝑥 + 3)2 + (𝑦 − 2)2 = (√10) From 𝐶2 :

cos 𝜃 =

𝑥−3 4

;

2

sin 𝜃 =

𝑦−2 4

𝑥−3 2 𝑦−2 2 ( ) +( ) =1 4 4 (𝑥 − 3)2 + (𝑦 − 2)2 = 42 2

(𝑥 + 3)2 + (𝑦 − 2)2 = (√10) ……….

ii. From (1)

(1)

2

(𝑦 − 2)2 = (√10) − (𝑥 + 3)2 ……….. (𝑥 − 3)2 + (𝑦 − 2)2 = 42

(2)

…….. (3)

(𝑦 − 2)2 = 42 − (𝑥 − 3)2 …………..

(4)

Equating (2) and (4) we have 2

(√10) − (𝑥 + 3)2 = 42 − (𝑥 − 3)2 10 − (𝑥 2 + 6𝑥 + 9) = 16 − (𝑥 2 − 6𝑥 + 9) 10 − 𝑥 2 − 6𝑥 − 9 = 16 − 𝑥 2 + 6𝑥 − 9 10 − 16 = 6𝑥 + 6𝑥 −6 = 12𝑥 𝑥=−

1 2

18

1

Substituting 𝑥 = − 2 into (4) we have 1 (𝑦 − 2)2 = 42 − ((− ) − 3) 2 𝑦 2 − 4𝑦 + 4 = 16 − 𝑦 2 − 4𝑦 +

2

49 4

1 =0 4

4𝑦 2 − 16𝑦 + 1 = 0 𝑦=

16 ± √162 − 16 8

𝑦=

16 ± 4√15 8

𝑦 = 3.94,

0.0635 1

1

Points of intersection are (− 2 , 3.94) and (− 2 , 0.0635 )

b.

If the point 𝑃(𝑥, 𝑦) moves so that its distance from a fixed point (0, 3) is two times the distance from the fixed point (5, 2) then; 𝑥 2 + (𝑦 − 3)2 = 4[(𝑥 − 5)2 + (𝑦 − 2)2 ] 𝑥 2 + 𝑦 2 − 6𝑦 + 9 = 4[𝑥 2 − 10𝑥 + 25 + 𝑦 2 − 4𝑦 + 4] 𝑥 2 + 𝑦 2 − 6𝑦 + 9 = 4𝑥 2 − 40𝑥 + 100 + 4𝑦 2 − 16𝑦 + 16 3𝑥 2 + 3𝑦 2 − 40𝑥 − 10𝑦 + 107 = 0 𝑥2 + 𝑦2 −

40 10 107 𝑥− 𝑦+ =0 3 3 3

20 2 400 5 2 25 107 (𝑥 − ) − + (𝑦 − ) − + =0 3 9 3 9 3 20 2 5 2 104 (𝑥 − ) + (𝑦 − ) = 3 3 9 20 5

This is the equation of a circle with centre ( 3 , 3) and radius

19

√104 . 3


5.

a.

Given 𝑓(𝑥) = {

sin(𝑎𝑥) if 𝑥 ≠ 0, 𝑎 ≠ 0 𝑥 4 if 𝑥 = 0

If f is continuous at 𝑥 = 0, then sin(𝑎𝑥) = lim 4 𝑥→0 𝑥→0 𝑥 lim

sin(𝑎𝑥) =4 𝑥→0 𝑥 lim

Multiplying numerator and denominator by a we have 𝑎 sin(𝑎𝑥) =4 𝑥→0 𝑎𝑥 lim

sin(𝑎𝑥) =4 𝑥→0 𝑎𝑥

𝑎 lim

sin(𝑎𝑥) =1 𝑥→0 𝑎𝑥 lim

𝑎=4 b.

Given 𝑓(𝑥) = sin(2𝑥) Differentiating from first principles we have 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→0 ℎ

𝑓 ′ (𝑥) = lim

sin 2(𝑥 + ℎ) − sin(2𝑥) ℎ→0 ℎ

= lim

= lim

2𝑥 + 2ℎ + 2𝑥 2𝑥 + 2ℎ − 2𝑥 ) sin ( ) 2 2 ℎ

2 cos (

ℎ→0

= lim

4𝑥 + 2ℎ 2 ) sin(ℎ) ℎ

2 cos (

ℎ→0

sin(ℎ) ℎ→0 ℎ

= lim 2 cos(2𝑥 + ℎ) × lim ℎ→0

= 2 cos 2𝑥

20

c.

Given 𝑦 = i.

2𝑥 √1 + 𝑥 2

Using the quotient and power rule: 𝑑𝑢 𝑑𝑣 𝑑𝑦 𝑣 𝑑𝑥 − 𝑢 𝑑𝑥 𝑑𝑦 = and = 𝑛(𝑓(𝑥))𝑛−1 × 𝑓′(𝑥) 2 𝑑𝑥 𝑣 𝑑𝑥 𝑑𝑦 = 𝑑𝑥

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

(√1 + 𝑥 2 )

2

2𝑥 2 √1 + 𝑥 2 1 + 𝑥2

2√1 + 𝑥 2 − = =

2(1 + 𝑥 2 ) − 2𝑥 2 (1 + 𝑥 2 )√1 + 𝑥 2

𝑑𝑦 2 2 = = 𝑑𝑥 (1 + 𝑥 2 )√1 + 𝑥 2 (1 + 𝑥 2 )3/2 Multiplying both sides by x we have 𝑥

𝑑𝑦 2𝑥 = 𝑑𝑥 (1 + 𝑥 2 )√1 + 𝑥 2 1 2𝑥 =( ) 2 1 + 𝑥 √1 + 𝑥 2

𝑥 ii.

𝑑𝑦 𝑦 = 𝑑𝑥 1 + 𝑥 2

From (i) 𝑑𝑦 2 = = 2(1 + 𝑥 2 )−3/2 𝑑𝑥 (1 + 𝑥 2 )3/2 𝑑2 𝑦 3 = (2) (− ) (1 + 𝑥 2 )−5/2 (2𝑥) 2 𝑑𝑥 2 =− =− =−

6𝑥 (1 + 𝑥 2 )5/2 6𝑥

(1 + 𝑥 2 )2 √1 + 𝑥 2 3(2𝑥)

(1 + 𝑥 2 )2 √1 + 𝑥 2 21


Since 𝑦 =

2𝑥 √1 + 𝑥 2

𝑑2 𝑦 3𝑦 = − (1 + 𝑥 2 )2 𝑑𝑥 2 𝑑2 𝑦 3𝑦 + =0 𝑑𝑥 2 (1 + 𝑥 2 )2 6.

Given 𝑦 = 3𝑥 − 7, 𝑦 + 𝑥 = 9 and 3𝑦 = 𝑥 + 3 i.

AB is the line 𝑦 = 3𝑥 − 7, AC is the line 𝑦 + 𝑥 = 9 and BC is the line 3𝑦 = 𝑥 + 3. The lines AB intersects AC at the point A therefore the coordinates of A is found by solving these equations simultaneously. 𝑦 = 3𝑥 − 7 ……. (1) 𝑦+𝑥 =9

……. (2)

Substituting (1) into (2) we have 3𝑥 − 7 + 𝑥 = 9 4𝑥 = 16 𝑥=4 From (2) 𝑦 + 4 = 9 𝑦=5 Therefore the coordinates of A is (4, 5) AB intersects BC at the point B therefore solving these equations gives the coordinates of the point B. 𝑦 = 3𝑥 − 7 …… (1) 3𝑦 = 𝑥 + 3 …… (2) Substituting (1) into (2) we have 3(3𝑥 − 7) = 𝑥 + 3 9𝑥 − 21 = 𝑥 + 3 8𝑥 = 24 𝑥=3 22

From (1) 𝑦 = 3(3) − 7 = 2 Therefore the coordinates of B is (3, 2) AC intersects BC at the point C therefore solving these equations gives the coordinates of the point C. 𝑦 + 𝑥 = 9 ……

(1)

3𝑦 = 𝑥 + 3 …… (2) From (1) 𝑦 = 9 − 𝑥 …. (3) Substituting (3) into (2) we have 3(9 − 𝑥) = 𝑥 + 3 27 − 3𝑥 = 𝑥 + 3 4𝑥 = 24 𝑥=6 𝑦 = 9−6 =3 Therefore the coordinates of C is (6, 3) ii.

The area bounded by these three lines is given by 6

4

6

∫ 3𝑥 − 7 𝑑𝑥 + ∫ 9 − 𝑥 𝑑𝑥 − ∫ 3

=[

4

3

𝑥+3 𝑑𝑥 3

3𝑥 2 𝑥2 6 1 𝑥2 4 6 − 7𝑥] + [9𝑥 − ] − [ + 3𝑥] 3 3 2 2 4 3 2

1 = [(−4) − (−7.5)] + [(36) − (28)] − [(36) − (13.5)] 3 = 4 sq. units b.

Given that 𝑓 ′ (𝑥) = 3𝑥 2 + 8𝑥 − 3 at the point (0, −6) i.

𝑓(𝑥) = ∫ 3𝑥 2 + 8𝑥 − 3 𝑑𝑥 = 𝑥 3 + 4𝑥 2 − 3𝑥 + 𝐶 The curve passes through the point (0, −6) therefore −6 = 𝐶 The equation of the curve is therefore 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 3𝑥 − 6 23


ii.

At the stationary point 𝑓 ′ (𝑥) = 0 therefore 3𝑥 2 + 8𝑥 − 3 = 0 (3𝑥 − 1)(𝑥 + 3) = 0 1

𝑥 = 3 , and − 3 1 3

1

1 2

1

When 𝑥 = 3 , 𝑦 = (3) + 4 (3) − 3 (3) − 6 𝑦=−

176 27

= −6.5

When 𝑥 = −3, 𝑦 = (−3)3 + 4(−3)2 − 3(−3) − 6 𝑦 = 12 1

Therefore the stationary points are (3 , −6.5) and (−3, 12) 𝑓 ′′ (𝑥) = 6𝑥 + 8 1

1

1

When 𝑥 = 3 , 𝑓 ′′ (3) = 6 (3) + 8 > 0 1

Therefore (3 , −6.5) is a minimum point When 𝑥 = −3, 𝑓 ′′ (−3) = 6(−3) + 8 < 0 Therefore (−3, 12) is a maximum point. iii.

𝑦

Max (−3, 12)

12

𝑥 3 + 4𝑥 2 − 3𝑥 − 6

−3

−2

−1

0

𝑥 1 3

1

−6

1 Min ( , −6.5) 3

24

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2014 EXAM Question 1

a. p T T T T F F F F b.

i.

q T T F F T T F F

r T F T F T F T F

𝑝→𝑞 T T F F T T T T

(𝑝 → 𝑞) ∧ (𝑟 → 𝑞) T T F F T T F T

𝑟→𝑞 T T F T T T F T

Given 𝑦 ⊕ 𝑥 = 𝑦 3 + 𝑥 3 + 𝑎𝑦 2 + 𝑎𝑥 2 − 5𝑦 − 5𝑥 + 16 𝑥 ⊕ 𝑦 = 𝑥 3 + 𝑦 3 + 𝑎𝑥 2 + 𝑎𝑦 2 − 5𝑥 − 5𝑦 + 16 𝑦 ⊕ 𝑥 = 𝑥 ⊕ 𝑦 therefore ⊕ is commutative in R

ii.

a.

We have 𝑓(𝑥) = 2 ⊕ 𝑥 = 23 + 𝑥 3 + 𝑎22 + 𝑎𝑥 2 − 5(2) − 5𝑥 + 16 𝑓(𝑥) = 8 + 𝑥 3 + 4𝑎 + 𝑎𝑥 2 − 10 − 5𝑥 + 16 𝑓(𝑥) = 𝑥 3 + 𝑎𝑥 2 − 5𝑥 + 4𝑎 + 14 If (𝑥 − 1) is a factor then 𝑓(1) = 0 Therefore 𝑓(1) = 13 + 𝑎(1)2 − 5(1) + 4𝑎 + 14 = 0 1 + 𝑎 − 5 + 4𝑎 + 14 = 0 5𝑎 + 10 = 0, 𝑎 = −2

b.

When 𝑎 = −2, 𝑓(𝑥) = 𝑥 3 + (−2)𝑥 2 − 5𝑥 + 4(−2) + 14 𝑓(𝑥) = 𝑥 3 − 2𝑥 2 − 5𝑥 + 6 (𝑥 − 1) is a factor of 𝑓(𝑥) therefore using long division we have.

25


𝑥2 − 𝑥 − 6

𝑥 − 1 𝑥 3 − 2𝑥 2 − 5𝑥 + 6 𝑥3 − 𝑥2 −𝑥 2 − 5𝑥 −𝑥 2 + 𝑥 −6𝑥 + 6 −6𝑥 + 6 0 3

2

2

𝑥 − 2𝑥 − 5𝑥 + 6 = (𝑥 − 1)(𝑥 − 𝑥 − 6) = (𝑥 − 1)(𝑥 + 2)(𝑥 − 3) Therefore factors are (𝑥 − 1), (𝑥 + 2) and (𝑥 − 3) c.

