2014-2-WPERSEKUTUAN-SMKMethodistKL_MATHS QA

Section A [45 marks]

Answer all questions in this section.

1 Evaluate the following limits.

(a) 2 6 8

lim44

x x

xx

[3 marks]

(b) 3

2

1lim

0 1

x

x

e

x e

[3 marks]

2 A curve 2

px ry

x x

where p and r are constants has the value of 3

dy

dx at the point (-1, 7).

(a) Calculate the values of p and r. [5 marks]

(b) Determine all the asymptotes of the curve. [2 marks]

(c) Sketch the graph of 2

px ry

x x

. [2 marks]

3 The volume of a solid cylinder of radius r cm is 250 cm.

(a) Show that the total surface area, S cm2, of the cylinder is given by

2 5002S r

r

. [3 marks]

(b) Given that r can vary, find the stationary value of S. [3 marks]

(c) Determine the nature of this stationary value. [3 marks]

4 The quantities x and y are related by the differential equation 2 24 cosdy

y xdx

. Find the general solution

of the equation. If y = 0 when x = 0, show that 2tan sin cosy x x x .

[6 marks]

5 (a) Given that ln(cos2 )y x , find 4

4

d y

dx. [4 marks]

(b) Use Maclaurin’s theorem to show that the first two non-zero terms in the expansion, in ascending

powers of x, of ln(cos2 )x are 2 442

3x x . [3 marks]

(c) Hence, find the first two terms in the expansion, in ascending powers of x, of 2ln(sec 2 )x .

[2 marks]

6 (a) Show that the root of the equation 1

sin2

x x lies between 1 and 2. [2 marks]

(b) Taking x = 1.4 as a first approximation, use the Newton-Raphson method to find this root, correct to

four decimal places. [4 marks]

Section B [15 marks]

Answer only one question in this section.

7 At time t, the population of Town A is x. The population increases by birth at a rate of bx and decreases by

death at a rate of 21

2mx fx , where b, m and f are constants.

Write down a differential equation that relates x and t. [2 marks]

By assuming that 3

2

mb and

mf

N , show that the differential equation can be written in the form of

21

dx xx

m dt N

. [3 marks]

Solve the equation, given that 1

4x N when t = 0. [8 marks]

Find, in terms of m, the value of t when 1

2x N . [2 marks]

8 A curve has the equation 4 cos2y x x .

(a) Find an exact equation of the tangent to the curve at the point on the curve where 4

x

.

[4 marks]

(b) The region shaded on the diagram below is bounded by the curve 4 cos2y x x and the

x-axis from x = 0 to 4

x

.

By using integration by parts, find the exact value of the area of the shaded region.

[5 marks]

(c) When the shaded region is rotated 2 radians about the x-axis, find the volume of the solid generated.

[6 marks]

STPM TRIAL SEMESTER 2 2014 MATHEMATICS T MARKING SCHEME

Section A

1. (a) 2 6 8

lim44

x x

xx

=

2 4lim

44

x x

xx

(M1)

= -(4 – 2) (M1)

= -2 (A1)

(b) 3

2

1lim

0 1

x

x

e

x e

=

21 1lim

0 1 1

x xx

x x

e e e

x e e

(M1)

= 1 1 1

1 1

(M1)

= 3

2 (A1) [Total = 6]

2. (a) At (-1, 7) 71(1)

p r

r = p – 7 (M1)

–

2

7pxy

p

x x

22

2 7 2 2

2

x x p px p xdy

dx x x

(M1)

22

1 1 2 ( 1) 7 2( 1) 23

( 1) 1 2

p p p

(M1)

p = 3 (A1)

r = 3 – 7 = 4 (A1)

(b) 0, 2, 0x x y (B2)

(c) 2

px ry

x x

(D1 shape)

(D1 asymptotes, x-int)

[Total = 9]

3. (a)

2

2

250

250

r h

hr

2

2

2

2

2

2 2

2502 2

5002

S r rh

r rr

rr

(b) 2

5004

dSr

dr r

(M1)

S is stationary when 0dS

dt

2

5004 r

r

-2 0

y

x

(M1)

(M1)

(A1)

r = 5 (M1)

