Set 17 - 2019 SEM 2 SMK ST PAUL, SEREMBAN (A)

954/2 STPM 2019

CONFIDENTIAL*

PEPERIKSAAN PERCUBAAN

SIJIL TINGGI PERSEKOLAHAN MALAYSIA BAHARU

SMK. ST. PAUL, SEREMBAN

Mathematics T, Paper 2

(Marking Scheme)

1. | + 2| − , < 0

Given ( ) = { − 2 , = 0 . Find the constant p and q such that f(-2) + f(2) =

2 + 3 , > 0 [4 marks]
4 and f is continuous from the left at x = 0.

|-2 + 2| - p + q(4) + 3 = 4 ; 4q – p = 1 ……..(1) M1
2- p = q – 2; p + q = 4 ……(2) M1

Solving (1) and (2) M1

p = 3, q = 1 A1 [4m]

2. The curve C has equation − = ( + )2. Given C has only one turning point

(a) Show that 1 + = 2 [3 marks]

2 +2 +1

(b) Hence, or otherwise, show that 2 = − (1 + )3. [3 marks]
2


(c) Hence state, with reason, whether the turning point is a maximum or minimum [2 marks]

(a) 1 − = 2( + ) (1 + ) differentiate w.r.t. x M1

M1

1−2 −2 = expand and factorise // rearrange A1
[3m]
1+2 +2

1 + = 2 ………..(1)

2 +2 +1

(b) 2 = −2(2+2 ) ………(2) differentiate (1) w.r.t. x M1
2 (2 +2 +1)2
M1
Substitute 2 + 2 + 1 = 2 into (2) and simplify
1+ A1
2 3 [3m]
2 = − (1 + )
2 M1
(c) 2 = −(1 + 0)3 < 0, substitute = 0 for stationary and explain. A1 [2m]



Turning point is a maximum.
3. (a) Find ∫1 (2 + 1) ln . Giving your answer in terms of e.
[3 marks]
(b) Show that ( 3 ) = 3 4 + 3 2 − 3. [5 marks]


Hence, determine the value of ∫04 4 .
(a) = ln , = 2 + 1


= 1 , = 2 + ∫1 (2 + 1) ln = [( 2 + ) ln ]1 − ∫1 (1) ( 2 + ) M1



= 2 + − [ 2 + ] = ….. (subs limit) M1
21
= 1 2 + 3
A1 [3m]
22

(b) ( 3 ) = 3 2 2 M1



= 3 2 (1 + 2 ) subs 2 = (1 + 2 ) or any suitable subs.

= 3 4 + 3 2 A1
= 3 4 + 3 2 − 3

4 =31 + 1 − 2 M1
M1
∫04 [ 3 ]04 ∫04 ∫04 A1 [5m]

1
= 3 [ 3 ]04 + [ ]04 − [tan ]04

= − 2
43

CONFIDENTIAL* 2

4. (a) Find ∫ sin cos . [2 marks]

(b) Show that the differential equation − = 2 sin cos may be reduce [6 marks]
M1
A1 [2m]
by means of substitution y = u sec x to = 2 sin cos cos . Hence, find M1
M1
the general solution of the differential equation, giving your answer in a form M1

expressing y in terms of x M1
A1
(a) ∫ sin cos = − cos + A1 [6m]

(b) 1 [3 marks]
= sec = ; = − + ; = ( + ) [4 marks]
1 ( + ) − ( ) (tan ) = 2 sin cos [3 marks]
M1

( + ) − ( ) (tan ) = 2 cos x sin cos M1

A1 [3m]
= 2 sin cos cos (Shown) M1 A1
M1
A1 [4m]
= ∫ 2 sin cos cos ; M1
M1
= [−2 ] − ∫ 2 A1 [3m]

= −2 + 2 cos +

y = -2 +2 sec cos + sec

5. Given that = [1 + sin ].

(a) Show that 2 = 1 − [( )2 + 1].
2


(b) By using result in (a), find the Maclaurin’s series for y, up to and including the

term in x2.

(c) Obtain the Maclaurin’s series for = [1 + sin ] , up to and including the

terms in x2, by using the standard series expansion for sin x and ln (1 + x).