𝑛

12 + 32 + 52 + ⋯ + (2𝑛 − 1)2 = 3 (4𝑛2 − 1) When 𝑛 = 1,

1

L.H.S = 12 = 1, and R.H.S = 3 (4(1)2 − 1) = 1

L.H.S = R.H.S therefore result is true for 𝑛 = 1 Assume result is true for 𝑛 = 𝑘, therefore 𝑘

12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 = 3 (4𝑘 2 − 1) When 𝑛 = 𝑘 + 1 R.H.S =

𝑘+1 3

(4[𝑘 + 1]2 − 1)

L.H.S = 12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 + (2(𝑘 + 1) − 1)2 𝑘

12 + 32 + 52 + ⋯ + (2𝑘 − 1)2 = 3 (4𝑘 2 − 1) 𝑘

L.H.S = 3 (4𝑘 2 − 1) + (2(𝑘 + 1) − 1)2 𝑘

= 3 (2𝑘 − 1)(2𝑘 + 1) + (2𝑘 + 1)2 𝑘

= (2𝑘 + 1) [3 (2𝑘 − 1) + (2𝑘 + 1)] = (2𝑘 + 1) [

𝑘(2𝑘 − 1) + 3(2𝑘 + 1) ] 3 26

L. H. S =

2𝑘 + 1 (2𝑘 2 + 5𝑘 + 3) 3

=

2𝑘 + 1 (2𝑘 + 3)(𝑘 + 1) 3

=

𝑘+1 (2𝑘 + 1)(2𝑘 + 3) 3

=

𝑘+1 (4𝑘 2 + 8𝑘 + 3) 3

=

𝑘+1 (4{𝑘 2 + 2𝑘} + 3) 3

=

𝑘+1 (4[𝑘 + 1]2 − 4 + 3) 3

=

𝑘+1 (4[𝑘 + 1]2 − 1) 3

L.H.S = R.H.S therefore result is true for 𝑛 = 𝑘 + 1 Since result is true for 𝑛 = 1, 𝑘, and 𝑘 + 1 it is true for all positive integer n.

Question 2 a.

𝑥−1

Given 𝑓(𝑥) = 2𝑥 2 + 1, 𝑔(𝑥) = √ i.

a.

2

𝑓𝑓(𝑥) = 2(𝑓(𝑥))2 + 1 = 2(2𝑥 2 + 1)2 + 1 = 2(4𝑥 4 + 4𝑥 2 + 1) + 1 = 8𝑥 4 + 8𝑥 2 + 2 + 1 = 8𝑥 4 + 8𝑥 2 + 3

b.

𝑓[𝑔(𝑥)] = 2[𝑔(𝑥)]2 + 1 2 𝑥−1

= 2 (√

2

) +1

𝑥−1

= 2(

2

)+1

=𝑥−1+1 =𝑥 ii.

𝑓 −1 (𝑥) = 𝑔(𝑥) Since 𝑓[𝑓 −1 (𝑥)] = 𝑥 27


b.

Given that 𝑎3 + 𝑏 3 + 3𝑎2 𝑏 = 5𝑎𝑏 2 We know that (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 3 Adding 3𝑎𝑏 2 to both sides we have 𝑎3 + 𝑏 3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 = 3𝑎𝑏 2 + 5𝑎𝑏 2 (𝑎 + 𝑏)3 = 8𝑎𝑏 2 (𝑎 + 𝑏)3 = 𝑎𝑏 2 8 𝑎+𝑏 3 ( ) = 𝑎𝑏 2 2 𝑎+𝑏 3 ) 2

log (

c.

i.

= log 𝑎𝑏 2

[Log both sides]

3 log (

𝑎+𝑏 ) = log 𝑎 + log 𝑏 2 2

3 log (

𝑎+𝑏 ) = log 𝑎 + 2 log 𝑏 2

𝑒𝑥 +

1 𝑒𝑥

−2=0

𝑒 2𝑥 − 2𝑒 𝑥 + 1 = 0

[Multiplying both sides by 𝑒 𝑥 ]

Let 𝑢 = 𝑒 𝑥 , 𝑢2 − 2𝑢 + 1 = (𝑢 − 1)2 𝑢 = 1, 𝑒 𝑥 = 1 , 𝑥 = 0 ii.

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

𝑥+1 = 2 log 2 2 3𝑥 + 1

log 2

𝑥+1 = log 2 22 3𝑥 + 1

𝑥+1 =4 3𝑥 + 1 𝑥 + 1 = 4(3𝑥 + 1) 𝑥 + 1 = 12𝑥 + 4 11𝑥 = −3 𝑥=−

3 11 28

d.

√3−1 √3+1

√3+1 √3−1

+

+

√2−1 √2+1

2

+

2

(√3 − 1) + (√3 + 1) (√3 + 1)(√3 − 1)

√2+1 √2−1 2

+

2

(√2 − 1) + (√2 + 1) (√2 + 1)(√2 − 1)

4 − 2√3 + 4 + 2√3 3 − 2√2 + 3 + 2√2 + 3−1 2−1 8 6 + = 4 + 6 = 10 2 1

Question 3 a.

i.

ii.

cot 𝑦−cot 𝑥 cot 𝑥+cot 𝑦

cot 𝑦−cot 𝑥 cot 𝑥+cot 𝑦

=

cos 𝑦 ⁄sin 𝑦−cos 𝑥⁄sin 𝑥 cos 𝑦 cos 𝑥⁄ ⁄sin 𝑦 sin 𝑥 +

=

sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦 / sin 𝑦 sin 𝑥 sin 𝑥 sin 𝑦

=

sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 sin 𝑥 sin 𝑦 × sin 𝑦 sin 𝑥 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦

=

sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦

=

sin(𝑥 − 𝑦) sin(𝑥 + 𝑦)

= 1, 0 ≤ 𝑦 ≤ 2𝜋, 1

1

𝜋

𝜋

When sin 𝑥 = 2 , sin−1 (2) = 6 , cos ( 6 ) =

√3 2

for 0 ≤ 𝑥 ≤

𝜋 2

cot 𝑦 − cot 𝑥 sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 = cot 𝑥 + cot 𝑦 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦 sin 𝑥 cos 𝑦 − sin 𝑦 cos 𝑥 =1 cos 𝑥 sin 𝑦 + sin 𝑥 cos 𝑦 1 2

√3 ) 2

( ) cos 𝑦−sin 𝑦(

1 √3 ) sin 𝑦+( ) cos 𝑦 2 2

(

cos 𝑦 − √3 sin 𝑦 √3 sin 𝑦 + cos 𝑦

= 1 Multiplying both numerator & denominator by 2

=1

cos 𝑦 − √3 sin 𝑦 = √3 sin 𝑦 + cos 𝑦 29


0 = 2√3 sin 𝑦 sin 𝑦 = 0 𝑦 = 0, 𝜋, 2𝜋

b.

i.

for 0 ≤ 𝑦 ≤ 2𝜋.

Given 𝑓(𝜃) = 3 sin 2𝜃 + 4 cos 2𝜃 to be written in the form 𝑟 sin(2𝜃 + 𝛼) We have 𝑟 sin(2𝜃 + 𝛼 ) = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃 𝑟 = √32 + 42 = 5 3 sin 2𝜃 + 4 cos 2𝜃 = 𝑟 sin 2𝜃 cos 𝛼 + 𝑟 sin 𝛼 cos 2𝜃 Comparing we have 𝑟 cos 𝛼 = 3, 𝑟 sin 𝛼 = 4 𝑟 sin 𝛼 4 = tan 𝛼 = 𝑟 cos 𝛼 3 4

𝛼 = tan−1 (3) = 0.927 rad 𝑓(𝜃) = 𝑟 sin(2𝜃 + 𝛼) = 5 sin(2𝜃 + 0.927) ii.

a.

𝑓(𝜃) is at a minimum when (2𝜃 + 0.927) = 2𝜃 = 𝜃=

3𝜋 2

3𝜋 − 0.927 2

3𝜋 0.927 − 4 2

𝜃 = 1.89 rad b.

The maximum value of

1 7−𝑓(𝜃)

is when 𝑓(𝜃) = 5 so

1 1 1 = = 7 − 𝑓(𝜃) 7 − 5 2

And the minimum value of 1 1 1 = = 7 − 𝑓(𝜃) 7 − (−5) 12

30

1 7−𝑓(𝜃)

is when 𝑓(𝜃) = −5

Question 4

a.

Given the equations of 𝐿1 and 𝐿2 are 𝑥 − 𝑦 + 1 = 0 and 𝑥 + 𝑦 − 5 = 0 i.

𝐿1 and 𝐿2 intersects at the centre of the circle therefore Solving the equations simultaneously we have 𝑥 − 𝑦 + 1 = 0 ------ (1) 𝑥 + 𝑦 − 5 = 0 ------ (2) 2𝑥 − 4 = 0

Adding (1) and (2)

𝑥=2 When 𝑥 = 2, from (2) 𝑦 = 5 − 𝑥 𝑦 = 5−2 =3 Therefore the coordinate of the centre of the circle is (2, 3)

ii.

Let A (1, 2) and B (a, b) be the coordinates of the endpoints of the diameter of the circle and the coordinates of the it’s centre (2, 3) is the midpoint of the line AB. Therefore in calculating the midpoint we have 𝑎+1 = 2, 2

𝑎=3

𝑏+2 =3 𝑏=4 2 Therefore B has coordinates (3, 4)

iii.

The point p moves in a circular path with centre (2, 3) and radius √2. The equation of the path of p is given by (𝑥 − 2)2 + (𝑦 − 3)2 = (√2)

2

𝑥 2 − 4𝑥 + 4 + 𝑦 2 − 6𝑦 + 9 = 2 𝑥 2 + 𝑦 2 − 4𝑥 − 6𝑦 + 11 = 0

31


b.

1

𝑡

Given 𝑥 = 1+𝑡, and 𝑦 = 1−𝑡 2 𝑥=

1 1+𝑡

𝑥(1 + 𝑡) = 1 𝑥 + 𝑥𝑡 = 1 𝑥𝑡 = 1 − 𝑥 𝑡=

1−𝑥 𝑥

𝑦=

𝑡 𝑡 = 2 (1 + 𝑡)(1 − 𝑡) 1−𝑡

𝑦=

1 𝑡 1 × , substituting 𝑥 = we have 1+𝑡 1−𝑡 1+𝑡

𝑦=

𝑥𝑡 , 1−𝑡

𝑦=

c.

i.

substituting 𝑡 =

1−𝑥 we have 𝑥

1−𝑥 𝑥( 𝑥 ) 1−𝑥 1−( 𝑥 )

𝑦=

1−𝑥 𝑥 − (1 − 𝑥) 𝑥

𝑦=

𝑥(1 − 𝑥) 2𝑥 − 1

Given 𝑃(3, −2, 1), 𝑄(−1, 𝜆, 5) and 𝑅(2, 1, −4) ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ 𝑃𝑄 = ⃗⃗⃗⃗⃗ 𝑃𝑂 + 𝑂𝑄 3 −1 −4 ⃗⃗⃗⃗⃗ = − (−2) + ( 𝜆 ) = (2 + 𝜆) 𝑃𝑄 1 5 4 ⃗⃗⃗⃗⃗ 𝑃𝑄 = −4𝒊 + (2 + 𝜆)𝒋 + 4𝒌 ⃗⃗⃗⃗⃗ = 𝑄𝑂 ⃗⃗⃗⃗⃗⃗ + 𝑂𝑅 ⃗⃗⃗⃗⃗ 𝑄𝑅

32

−1 2 3 ⃗⃗⃗⃗⃗ = − ( 𝜆 ) + ( 1 ) = (1 − 𝜆) 𝑄𝑅 5 −4 −9 ⃗⃗⃗⃗⃗ = 3𝒊 + (1 − 𝜆)𝒋 − 9𝒌 𝑄𝑅 ⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 𝑅𝑃 = 𝑅𝑂 𝑂𝑃 2 3 1 ⃗⃗⃗⃗⃗ 𝑅𝑃 = − ( 1 ) + (−2) = (−3) −4 1 5 ⃗⃗⃗⃗⃗ = 𝒊 − 3𝒋 + 5𝒌 𝑅𝑃

ii.

Given PQ is the hypotenuse therefore RQ and RP are perpendicular to each other. (𝑅𝑄) ∙ (𝑅𝑃) = 0 −(3𝒊 + (1 − 𝜆)𝒋 − 9𝒌) ∙ (𝒊 − 3𝒋 + 5𝒌) = −3 − 3(𝜆 − 1) + 9(5) = 0 −3 − 3𝜆 + 3 + 45 = 0 −3𝜆 = −45, 𝜆=

Question 5

a.

Given i.

𝑓(𝑥) = {

𝑎𝑥 + 2, 𝑎𝑥 2 ,

𝑥<3 . 𝑥≥3

lim 𝑓(𝑥) = 𝑎(32 ) = 9𝑎

𝑥→3+

lim 𝑓(𝑥) = 3𝑎 + 2

𝑥→3−

If 𝑓(𝑥) is continuous at 𝑥 = 3, then lim 𝑓(𝑥) = lim− 𝑓(𝑥) Therefore

𝑥→3+

𝑥→3

9𝑎 = 3𝑎 + 2 6𝑎 = 2, 𝑎=

1 3

33

−45 = 15 −3


ii.