S = 2 5002 5

5

= 150 cm2 (A1)

(c) 2

2 3

10004

d S

dr r

(M1)

When r = 5, 2

2 3

10004 12 0

5

d S

dr

(M1)

S is a minimum value. (A1) [Total = 9]

4. 2 24 cosdy

y xdx

2

2cos ( 1)

4M

dyx dx

y

2

2

2sec 1cos2 1 ( 1)

24secM

dx dx

1 1 sin 2( 1)

2 2 2M

xx b

1 sin 2tan ( 2 )

2 2

y xx c c b

(M1)

Given x = 0, y = 0, c = 0 (M1)

2sin costan

2 2

y x xx

2tan sin cosy x x x (A1) [Total = 6]

5. (a) ln cos2y x

2sin 2

2 tan 2cos2

dy xx

dx x

(M1)

2

2 2

22 2sec 2 4sec 2

d yx x

dx (M1)

3

2

34 2sec2 (2sec2 tan 2 ) 16sec 2 tan 2

d yx x x x x

dx (M1)

4

2 2

416 sec 2 2sec 2 tan 2 2sec2 2sec2 tan 2

d yx x x x x x

dx

(M1)

2 2 232sec 2 sec 2 2tan 2x x x

(b) x = 0, y = 0

2 3 4

2 3 40, 4, 0, 32

dy d y d y d y

dx dx dx dx (M1)

2 4

2 4

4 32ln cos2 ... (M1)

2! 4!

42 ... (A1)

3

x x x

x x

(c) 2 2 44ln sec 2 2ln cos2 2 2 ... ( 1)

3Mx x x x

= 2 48

4 ...3

x x (A1) [Total = 9]

Let

2

2 2 2

2tan

2sec

4 4 1 tan 4sec

y

dy d

y

6. (a) Let f(x) = sin x +1

2 x .

f(1) = 0.341 > 0

f(2) = -0.591 < 0 (M1) There is a change in signs. The root lies between 1 and 2. (A1)

(b) 1.4 ' cos 1ox f x x (B1)

1

1.41.4

' 1.4

fx

f = 1.502947 (M1)

x2 = 1.497317

x3 = 1.497300 x4 = 1.497300 (M1)

The root = 1.4973 (4 d.p.) (A1) [Total = 6]

7. 21

62

dxx mx fx

dt

(M1)

216

2m x fx (A1)

3,

2

m mb f

N

23 1

2 2

dx m mm x x

dt N

(M1)

21 1

2 2

mmx x

n

1

12

xmx

N

(M1)

2

1dx x

xm dt N

(A1)

2

N mdx dt

x N x

(M1 – isolate variables)

1 1

2

mdx t c

x N x

(M1- partial fractions)

ln2

x mt c

N x

(M1A1)

1, 0 ln3

4x N t c

(M1A1)

2

2

2 3ln or

3

mt

mt

x Net x

m N xe

(M1A1)

3

1 2 2ln12

2

N

x N tm

N

(M1)

2ln3t

m

(A1) [Total = 15]

8. (a) 4 cos2y x x

4 2sin 2 4cos2dy

x x xdx

(M1)

4 cos2 2 sin 2x x x

4 0 24 2

dyx

dx

(A1)

Equation of tangent is 0 24

y x

(M1)

24

y x

(A1)

(b) A = 4

4 cos2o

x x dx

(M1)

= 44

00

sin2 sin24

2 2

x xx dx

(M1)

= 4

0

1 cos24 2

4 2 2

x

(M1)

= 1

2 2 02

(M1)

= 1 or 0.5708 unit2 (A1)

(c) V = 4 2

4 cos2o

x x dx

(M1)

= 4

2 116 cos4 1

2o

x x dx

(M1)

= 4

2 28 cos4o

x x x dx

= 3 4442

00 0

sin 4 18 2 sin 4

4 4 3

x xx x x dx

(M1)

= 344

00

cos4 cos40 2 2 2 8

4 4 192

x xx dx

(M1)

= 4

4

0

sin 4( 1)

4 4 24

x

(M1)

= 2 4

4 24

= 4 2 2 2

124 4 4 6

or 1.591 cm3 (A1)