PSPM 1 Q&A

PSPM I QS 015/1 Session
2014/2015

3 CHOW CHOON WOOI

5. (a) State the interval for x such that the expansion for (4 + 3 )2 is valid.

3

(b) Expand (4 + 3 )2 in ascending power of x up to the term in 3.

3

(c) Hence, by substituting an appropriate value of x, evaluate (5)2 correct to three decimal
places.

SOLUTION

a) (4 + 3 = [4 (1 + 3 3

3 )2 4 )]2

ℎ
3

−1 < 4 < 1

44
− 3 < < 3

b) (4 + 3 = [4 (1 + 3 3

3 )2 4 )]2

3
3 32
= 42 (1 + 4 )
3
32
= 8 (1 + 4 )

[1 (23) 3 (23) (12) 3 2 (23) (21) (− 21) 3 3 ]
1! (4 2! (4 3! (4
= 8 + ) + ) + ) + ⋯

= 8 [1 + 9 + 3 9 2 ) − 3 27 3) + ⋯ ]
8 8 (16 48 (64

= 8 [1 + 9 + 27 2 − 81 3 + ⋯ ]
8 128 3072

9 27 2 − 27 3 + ⋯ ]
= 8 [1 + +
8 128 1024
27 27
= 8 + 9 + 16 2 − 128 3 + ⋯

c) (4 + 3 = 8 + 9 + 27 2 − 27 3 +⋯

3 )2 16 128

33

(4 + 3 )2 = (5)2
4 + 3 = 5

Page 9

101

PSPM I QS 015/1 Session
2014/2015

3 = 1
1

= 3

(4 + 3 = 8 + 9 + 27 2 − 27 3 + ⋯
16 128
3 )2

3 1 27 1 2 27 1 3 CHOW CHOON WOOI

12
[4 + 3 (3)] = 8 + 9 (3) + 16 (3) − 128 (3) + ⋯

3 1 27 1 2 27 1 3
[5]2 = 8 + 9 (3) + 16 (3) − 128 (3) + ⋯

= 11.180

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102

PSPM I QS 015/1 Session
2014/2015
6. (a) Given ( ) = 2 + 1 and ( ) = 2 + 2x − 1.
(i) Find ( − )(x).
(ii) Evaluate (3 − 2 )(1).

(b) Given ( ) = √2 + 1. State the domain and range of ( ). Hence, on the same axes, CHOW CHOON WOOI

2

sketch the graph of ( ) and −1(x).

SOLUTION

a) ( ) = 2 + 1 and ( ) = 2 + 2x − 1

i. ( − )(x) = ( ) − ( )
= (2 + 1 ) − ( 2 + 2x − 1)
= 2 + 1 − 2 − 2x + 1
= − 2 + 2

ii. (3 − 2 )( ) = 3 ( ) − 2 ( )
= 3( 2 + 2x − 1) − 2( 2 + 1)
= 3 2 + 6x − 3 − 4 − 2
= 3 2 + 2x − 5

(3 − 2 )(1) = 3(1)2 + 2(1) − 5
=3+2−5
=0

b) ( ) = √2 + 1

2

1
: 2 + 2 ≥ 0

1
2 ≥ − 2

1
≥ − 4

{ : 1
= ≥ − }
4

= { : ≥ 0}

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103

PSPM I QS 015/1 Session
2014/2015
CHOW CHOON WOOI
y −1( )
( )

1 x
−4

1
−4

Page 12

104

PSPM I QS 015/1 Session
2014/2015
7. Let = + be a nonzero complex number.

a. Show that 1 = |zz̅|2.


b. Show that if z̅ = −z, then z is a complex number with only an imaginary part.

c. Find the value of a and b if (2 − ) = (̅z + 1)(1 + i). CHOW CHOON WOOI

SOLUTION

a) = +

z̅ = −
|z|2 | + |2

−
= (√a2 + 2)2

−
= a2 + 2

− +
= a2 + 2 ( + )

2 + − − ( )2
= (a2 + 2)( + )

2 − 2 2
= (a2 + 2)( + )

2 − 2(−1)
= (a2 + 2)( + )

2 + 2
= (a2 + 2)( + )

1
= ( + )

1
=z

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PSPM I QS 015/1 Session
2014/2015

b) z̅ = −z

− = −( + )

− = − −

= − CHOW CHOON WOOI

2 = 0

= 0

= +

= 0 +

=

∴ z is a complex number with only an imaginary part when z̅ = −z

c) (2 − ) = (̅z + 1)(1 + i)
( + )(2 − ) = [(a − bi) + 1](1 + i)
2 + 2 − − 2 = (a − bi)(1 + i) + (1 + i)
2 + 2 − + = ( − + − 2) + (1 + i)
2 + + (2 − ) = − + + + 1 +
2 + (2 − ) = ( + 1) + ( − + 1)
By equating real part
2 = + 1
= 1

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PSPM I QS 015/1 Session
2014/2015
By equating imaginary part
(2 − ) = ( − + 1) CHOW CHOON WOOI
3 − 2 = 1
3 − 2(1) = 1
3 − 2 = 1
3 = 3
= 1

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PSPM I QS 015/1 Session
2014/2015
8. (a)
Solve the for the following equation |6 2 + − 11| = 4.