𝑔(𝑥) =

𝑥2 + 2 𝑏𝑥 2 + 𝑥 + 4

Given that lim 2𝑔(𝑥) = lim 𝑔(𝑥) we have 𝑥→1

2(

𝑥→0

12 + 2 02 + 2 = ) 𝑏(1)2 + 1 + 4 𝑏(0)2 + 0 + 4

2(

3 2 )= 𝑏+5 4 6 1 = 𝑏+5 2

12 = 𝑏 + 5 𝑏=7

b.

Let 𝑓(𝑥) =

1 √𝑥

,

𝑓(𝑥 + ℎ) =

1 √𝑥 + ℎ

Using differentiation from first principle, we have 𝑑𝑦 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = lim 𝑑𝑥 ℎ→0 ℎ 1 1 − 𝑑𝑦 + ℎ √𝑥 = lim √𝑥 𝑑𝑥 ℎ→0 ℎ √𝑥 − √𝑥 + ℎ (√𝑥 + ℎ)(√𝑥) = lim ℎ→0 ℎ = lim

√𝑥 − √𝑥 + ℎ

ℎ→0 ℎ(√𝑥

+ ℎ)(√𝑥)

√𝑥 − √𝑥 + ℎ √𝑥 + √𝑥 + ℎ = lim [ × ] ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥) √𝑥 + √𝑥 + ℎ 𝑥 − (𝑥 + ℎ) = lim [ ] ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ) −ℎ = lim [ ] ℎ→0 ℎ(√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ) 34

𝑑𝑦 −1 = lim [ ] 𝑑𝑥 ℎ→0 (√𝑥 + ℎ)(√𝑥)(√𝑥 + √𝑥 + ℎ) =

−1 (√𝑥)(√𝑥)(√𝑥 + √𝑥)

=−

ii.

1

1 = − 𝑥 −3/2 2 2𝑥√𝑥

Given 𝑦 =

𝑥 √1+𝑥 𝑢

Using the quotient rule when 𝑦 = 𝑣 , We have 𝑢 = 𝑥, 𝑢′ = 1

𝑑𝑦 = 𝑑𝑥

𝑑𝑦 𝑑𝑥

=

1 1 𝑣 ′ = (1 + 𝑥)−1/2 = 2 2√1 + 𝑥

1 ) (√1 + 𝑥)(1) − 𝑥 ( 2√1 + 𝑥 2

(√1 + 𝑥)

𝑑𝑦 = 𝑑𝑥

𝑥 √1 + 𝑥 1 − 2√1 + 𝑥 1+𝑥

𝑑𝑦 = 𝑑𝑥

(√1 + 𝑥)(2√1 + 𝑥) − 𝑥 2√1 + 𝑥 1+𝑥

𝑑𝑦 2(1 + 𝑥) − 𝑥 = 𝑑𝑥 2√(1 + 𝑥)3 𝑑𝑦 𝑥+2 = 𝑑𝑥 2√(1 + 𝑥)3

Given 𝑥 = cos 𝜃,

𝑦 = sin 𝜃

𝑑𝑥 𝑑𝑦 = − sin 𝜃 , = cos 𝜃 𝑑θ 𝑑𝜃 𝑑𝑦 𝑑𝑦/𝑑𝜃 = 𝑑𝑥 𝑑𝑥/𝑑𝜃 =

𝑣2

𝑣 = √1 + 𝑥

𝑑𝑦 (√1 + 𝑥)(2√1 + 𝑥) − 𝑥 = 𝑑𝑥 2√1 + 𝑥(1 + 𝑥)

c.

𝑣𝑢′ −𝑢𝑣′

cos 𝜃 = − cot 𝜃 − sin 𝜃 35


Question 6

a.

i.

a.

Given

𝑑𝑦 𝑑𝑥

= 3𝑥 2 − 4𝑥 + 1

Integrating both sides we have 𝑦 = ∫ 3𝑥 2 − 4𝑥 + 1 𝑑𝑥 𝑦=3

𝑥 2+1 𝑥 1+1 −4 +𝑥+𝐶 2+1 1+1

𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥 + 𝐶 When 𝑥 = −1, 𝑦 = −4 −4 = (−1)3 − 2(−1)2 + (−1) + 𝐶 −4 = −4 + 𝐶 𝐶=0 𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥

b.

𝑑𝑦

At the stationary points 𝑑𝑥 = 0 therefore 3𝑥 2 − 4𝑥 + 1 = 0 (3𝑥 − 1)(𝑥 − 1) = 0 1

𝑥 = 3 , or 1 1

1 3

1 2

1

When 𝑥 = 3 , 𝑦 = (3) − 2 (3) + 3 𝑦=

1 2 1 1−6+9 4 − + = = 27 9 3 27 27 1

4

Therefore coordinate of the stationary point is (3 , 27) When 𝑥 = 1, 𝑦 = 13 − 2(1)2 + 1 = 0 Therefore coordinate of the stationary point is (1, 0)

36

𝑑2 𝑦 = 6𝑥 − 4 𝑑𝑥 2 1

When 𝑥 = 3 , 𝑑2 𝑦 𝑑𝑥 2

𝑑𝑥 2

ii.

𝑑𝑥 2

1

= 6 (3) − 4 = −2 1

4

< 0 Therefore (3 , 27) is a maximum

When 𝑥 = 1, 𝑑2 𝑦

𝑑2 𝑦

𝑑2 𝑦 𝑑𝑥 2

= 6(1) − 4 = 2

> 0 Therefore (1, 0) is a minimum

y-intercept when 𝑥 = 0, 𝑦 = 0 (0, 0) x-intercept when 𝑦 = 0, 𝑥 3 − 2𝑥 2 + 𝑥 = 0 𝑥(𝑥 2 − 2𝑥 + 1 = 0) 𝑥(𝑥 − 1)2 = 0 When 𝑦 = 0, 𝑥 = 0, 1 x-intercepts (0, 0) and (1, 0) 𝑦

1 4 max ( , ) 3 27

4 27

min(1, 0) 0

𝑥 1 3

2 3

37

1


b.

i.

3

∫0 2𝑥√1 + 𝑥 2 𝑑𝑥 Using the substitution method with 𝑢 = √1 + 𝑥 2 We have

𝑑𝑢 𝑑𝑥

1

𝑥

2

√1+𝑥 2

= (1 + 𝑥 2 )−1/2 × 2𝑥 = 𝑑𝑢

𝑢 = √1 + 𝑥 2 ,

𝑑𝑥

𝑥

𝑢

𝑢

𝑥

= , 𝑑𝑥 = 𝑑𝑢

When 𝑥 = 3, 𝑢 = √1 + 32 = √10 When 𝑥 = 0, 𝑢 = √1 + 0 = 1

𝑢 𝑥

√10

2𝑥(𝑢) 𝑑𝑢

∫ 1

2𝑢3 √10 2𝑢 𝑑𝑢 = [ ] 3 1

√10

∫ 1

2

3

2(√10) 2(1)3 =[ − ] 3 3 2 = [√103 − 1] = 20.42 3

ii.

𝑏

Volume of revolution about the x-axis is given by 𝜋 ∫𝑎 𝑦 2 𝑑𝑥 therefore from b. (i) 2

2

Volume = 𝜋 ∫0 (2𝑥√1 + 𝑥 2 ) 𝑑𝑥 2

= 𝜋 ∫ 4𝑥 2 (1 + 𝑥 2 ) 𝑑𝑥 0 2

= 𝜋 ∫ 4𝑥 2 + 4𝑥 4 𝑑𝑥 0

4𝑥 3 4𝑥 5 2 = 𝜋[ + ] 3 5 0 = 𝜋 [( = 𝜋[

4(2)3 4(2)5 + ) − 0] 3 5

32 128 544 + 𝜋 cubic units ]= 3 5 15 38

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS FOR 2013 EXAM

Question 1

a.

i. p T T F F

q T F T F

𝑝→𝑞 T F T T

p T T F F

Q T F T F

𝑝∧𝑞 T F F F

ii.

b.

~(𝑝 ∧ 𝑞) F T T T

Given 𝑦 ⊕ 𝑥 = 𝑦 2 + 𝑥 2 + 2𝑦 + 𝑥 − 5𝑥𝑦 2 ⊕ 𝑥 = 22 + 𝑥 2 + 2(2) + 𝑥 − 5𝑥(2) 2 ⊕ 𝑥 = 4 + 𝑥 2 + 4 + 𝑥 − 10𝑥 2 ⊕ 𝑥 = 𝑥 2 − 9𝑥 + 8 2⊕𝑥 = 0 𝑥 2 − 9𝑥 + 8 = 0 (𝑥 − 1)(𝑥 − 8) = 0 𝑥 = 1, 8

39


c.

When 𝑛 = 1, 51 + 3 = 8 which is divisible by 2 Therefore statement is true for 𝑛 = 1 Assume statement is true when 𝑛 = 𝑘 Therefore 5𝑘 + 3 is divisible by 2 When 𝑛 = 𝑘 + 1, we have 5𝑘+1 + 3 = 5(5𝑘 + 3) − 12 5𝑘 + 3 is assumed to be divisible by 2 and 12 is divisible by 2 Therefore 5𝑘+1 + 3 is divisible by 2 Since the statement is true for 𝑛 = 1, 𝑘, and 𝑘 + 1 it is true for all positive integer n.

d.

Given 𝑓(𝑥) = 𝑥 3 − 9𝑥 2 + 𝑝𝑥 + 16 i.

If (𝑥 + 1) is a factor then 𝑓(−1) = 0 therefore 𝑓(−1) = (−1)3 − 9(−1)2 + 𝑝(−1) + 16 = 0 −𝑝 + 6 = 0, 𝑝 = 6

ii.

𝑥 2 − 10𝑥 + 16 (𝑥 + 1)

𝑥 3 − 9𝑥 2 + 6𝑥 + 16 𝑥3 + 𝑥2 −10𝑥 2 + 6𝑥 −10𝑥 2 − 10𝑥 16𝑥 + 16 16𝑥 + 16 0

𝑥 3 − 9𝑥 2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥2 − 10𝑥 + 16) = (𝑥 + 1)(𝑥 − 2)(𝑥 − 8) iii.

𝑥 3 − 9𝑥 2 + 6𝑥 + 16 = (𝑥 + 1)(𝑥 − 2)(𝑥 − 8) = 0 Therefore 𝑥 = −1, 2, 8

40

Question 2 a.

Given 𝑓(𝑥) = 𝑥 2 − 𝑥,

𝑥≥1

By completing the square we have 1

1

𝑥 2 − 𝑥 = (𝑥 2 − 𝑥 + 4) − 4 1 2

1

= (𝑥 − 2) − 4 1

The function has a parabolic shape with axis of symmetry 𝑥 = 2. The domain given is 𝑥 ≥ 1, and this section of the graph is a one to one function due to it passing the horizontal line test.

𝑦 𝑥=

1 2

𝑓(𝑥) = 𝑥 2 − 𝑥

𝑦=𝑐

1 𝑥

0 1 2

Alternatively, if we assume that 𝑓(𝑥) is not a one to one function when 𝑥 = 𝑎, or 𝑏 Then 𝑓(𝑎) = 𝑓(𝑏) where 𝑎 ≠ 𝑏. So

𝑓(𝑎) = 𝑓(𝑏) 𝑎2 − 𝑎 = 𝑏 2 − 𝑏 𝑎2 − 𝑏 2 − 𝑎 + 𝑏 = 0 (𝑎 + 𝑏)(𝑎 − 𝑏) − (𝑎 − 𝑏) = 0 (𝑎 − 𝑏)(𝑎 + 𝑏 + 1) = 0

Therefore 𝑎 = 𝑏, or 𝑎 = −𝑏 − 1 For 𝑥 ≥ 1 which eliminates 𝑎 = −(𝑏 + 1), 𝑓(𝑎) = 𝑓(𝑏) is only true if 𝑎 = 𝑏. Therefore the function is one to one for 𝑥 ≥ 1.

41


b.

i.

Given 𝑓(𝑥) = 3𝑥 + 2, and 𝑔(𝑥) = 𝑒 2𝑥 a.

𝑦 = 3𝑥 + 2 interchanging 𝑥 and 𝑦 we have

Let

𝑥 = 3𝑦 + 2, therefore 𝑦 = 𝑓 −1 (𝑥) =

𝑥−2 3

𝑥−2 3

Let 𝑦 = 𝑒 2𝑥 interchanging 𝑥 and 𝑦 we have 𝑥 = 𝑒 2𝑦

Natural log of both sides gives.

ln 𝑥 = ln 𝑒 2𝑦 ln 𝑥 = 2𝑦 1

1

𝑦 = 2 ln 𝑥 Therefore 𝑔−1 (𝑥) = 2 ln 𝑥

𝑓[𝑔(𝑥)] = 3[𝑔(𝑥)] + 2

b.

= 3𝑒 2𝑥 + 2

ii.

𝑦 = 3𝑒 2𝑥 + 2 interchanging 𝑥 and 𝑦 we have 𝑥 = 3𝑒 2𝑦 + 2

𝑒 2𝑦 =

𝑥−2

Natural log of both sides.

