2014 2 JOHOR SMK Dato Jaafar JohorBahru MATHS A

1) a) f(x) = 1 – x3 f ‫(׳‬x) = - 3x2 =− = −

dy = −4 dx Further differentiation

2 3x 2 dx − 3 ∫ 1 − x3

 2  d 2 y dy  dy   1 −  2 +  −2 ln y  = 2 + 2 ln x y  dx dx  dx   when x = y = 1

1 4 1 − x3 ) 2 + C ( 3

d2y = 32 dx 2

dV b) =10 dt

dA dr dA = × dV dV dr 1 = ( 8π r ) 4π r 2 1 = 4

A = 4πr2

dA = 8 πr dr

4 V = π r3 3 dA = 4π r 2 dr

3. 6x + 16 = A(2x + 4 ) + B A= 3, B = 4 6x + 16 = 3( 2x + 4 ) + 4 1

∫x

=

2

−2 1

∫

=

−2

6 x + 16 dx + 4 x + 13

3( 2x + 4) + 4 x 2 + 4 x + 13

dx

1

=

dA dV dA = × dt dt dV 1 = 10 × = 2.5m 2 s −1 4

1

3(2 x + 4) 4 ∫−2 x 2 + 4 x + 13 dx + −∫2 x 2 + 4 x + 13 dx 1

1  x + 2  = 3[ln( x 2 + 4 x + 13)]1−2 + 4  tan −1   3  3   −2  4π  = 3ln 2 +   3 4 

c)

= 3ln 2 + 4

π 3

4. y = ( tan x + sec x )

2

-5

5

=

10

-2

2

dy = 2 ( tan x + sec x ) ( sec2 x + tan x sec x ) dx

= 2sec x  2 tan x sec x + tan 2 x + sec 2 x 

-4

= 2sec x ( tan x + sec x )

Area= ∫ xdy

dy 2y = dx cos x

 y + 4 y2  − dy =∫ 2 4 −2  4

4

=

= 9 unit 2. y = x2 + 2 ln xy

dy 2  dy  = 2x + x + y  dx xy  dx   2  dy 2 = 2x + 1 −  y  dx x  1  2 x +  dy x =  2 dx 1− y when x = y =1

dy = 2 y (shown) dx From * cos x

4

 1  y2 1  y3   + 4y −   2 2  −2 4  3  −2 2

2

*

cos x

d 2 y dy dy + ( − sin x ) = 2 dx 2 dx dx

cos x

d2y  d3y d2 y dy d2y sin x sin x cos x 2 + − − − + − = ( ) ( ) ( )   2 dx dx 3 dx 2 dx 2  dx 

d 3 y dy d2y − cos x = 2 dx 3 dx dx 2 y(0) = 1 y‫(׳‬0) = 2 y‫(׳׳‬0) = 4 y‫(׳׳׳‬0) = 10 cos x

f(x) = 1 + 2x + 2x2 +

5 3 x +… 3

7.

5. The vertical asymptote, x = -1 2x + 1 lim x →±∞ x + 1 1 2+ x = lim x →±∞ 1 1+ x =2 The horizontal asymptote, y = 2.

dx = −k ( x − x0 ) dt dx ∫ x − x0 = − ∫ kdt ln ( x − x0 ) = −kt + c ln ( x − 27 ) = −kt + c

6

4

y=-1

2

-5

5

-2

-4

x=-1 when x = 0, y = 1 When y = 0, x = - ½ (0, 1), and ( - ½ , 0 ) Let y =

2x + 1 x +1

x=

2 y +1 y +1

x − 27 = 36e x = 27 + 36

1 − t 6

1 − t 6

( proven)

(a)

x – 27 −

7

= 36 6 = 11.2°C The fall of temperature after being left in the room for 7 minutes is 11.2°C t

−

(b) 36e 6 = 2 1

1− x y= x−2 ∴ f −1 : x →

When t = 0, x = 63 c = ln36 when t = 6ln 2, x = 45, 1 k= 6 1 ln ( x − 27 ) = − t + ln 36 6

1− x ,x ≠ 2 x−2

t

e 6 = 18 t = 6 ln 18 t = 17.3 minutes The time that elapses is 17.3 minutes.

D f −1 = { x : x ∈ R, x ≠ 2} R f −1 = { y : y ∈ R, y ≠ −1}

8. y =

25 − x 2 x

6. y = ex dy = ex dx dy = ea dx when x = a, y = ea The equation of the tangent at the point where x = a is, y – ea = ea ( x – a ) y = ea( 1+x – a ) At (0,0), ea ( 1 – a ) = 0 a=1 The equation of tangent which passes through origin is y=e(1+x–1) y = ex For y = mx to intersect y = ex at two distinct points, m > e. when x = a,

0 0.5 1.0

25 − x 2 5 4.9749

24

1.5

91 4

2.0

21

2.5

75 4 4

3.0 3.5 4.0 4.5 5.0

51 4 3 19 4 0

5−0 1 = 10 2

h=

11   5 + 2 ( 36.30644 )  2 2 = 19.40322 unit2 = 19.4 b) underestimated

Area =

5

c) Area =

∫

25 − x 2 dx

0

π 2

=

∫5

1 − sin 2 θ ( 5cos θ dθ )

0

π 2

= 25 ∫ cos 2 θ dθ 0

π

25 2 (cos 2θ + 1)dθ = 2 ∫0 π

25  sin 2θ 2 +θ  =  2  2 0 = 6.25π unit2 (d)

1 2 π ( 5 ) ≈ 19.4032 4 π ≈ 3.105

π ≈ 3.11