Pentaksiran Semester 2 STPM 2021 Kali Kedua

QUESTION 1

No Working Marks

1 (a) M1

lim 2 − x2 − 5 = lim 2 − x2 − 5  2 + x2 − 5 M1
x→−3 x + 3 x→−3 x + 3 2 + x2 − 5 (Subst x =-3)

4 − x2 + 5 A1
( )= lim
x→−3 ( x + 3) 2 + x2 − 5 M1
(Using laws of
(3+ x)(3− x)
limit)
( )= lim M1
x→−3 ( x + 3) 2 + x2 − 5 (Subst x = 4)
A1
=6 6 marks
2+ 4

=3
2

(b)

f (x)−5

lim = 1
x→4 x − 2

lim f ( x) − lim5

x→4 x→4 = 1
lim x − 2

x→4

lim f ( x) − 5

x→4 = 1
4−2

lim f ( x) = 2 + 5

x→4

=7

QUESTION 2 M1
A1
2(a). y = −3 (2 cos 2x + 3 sin 2x) [2]
= −3 (−4 sin 2 + 6 cos 2 ) − 3 −3 (2 cos 2 + 3 sin 2 ) either one M1

M1

= −13 −3 sin 2 A1
M1
(b). x = t − 1 y= 2 + 1
[6]

= 1 + 1 = 2 − 1
2 2

= 2+1 = 2 2−1
2 2

= 2 2−1

2
2+1

2

= 2 2−1
2+1

= 2( 2+1)−3
2+1

= 2 - 3
2+1

Since 0 < 3 < 3
2+1

-3 < - 3 < 0
2+1

-1 < 2 - 3 < 2
2+1

∴ −1 < < 2, ∴ a = −1, b = 2 A1A1


QUESTION 3

3. Given = 1, show that = 32 3 . [3 marks]
[4 marks]
( 2+32)2 ( 2+32)2 1
1
Find the exact value of the area of the region bounded by the curve = 1 3 , the x-axis

( 2+32)2

and the lines x = 2, x = 7.

A. ( 2 + 1 − ( )( 2 + 32)−21( )

= 32)2

2 + 32

= 2+32− 2
1
( 2+32)( 2+32)2

∴ = 32 1 [3m]
3
( 2+32)2

Area =∫27 1 1
3 1
1
( 2+32) 2

7

= 1 [ 1]
32
( 2+32)2 2

= 1 [( 7 1) − ( 2 1)]
32
(49+32)2 (4+32)2

= 1 units2 1 [4m]

72

QUESTION 4
4.

1 + 1 M1
∫ 2 = ∫ M1
− −1 = − − + ( )(− − ) − ∫ − − (1) M1
M1
Integration by parts correctly M1
A1
− −1 = − − − − − − + 1 A1
1 = − (2 + ) +
where = − 1


Find correctly when = 0, = 1. Where = −1

= 2 + −

QUESTION 5

5 f( ) = ln(1 + sin ), (0) = 0 5
f ′( ) = cos , f ′(0) = 1
M1
1+sin M1
(2nd and 3rd
f ′′( ) = (1+sin )(− sin )− cos (cos ) derivatives)
(1+sin )2
A1
f ′′( ) = − 1 , ′′(0) = −1 [for all
f(0),f ’(0),
1+sin x f ”(0),f ”’(0)]

f ′′′( ) = (1 cos x , ′′′(0) = 1
+ sin )2

f( ) = f(0) + f ′(0) + f ′′(0) 2 + f ′′′(0) 3 + ⋯ M1 [subs
2! 3! f(0),f’(0),
f’’(0),f’’’(0)]
( ) = 0 + − 1 2 + 1 3 + ⋯
2 6 A1

lim 2 −ln(1+sin )2= lim 2 −2ln(1+sin ) M1 3
2 2− 4 2 2− 4 M1
→0 →0 A1

=lim 2 −2(x−12x2+16x3+⋯ )
2 2− 4
→0

lim 2(1−13 +⋯ ) = lim 1−31 +⋯
2(2− 2) 2− 2
→0 →0

=12

QUESTION 6

Q6 By using a graphical method, show that equation 2x − 2x + 1 = 0 D1 correct graph
has only one real root. Show also that this equation can also be
written in the form (i) x = 1 2x + 1 and (ii) x = 1 (4x2 −1) .
22
Determine which of the two form would be suitable for the use of
iterative formula. Hence, estimate this root correct to three decimal
places given that xo = 1.
[9 marks]

y y= 2x

y= 2x + 1

−1 x
2
The graphs = 2 and = 2x + 1 have
only 1 point of intersection so thAe 1equation

2x − 2x + 1 = 0 has only one real root .