3

ln 𝑒 2𝑦 = ln ( 2𝑦 = ln (

𝑥−2 ) 3

𝑥−2 1 𝑥−2 ) , therefore 𝑦 = ln ( ) 3 2 3

[𝑓[𝑔(𝑥)]]

−1

=

𝑔−1 [𝑓 −1 (𝑥)] =

1 𝑥−2 ln ( ) 2 3

1 ln(𝑓 −1 (𝑥)) 2

1 𝑥−2 = ln ( ) 2 3 −1

Therefore [𝑓[𝑔(𝑥)]]

= 𝑔−1 [𝑓 −1 (𝑥)] 42

c.

i.

3𝑥 2 + 4𝑥 + 1 ≤ 5 3𝑥 2 + 4𝑥 − 4 ≤ 0 (3𝑥 − 2)(𝑥 + 2) ≤ 0 Critical points 𝑥 = −2,

2 3

−2

𝑥 ≤ −2 3𝑥 − 2 𝑥+2 (3𝑥 − 2)(𝑥 + 2)

−2 ≤ 𝑥 ≤

− − +

2 3

𝑥≥

− + −

2 3

2 3

+ + +

2

2

Therefore −2 ≤ 𝑥 ≤ 3 from the table is negative or from the graph −2 ≤ 𝑥 ≤ 3 is the part of the graph that is below the x-axis.

ii.

|𝑥 + 2| = 3𝑥 + 5 Squaring both sides we have (𝑥 + 2)2 = (3𝑥 + 5)2 𝑥 2 + 4𝑥 + 4 = 9𝑥 2 + 30𝑥 + 25 8𝑥 2 + 26𝑥 + 21 = 0 8𝑥 2 + 12𝑥 + 14𝑥 + 21 = 0 4𝑥(2𝑥 + 3) + 7(2𝑥 + 3) = 0 (4𝑥 + 7)(2𝑥 + 3) = 0 7 3 𝑥 = − , not possible − only answer 4 2

Alternatively, for (𝑥 + 2) ≥ 0 we have 𝑥 + 2 = 3𝑥 + 5 2𝑥 = −3, 𝑥 = −

3 2

For (𝑥 + 2) < 0, we have −(𝑥 + 2) = 3𝑥 + 5 −𝑥 − 2 = 3𝑥 + 5 4𝑥 = −7, 𝑥 = − 43

7 not possible 4


Question 3

a.

i.

L.H.S

2 tan 𝜃 1+tan2 𝜃

substituting tan 𝜃 =

2 tan 𝜃 We have = 1 + tan2 𝜃

sin 𝜃 cos 𝜃

sin 𝜃 2 cos 𝜃 sin 𝜃 2 1 + (cos 𝜃 )

Multiplying denominator and numerator by cos2 𝜃 2 tan 𝜃 2 sin 𝜃 cos 𝜃 = 2 1 + tan 𝜃 cos 2 𝜃 + sin2 𝜃

cos2 𝜃 + sin2 𝜃 = 1

2 tan 𝜃 = 2 sin 𝜃 cos 𝜃 1 + tan2 𝜃

2 sin 𝜃 cos 𝜃 = sin 2𝜃

ii.

Given sin 2𝜃 − tan 𝜃 = 0 2 tan 𝜃 − tan 𝜃 = 0 1 + tan2 𝜃

2 tan 𝜃 − tan 𝜃 (1 + tan2 𝜃) = 0 2 tan 𝜃 − tan 𝜃 − tan3 𝜃 = 0 tan 𝜃 − tan3 𝜃 = 0 tan 𝜃 (1 − tan2 𝜃) = 0 tan 𝜃(1 − tan 𝜃)(1 + tan 𝜃) = 0 tan 𝜃 = 0, 1, −1 𝜃 = tan−1 (0) = 0, 𝜋, 2𝜋 Acute angle for 𝜃 = tan−1 (1) = 𝜋 3𝜋 5𝜋 7𝜋 𝜃 = 0, , , , 4 4 4 4

44

𝜋 4

b.

i.

Given 𝑓(𝜃) = 3 cos 𝜃 − 4 sin 𝜃 𝑟 cos(𝜃 + 𝛼) = 𝑟 cos 𝜃 cos 𝛼 − 𝑟 sin 𝜃 sin 𝛼 𝑟 cos 𝛼 = 3,

𝑟 sin 𝛼 = 4

𝑟 sin 𝛼 4 = tan 𝛼 = 𝑟 cos 𝛼 3 4 𝛼 = tan−1 ( ) = 0.927 3

𝑟 = √32 + 42 = 5 3 cos 𝜃 − 4 sin 𝜃 = 5 cos(𝜃 + 0.927)

ii.

a.

𝑓(𝜃) = 5 cos(𝜃 + 0.927) Therefore maximum value of 𝑓(𝜃) is 5 (−1 ≤ cos 𝜃 ≤ 1)

b.

Minimum value of Therefore

iii.

a.

1 8+𝑓(𝜃)

is when 𝑓(𝜃) is maximum

1 1 1 = = 8 + 𝑓(𝜃) 8 + 5 13

Given that A, B and C are the angles of a triangle where their sum is π. 𝐴+𝐵+𝐶 =𝜋 𝐴 = 𝜋 − (𝐵 + 𝐶) Taking sine of the angles sin 𝐴 = sin[𝜋 − (𝐵 + 𝐶)] sin[𝜋 − (𝐵 + 𝐶)] = sin 𝜋 cos(𝐵 + 𝐶) − sin(𝐵 + 𝐶) cos 𝜋 sin 𝜋 = 0,

cos 𝜋 = −1 Therefore

sin[𝜋 − (𝐵 + 𝐶)] = 0 − sin(𝐵 + 𝐶) (−1) sin[𝜋 − (𝐵 + 𝐶)] = sin(𝐵 + 𝐶) sin 𝐴 = sin(𝐵 + 𝐶)

b.

from (iii).a. sin 𝐴 = sin(𝐵 + 𝐶) therefore sin 𝐵 = sin(𝐴 + 𝐶) and sin 𝐶 = sin(𝐴 + 𝐵) so sin 𝐴 + sin 𝐵 + sin 𝐶 = sin(𝐵 + 𝐶) + sin(𝐴 + 𝐶) + sin(𝐴 + 𝐵) 45


Question 4 a.

i.

Given 𝑥 2 + 𝑦 2 − 6𝑥 − 4𝑦 + 4 = 0 By completing the square we have −6 2 −6 2 −4 2 −4 2 (𝑥 2 − 6𝑥 + ( ) ) − ( ) + (𝑦 2 − 4𝑦 + ( ) ) − ( ) + 4 = 0 2 2 2 2

(𝑥 2 − 6𝑥 + 9) − 9 + (𝑦 2 − 4𝑦 + 4) − 4 + 4 = 0 (𝑥 − 3)2 + (𝑦 − 2)2 = 9 = 32 The equation of a circle is given by (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 Where (𝑎, 𝑏) is the centre and r its radius. Therefore the circle has centre (3, 2) and radius √9 = 3 ii.

The gradient of the line between the centre (3, 2) and a point on the

a.

2−2

circumference (6, 2) is given by 6−3 = 0, Therefore equation of the normal to the circle at (6, 2) is given by 𝑦 = 2 The tangent is perpendicular to the normal therefore the line 𝑦 = 2 at

b.

(6, 2) is perpendicular to 𝑥 = 6, which is a vertical line parallel to the y-axis. b.

Given 𝑥 = 𝑡 2 + 𝑡, 𝑦 = 2𝑡 − 4 From 𝑦 = 2𝑡 − 4, 𝑡 =

Substituting 𝑡 =

𝑦+4 2

𝑦+4 2

into 𝑥 = 𝑡 2 + 𝑡

𝑦+4 2 𝑦+4 We have 𝑥 = ( ) + 2 2 𝑥=

𝑦 2 + 8𝑦 + 16 𝑦 + 4 + 4 2

𝑥=

𝑦 2 + 8𝑦 + 16 + 2(𝑦 + 4) 4

𝑥=

𝑦 2 + 10𝑦 + 24 4

4𝑥 = 𝑦 2 + 10𝑦 + 24 46

c.

i.

Given 𝐴(3, −1, 2), 𝐵(1, 2, −4) and 𝐶(−1, 1, −2) ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ 𝐴𝐵 = ⃗⃗⃗⃗⃗ 𝐴𝑂 + 𝑂𝐵 3 1 = − (−1) + ( 2 ) 2 −4 −2 =( 3 ) −6 ⃗⃗⃗⃗⃗ 𝐴𝐵 = −2𝒊 + 3𝒋 − 6𝒌 ⃗⃗⃗⃗⃗ = 𝐵𝑂 ⃗⃗⃗⃗⃗ + 𝑂𝐶 ⃗⃗⃗⃗⃗ 𝐵𝐶 1 −1 = −( 2 )+( 1 ) −4 −2 −2 = (−1) 2 ⃗⃗⃗⃗⃗ = −2𝒊 − 𝒋 + 2𝒌 𝐵𝐶

ii.

Given 𝒓 = −16𝒋 − 8𝒌, if 𝒓 is perpendicular to the plane through A, B, and C Then 𝒓 ∙ 𝑨𝑩 = 0 and 𝒓 ∙ 𝑩𝑪 = 0 0 −2 𝒓 ∙ 𝑨𝑩 = (−16) ∙ ( 3 ) −8 −6 = (0 × −2) + (−16 × 3) + (−8 × −6) = 0 − 48 + 48 = 0 Therefore r is perpendicular to ⃗⃗⃗⃗⃗ 𝐴𝐵 0 −2 𝒓 ∙ 𝑩𝑪 = (−16) ∙ (−1) −8 2 = (0 × −2) + (−16 × −1) + (−8 × 2) = 0 + 16 − 16 = 0 ⃗⃗⃗⃗⃗ Therefore r is perpendicular to 𝐵𝐶

47


iii.

The vector equation of a plane is given by 𝒓 ∙ 𝒏 = 𝒂 ∙ 𝒏 where r is any vector (𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌) on the plane, n is a vector normal to the plane and a is the position vector for a point on the plane. Using 𝒏 = −16𝒋 − 8𝒌 and 𝒂 = 3𝒊 − 𝒋 + 2𝒌 we have 𝒓∙𝒏=𝒂∙𝒏 (𝑥𝒊 + 𝑦𝒋 + 𝑧𝒌) ∙ (−16𝒋 − 8𝒌) = (3𝒊 − 𝒋 + 2𝒌) ∙ (−16𝒋 − 8𝒌) −16𝑦 − 8𝑧 = −16 + 16 −16𝑦 − 8𝑧 = 0 2𝑦 + 𝑧 = 0

Question 5

a.

Given 𝑓(𝑥) = { i.

𝑥 + 2, 𝑥2,

𝑥<2 . 𝑥>2

lim 𝑓(𝑥) = lim+ 𝑥 2

𝑥→2+

𝑥→2

= 22 = 4 lim 𝑓(𝑥) = lim− 𝑥 + 2

𝑥→2−

𝑥→2

=2+2=4 lim 𝑓(𝑥) = lim− 𝑓(𝑥)

𝑥→2+

𝑥→2

lim 𝑓(𝑥) = 4

𝑥→2

ii.

𝑓(𝑥) is not continuous at 𝑥 = 2 because 𝑓(2) is undefined.

48

b.

𝑥 2 + 2𝑥 + 3 Let 𝑦 = (𝑥 2 + 2)3 Using the quotient rule 𝑦 = Let 𝑢 = 𝑥 2 + 2𝑥 + 3 𝑣 = (𝑥 2 + 2)3

𝑢 𝑣

,

𝑑𝑦 𝑑𝑥

=

𝑣𝑢′ −𝑢𝑣′ 𝑣2

𝑢′ = 2𝑥 + 2

𝑣 ′ = 3(𝑥 2 + 2)2 × 2𝑥 𝑣 ′ = 6𝑥(𝑥 2 + 2)2

𝑑𝑦 (𝑥 2 + 2)3 × (2𝑥 + 2) − (𝑥 2 + 2𝑥 + 3) × (6𝑥(𝑥 2 + 2)2 ) = ((𝑥 2 + 2)3 )2 𝑑𝑥 𝑑𝑦 (𝑥 2 + 2)2 [(𝑥 2 + 2)(2𝑥 + 2) − 6𝑥(𝑥 2 + 2𝑥 + 3)] = (𝑥 2 + 2)6 𝑑𝑥 𝑑𝑦 2𝑥 3 + 2𝑥 2 + 4𝑥 + 4 − 6𝑥 3 − 12𝑥 2 − 18𝑥 = (𝑥 2 + 2)4 𝑑𝑥 𝑑𝑦 −4𝑥 3 − 10𝑥 2 − 14𝑥 + 4 = (𝑥 2 + 2)4 𝑑𝑥

c.

Given 𝑥 = 1 − 3 cos 𝜃 ,

𝑦 = 2 sin 𝜃

𝑑𝑥 = 0 − 3(− sin 𝜃) = 3 sin 𝜃 𝑑𝜃 𝑑𝑦 = 2 cos 𝜃 𝑑𝜃 𝑑𝑦 𝑑𝑦/𝑑𝜃 = 𝑑𝑥 𝑑𝑥/𝑑𝜃 𝑑𝑦 2 cos 𝜃 = 𝑑𝑥 3 sin 𝜃 𝑑𝑦 2 = cot 𝜃 𝑑𝑥 3

d.

i.