(a) = 1 + sin , = cos

( )2
2 + = − sin …………….(1)
2
( )2
2 + = −( − 1) Subs sin x = ey – 1 into (1) and rearrange
2
[( )2
2 = 1 − + 1]
2

(b) let y = f(x), f(0) = 0, f ´(0) = 1, f ´´ (0) = -1 Subs x = 0 to find derivatives

= 0 + 1 + (−1) 2 + ⋯
2!
= − 1 2 + ⋯
2
(c) ≈ ln [1 + ( − 3)]
6
= ( − 3) − 1 ( − 3)2
62 6
= − 1 2
2

6. The equation x3-3x + 1 = 0 has 3 real roots. [2 marks]
(a) Show that one of the root lies between -2 and -1.
(b) By taking xo = -2 as the first approximation to one of the roots, use Newton- [4 marks]
Raphson method to find the root correct to 3 decimal places.
(c) Explain why Newton-Raphson method fails in the case where the first [1 mark]
approximation, xo = -1 M1
A1 [2m]
(a) let f(x) = x3-3x + 1, f(-2) = -1, f(-1) = 3 > 0, M1
f(-2).f(-1) < 0, f(x) continuous, therefore one of the root lies between -2 and -1

(b) f ´(x) = 3x2 - 3

CONFIDENTIAL* 3

x0 = -2, 1 = −2 − −23−3(−2)+1 = −1.8889 M1
3(−2)2−3 M1A1[4m]

2 = −1.8889 − −1.88893 − 3(−1.8889) + 1 = −1.8795

3(−1.8889)2 − 3

3 = −1.8794, 4 = −1.8794, ∴ the root is -1.879 (3 decimal places)

Section B [15 marks] Answer any one question in this section.

7. (a) (i) If = ∫01( 2 + 1)−23 , use trapezium rule with 3 ordinates to estimate
the value of I, giving your answer correct to 2 significant figures.
[4 marks]

(ii) By using trapezium with same ordinates as (i), estimate the volume of the
solid formed when the region bounded by the curve y = ( 2 + 1)−23, the axes

and the line x = 1 is rotated completely about x-axis, giving your answer correct

to 2 significant figures. [4 marks]

(b) A cylinder of radius r and height h is inscribed in a sphere of fixed radius a

such that all the points of the circumference of base and top of the cylinder are

in contact with the inner surface of the sphere.

Show that the volume, V, of the cylinder can be expressed as

= [ 2ℎ − 1 ℎ3] [7 marks]

4

Hence, find , in terms of a, the exact maximum volume as h varies.

(a) ≈ 1 (1) [1 + 2(0.716) + 0.354] find d, any two ordinates correct, correct formula M1A1M1

22

= 0.70 (2 sig. fig) A1 [4m]

(ii) = ∫01 ( 2 + 1)−3 correct formula for volume M1

≈ 1 (1) [1 + 2(0.512) + 0.125] any two ordinates correct, correct formula A1 M1

22

= 1.7 (2 sig. fig) A1 [4m]

(b) 2 = 2 − ℎ2 M1

4

= 2ℎ

= ( 2 − ℎ2) ℎ = ( 2ℎ − 1 ℎ3) M1 A1
44 M1
= ( 2 − 3ℎ2) Find first derivative and equate to 0 A1
ℎ 4
( 2 − 3ℎ2) = 0; ℎ = 2√3 M1A1[7m]
, find h in terms of a
43
2
ℎ2 = (− 3ℎ) < 0 , maximum volume.

2 3

= ( 2 2√3 − 1 (2√3 ) ) = 4√3 3

3 43 9

8. (a) A child was blowing bubbles and noticed that if he blows too hard, the bubble

would burst immediately. However, if he were to blow gently, the bubble

increases in size till its optimal volume and detaches itself off the blowing stick.

At time t minutes, the radius of the bubble blown is r cm, assuming that all

bubbles blown are spherical in shape
(i) If the child blows gently at a rate of ( − 3) 3 −1 , where λ is a

constant. Show that 4 2 = − 3. [3 marks]
[4 marks]


(ii) Find the general solution for the differential equation derived in (i)

(iii) Given that the initial volume of a bubble is negligible and the optimal

volume of a bubble is 3, find the time taken for a bubble to detach itself

6

off the blowing stick. [4 marks]
[4 marks]
(b) Show that √1 − 2 = √1−− 2. Hence, show that

3
√1− 2
∫ = − 1 (2 + 2)√1 − 2 + .

3

CONFIDENTIAL* 4

(a)i = 4 3; = 4 2 M1
3 M1
= × ; ( − 3) = 4 2 × A1 [3m]
M1
4 2 = − 3 M1 M1
A1 [4m]
4 2 B1
(ii) ∫ − 3 = ∫ 1 B1
M1
− 4 ln( − 3) = + ; simplify
A1 [4m]
3
− 3 = −34 B1

(iii) t = 0, r = 0, A = λ M1
M1
− 3 = −43 A1 [4m]

max = ; 4 3 = ; 3 =
6
36 8
− = −43 ; − 3 = ln 7

8 48

= − 4 ln 7 = 0.178 min(3 . . )

38
(b) √1 − 2 = 1 (1 − 2)−12(−2 )
2
−
= √1− 2

= − 2; = − ; = −2 ; = √1 − 2
√1− 2
3
∫ √1− 2 = ∫(− 2) (√1−− 2)

= [− 2√1 − 2] − ∫(−2 )√1 − 2

= [− 2√1 − 2] − 2 (1 − 3 +

3 2)2

= − 1 √1 − 2(2 + 2) +
3