(b) Find the solution set for the inequality

+ 2
2 − ( − 4) < 5

SOLUTION | | = ⟺ = = − CHOW CHOON WOOI

a) |6 2 + − 11| = 4

6 2 + − 11 = 4 or 6 2 + − 11 = −4
= 1 6 2 + − 7 = 0
6 2 + − 15 = 0 (6 + 7)( − 1) = 0
6 + 7 = 0 − 1 = 0
(3 + 5)(2 − 3) = 0 = − 7 , = 1

3 + 5 = 0 2 − 3 = 0 6

= − 5 , = 3

32

5 3 7
∴= − 3 , = 2 , = − 6 ,

b) 2 − ( +2) < 5

−4

+ 2
2 − ( − 4) − 5 < 0

2( − 4) − ( + 2) − 5( − 4) <0

− 4

2 − 8 − − 2 − 5 + 20 <0

− 4

−4 + 10
− 4 < 0

−4 + 10 = 0 − 4 = 0

= 10 = 5 = 4

42

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PSPM I QS 015/1 Session
2014/2015
−4 + 10 5 5
(−∞, 2) (2 , 4) (4, ∞)
-
+ - +
- -
− 4 - + CHOW CHOON WOOI
-
−4 + 10
− 4

5
ℎ : { : < 2 ∪ > 4}

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109

PSPM I QS 015/1 Session
2014/2015

9. Two companies P and Q decided to award prizes to their employees for three work ethical
values, namely punctuality (x), creativity (y) and efficiency (z). Company P decided to award
a total of RM3850 for the three values to 6, 2 and 3 employees respectively, while company
Q decided to award RM3200 for the three values to 4, 1 and 5 employees respectively. The
total amount for all the three prizes is RM1000.

(a) Construct a system of linear equations to represent the above situation. CHOW CHOON WOOI

(b) By forming a matrix equation, solve this equation system using the elimination
method.

(c) With the same total amount of money spent by company P and Q, is it possible for
company P to award 15 employees for their creativity instead of 2 employees? Give
your reason.

SOLUTION

a) 6 + 2 + 3 = 3850
4 + + 5 = 3200
+ + = 1000

b) =

6 2 3 3850

(4 1 5) ( ) = (3200)

1 1 1 1000

3 ↔ 3 6 2 3 3850
2 = 4 1 − 2 (4 1 5|3200)
3 = 6 1 − 3
3 = 4 2 − 3 3 1 1 1 1000

1 1 1 1000
(4 1 5|3200)

6 2 3 3850

1 1 1 1000
(0 3 −1| 800 )

6 2 3 3850

1 1 1 1000
(0 3 −1| 800 )

0 4 3 2150

1 1 1 1000
(0 3 −1 | 800 )

0 0 −13 −3250

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110

PSPM I 1 1 1 1000 QS 015/1 Session CHOW CHOON WOOI
(0 3 −1| 800 ) 2014/2015
1
3 = −13 3 0 0 1 250 Page 19
2 = 2 + 3
1 1 1 1000
1 = 1 − 3 (0 3 0|1050)

1 0 0 1 250
2 = 3 2
1 = 1 − 2 1 1 0 750
(0 3 0|1050)

0 0 1 250

1 1 0 750
(0 1 0|350)

0 0 1 250

1 0 0 400
(0 1 0|350)

0 0 1 250

∴ = 400, = 350, = 250

c) 6 + 15 + 3 = 3850
4 + + 5 = 3200
+ + = 1000

6 15 3 3850

(4 1 5) ( ) = (3200)

1 1 1 1000

6 15 3
= (4 1 5)

111

6 15 3
| | = |4 1 5|

111

= (1) |115 53| − (1) |64 53| + (1) |46 115|

= (75 − 3) − (30 − 12) + (6 − 60)

=0

111

PSPM I QS 015/1 Session
2014/2015

| | = 0, . ℎ ℎ ℎ .
∴ 15 ℎ
2 ℎ ℎ .

CHOW CHOON WOOI

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112

PSPM I QS 015/1 Session
2014/2015

10. (a) Determine wheather ( ) = 1 and ( ) = 4 +1 are inverse function of each other
−4

by computing their composite functions.

(b) Given ( ) = ln(1 − 3 ).

(i) Determine the domain and range of ( ). Then sketch the graph of ( ). CHOW CHOON WOOI
(ii) Find −1(x), if it exists. Hence, state the domain and range of −1(x).