𝑦 = 𝑥 2 + 3 ------ (1) 𝑦 = 4𝑥

------- (2)

4𝑥 = 𝑥 2 + 3

Substitute 𝑦 = 4𝑥 into (1)

𝑥 2 − 4𝑥 + 3 = 0 49


(𝑥 − 1)(𝑥 − 3) = 0 𝑥 = 1, or 3 When 𝑥 = 1, 𝑦 = 4(1) = 4 𝑥 = 3, 𝑦 = 4(3) = 12 𝑃(1, 4) and 𝑄(3, 12)

ii.

Area of the shaded region is given by A 3

3

𝐴 = ∫ 4𝑥 𝑑𝑥 − ∫ 𝑥 2 + 3 𝑑𝑥 1

1

𝑥3 3 3 𝐴 = [2𝑥 2 ] − [ + 3𝑥] 1 1 3 𝐴 = [2(32 ) − 2] − [(

33 1 + 3(3)) − ( + 3)] 3 3

1

𝐴 = 16 − 18 + 3 3 4

= 3 sq. units

Question 6

a.

i.

∫ 𝑥(1 − 𝑥)2 𝑑𝑥 Let 𝑢 = 1 − 𝑥, 𝑥 = 1 − 𝑢,

𝑑𝑢 𝑑𝑥

= −1, 𝑑𝑥 = −𝑑𝑢

∫(1 − 𝑢)𝑢2 (−1)𝑑𝑢 ∫ 𝑢3 − 𝑢2 𝑑𝑢 =

𝑢4 𝑢3 − +𝐶 4 3

∫ 𝑥(1 − 𝑥)2 𝑑𝑥 =

(1 − 𝑥)4 (1 − 𝑥)3 − +𝐶 4 3

=−

1 [(1 − 𝑥)3 (1 + 3𝑥)] + 𝐶 12 50

ii.

Given 𝑓(𝑡) = 2 cos 𝑡 , 𝑔(𝑡) = 4 sin 5𝑡 + 3 cos 𝑡 ∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = ∫ 2 cos 𝑡 + 4 sin 5𝑡 + 3 cos 𝑡 𝑑𝑡 = ∫ 5 cos 𝑡 + 4 sin 5𝑡 𝑑𝑡 4 ∫[𝑓(𝑡) + 𝑔(𝑡)] 𝑑𝑡 = 5 sin 𝑡 − cos 5𝑡 + 𝐶 5 ∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡 = ∫ 2 cos 𝑡 𝑑𝑡 + ∫ 4 sin 5𝑡 + 3 cos 𝑡 𝑑𝑡 4 = 2 sin 𝑡 + 𝐴 + (− cos 5𝑡) + 3 sin 𝑡 + 𝐵 5 4 ∫ 𝑓(𝑡) 𝑑𝑡 + ∫ 𝑔(𝑡) 𝑑𝑡 = 5 sin 𝑡 − cos 5𝑡 + 𝐶 5

A, B, and C are merely constants of the integrals therefore 𝐶 = 𝐴 + 𝐵

b.

i.

Length of rectangle is x, width of rectangle is 2r and length of semi-circle is

2𝜋𝑟 2

= 𝜋𝑟

Perimeter of track is given by 2𝑥 + 2𝑟 + 𝜋𝑟 = 600 𝑟(2 + 𝜋) = 600 − 2𝑥 𝑟=

ii.

600 − 2𝑥 2+𝜋

Area of track is given by 1

𝐴 = 𝑥(2𝑟) + 2 (𝜋𝑟 2 ) 𝜋

𝐴 = 2𝑥𝑟 + 2 𝑟 2 600 − 2𝑥 𝜋 600 − 2𝑥 2 𝐴 = 2𝑥 ( )+ ( ) 2+𝜋 2 2+𝜋 1200𝑥 − 4𝑥 2 𝜋 600 − 2𝑥 2 𝐴=( ) )+ ( 2+𝜋 2 2+𝜋 𝑑𝐴 1 𝜋 = (2)(600 − 2𝑥)(−2)] [1200 − 8𝑥 + 𝑑𝑥 2 + 𝜋 2(2 + 𝜋) 𝑑𝐴 1 2𝜋 (600 − 2𝑥)] = [1200 − 8𝑥 − (2 + 𝜋) 𝑑𝑥 2 + 𝜋 51


When

𝑑𝐴 1 2𝜋 (600 − 2𝑥)] = 0 = 0, [1200 − 8𝑥 − (2 𝑑𝑥 2+𝜋 + 𝜋)

1200 − 8𝑥 −

2𝜋 (600 − 2𝑥) = 0 (2 + 𝜋)

(2 + 𝜋)(1200 − 8𝑥) − 2𝜋(600 − 2𝑥) = 0 2400 − 16𝑥 + 1200𝜋 − 8𝜋𝑥 − 1200𝜋 + 4𝜋𝑥 = 0 2400 − 16𝑥 − 4𝜋𝑥 = 0 16𝑥 + 4𝜋𝑥 = 2400 4𝑥(4 + 𝜋) = 2400 4𝑥 =

2400 4+𝜋 600

𝑥 = 4+𝜋 ≈ 84 metres 𝑑2 𝐴 1 2𝜋 1 4𝜋 = ) (−2)] = − 8] [−8 − ( [ 2 (2 + 𝜋) 𝑑𝑥 2+𝜋 2+𝜋 2+𝜋 8>

4𝜋 2+𝜋

Therefore

𝑑2 𝐴 <0 𝑑𝑥 2

600

Therefore 𝑥 = 4+𝜋 give the maximum area.

c.

i.

Let 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵 𝑦 ′ = −[𝑥 cos 𝑥 + sin 𝑥] − 2(− sin 𝑥) + 𝐴 = −𝑥 cos 𝑥 − sin 𝑥 + 2 sin 𝑥 + 𝐴 = sin 𝑥 − 𝑥 cos 𝑥 + 𝐴 𝑦 ′′ = cos 𝑥 − [cos 𝑥 + 𝑥(− sin 𝑥)] = cos 𝑥 − cos 𝑥 + 𝑥 sin 𝑥 𝑦′′ = 𝑥 sin 𝑥

52

ii.

𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝐴𝑥 + 𝐵 When 𝑥 = 0, 𝑦 = 1, We have 1 = −2 + 𝐵 𝐵=3 When 𝑥 = 𝜋, 𝑦 = 6, 𝐵 = 3 we have 6 = −𝜋 sin 𝜋 − 2 cos 𝜋 + 𝜋𝐴 + 3 6 = 2 + 𝜋𝐴 + 3 𝜋𝐴 = 1 1

𝐴=𝜋 1

The specific solution is 𝑦 = −𝑥 sin 𝑥 − 2 cos 𝑥 + 𝜋 𝑥 + 3

53

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS TO 2012 EXAM Question 1

a.

Given 𝑓(𝑥) = 2𝑥 3 − 𝑝𝑥 2 + 𝑞𝑥 − 10

i.

𝑥 − 1 is a factor of 𝑓(𝑥) therefore 𝑓(1) = 0 𝑓(1) = 2(1)3 − 𝑝(1)2 + 𝑞(1) − 10 = 0 2 − 𝑝 + 𝑞 − 10 = 0 𝑝 − 𝑞 = −8 ------- (1) When 𝑓(𝑥) is divided by 𝑥 + 1 it gives a remainder of −6 therefore 𝑓(−1) = 2(−1)3 − 𝑝(−1)2 + 𝑞(−1) − 10 = −6 −2 − 𝑝 − 𝑞 − 10 = −6 𝑝 + 𝑞 = −6 -------- (2) Adding equations (1) and (2) we have 2𝑝 = −14,

therefore 𝑝 = −7

Substituting 𝑝 = −7 into (2) give −7 + 𝑞 = −6, therefore 𝑞 = 1 𝑓(𝑥) = 2𝑥 3 + 7𝑥 2 + 𝑥 − 10

ii.

2𝑥 2 + 9𝑥 + 10 𝑥 − 1 2𝑥 3 + 7𝑥 2 + 𝑥 − 10

2𝑥 3 − 2𝑥 2 9𝑥 2 + 𝑥 9𝑥 2 − 9𝑥 10𝑥 − 10 10𝑥 − 10 0 54

2𝑥 3 + 7𝑥 2 + 𝑥 − 10 = (𝑥 − 1)(2𝑥2 + 9𝑥 + 10) = (𝑥 − 1)(𝑥 + 2)(2𝑥 + 5) Therefore the factors of 𝑓(𝑥) are (𝑥 − 1), (𝑥 + 2), and (2𝑥 + 5)

b.

2

Given (√𝑥 + √𝑦) = 16 + √240 2

(√𝑥 + √𝑦) = 𝑥 + 2√𝑥𝑦 + 𝑦 𝑥 + 𝑦 + 2√𝑥𝑦 = 16 + √240 Therefore 𝑥 + 𝑦 = 16 --------

(1)

𝑦 = 16 − 𝑥 ------- (2) 2√𝑥𝑦 = √240 --------

(3)

√240 = √4 × 60 = 2√60 2√𝑥𝑦 = 2√60 -----Therefore

𝑥𝑦 = 60

(4)

--------

(5)

Substituting (2) into (5) we have 𝑥(16 − 𝑥) = 60 16𝑥 − 𝑥 2 = 60 𝑥 2 − 16𝑥 + 60 = 0 (𝑥 − 6)(𝑥 − 10) = 0 𝑥 = 6, 10 When 𝑥 = 6, 𝑦 = 16 − 6 = 10 When 𝑥 = 10, 𝑦 = 16 − 10 = 6 𝑥 = 6, 𝑦 = 10 and 𝑥 = 10, 𝑦 = 6

55


c.

Given |3𝑥 − 7| ≤ 5 Squaring both sides we have (3𝑥 − 7)2 ≤ 25

|3𝑥 − 7| ≤ 5

9𝑥 2 − 42𝑥 + 49 ≤ 25 9𝑥 2 − 42𝑥 + 24 ≤ 0 2 3

3(3𝑥 2 − 14𝑥 + 8) ≤ 0

4

3(3𝑥 − 2)(𝑥 − 4) ≤ 0 2

Critical points 𝑥 = 3 , 4 𝑥≤ 3𝑥 − 2 𝑥−4 (3𝑥 − 2)(𝑥 − 4)

2 3

− − +

2 ≤𝑥≤4 3 − + −

𝑥≥4 + + +

2

Therefore 3 ≤ 𝑥 ≤ 4 Alternatively, when (3𝑥 − 7) ≥ 0 we have 3𝑥 − 7 ≤ 5 3𝑥 ≤ 12, 𝑥 ≤ 4 When (3𝑥 − 7) ≤ 0 we have −(3𝑥 − 7) ≤ 5 −3𝑥 + 7 ≤ 5 −3𝑥 ≤ −2 2

𝑥≥3 2

Therefore 3 ≤ 𝑥 ≤ 4 ii.

|3𝑥 − 7| + 5 ≤ 0 |3𝑥 − 7| ≥ 0 Modulus always give the value of the function as positive and 5 is also greater than zero, therefore if we add the two together a number greater than zero will be the result. Therefore |3𝑥 − 7| + 5 cannot be less than zero for any real value of x and this function will not intersect the x-axis resulting in us having no real solution. 56

Question 2 a.

Given 𝑓(𝑥) → 𝑥 2 − 3 i.

𝑓(𝑓(𝑥)) = [𝑓(𝑥)]2 − 3 = (𝑥 2 − 3)2 − 3 = 𝑥 4 − 6𝑥 2 + 9 − 3 = 𝑥 4 − 6𝑥 2 + 6

ii.

𝑓(𝑓(𝑥)) = 𝑓(𝑥 + 3) 𝑥 4 − 6𝑥 2 + 6 = (𝑥 + 3)2 − 3 𝑥 4 − 6𝑥 2 + 6 = 𝑥 2 + 6𝑥 + 9 − 3 𝑥 4 − 6𝑥 2 + 6 = 𝑥 2 + 6𝑥 + 6 𝑥 4 − 7𝑥 2 − 6𝑥 = 𝑥(𝑥 3 − 7𝑥 − 6) 𝑥 + 1 is a factor of 𝑥 3 − 7𝑥 − 6 (−1)3 − 7(−1) − 6 = −1 + 7 − 6 = 0 𝑥 + 2 is a factor of 𝑥 3 − 7𝑥 − 6 (−2)3 − 7(−2) − 6 = −8 + 14 − 6 = 0 𝑥 − 3 is a factor of 𝑥 3 − 7𝑥 − 6 (3)3 − 7(3) − 6 = 27 − 21 − 6 = 0 Therefore 𝑥 4 − 7𝑥 2 − 6𝑥 = 𝑥(𝑥 + 1)(𝑥 + 2)(𝑥 − 3) 𝑥 = 0, −1, −2, 3

b.

i.

Given 𝛼 𝑎𝑛𝑑 𝛽 are the roots the equation 4𝑥 2 − 3𝑥 + 1 = 0 (𝑥 − 𝛼)(𝑥 − 𝛽) = 𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 4𝑥 2 − 3𝑥 + 1 = 0 Dividing both sides by 4 we have 3

1

3

1

𝑥 2 − 4 𝑥 + 4 = 0 Therefore 𝛼 + 𝛽 = 4 𝑎𝑛𝑑 𝛼𝛽 = 4 ii.

(𝛼 + 𝛽)2 = 𝛼 2 + 2𝛼𝛽 + 𝛽 2 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽 3 2

1

= (4) − 2 (4) 1

= 16 57


iii.