2x − 2x + 1 = 0 2x − 2x + 1 = 0

2x = 2x + 1 2x = 2x + 1
4x2 = 2x +1
(square both sides) B1 both shown with
correct working
x = 1 (4x2 – 1)
1 2x +1 2

x=
2

1 2x +1 F2= 1 (4x2 – 1)
2
Let F1=
2 F2’= 1 (8x)
2
F1’ = 11 −1
F2’= 4x
. (2x + 1) 2 (2)
22

F1’ = 1

2 2x +1

Substitute x=1

F1’ = 1 F2’= 4(1) = 4

=0.2887 1< F2’
2 2(1) + 1 The iteration diverges

0 < F1’< 1 M1 find correct
The iteration converges iterative

1 2x + 1 as the iterative formula A1

Choose x =
2

ALTERNATIVE METHOD

1 2(1) + 1 =0.8660 1

x0 = 1, x1 = x0 = 1, x1 =
2
2

(4(1)2 – 1) =1.5

1 2(0.8660) + 1 =0.8264 x2 = 1 (4(1.5)2 – 1) = 4
2
x2 =
2

1 2(0.8264 ) + 1 =0.8144 x3 = 1 (4(4)2 – 1) = 31.5
2
x3 =
2
M1 find correct
x4 =0.8107 , x5 =0.8095 , x4 =1984 , iterative
x5 =7872511.5 The iteration diverges A1
x6 =0.8092 , x7 =0.8091 ,
x8 =0.8090 , x9 =0.8090 M1 A1
The iteration converges
M1 stopping
1 2x + 1 as the iterative formula A1

Choose x =
2

Iterative formula 1 2xn + 1

xn+1 =

2

1 2(1) + 1 =0.8660

x0 = 1, x1 =
2

1 2(0.8660) + 1 =0.8264

x2 =
2

1 2(0.8264 ) + 1 =0.8144

x3 =
2

x4 =0.8107 , x5 =0.8095 , x6 =0.8092 , x7 =0.8091 ,
x8 =0.8090
x9 =0.8090

So, the root x = 0.809 (3d.p)

QUESTION 7

7 = − 2+ 2



Substitute = , function

= + B1
M1
2+ 2 2 A1
∴ + = − 2 M1
A1
2(1+ 2)
+ = − 2 M1
A1
+ = − (1+ 2) M1

+ = − 1 − A1
M1
∴ = − (1+2 2)
A1

∫ d = − ∫
1+2 2
1 4
4 ∫ 1+2 2 = − ∫

1 ln(1 + 2 2) = − ln + ln ;
4
2 2
1 ln (1 + 2 ) = − ln + ln
4

Given that = 2, = 4

1 ln(1 + 8) = − ln 2 + ln
4
ln = 1 ln 9 + ln 2
4
= 1 ln 3 + 1 ln 4
22
ln = 1 ln 12
2
2 2
1 ln (1 + 2 ) = − ln + 1 ln 12
4 2
ln (1 + 2 22) = −4 ln + 2ln 12
2 22)
ln (1 + = ln 144
4
2 2
1 + 2 = 144
4
4 + 2 2 2 = 144

2( 2 + 2 2) = 144

When = ; 2( 2 + 2 2) = 144
2(3 2) = 144
4 = 144

3

2 = 12

√3

1

= 2(3)4

11 M1

= ∫22(3)4 (1424 − 2 4) − ∫22(3)4 2 A1
M1
1 2
2 A1
= ∫22(3)4 ( 7 22 − − 2)

1 3 2)

= ∫22(3)4 ( 7 22 − 2

1

= (− 72 − 3)2(3)4

2 2

3

= {[− 72 − 8(3)4] − [−36 − 4]}

1 2

2(3)4

= {− 36 − 4 3 + 40}

1 (34)

34 3 3

= {− 36 (34) − 4 (34) + 40}

3 3

= {40 − 16 (34)}

= 8 [5 − 2 3 unit3

(34)]

QUESTION 8

Q8 Given that y = ln (1 + tan x).