SOLUTION

a) ( ) = 1 ( ) = 4 +1

−4

[ ( )] = 4 + 1
[ ]

1
= (4 + 1) − 4

1
= 4 + 1 − 4



=

[ ( )] = , ℎ ℎ ℎ .

b) ( ) = ln(1 − 3 )

i. : 1 − 3 > 0

1 > 3

3 < 1

1
< 3

{ : 1
= < }
3

= { ∈ ℝ}

( )

( ) = ln(1 − 3 )


1
3

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PSPM I QS 015/1 Session
2014/2015

ii. [ −1( )] =

ln[1 − 3 −1( )] =

1 − 3 −1( ) =

3 −1( ) = 1 −

−1 ( ) = 1 −
3
CHOW CHOON WOOI
−1 = { ∈ ℝ}
1

−1 = { : < 3}

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114

QS 015/2 CHOW CHOON WOOI
Matriculation Programme Examination

Semester I
Session 2014/2015

115

PSPM I QS 015/2 Session
2014/2015

1. Given that ( − 2) is a factor of the polynomial ( ) = 3 − 10 2 + − 2 where and CHOW CHOON WOOI
are real numbers. If ( ) is divided by ( + 1) the remainder is −24, find the values of and
. Hence, find the remainder when ( ) is divided by (2 + 1).

2. Solve the equation 2 2 − 1 = sin 0 ≤ ≤ 2 . Give your answer in terms of .
3. Find the relative extremum of the curve = 3 − 4 2 + 4 .
4. Car X is travelling east at a speed of 80km/h and car Y is travelling north at 100km/h as shown

in the diagram below. Obtain an equation that describes the rate of change of the distance
between the two cars.
Hence, evaluate the rate of change of the distance between the two cars when car X is 0.15km
and car Y is 0.08km from P.

Car X P

Car Y

5. Expand ( + )( + )2, and are real numbers with > 0. Hence, find the values of and
if ( + )( + )2 = 3 − 3 − 2.

Express 4 −4 2 +5 −1 in the form of partial fractions.
3 −3 −2

6. (a) Express sin 6 − sin 2 in product form. Hence, show that

sin 6 − sin 2 + sin 4 = 4 cos 3 sin 2 cos .

(b) Use the result in (a) to solve

sin 6 − sin 2 + sin 4 = sin 2 cos for 0 ≤ ≤ 180°.

7. Find the limit of the following, if it exists.

a. lim +3
3+27
→−3

b. lim 2 −1
√ 2 −9
→−∞

c. lim 2−3 −4

→4 √ −2

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PSPM I QS 015/2 Session
2014/2015

1 + ≤ 0

8. Given that ( ) = { +6 0 < ≤ 4

3−

> 4

Where C is a constant.

a) Determine whether ( ) is continuous at = 0. CHOW CHOON WOOI
b) Given that ( ) is discontinuous at = 4, determine the values of .
c) Find the vertical asymptote of ( ).
9. Consider the parametric equations of the curve

= 3 and = 3 , 0 < ≤ 2 .

a) Find and express your answer in terms of .



b) Find the value of if = √2
4

c) Show that 2 = 1 .
2 3 4
sin

Hence, calculate 2 at = .
2
3

10. (a) Use the first principle to find the derivative of ( ) = √1 − .

(b) Given that + + ln(1 + 2 ) = 1, ≥ 0

Show that ( + ) 2 + ( 2 + 2 − 4 = 0.
2 (1+2 )2
)

Hence, find the value of 2 at the point (0,0).
2

END OF QUESTION PAPER

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117

PSPM I QS 015/2 Session
2014/2015

1. Given that ( − 2) is a factor of the polynomial ( ) = 3 − 10 2 + − 2 where and
are real numbers. If ( ) is divided by ( + 1) the remainder is −24, find the values of and
. Hence, find the remainder when ( ) is divided by (2 + 1).

SOLUTION

( ) = 3 − 10 2 + − 2 ( − 2) is a factor of ( ) CHOW CHOON WOOI
(2) = 0

(2)3 − 10(2)2 + (2) − 2 = 0

8 − 40 + 2 − 2 = 0

8 + 2 = 42

4 + = 21 …………………………. (1)

(−1) = −24 ( ) is divided by ( + 1) the remainder is −24

(−1)3 − 10(−1)2 + (−1) − 2 = −24

− − 10 − − 2 = −24

+ = 12 …………………………. (2)

(1) − (2)

3 = 9

= 3

= 9

( ) = 3 3 − 10 2 + 9 − 2
1 13 12 1

(− 2) = 3 (− 2) − 10 (− 2) + 9 (− 2) − 2
1 19

= 3 (− 8) − 10 (4) − 2 − 2
3 10 9

= −8− 4 −2−2
75

=− 8

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PSPM I QS 015/2 Session
2014/2015

2. Solve the equation 2 2 − 1 = sin 0 ≤ ≤ 2 . Give your answer in terms of .

SOLUTION CHOW CHOON WOOI
0 ≤ ≤ 2
2 2 − 1 = sin
2 (1 − 2 ) − 1 = sin
2 − 2 2 − 1 = sin
2 2 + sin − 1 = 0
=
2 2 + − 1 = 0
(2 − 1)( + 1) = 0

(2 − 1) = 0 ( + 1) = 0
= −1
1 sin = −1
= 2

sin = 1

2

= −1 1 =
(2) 6


66

3
= 6 , − 6 = 2

5
= 6 , 6

5 3
∴ = 6 , 6 , 2

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PSPM I QS 015/2 Session
2014/2015
3. Find the relative extremum of the curve = 3 − 4 2 + 4 .