Given

2 𝛼2

and

2 𝛽2

are the roots of a quadratic equation we have 2 2 2𝛽 2 + 2𝛼 2 + = 𝛼 2 𝛽2 𝛼 2 𝛽2

Sum of roots

=

2(𝛽 2 + 𝛼 2 ) (𝛼𝛽)2

1 1 ) 16 = = 8 1 1 2 ( ) ( ) 16 4 2(

=2 Product of roots

2 2 4 × 2= 2 (𝛼𝛽)2 𝛼 𝛽 =

4 1 2 ( ) 4

= 4 × 16 = 64 Therefore the quadratic equation is 𝑥 2 − 2𝑥 + 64 = 0 c.

i.

1

3

5

7

9

log10 (3) + log10 (5) + log10 (7) + log10 (9) + log10 (10) 1

3

5

7

9

log10 [(3) × (5) × (7) × (9) × (10)] 1

log10 (10) = log10 10−1 = −1 99

ii.

𝑟 1 2 98 99 ∑ log10 ( ) = log10 ( ) + log10 ( ) + ⋯ + log10 ( ) + log10 ( ) 𝑟+1 2 3 99 100

𝑟=1

1 2 98 99 = log10 [( ) × ( ) × … × ( ) ( )] 2 3 99 100 = log10 (

1 ) 100

= log10 10−2

= −2 58

Question 3 a.

i.

Given cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 and cos 2𝜃 = 2 cos2 𝜃 − 1 cos 3𝜃 = cos(2𝜃 + 𝜃) = cos 2𝜃 cos 𝜃 − sin 2𝜃 sin 𝜃 = (2 cos2 𝜃 − 1) cos 𝜃 − (2 sin 𝜃 cos 𝜃) sin 𝜃

[sin 2𝜃 = 2 sin 𝜃 cos 𝜃]

= (2 cos2 𝜃 − 1) cos 𝜃 − 2 sin2 𝜃 cos 𝜃 = cos 𝜃 [(2 cos2 𝜃 − 1) − 2 sin2 𝜃] 1

= 2 cos 𝜃 [cos2 𝜃 − sin2 𝜃 − 2] ii.

1

L.H.S = 2 [sin 6𝜃 − sin 2𝜃] Using the factor formulae sin 𝐴 − sin 𝐵 = 2 cos ( 1 2

1

[sin 6𝜃 − sin 2𝜃] = [2 cos ( 2

6𝜃+2𝜃 2

6𝜃−2𝜃

) sin (

2

= (2 cos 2 2𝜃 − 1) sin 2𝜃 sin 6𝜃 − sin 2𝜃 = 0 (2 cos 2 2𝜃 − 1) sin 2𝜃 = 0 2 cos 2 2𝜃 − 1 = 0 cos2 2𝜃 =

1 2

cos 2𝜃 = ± 2𝜃 = 𝜃=

1 √2

𝜋 , 4

0≤𝜃≤

𝜋 2

3𝜋 4

𝜋 3𝜋 , 8 8

sin 2𝜃 = 0

0≤𝜃≤

2𝜃 = 0, 𝜋 𝜃 = 0,

𝜋 2

𝜃 = 0,

𝜋 , 2

𝜋 , 8

3𝜋 8 59

2

) sin (

𝐴−𝐵 2

)

)]

[cos 4𝜃 = 2 cos2 2𝜃 − 1]

= cos 4𝜃 sin 2𝜃

iii.

𝐴+𝐵

𝜋 2


b.

Given cot 2 𝜃 + cos 𝜃 = 0 cos 2 𝜃 2 + cos 𝜃 = 0 sin2 𝜃 2 cos 2 𝜃 + sin2 𝜃 cos 𝜃 = 0 2 cos 2 𝜃 + (1 − cos 2 𝜃) cos 𝜃 = 0 2 cos 2 𝜃 + cos 𝜃 − cos3 𝜃 = 0 cos3 𝜃 − 2 cos2 𝜃 − cos 𝜃 = 0 cos 𝜃 (cos 2 𝜃 − 2 cos 𝜃 − 1) = 0 cos 𝜃 = 0 cos2 𝜃 − 2 cos 𝜃 − 1 = 0 Using the quadratic formula 𝑥 =

−𝑏±√𝑏2 −4𝑎𝑐

cos 𝜃 =

−(−2) ± √(−2)2 − 4(1)(−1) 2

cos 𝜃 =

2 ± √8 , 2

cos 𝜃 =

2 ± 2√2 = 1 ± √2 2

cos 𝜃 = 1 − √2,

√8 = √2 × 4 = 2√2

cos 𝜃 ≠ 1 + √2

cos 𝜃 = 0, 1 − √2 Question 4 a.

i.

2𝑎

Given 𝑦 = 3 sec 𝜃 , and 𝑥 = 3 tan 𝜃 sec 2 𝜃 = 1 + tan2 𝜃 𝑦 𝑥 sec 𝜃 = , tan 𝜃 = 3 3 𝑦 2 𝑥 2 ( ) =1+( ) 3 3 𝑦2 𝑥2 =1+ 9 9

𝑦2 = 9 + 𝑥2

60

ii.

𝑦 2 = 9 + 𝑥 2 ------ (1) 𝑦 = √10𝑥

------ (2)

𝑦 2 = 10𝑥

--------- (3)

10𝑥 = 9 + 𝑥 2

squaring (2)

Substitute (3) into (1)

𝑥 2 − 10𝑥 + 9 = 0 (𝑥 − 1)(𝑥 − 9) = 0 𝑥 = 1, 9 When 𝑥 = 1, 𝑦 = √10 point (1, √10) When 𝑥 = 9, 𝑦 = √90 = 3√10 point (9, 3√10)

b.

i.

𝒑 = −3𝒊 + 4𝒋 and 𝒒 = −𝒊 + 6𝒋

ii.

𝒑 − 𝒒 = (−3𝒊 + 4𝒋) − (−𝒊 + 6𝒋) = −2𝒊 − 2𝒋

iii.

𝒑 ∙ 𝒒 = (−3𝒊 + 4𝒋) ∙ (−𝒊 + 6𝒋) = 3 + 24 = 27

iv.

𝒑 ∙ 𝒒 = |𝒑| × |𝒒| cos 𝜃 cos 𝜃 =

𝒑∙𝒒 |𝒑| × |𝒒|

|𝒑| = √(−3)2 + (4)2 = √25 = 5 |𝒒| = √(−1)2 + (6)2 = √37 cos 𝜃 =

27 5 × √37

cos 𝜃 = 0.888 𝜃 = cos −1 0.888 = 27.4𝑜

61


Question 5

a.

i.

Let 𝑓(𝑥) =

𝑥 3 +8 𝑥 2 −4

𝑓(𝑥) is discontinuous when 𝑥 2 − 4 = 0 Therefore 𝑥 = 2, −2

ii.

lim

𝑥 3 +8

𝑥→−2 𝑥 2 −4

𝑥 3 + 8 = (𝑥 + 2)(𝑥 2 − 2𝑥 + 4) (𝑥 + 2)(𝑥 2 − 2𝑥 + 4) lim 𝑥→−2 (𝑥 + 2)(𝑥 − 2) (𝑥 2 − 2𝑥 + 4) (−2)2 − 2(−2) + 4 = 𝑥→−2 (𝑥 − 2) −2 − 2 lim

=

ii.

lim

12 = −3 −4

2𝑥 3 +4𝑥

𝑥→0 sin 2𝑥

2𝑥 3 + 4𝑥 2𝑥 lim 𝑥→0 sin 2𝑥 2𝑥 𝑥+2 𝑥→0 sin 2𝑥 2𝑥 lim

lim 𝑥 + 2

𝑥→0

sin 2𝑥 lim 2𝑥

=

0+2 1

𝑥→0

2𝑥 3 + 4𝑥 =2 𝑥→0 sin 2𝑥 lim

62

b.

Given 𝑓(𝑥) = { i.

a. b.

𝑥 2 + 1, 𝑥 > 1, 4 + 𝑝𝑥, 𝑥 < 1.

lim 𝑓(𝑥) = 12 + 1 = 2

𝑥→1+

lim 𝑓(𝑥) = 4 + 𝑝(1)

𝑥→1−

When lim 𝑓(𝑥) exist lim+ 𝑓(𝑥) = lim− 𝑓(𝑥) 𝑥→1

𝑥→1

𝑥→1

Therefore 4 + 𝑝 = 2 𝑝 = −2 ii.

c.

𝑓(1) = 2 For 𝑓 to be continuous at 𝑥 = 1.

𝑣

Given 𝑀 = 𝑢𝑡 2 + 𝑡 2 𝑑𝑀 = 2𝑢𝑡 + (−2𝑣𝑡 −3 ) 𝑑𝑡 𝑑𝑀 2𝑣 = 2𝑢𝑡 − 3 𝑑𝑡 𝑡

When 𝑀 = −1, 𝑡 = 1 therefore −1 = 𝑢 + 𝑣 When

𝑑𝑀 𝑑𝑡

=

------ (1) 35 4

, 𝑡 = 2 therefore

35 2𝑣 = 2𝑢(2) − 3 4 2 35 𝑣 = 4𝑢 − 4 4

35 = 16𝑢 − 𝑣 ----- (2) Add eq. (1) and (2) we have 34 = 17𝑢, 𝑢 = 2 From (1) when 𝑢 = 2 we have −1 = 2 + 𝑣, 𝑣 = −3 Therefore 𝑢 = 2, 𝑣 = −3

63


Question 6

a.

i.

𝑦 = √4𝑥 2 − 7

Given

𝑑𝑦 1 = (4𝑥 2 − 7)−1/2 × 8𝑥 𝑑𝑥 2 𝑑𝑦 4𝑥 = 𝑑𝑥 √4𝑥 2 − 7 √4𝑥 2 − 7

𝑑𝑦 = 4𝑥 𝑑𝑥

𝑦 = √4𝑥 2 − 7 𝑦

ii.

𝑑𝑦 𝑑𝑥

=

𝑑𝑦 = 4𝑥 𝑑𝑥 4𝑥

√4𝑥 2 −7 𝑢

𝑑𝑦

𝑣

𝑑𝑥

using the quotient rule 𝑦 = ,

𝑑𝑢 𝑑𝑣 −𝑢 𝑑𝑥 𝑑𝑥 𝑣2

𝑑𝑢 =4 𝑑𝑥

𝑢 = 4𝑥, 𝑣 = √4𝑥 2 − 7 𝑑2 𝑦 = 𝑑𝑥 2

=

𝑣

𝑑𝑣 4𝑥 = 𝑑𝑥 √4𝑥 2 − 7

√4𝑥 2 − 7(4) − 4𝑥 (

4𝑥 ) √4𝑥 2 − 7 2

(√4𝑥 2 − 7)

4(√4𝑥 2 − 7)(√4𝑥 2 − 7) − (4𝑥)(4𝑥) √4𝑥 2 − 7 = 4𝑥 2 − 7 =

= =

4(4𝑥 2 − 7) − 16𝑥 2 (4𝑥 2 − 7)√4𝑥 2 − 7 16𝑥 2 − 28 − 16𝑥 2 (4𝑥 2 − 7)√4𝑥 2 − 7 −28 (4𝑥 2 − 7)√4𝑥 2 − 7 64

(√4𝑥 2 − 7) 𝑦

𝑑2 𝑦 28 =− 2 2 𝑑𝑥 4𝑥 − 7

𝑑2 𝑦 28 =− 2 2 𝑑𝑥 4𝑥 − 7

2 𝑑𝑦 2 4𝑥 ( ) =( ) 𝑑𝑥 √4𝑥 2 − 7

=

16𝑥 2 4𝑥 2 − 7

𝑑2 𝑦 𝑑𝑦 2 28 16𝑥 2 𝑦 2+( ) =− 2 + 2 𝑑𝑥 𝑑𝑥 4𝑥 − 7 4𝑥 − 7

Therefore 𝑦

b.

i.

𝑑𝑦 𝑑𝑥

=

16𝑥 2 − 28 4𝑥 2 − 7

=

4(4𝑥 2 − 7) =4 4𝑥 2 − 7

𝑑2𝑦

𝑑𝑦 2

+( ) =4 𝑑𝑥 2 𝑑𝑥

= 3𝑥 2 − 6𝑥

Integrating both sides we have 𝑦 = ∫ 3𝑥 2 − 6𝑥 𝑑𝑥 𝑦 = 𝑥 3 − 3𝑥 2 + 𝐶 When 𝑥 = −1, 𝑦 = 0 0 = (−1)3 − 3(−1)2 + 𝐶 0 = −1 − 3 + 𝐶 𝐶=4 𝑦 = 𝑥 3 − 3𝑥 2 + 4

65


ii.

𝑑𝑦 𝑑𝑥

= 3𝑥 2 − 6𝑥

At the stationary points

𝑑𝑦 𝑑𝑥

= 0 therefore

3𝑥 2 − 6𝑥 = 0 3𝑥(𝑥 − 2) = 0 𝑥 = 0, 2 When 𝑥 = 0, 𝑦 = 4 (0, 4) When 𝑥 = 2, 𝑦 = 23 − 3(22 ) + 4 𝑦 = 8 − 12 + 4 = 0

(2, 0)

Therefore the stationary points are (0, 4) and (2, 0)

iii.