(a) Show that dy = ey + 2(e−y −1) and
dx

d3y = (e y − 2e−y ) d2y + (e y + 2e−y ) dy 2 . [5 marks]
dx3 dx2  dx 

(b) Hence, find the Maclaurin series of ln (1 + tan x) in ascending powers of x up to

and [3 marks]
including the term in x3.

(c) Use the Maclaurin series in part(b) to estimate an approximate value for

01ln(1+ tan x) dx [2 marks]

(d) Use the trapezium rule with five ordinates to find another approximate value for

10 + tan x) dx , giving your answer correct to three decimal places. [3 marks]

ln(1

(e) If x is sufficient small for x4 and higher powers of x to be neglected, show

that

ln( 1+ 2x )2  2x − 3x2 + 4x3. [2 marks]
1+ tan x

8(a) y = ln (1 + tan x) ➔ dy = 1 (sec2 x) M1
dx 1 + tan x differentiate

dy = 1 (1 + tan2 x) e y = 1 + tan x A1
dx 1 + tan x e y −1 = tan x M1
differentiate
dy = 1 (1 + [e y −1]2 ) M1
dx e y A1

dy = 1 (1 + e2y − 2e y + 1)
dx e y

dy = e y + 2e−y − 2
dx
dy = ey + 2(e−y −1)
dx

d 2 y = e y dy + 2(e−y[− dy ])
dx2 dx dx

d 2 y = dy (e y − 2e−y )
dx2 dx

d 3 y = (e y − 2e−y ) d 2 y + dy (e y dy − 2e−y[− dy ])
dx3 dx2 dx dx dx

d3y = (e y − 2e−y ) d2y + (e y + 2e−y ) dy 2
dx3 dx2  dx 

8(b) dy = 1 (1 + tan2 0) = 1 M1 find
Substitute x=0 : y = ln (1 + tan 0)= 0 ; dx 1 + tan 0 coefficient ,
at least 3
; d 2 y = (1)(e0 − 2e−0 ) = –1 ; correct
dx2
M1 sub his
d 3 y = (e0 − 2e−0 )(−1) + (e0 + 2e−0 )(1)2 = 4 coef in
correct MS
dx3 formula
A1
Maclaurin’s Series : y = f (0) + f ’(0).x + f ' ' (0) x2 + f ' ' ' (0) x3 + ….
2! 3!

y = ln (1 + tan x) =0 +(1).x + (−1) x2 + (4) x3 + ….
2! 3!

= x − 1 x2 + 2 x3 + ….
23

8c 01ln(1 + tan x) dx = 1  x − 1 x2 + 2 x3 + .....dx
 2 3 
0

 x2 x3 x4  1 M1
 2 6 6 ..... his correct
= − + + integration
A1
0 B1

 (1) 2 − (1)3 + (1)4 +  − (0)  1 M1 correct
=  2 6 6 ..... 2 TR
A1 CAO
8d h = 1 − 0 = 1 =0.25
44 M1
Expansion
x y = ln (1 + tan x) ln(1+2x)
0 y0= 0
0.25 y1= 0.22741
0.5 y2= 0.43587
0.75 y3= 0.65835
1 y4= 0.93899

01ln(1+ tan x) dx

 1  1 0 + 2(0.22741+ 0.43587 + 0.65835) + 0.93899

2 4

= 0.448 (3d.p.)

8e ln( 1+ 2x )2 = 2ln( 1+ 2x ) = 2 [ln(1+ 2x) − ln(1+ tan x)]
1+ tan x 1+ tan x

= 2 [(2x) − (2x)2 + (2x)3 + ....)− (x − 1 x2 + 2 x3 + ....)]
23 23

= 2 [2x − 2x2 + 8 x3 + ....− x + 1 x2 − 2 x3 + ....] A1
3 23

= 2 [x − 3 x2 + 2x3 + ....]
2

 2x − 3x2 + 4x3