SOLUTION

= 3 − 4 2 + 4 CHOW CHOON WOOI

= 3 2 − 8 + 4



= 0

3 2 − 8 + 4 = 0

(3 − 2)( − 2) = 0

2
= 3 = 2

2
2 = 6 − 8

ℎ = 2

3

23 22 2
= (3) − 4 (3) + 4 (3)

8 16 8
= 27 − 9 + 3

32
= 27
2 2
2 = 6 (3) − 8

= −4 < 0 ( )
2 32

∴ ℎ (3 , 27 ) .

ℎ = 2
= (2)3 − 4(2)2 + 4(2)

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PSPM I QS 015/2 Session
2014/2015
= 8 − 16 + 8
=0 CHOW CHOON WOOI
2
2 = 6(2) − 8

= 4 > 0 ( )
∴ ℎ (2, 0) .

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121

PSPM I QS 015/2 Session
2014/2015

4. Car X is travelling east at a speed of 80km/h and car Y is travelling north at 100km/h as shown CHOW CHOON WOOI
in the diagram below. Obtain an equation that describes the rate of change of the distance
between the two cars.
Hence, evaluate the rate of change of the distance between the two cars when car X is 0.15km
and car Y is 0.08km from P.

Car X P

Car Y

SOLUTION Negative sign as the distance is decreasing.

= −80

= −100

Car X P

Z

Car Y

2 = 2 + 2


2 = 2 + 2

1
= 2 (2 + 2 )

1
= ( + )

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PSPM I QS 015/2 Session
2014/2015


ℎ = 0.15, = 0.08, = −80, = −100

2 = 2 + 2

2 = (0.15)2 + (0.08)2

2 = (0.15)2 + (0.08)2 CHOW CHOON WOOI

= 0.17

1
= ( + )
1
= 0.17 [(0.15)(−80) + (0.08)(−100)]

= −117.647

The rate of change of the distance between the two cars is 117.65km/h

Page 9

123

PSPM I QS 015/2 Session
2014/2015

5. Expand ( + )( + )2, and are real numbers with > 0. Hence, find the values of and
if ( + )( + )2 = 3 − 3 − 2.

Express 4 −4 2 +5 −1 in the form of partial fractions.
3 −3 −2

SOLUTION CHOW CHOON WOOI

( + )( + )2 = ( + )( 2 + 2 + 2)

= 3 + 2 2 + 2 + 2 + 2 + 2

= 3 + ( + 2 ) 2 + ( 2 + 2 ) + 2

( + )( + )2 = 3 − 3 − 2
3 + ( + 2 ) 2 + ( 2 + 2 ) + 2 = 3 − 3 − 2

+ 2 = 0 ……………… (1)

2 + 2 = −3 ……………… (2)

2 = −2 ……………… (3)

(1)

= −2

2 + 2(−2 ) = −3
2 − 4 2 = −3
−3 2 = −3
2 = 1

= ±1

> 0, = 1

ℎ = 1

= −2(1)

= −2

= −2, = 1

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PSPM I QS 015/2 Session
2014/2015
4 − 4 2 + 5 − 1
3 − 3 − 2

CHOW CHOON WOOI
3 − 3 − 2√ 4 + 0 3 − 4 2 + 5 − 1

4 − 0 3 − 3 2 − 2

− 2 + 7 − 1

4 − 4 2 + 5 − 1 − 2 + 7 − 1
3 − 3 − 2 = + 3 − 3 − 2

2 − 7 + 1
= − ( − 2)( + 1)2

2 − 7 + 1
( − 2)( + 1)2 = − 2 + + 1 + ( + 1)2

( + 1)2 + ( − 2)( + 1) + ( − 2)
= ( − 2)( + 1)2

2 − 7 + 1 = ( + 1)2 + ( − 2)( + 1) + ( − 2)

ℎ = −1

2 − 7 + 1 = ( + 1)2 + ( − 2)( + 1) + ( − 2)

(−1)2 − 7(−1) + 1 = (−1 − 2)

9 = −3

= −3

ℎ = 2

2 − 7 + 1 = ( + 1)2 + ( − 2)( + 1) + ( − 2)

(2)2 − 7(2) + 1 = (2 + 1)2

−9 = 9

= −1

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PSPM I QS 015/2 Session
2014/2015