𝑑2𝑦 𝑑𝑥 2

= 6𝑥 − 6

When 𝑥 = 0,

𝑑2 𝑦 𝑑𝑥 2

= −6

Therefore (0, 4) is a maximum When 𝑥 = 2,

𝑑2𝑦 𝑑𝑥 2

= 6(2) − 6 = 6

Therefore (2, 0) is a minimum

66

iv.

The curve meets the x-axis when 𝑦 = 0 therefore 𝑥 3 − 3𝑥 2 + 4 = 0 The minimum point has 𝑦 = 0, 𝑥 = 2 Therefore (𝑥 − 2)2 is a factor (𝑥 − 2)2 (𝑥 − 𝑎) = 𝑥 3 − 3𝑥 2 + 4 Equating the constants we have −4𝑎 = 4, 𝑎 = −1 𝑥 3 − 3𝑥 2 + 4 = (𝑥 − 2)2 (𝑥 + 1) Therefore the curve meets the x-axis at 𝑥 = 2, and − 1 𝑃(−1, 0) and 𝑄(2, 0)

v. 𝑦 4

𝑚𝑎𝑥 (0,4)

2 𝑃(−1, 0)

𝑚𝑖𝑛 (2,0) −1

−2

0

2 𝑄(2, 0)

67

𝑥

CAPE PURE MATHEMATICS UNIT 1 SOLUTIONS TO 2011 EXAM

Question 1

a.

i.

2

(√75 + √12) − (√75 − √12)

2

[(√75 + √12) + (√75 − √12)][(√75 + √12) − (√75 − √12)] (2√75)(2√12) (2√25 × 3)(2√4 × 3) 2 × 5√3 × 2 × 2√3 40 × 3 = 120

ii.

271/4 × 93/8 × 811/8 1

3

1

(33 )4 × (32 )8 × (34 )8 3

3

1

34 × 34 × 32 3 3 1

3(4+4+2) 32 = 9

b.

𝑓(𝑥) = 𝑥 3 + 𝑚𝑥 2 + 𝑛𝑥 + 𝑝 f(x)

(0, 4)

Q

0

1

68

2

x

i.

When 𝑥 = 0, 𝑓(0) = 𝑝, therefore 𝑝 is the y-intercept From the graph the y-intercept is where the curve cuts the y-axis. Therefore 𝑝 = 4

ii.

From the graph when 𝑦 = 0, 𝑥 = 1, 2 therefore 𝑓(1) = 1 + 𝑚 + 𝑛 + 4 = 0 𝑚 + 𝑛 = −5 ------ (1) 𝑓(2) = 23 + 𝑚(22 ) + 𝑛(2) + 4 = 0 4𝑚 + 2𝑛 = −12 ----- (2) 2𝑚 + 𝑛 = −6 ---- (3)

dividing (2) by 2

−𝑚 = 1 (1) Subtract (2) 𝑚 = −1 From (1) when 𝑚 = −1 we have −1 + 𝑛 = −5 𝑛 = −4 So 𝑓(𝑥) = 𝑥 3 − 𝑥 2 − 4𝑥 + 4

iii.

𝑥 3 − 𝑥 2 − 4𝑥 + 4 = (𝑥 − 1)(𝑥 − 2)(𝑥 + 𝑎) Equating the constants we have 4 = (−1)(−2) × 𝑎 4 = 2𝑎 𝑎=2 Therefore the third factor is 𝑥 + 2 𝑥+2=0 𝑥 = −2 The x coordinate of the point Q is −2.

69


c.

i.

Given √log 2 𝑥 = log 2 √𝑥 1

√log 2 𝑥 = log 2 (𝑥)2 1 √log 2 𝑥 = log 2 𝑥 2 Let 𝑦 = log 2 𝑥 therefore 1

√𝑦 = 2 𝑦 Squaring both sides we have 𝑦=

1 2 𝑦 4

4𝑦 = 𝑦 2 𝑦 2 − 4𝑦 = 0 𝑦(𝑦 − 4) = 0 𝑦 = 0, 4 When 𝑦 = 0, log 2 𝑥 = 0 𝑥 = 20 = 1 When 𝑦 = 4, log 2 𝑥 = 4 𝑥 = 24 = 16 Therefore 𝑥 = 1, 16

ii.

Given 𝑥 2 − |𝑥| − 12 < 0 Because of |𝑥| = {

𝑥, 𝑥 ≥ 0 −𝑥, 𝑥 < 0

We have 𝑥 2 − 𝑥 − 12 < 0, for 𝑥 ≥ 0 (𝑥 + 3)(𝑥 − 4) < 0 Critical points 𝑥 = −3, 4

70

𝑥 2 − 𝑥 − 12

−3

𝑥 2 + 𝑥 − 12 4

−4

3

𝑥 2 − 𝑥 − 12 < 0, has inequality for values of 𝑥 between −3 < 𝑥 < 4

We have 𝑥 2 + 𝑥 − 12 < 0, for 𝑥 < 0 (𝑥 − 3)(𝑥 + 4) < 0 Critical points 𝑥 = 3, −4, therefore for 𝑥 2 + 𝑥 − 12 < 0, has inequality for values of 𝑥, −4 < 𝑥 < 3

Taking the union of both sets we have inequality for values of 𝑥, −4 < 𝑥 < 4

Question 2

a.

i.

Given 𝛼 and 𝛽 are the roots of 𝑥 2 − 𝑝𝑥 + 24 = 0 (𝑥 − 𝛼)(𝑥 − 𝛽) = 0, 𝑥 2 − (𝛼 + 𝛽)𝑥 + 𝛼𝛽 a.

𝛼+𝛽 =𝑝

b.

𝛼𝛽 = 24, 𝛼 2 + 𝛽 2 = (𝛼 + 𝛽)2 − 2𝛼𝛽 = 𝑝2 − 2(24) = 𝑝2 − 48

ii.

Given 𝛼 2 + 𝛽 2 = 33, 𝑝2 − 48 = 33 𝑝2 − 81 = 0 (𝑝 − 9)(𝑝 + 9) = 0 𝑝 = 9, or − 9 71


b.

Given 𝑓(2𝑥 + 3) = 2𝑓(𝑥) + 3 and 𝑓(0) = 6 i.

When 𝑥 = 0, we have 𝑓(3) = 2𝑓(0) + 3 = 2(6) + 3 = 12 + 3 = 15

ii.

When 𝑥 = 2, we have 𝑓(2(3) + 3) = 2𝑓(3) + 3 𝑓(6 + 3) = 2(15) + 3 𝑓(9) = 30 + 3 = 33

iii.

When 𝑥 = −3 we have 𝑓(2(−3) + 3) = 2𝑓(−3) + 3 𝑓(−6 + 3) = 2𝑓(−3) + 3 𝑓(−3) = 2𝑓(−3) + 3 −3 = 2𝑓(−3) − 𝑓(−3) 𝑓(−3) = −3

c.

An even number can be express as 2𝑛 where 𝑛 is an integer. A odd number can be express as 2𝑛 − 1 where 𝑛 is an integer. For two consecutive numbers one must be even and the other odd, therefore The product of two consecutive integers can be 𝑘(𝑘 + 1) = 2𝑛(2𝑛 − 1). Where 2𝑛(2𝑛 − 1) = 2[𝑛(𝑛 − 1)]. Two times any number makes it even. Therefore the product of two consecutive integers is an even integer.

72

d.

Given to prove that 𝑛(𝑛2 + 5) is divisible by 6 When 𝑛 = 1 we have 1(12 + 5) = 6 which is divisible by 6 Therefore the statement is true for 𝑛 = 1 Assume statement is true when 𝑛 = 𝑘, therefore 𝑘(𝑘 2 + 5) is divisible by 6 When 𝑛 = 𝑘 + 1 we have (𝑘 + 1)[(𝑘 + 1)2 + 5] (𝑘 + 1)(𝑘 2 + 2𝑘 + 1 + 5) (𝑘 + 1)(𝑘 2 + 2𝑘 + 6) 𝑘 3 + 2𝑘 2 + 6𝑘 + 𝑘 2 + 2𝑘 + 6 𝑘 3 + 3𝑘 2 + 8𝑘 + 6 𝑘 3 + 5𝑘 + 3𝑘 2 + 3𝑘 + 6 𝑘(𝑘 2 + 5) + 3𝑘(𝑘 + 1) + 6 We assumed 𝑘(𝑘 2 + 5) is divisible by 6, 𝑘(𝑘 + 1) is an even integer which when multiplied by 3 is divisible by 6 and 6 is divisible by 6. Therefore when 𝑛 = 𝑘 + 1 the statement is true. Since the statement is true for 𝑛 = 1, 𝑘 and 𝑘 + 1, it is true for all positive integer n.

Question 3

a.

𝒂 = 𝑎1 𝒊 + 𝑎2 𝒋 and 𝒃 = 𝑏1 𝒊 + 𝑏2 𝒋 with |𝒂| = 13 and |𝒃| = 10 i.

|𝒂| = √(𝑎1 )𝟐 + (𝑎2 )𝟐 = 13 and |𝒃| = √(𝑏1 )𝟐 + (𝑏2 )𝟐 = 10 (𝒂 + 𝒃) ∙ (𝒂 − 𝒃) = [(𝑎1 + 𝑏1 )𝒊 + (𝑎2 + 𝑏2 )𝒋] ∙ [(𝑎1 − 𝑏1 )𝒊 + (𝑎2 − 𝑏2 )𝒋] = ((𝑎1 )2 − (𝑏1 )2) + ((𝑎2 )2 − (𝑏2 )2 ) = ((𝑎1 )2 + (𝑎2 )2 ) − ((𝑏1 )2 + (𝑏2 )2 ) = 132 − 102 = 169 − 100 = 69 73


ii.

2𝒃 − 𝒂 = 11𝒊 and so 𝒂 = 2𝒃 − 11𝒊 (𝒂 + 𝒃) = 2𝒃 − 11𝒊 + 𝒃 and (𝒂 − 𝒃) = 2𝒃 − 11𝒊 − 𝒃 (𝒂 + 𝒃) = 3𝒃 − 11𝒊 𝑎𝑛𝑑 (𝒂 − 𝒃) = 𝒃 − 11𝒊 (𝒂 + 𝒃) = 3( 𝑏1 𝒊 + 𝑏2 𝒋) − 11𝒊 and (𝒂 − 𝒃) = ( 𝑏1 𝒊 + 𝑏2 𝒋) − 11𝒊 (𝒂 + 𝒃) = (3𝑏1 − 11)𝒊 + 3𝑏2 𝒋 and (𝒂 − 𝒃) = (𝑏1 − 11)𝒊 + 𝑏2 𝒋 (𝒂 + 𝒃) ∙ (𝒂 − 𝒃) = [(3𝑏1 − 11)(𝑏1 − 11)] + [3𝑏2 × 𝑏2 ] = 69 = 3𝑏1 2 − 44𝑏1 + 121 + 3𝑏2 2 = 69 = 3(𝑏1 2 + 𝑏2 2 ) − 44𝑏1 + 52 = 0 |𝑏| = √(𝑏1 2 + 𝑏2 2 ) = 10 𝑠𝑜 𝑏1 2 + 𝑏2 2 = 100 = 3(100) − 44𝑏1 + 52 = 0 44𝑏1 = 352 𝑏1 = 8 𝑏1 2 + 𝑏2 2 = 100 𝑏2 = √100 − 82 = ±6 Therefore 𝑏 = 8𝒊 + 6𝒋 or 8𝒊 − 6𝒋 𝒂 = 2𝒃 − 11𝒊 𝒂 = 2(8𝒊 + 6𝒋) − 11𝒊 = 5𝒊 + 12𝒋 𝒂 = 2(8𝒊 − 6𝒋) − 11𝒊 = 5𝒊 − 12𝒋

b.

i.

Given the line L has equation 𝑥 − 𝑦 + 1 = 0 and the circle C has equation 𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0 The general equation of a circle is given by 𝑥 2 + 𝑦 2 − 2𝑓𝑥 − 2𝑔𝑦 + 𝑐 = 0 where f and g are the coordinates of the centre of the circle. From the equation of the circle the coordinate of the centre is (0, 1) Therefore from the line equation 𝑥 − 𝑦 + 1 = 0 when 𝑥 = 0 and 𝑦 = 1 74

We have 0 − 1 + 1 = 0 therefore it is shown that the line L passes through the centre of the circle.

ii.

L intersects C at P and Q therefore we solve simultaneously the equations of L and C. 𝑥 2 + 𝑦 2 − 2𝑦 − 15 = 0 -------- (1) 𝑥 − 𝑦 + 1 = 0 -------- (2) From (2) 𝑦 = 𝑥 + 1 ------- (3) Substitute (3) into (1) we have 𝑥 2 + (𝑥 + 1)2 − 2(𝑥 + 1) − 15 = 0 𝑥 2 + 𝑥 2 + 2𝑥 + 1 − 2𝑥 − 2 − 15 = 0 2𝑥 2 − 16 = 0 𝑥2 = 8 𝑥 = ±√8 = ±2√2 𝑦 = 1 ± 2√2 The coordinates of P and Q are (2√2, 1 + 2√2) and (−2√2, 1 − 2√2)

iii.

Given the parametric equations 𝑥 = 𝑏 + 𝑎 cos 𝜃 and 𝑦 = 𝑐 + 𝑎 sin 𝜃. We have cos 𝜃 =

𝑥−𝑏 𝑎

and sin 𝜃 = 𝑥−𝑏 2

cos2 𝜃 + sin2 𝜃 = 1 so (

𝑎

𝑦−𝑐 𝑎 𝑦−𝑐 2

) +(

𝑎

) =1

(𝑥 − 𝑏)2 + (𝑦 − 𝑐)2 = 𝑎2 The equation of C written in this form is given by 𝑥 2 + (𝑦 − 1)2 − 1 − 15 = 0 𝑥 2 + (𝑦 − 1)2 = 16 = 42 Therefore 𝑏 = 0, 𝑐 = 1, and 𝑎 = 4

75


iv.

Let the circle 𝐶2 has centre (𝑓, 𝑔) and radius 4 therefore 𝐶2 has equation (𝑥 − 𝑓)2 + (𝑦 − 𝑔)2 = 16 The circle 𝐶2 touches the line L at the centre of C (0, 1) and has the same radius Therefore (0 − 𝑓)2 + (1 − 𝑔)2 = 16 𝑓 2 + (1 − 𝑔)2 = 16 Also the line through the centres of the circle is perpendicular to the line L therefore it has gradient −1. 𝑔−1 = −1 𝑓−0

𝑔 − 1 = −𝑓 𝑓 =1−𝑔 Substituting 𝑓 = 1 − 𝑔 into

𝑓 2 + 𝑔2 − 2𝑔 − 15 = 0 we have

(1 − 𝑔)2 + (1 − 𝑔)2 = 16 2(1 − 𝑔)2 = 16 (1 − 𝑔)2 = 8 1 − 𝑔 = ±√8 = ±2√2 𝑔 = 1 − 2√2, 𝑜𝑟 1 + 2√2 𝑓 =1−𝑔 𝑓 = 1 − (1 − 2√2) = 2√2 𝑓 = 1 − (1 + 2√2) = −2√2 Therefore the centres are (2√2, −1 − 2√2) and (−2√2, 1 + 2√2 ) 2

2

The possible equations are (𝑥 − 2√2) + (𝑦 − (1 − 2√2)) = 16 2

2

(𝑥 − 2√2) + (𝑦 − 1 + 2√2) = 16 2

and

2

(𝑥 − (−2√2)) + (𝑦 − (1 + 2√2)) = 16 2

2

(𝑥 + 2√2) + (𝑦 − 1 − 2√2) = 16 76

Question 4

a.

i.

Given 8 cos 4 𝜃 − 10 cos2 𝜃 + 3 = 0 Let 𝑥 = cos 2 𝜃 then 8𝑥 2 − 10𝑥 + 3 = 0 8𝑥 2 − 6𝑥 − 4𝑥 + 3 = 0 2𝑥(4𝑥 − 3) − (4𝑥 − 3) = 0 (2𝑥 − 1)(4𝑥 − 3) = 0 1

𝑥 = 2 or

3 4 1

cos2 𝜃 = 2 , so cos 𝜃 = ±

1 √2 1

𝜋

√2

4

The acute angle is 𝜃 = cos −1 ( ) =

0 ≤ 𝜃 ≤ 𝜋 therefore 𝜃 is in the first and second quadrants . The angles are 𝜃 =

𝜋 4

𝜋

or 𝜃 = 𝜋 − 4 =

3

cos2 𝜃 = 4 , so cos 𝜃 = ±

√3

𝜃=

b.

i.

𝜋 6

4

for the second quadrant.

√3 2

The acute angle is 𝜃 = cos −1 ( 2 ) = The angles are 𝜃 =

3𝜋

𝜋

𝜋 3

or 𝜃 = 𝜋 − 6 =

5𝜋 6

for the second quadrant.

𝜋 3𝜋 𝜋 5𝜋 , , or 4 4 6 6

From the diagram angle 𝑄𝑅𝑆 𝑖𝑠 90𝑜 therefore triangle 𝑆𝑅𝐶 is similar to triangle 𝑅𝑄𝐵 and as a result angle 𝑅𝑆𝐶 is equal to angle 𝑄𝑅𝐵. 𝐵𝐶 = 𝐵𝑅 + 𝑅𝐶 𝐵𝑅 = cos 𝜃 , 6 𝑅𝐶 = sin 𝜃, 8

so 𝐵𝑅 = 6 cos 𝜃 so 𝑅𝐶 = 8 sin 𝜃

𝐵𝐶 = 6 cos 𝜃 + 8 sin 𝜃 77


ii.

Given |𝐵𝐶| = 7 then 6 cos 𝜃 + 8 sin 𝜃 = 7 Using 𝑅 sin(𝜃 + 𝛼) = 6 cos 𝜃 + 8 sin 𝜃 We have 𝑅 sin(𝜃 + 𝛼) = 𝑅 sin 𝜃 cos 𝛼 + 𝑅 sin 𝛼 cos 𝜃 Therefore 8 = 𝑅 cos 𝛼 𝑎𝑛𝑑 6 = 𝑅 sin 𝛼 𝑅 sin 𝛼 6 = tan 𝛼 = 𝑅 cos 𝛼 8 tan 𝛼 =

6 3 = 8 4 3

𝛼 = tan−1 (4) = 36.87𝑜 or 0.644 rad 𝑅 = √62 + 82 = 10 6 cos 𝜃 + 8 sin 𝜃 = 10 sin(𝜃 + 0.644) 10 sin(𝜃 + 0.644) = 7 7

sin(𝜃 + 0.644) = 10 7

𝜃 + 0.644 = sin−1 (10) 𝜃 + 0.644 = 44.42𝑜 or 0.775 rad 𝜃 = 0.775 − 0.644 𝜃 = 0.131 rad or 7.55𝑜

iii.

𝐵𝐶 = 6 cos 𝜃 + 8 sin 𝜃 = 10 sin(𝜃 + 0.644) Therefore the maximum value of BC is 10 because sin(𝜃 + 0.644) has a maximum value of 1. So |𝐵𝐶| = 15 is NOT possible

c.

i.

1−cos 2𝜃 sin 2𝜃

=

=

2 sin2 𝜃 2 sin 𝜃 cos 𝜃

sin 𝜃 = tan 𝜃 cos 𝜃 78

ii.

a.

1−cos 4𝜃

=

sin 4𝜃

=

b.

1−cos 6𝜃

=

iii.

2 sin 2𝜃 cos2 𝜃

sin 2𝜃 = tan 2𝜃 cos 2𝜃

=

sin 6𝜃

2 sin2 2𝜃

2 sin2 3𝜃 2 sin 3𝜃 cos 3𝜃

sin 3𝜃 = tan 3𝜃 cos 3𝜃

From the above identities it can be seen that 1 − cos 2𝑟𝜃 = tan 𝑟𝜃 sin 2𝑟𝜃

Therefore 1 − cos 2𝑟𝜃 = tan 𝑟𝜃 sin 2𝑟𝜃 1 = tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃 𝑛

𝑛

∑ tan 𝑟𝜃 sin 2𝑟𝜃 + cos 2𝑟𝜃 = ∑ 1 = 𝑛 𝑟=1

𝑟=1

Question 5 a.

lim

𝑥 2 +5𝑥+6

𝑥→−2 𝑥 2 −𝑥−6

(𝑥 + 2)(𝑥 + 3) 𝑥→−2 (𝑥 + 2)(𝑥 − 3) lim

(𝑥 + 3) −2 + 3 1 = =− 𝑥→−2 (𝑥 − 3) −2 − 3 5 lim

b.

2 Given 𝑓(𝑥) = {𝑥 + 1 if 𝑥 ≥ 2 𝑏𝑥 + 1 if 𝑥 < 2

i. ii. iii.

𝑓(2) = 22 + 1 = 5 lim 𝑓(𝑥) = lim+(𝑥 2 + 1) = 22 + 1 = 5

𝑥→2+

𝑥→2

lim 𝑓(𝑥) = lim−(𝑏𝑥 + 1 ) = 2𝑏 + 1

𝑥→2−

𝑥→2

79


iv.

if f is continuous at 𝑥 = 2 then lim 𝑓(𝑥) = lim− 𝑓(𝑥)

𝑥→2+

𝑥→2

5 = 2𝑏 + 1 4 = 2𝑏 => 𝑏 = 2 c.

Given 𝑦 = 𝑝𝑥 3 + 𝑞𝑥 2 + 3𝑥 + 2 and at 𝑇(1, 2)

𝑑𝑦 𝑑𝑥

=7

𝑑𝑦

We have 𝑑𝑥 = 3𝑝𝑥 2 + 2𝑞𝑥 + 3 When 𝑥 = 1,

𝑑𝑦

7 = 3𝑝 + 2𝑞 + 3 ------

(1)

4 = 3𝑝 + 2𝑞 ---------

(2)

2 = 𝑝 + 𝑞 + 3 + 2 -----

(3)

−3 = 𝑝 + 𝑞

--------

(4)

Multiple (4) by 2

−6 = 2𝑝 + 2𝑞 ------

(5)

Subtract (5) from (2)

10 = 𝑝

𝑑𝑥

= 7,

When 𝑥 = 1, 𝑦 = 2

−3 = 𝑝 + 𝑞 −3 = 10 + 𝑞 =>

𝑞 = −13

Therefore the equation is 𝑦 = 10𝑥 3 − 13𝑥 2 + 3𝑥 + 2 ii.

The gradient of the tangent at T is

𝑑𝑦 𝑑𝑥

=7

Therefore the gradient of the normal is −7 1

The equation of the normal is given by 𝑦 − 2 = − 7 (𝑥 − 1) 7𝑦 − 14 = 1 − 𝑥 7𝑦 + 𝑥 = 15 iii.

The line 𝑥 = 1 cuts the x-axis at 𝑥 = 1, therefore coordinates of M is (1, 0) the normal has equation 7𝑦 + 𝑥 = 15 therefore coordinates for N when 𝑦 = 0, 𝑥 = 15, M and N is on the x-axis therefore length of 𝑀𝑁 = 15 − 1 = 14

80

Question 6

a.

i.

Given 𝑦 = 𝑥(𝑥 2 − 12), 𝑦 = 𝑥 3 − 12𝑥 We have

𝑑𝑦 𝑑𝑥

= 3𝑥 2 − 12 𝑑𝑦

At the stationary points 𝑑𝑥 = 0, 3𝑥 2 − 12 = 0 3(𝑥 2 − 4) = 0 𝑥 = 2, −2 When 𝑥 = 2, 𝑦 = 2(22 − 12) = −16 When 𝑥 = −2, 𝑦 = −2((−2)2 − 12) = 16 Therefore the stationary points have coordinates (2, −16) and (−2, 16) ii.

𝑑𝑦

At the origin 𝑥 = 0,

𝑑𝑥

= −12 this is the gradient of the tangent. 1

The gradient of the normal is therefore − 12, and the equation of the normal at the origin is given by 𝑦 − 0 = − iii.

1 12

(𝑥 − 0), 𝑦 = −

1 12

𝑥.

The curve 𝑦 = 𝑥(𝑥 2 − 12) cuts the x-axis when 𝑦 = 0 therefore 𝑥(𝑥 2 − 12) = 0, and 𝑥 = 0, ±√12 = ±2√3 The area between the curve and the positive x-axis is given by √12

𝐴 = − ∫ 𝑥(𝑥 2 − 12)𝑑𝑥 0 √12

𝐴 = ∫ (12𝑥 − 𝑥 3 ) 𝑑𝑥 0

𝐴 = [6𝑥 2 −

𝑥 4 √12 ] 4 0 4

(√12) 𝐴 = 6(√12) − 4 2

= 6(12) −

144 = 36 sq. units 4 81


𝑎

b.

i.

𝑎

Using the result ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑎 − 𝑥) 𝑑𝑥 0

0

𝜋

𝜋

We have ∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin(𝜋 − 𝑥) 𝑑𝑥 0

0 𝜋

𝜋

sin(𝜋 − 𝑥) = sin 𝑥 therefore ∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥 𝑑𝑥 0 𝜋

ii.

a.

0

𝜋

∫ 𝑥 sin 𝑥 𝑑𝑥 = ∫(𝜋 − 𝑥) sin 𝑥 𝑑𝑥 0

0 𝜋

= ∫(𝜋 sin 𝑥 − 𝑥 sin 𝑥) 𝑑𝑥 0 𝜋

𝜋

= ∫ πsin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥 0

0 𝜋

𝜋

= 𝜋 ∫ sin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥 0 𝜋

𝑏.

0

𝜋

𝜋

∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫ sin 𝑥 𝑑𝑥 − ∫ 𝑥 sin 𝑥 𝑑𝑥 0

0

𝜋

0

𝜋

2 ∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 ∫ sin 𝑥 𝑑𝑥 0

0

= 𝜋[− cos 𝑥]

𝜋 0

= 𝜋[(− cos 𝜋) − (− cos(0))] = 𝜋[(−(−1)) − (−1)] 𝜋

2 ∫ 𝑥 sin 𝑥 𝑑𝑥 = 2𝜋 0 𝜋

∫ 𝑥 sin 𝑥 𝑑𝑥 = 𝜋 0

82

Cape Pure Maths Unit 1 Solutions 2011-2016.PDF

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