ℎ = 0, = −1, = −3 CHOW CHOON WOOI
2 − 7 + 1 = ( + 1)2 + ( − 2)( + 1) + ( − 2)
1 = − 2 − 2
1 = −1 − 2 − 2(−3)
1 = 5 − 2
2 = 4
= 2

4 − 4 2 + 5 − 1 2 − 7 + 1
3 − 3 − 2 = − ( − 2)( + 1)2


= − [ − 2 + + 1 + ( + 1)2]

−1 2 −3
= − [ + + 1)2]
− 2 + 1 ( +

12 3
= + − 2 − + 1 + ( + 1)2

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126

PSPM I QS 015/2 Session
2014/2015
6. (a) Express sin 6 − sin 2 in product form. Hence, show that
sin 6 − sin 2 + sin 4 = 4 cos 3 sin 2 cos .

(b) Use the result in (a) to solve
sin 6 − sin 2 + sin 4 = sin 2 cos for 0 ≤ ≤ 180°.

SOLUTION CHOW CHOON WOOI
(a) Express sin 6 − sin 2 in product form. Hence, show that

sin 6 − sin 2 + sin 4 = 4 cos 3 sin 2 cos .

+ −
6 + 2 6 − 2 sin − sin = 2 cos ( 2 ) sin ( 2 )
sin 6 − sin 2 = 2 cos ( 2 ) sin ( 2 )

= 2 cos 4 sin 2

sin 6 − sin 2 + sin 4

= 2 cos 4 sin 2 + sin 4 sin 2 = 2 sin cos
sin 2(2 ) = 2 sin(2 ) cos(2 )
= 2 cos 4 sin 2 + sin 2(2 )

= 2 cos 4 sin 2 + 2 sin 2 cos 2 + −

= 2 sin 2 (cos 4 + cos 2 ) cos + cos = 2 cos ( 2 ) cos ( 2 )

= 2 sin 2 [2 cos ( 4 + 2 ) cos ( 4 − 2 )]

22

= 2 sin 2 [2 cos 3 cos ]

= 4 cos 3 sin 2 cos

(b) Use the result in (a) to solve sin 6 − sin 2 + sin 4 = sin 2 cos .
sin 6 − sin 2 + sin 4 = sin 2 cos
4 cos 3 sin 2 cos = sin 2 cos
4 cos 3 sin 2 cos − sin 2 cos = 0
sin 2 cos (4 cos 3 − 1) = 0

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PSPM I QS 015/2 Session
2014/2015
0 ≤ ≤ 180° 0 ≤ ≤ 180°.
0 ≤ 2 ≤ 360° cos = 0 0 ≤ ≤ 180°
0 ≤ 3 ≤ 540°

sin 2 = 0 4 cos 3 − 1 = 0 CHOW CHOON WOOI
1

cos 3 = 4

90° 75.5°

180° 360° 75.5°

2 = 0°, 180°, 360° = 90° = 75.5°
= 0°, 90°, 180° 3 = 75.5°, 284.5°, 435.5°

= 25.2°, 94.8°, 145.2°

∴ = 0°, 25.2°, 90°, 94.8°, 145.2°, 180°

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128

PSPM I QS 015/2 Session
2014/2015

7. Find the limit of the following, if it exists.

a. lim +3
3+27
→−3

b. lim 2 −1
√ 2 −9
→−∞

c. lim 2−3 −4 CHOW CHOON WOOI

→4 √ −2

SOLUTION

a. lim +3 = lim ( +3) +3 2 − 3 + 9
3 +27 ( 2−3 +9)
→−3 →−3 + 3√ 3 + 0 2 + 0 + 27
3 + 3 2 + 0
= lim 1
−3 2 + 0 + 27
→−3 ( 2 − 3 + 9) −3 2 − 9

1 9 + 27
= (−3)2 − 3(−3) + 9 9 + 27

1
= 27

b. lim 2 −1 = lim −2 −−1
√ 2 −9 √ 22− 92
→−∞ →−∞

−2 + 1

= lim
→−∞ 9
√1 − 2

−2
=

√1

= −2

c. lim 2−3 −4 = lim ( −4)( +1) 2 − 2 = ( − )( + )
→4 √ −2 →4 √ −2 − 4 = √ 2 − 22

(√ − 2)(√ + 2)( + 1) = (√ − 2)(√ + 2)
= lim
Page 15
→4 √ − 2

= lim(√ + 2)( + 1)

→4

= (√4 + 2)(4 + 1)

= 20

129

PSPM I QS 015/2 Session
2014/2015

1 + ≤ 0

8. Given that ( ) = { +6 0 < ≤ 4

3−

> 4

Where C is a constant.

a) Determine whether ( ) is continuous at = 0. CHOW CHOON WOOI
b) Given that ( ) is discontinuous at = 4, determine the values of .
c) Find the vertical asymptote of ( ).

SOLUTION

1 + ≤ 0
( ) = { + 6 0 < ≤ 4

3 − > 4


a) Determine whether ( ) is continuous at = 0 Continuity at a point

(0) = 1 + 0 = 2 ( ) = ℎ

lim ( ) = lim 1 + 1. ( )
→0− →0− 2. lim ( )

= 1 + 0 →

3. lim ( ) = ( )

→

=2

lim ( ) = lim + 6

→0+ →0+ 3 −

0+6
=3−0

=2

lim ( ) = 2

→0

(0) = lim ( ) = 2, ℎ ( ) = 0.

→0

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PSPM I QS 015/2 Session
2014/2015

b) Given that ( ) is discontinuous at = 4, determine the values of .

= 4

lim ( ) = lim + 6

→4− →4− 3 − CHOW CHOON WOOI

4+6
= 3−4

10
= −1

= −10

lim ( ) = lim
→4+ →4+

=

( ) = 4, ℎ ≠ −10.

{ : < −10 ∪ > −10}

c) Find the vertical asymptote of ( ).

:

3 − = 0

= 3

+ 6
lim = −∞
→3+ 3 −

lim + 6 = ∞

→3− 3 −

∴ ℎ = 3.

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131

PSPM I QS 015/2 Session
2014/2015

9. Consider the parametric equations of the curve

= 3 and = 3 , 0 < ≤ 2 .

a) Find and express your answer in terms of .


b) Find the value of if = √2 CHOW CHOON WOOI

4

c) Show that 2 = 1 .
2 3 4
sin

Hence, calculate 2 at = .
2
3

SOLUTION
a) Find and express your answer in terms of .



= 3 = 3

= 3 cos2 (cos ) = 3 sin2 (sin )


= 3 cos2 (− sin ) = 3 sin2 (cos )

= −3 sin cos2 = 3 cos sin2


= .

= (3 cos sin2 ) (−3 sin 1 cos2 )


= (sin ) (− 1 )
cos

sin
= − cos

= − tan

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132

PSPM I QS 015/2 Session
2014/2015

b) Find the value of if = √2

4

√2
ℎ = 4

= 3

3 = √2 CHOW CHOON WOOI
4

1

√2 3
cos = ( 4 )

≈ 0.7071


4 4


, = 4


= 4 , 2 − 4

7
= 4 , 4

ℎ = √2 , =

44


= − tan


= − tan ()
4

= −1

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PSPM I QS 015/2 Session CHOW CHOON WOOI
2014/2015
√2 7
ℎ = 4 , = 4 Page 20


= − tan

7
= − tan ( 4 )
= −(−1)
=1

c) Show that 2 = 1 sin . Hence, calculate 2 at = .
2 3 4 2
3

2
2 = [ ] .

1
= [ (− tan )] . (−3 )
sin cos2

= (− 2 ). (−3 sin 1 cos2 )


2
= 3 sin cos2

1

= cos2
3 sin cos2

1
= 3 cos4 sin


= 3

2 1
2 = 3 cos4 sin

1
= 3 cos4 ( 3 ) sin ( 3 )
= 6.16

134

PSPM I QS 015/2 Session
2014/2015

10. (a) Use the first principle to find the derivative of ( ) = √1 − .

(b) Given that + + ln(1 + 2 ) = 1, ≥ 0

Show that ( + ) 2 + ( 2 + 2 − 4 = 0.
2 (1+2 )2
)

Hence, find the value of 2 at the point (0,0). CHOW CHOON WOOI
2

SOLUTION

(a) ( ) = √1 −

( + ℎ) = √1 − ( + ℎ)

′( ) = lim ( + ℎ) − ( )
ℎ→0 ℎ

= lim √1 − ( + ℎ) − √1 −

ℎ→0 ℎ

= lim √1 − ( + ℎ) − √1 − √1 − ( + ℎ) + √1 −
.
ℎ→0 ℎ √1 − ( + ℎ) + √1 −

[1 − ( + ℎ)] − (1 − )
= lim

ℎ→0 ℎ (√1 − ( + ℎ) + √1 − )

1 − − ℎ − 1 +
= lim

ℎ→0 ℎ (√1 − ( + ℎ) + √1 − )

−ℎ
= lim

ℎ→0 ℎ (√1 − ( + ℎ) + √1 − )

−1
= lim

ℎ→0 (√1 − ( + ℎ) + √1 − )

−1
=

(√1 − ( + 0) + √1 − )

−1
=

(√1 − + √1 − )
−1

=
2√1 −

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PSPM I QS 015/2 Session
2014/2015

(b) + + ln(1 + 2 ) = 1

+ [ + ] + (1 1 (1 + 2 ) = 0
+ 2 )

+ + + (1 1 (2) = 0
+ 2 )
CHOW CHOON WOOI
+ + + (1 2 = 0
+ 2 )

( + ) + + (1 2 = 0
+ 2 )

( + ) + + 2(1 + 2 )−1 = 0


[( + ) 2 + ( + 1)] + − 2(1 + 2 )−2 (1 + 2 ) = 0
2

[( + ) 2 + 2 + + − (1 2 (2) = 0
2 ( ) ] + 2 )2

( + ) 2 + 2 + 2 − (1 4 = 0
2 ( ) + 2 )2

ℎ (0, 0) ➔ = 0, = 0

( 0 + 0) + 0 + (1 2 = 0
+ 2(0))

(1 + 0) + 2 = 0
(1)


= −2

ℎ = 0, = 0,
= −2

( 0 + 0) 2 + 0(−2)2 + 2(−2) − (1 + 4 = 0
2 2(0))2

2 4
(1 + 0) 2 + 4 − 4 − 1 = 0
2
2 = 4

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2015/2016

137

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2015/2016

138

PSPM I QS 015/1
Session 2015/2016

1. Evaluate the solution of 4 −2 = 1 up to three decimal places.
3−y

2. The first three terms of a geometric sequence are (4 − 2) , (2m − 1) and 12. CHOW CHOON WOOI

3

Determine the value of m. Hence, find the sixth term for this sequence.

3. Solve the equation
2 + log2 = 15 logx 2 .

1 −1
4. (a) Determine the values of x so that [ 0 1 ] is singular.

1 3 −1

(b) If = [13 25] and = [01 −41], find C when = −1.

5. (a) Expand (2 + )−12 in ascending powers of x, up to the term 3.
(b) Use the expansion in (a) to approximate √2.

3

6. Given 1 = 3 − 3i and 2 = 3 + 2i.
(a) Write ̅ 1 in polar form.
(b) Express ( ̅ ̅1̅ 2) + [̅̅̅ 3̅̅̅] in the form of + , , ∈ .

13 − 2

7. A curve = 2 + bx + c where a, b and c are constants, passes through the points (2,11),
(-1, -16) and (3,28).
(a) By using the above information, construct a system containing three linear equations.

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PSPM I QS 015/1
Session 2015/2016

(b) Express the above system as a matrix equation AX=B.
(c) Find the inverse of matrix A by using the adjoint matrix method. Hence, obtain the

values of a, b and c.

8. Given a function ( ) = √3 − 2 . CHOW CHOON WOOI
(a) Show that is a one to one function.
(b) Find the domain and range of .
(c) Determine the inverse function of and state its domain and range.
(d) Sketch the graphs of and −1 on the same axis.

9. (a) The function is given as ( ) = +2 , x ≠ 4. If ( )(x) = x, find the value of a.
3 −4 3

(b) Let ( ) = |3 + 2| and ( ) = − + 2 be two functions. Evaluate ( )−1(3).

10. (a) Solve the inequality | −1| > 2.

+3

(b) Show that 2 42 = 22 .
8

Hence, find the interval for x so that 2 x 42 − 13(2 ) + 36 ≥ 0.
8

END OF QUESTION PAPER

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PSPM I QS 015/1
Session 2015/2016

1. Evaluate the solution of 4 −2 = 1 up to three decimal places.
3−y

SOLUTION

4 −2 = 1 CHOW CHOON WOOI
3−y

4 −2 = 3y

ln 4 −2 = ln 3y

( − 2) ln 4 = ln 3

ln 4 − 2 ln 4 = ln 3

ln 4 − ln 3 = 2 ln 4

(ln 4 − ln 3) = 2 ln 4

2 ln 4
= (ln 4 − ln 3)

= 9.638

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PSPM I QS 015/1
Session 2015/2016

2. The first three terms of a geometric sequence are (4 − 2) , (2m − 1) and 12.

3

Determine the value of m. Hence, find the sixth term for this sequence.

SOLUTION

Alternative CHOW CHOON WOOI

, , ℎ , = =

= ±√
2 − 1 12
4 =
(3 − 2) , (2m − 1), 12 4 2 − 1
(2m − 1) = √(43 − 2) ( 12) 3 − 2

(2m − 1) = √16 − 24 (2m − 1)2 = 16 − 24
(2m − 1)2 = 16 − 24
4 2 − 4 + 1 = 16 − 24 4 2 − 20 + 25 = 0
4 2 − 20 + 25 = 0
(2 − 5)(2 − 5) = 0 (2 − 5)(2 − 5) = 0

5 5
= 2 = 2

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PSPM I QS 015/1
Session 2015/2016
When = 5,

2

4 45 4
1 = = 3 − 2 = 3 (2) − 2 = 3

5 CHOW CHOON WOOI
2 = 2 − 1 = 2 (2) − 1 = 4

=


4
= 4 = 3

3

= −

6 = 4 (3)5
(3)

= 324

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PSPM I QS 015/1
Session 2015/2016
3. Solve the equation

2 + log2 = 15 logx 2.

SOLUTION

2 + log2 = 15 logx 2 CHOW CHOON WOOI

2 + log2 = 15 (lloogg22 2
x)
logc a
logb a = logc b

1
2 + log2 = 15 (log2 x)

15
2 + log2 = log2 x

= log2

15
2 + = u

2 + 2 = 15

2 + 2 − 15 = 0

( − 3)( + 5) = 0

= 3 = −5 logb a = logc a
log2 = −5 logc b
= 2−5
log2 = 3 = 1
= 23
32
= 8

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PSPM I QS 015/1
Session 2015/2016
1 −1
4. (a) Determine the values of x so that [ 0 1 ] is singular.

1 3 −1

(b) If = [13 52] and = [10 −41], find C when = −1.

SOLUTION CHOW CHOON WOOI

(a) Singular Matrix

1 −1 Singular matrix is square matrix
| 0 1 | = 0 whose determinant is zero.
1 3 −1

(1) |30 −11| − ( ) |1 −11| + (−1) |1 30| = 0

(1)[0 − 3] − ( )[− − 1] + (−1)[3 − 0] = 0

−3 + 2 + − 3 = 0
2 − 2 − 3 = 0

( − 3)( + 1) = 0

= 3 = −1

(b)

= [31 52] , = [10 −41]

= −1

−1( ) = −1( −1)

−1( ) = −1

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PSPM I QS 015/1
Session 2015/2016

−1 = −1

( −1 ) = ( −1)

−1 =

= −1 CHOW CHOON WOOI

= [31 25] , = [01 −41]

−1 = 1 0 [−01 −14]
−1 −

= −1 [−01 −14]

= [01 −41]

= −1
= [10 −41] [31 52] [01 −41]
= [−71 −132] [10 −41]
= [−71 −152]

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PSPM I QS 015/1
Session 2015/2016

5. (a) Expand (2 + )−21 in ascending powers of x, up to the term 3.

(b) Use the expansion in (a) to approximate √2.

3

SOLUTION CHOW CHOON WOOI

(a)

(2 + )−12 = [2 (1 + )]−12
2

= 2−12 (1 + )−12
2

1 [1 (− 21) 1 (− 21) (− 23) 2 (− 21) (− 23) (− 25) 3 ]
√2 1! () 2! () 3! ()
= + + 2 + 2 + ⋯
2

1 3 2 15 3
= [1 − + 8 ( 4 ) − 48 ( 8 ) + ⋯ ]
√2 4

1 3 2 5 3
= [1 − + − + ⋯ ]

√2 4 32 128

11 3 2 − 5 3 + ⋯
= − +
√2 4√2 32√2 128√2

(b)

√2 1
(32)2
3 =

3 −12
= (2)

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PSPM I QS 015/1
Session 2015/2016

(2 + )−21 = 3 −12
(2)

3
2 + = 2

1 CHOW CHOON WOOI
= − 2

(2 + )−12 = 1 − 1 + 3 2 − 5 3
√2 4√2 32√2 128√2

1 −12 1 1 1 3 12 5 13
[2 + (− 2)] = √2 − 4√2 (− 2) + 32√2 (− 2) − 128√2 (− 2)

3 −12
( 2 ) = 0.8155

1

22
( 3 ) = 0.8155

√2 = 0.8155
3

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PSPM I QS 015/1
Session 2015/2016
6. Given 1 = 3 − 3i and 2 = 3 + 2i.
(a) Write ̅ 1 in polar form. CHOW CHOON WOOI
(b) Express (̅ ̅1̅ 2) + [̅̅̅ 3̅̅̅] in the form of + , , ∈ .

13 − 2

SOLUTION

(a)
1 = 3 − 3i and 2 = 3 + 2i
̅ 1 = 3 + 3

= | ̅ 1| = √32 + 32
= √18

= 3√2

̅ 1, = −1
( )

= −1 3
(3)

= −1(1)

= 0.785

̅ 1 in polar form
̅ 1 = [cos + ]
= 3√2[cos 0.785 + 0.785]

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PSPM I QS 015/1
Session 2015/2016
(b)

( ̅ 1 2) + ̅̅̅̅ ̅3̅̅̅̅ = (3 + 3 )(3 + 2i) + ̅̅̅̅̅̅̅−̅̅ ̅̅̅̅̅̅
13 (− 2) 13 [−(3 + 2i)]

9 + 6 + 9 + 6 2 ̅̅̅̅̅̅̅ ̅̅̅̅̅̅ CHOW CHOON WOOI
= 13 + [(3 + 2i)]

3 + 15 ̅̅̅̅̅̅ ̅ (̅̅3̅̅−̅̅̅2̅̅i̅)̅̅̅̅̅̅
= 13 + [(3 + 2i)(3 − 2i)]

3 + 15 ̅̅3̅ ̅ ̅+̅̅̅2̅̅
= 13 + [ 9 + 4 ]

3 + 15 ̅̅2̅̅+̅̅̅3̅̅ ̅
= 13 + [ 13 ]

3 + 15 2 − 3
= 13 + 13

3 + 15 + 2 − 3
= 13

5 + 12
= 13

5 12
= 13 + 13

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