PSPM 1 Q&A

PSPM I QS 015/1 Session
2016/2017

4. If A and B are 2 2 matrices where ≠ 2, such that ( + )2 = 2 + 2 + 2,
deduce that = −1 [Assume = = ]. If = [19 −21], find .

SOLUTION

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

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

2 + + + 2 = 2 + 2 + 2

+ = 2

= 2 −

= = = ℎ −1 =
∴ = −1

= [19 −21]

= −1 = 1 (2)(9) [−−91 −12]
(1)(−1) −

= 1 [−−91 −12]
−19

= 1 [−−19 −12]
−19

1 2
= [199 119]
19
19

Page 7

201

PSPM I QS 015/1 Session
2016/2017

1

5. (a) Obtain the expansion for (1 − )4 in ascending powers of x up to the term 3.

4

1

State the interval for x such that the expansion (1 − )4 is valid. Hence, obtain

4

1

the simplest form of the expansion (16 − 4 )4.

1 CHOW CHOON WOOI

(b) Write 4√12 in the form of [(1 − )4]. Hence, approximate 4√12 correct to

4

three decimal places.

SOLUTION

a) (1 − 1 = 1 + (1) (− 1 + (41)(14−1) (− )2 + (41)(41−1)(14−2) (− )3 + ⋯

)4 4 )

4 1! 4 2! 4 3! 4

1 (− (14) (− 43) 2 (41) (− 43) (− 74) (− 3
() ) () )
= 1 + 44 + 2 1 16 + 3 2 1 64 + ⋯

3 2 21 3
= 1 − 16 − 32 (16) − 384 (64) + ⋯

= 1 − 1 − 3 2 − 7 3 + ⋯
16 512 8192



The interval for x such that the expansion ( − ) is valid




|4| < 1


−1 < 4 < 1

−4 < < 4

Page 8

202

PSPM I QS 015/1 Session
2016/2017



The simplest form of the expansion ( − )

1

1 4 4
(16 − 4 )4 = [16 (1 − 16)]

= 1 (1 − 1 CHOW CHOON WOOI

164 4 )4

= 2 (1 − )14
4

= 2 [1 − 1 − 3 2 − 7 3 + ⋯ ]
16 512 8192

= 2 − 1 − 3 2 − 7 3 + ⋯
8 256 4096



b) √ in the form of [( − ) ]



4√12 = 1

124

1

= (16 − 4)4

1

44
= [16 (1 − 16)]

1 Compare 2 (1 − 4 )41 = 2 − 1 − 3 2 − 7 3
8 256 4096
1 14
= 164 (1 − 4) +⋯

1

14
= 2 (1 − 4)

= 2 − 1 (1) − 3 (1)2 − 7 (1)3 + ⋯
8 256 4096

≈ 1.862

Page 9

203

PSPM I QS 015/1 Session
2016/2017

6. Given ( ) = 2+1, ≥ 0. CHOW CHOON WOOI

5

(a) Determine −1( ). Hence, if ( ( )) = 1 ( 2(3 −1) + 1), show that ( ) =

5

3 −1

(b) Evaluate ( (2)) correct to three decimal places.
(c) Assume that the domain for ( ) is ≥ 0, determine −1( ) and state its

domain and range.

SOLUTION
a) ( ) = 2+1, ≥ 0

5

Method I Method II

2 + 1 2 + 1
( ) = 5 = 5
[ −1( )] = 5 = 2 + 1
[ −1( )]2 + 1 2 = 5 − 1
= √5 − 1
5 = −1( ) = √5 − 1
[ −1( )]2 + 1 = 5
[ −1( )]2 = 5 − 1
−1( ) = √5 − 1

( ( )) = 1 ( 2(3 −1) + 1)
5

[ ( )]2 + 1 = 1 ( 2(3 −1) + 1)
55

[ ( )]2 + 1 = 2(3 −1) + 1

[ ( )]2 = 2(3 −1)

Page 10

204

PSPM I QS 015/1 Session
2016/2017
1

( ) = [ 2(3 −1)]2
( ) = 3 −1

b) Evaluate ( (2)) CHOW CHOON WOOI
22 + 1

(2) = 5
=1

( (2)) = (1)
= 3(1)−1
= 2

c) ( ) = 3 −1 Method II
Method I

( ) = 3 −1 ( ) = 3 −1

[ −1( )] = = 3 −1

3[ −1( )]−1 = ln = ln 3 −1

ln 3[ −1( )]−1 = ln ln = 3 − 1

3[ −1( )] − 1 = ln 3 = ln + 1

3[ −1( )] = ln + 1 ln + 1
= 3

−1( ) = ln + 1 ln + 1
3 3
−1( ) =

Page 11

205

PSPM I QS 015/1 Session
2016/2017

( ) = 3 −1; ≥ 0 ( ): ≥ 3(0)−1
( ): ≥ 0 ≥ −1
1
≥ CHOW CHOON WOOI

−1( ) = ( ) −1( ) = ( )
−1( ): ≥ 0
−1( ): ≥ 1


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206

PSPM I QS 015/1 Session
2016/2017

7. (a) If √7 − 3√5 = √ − √ , determine the values of and .

(b) Solve the equation 2 − 4(3 + 4) = 0.

SOLUTION CHOW CHOON WOOI

a) √7 − 3√5 = √ − √

2

[√7 − 3√5] = [√ − √ ]2

7 − 3√5 = √ 2 + √ 2 − 2√ √

7 − 3√5 = + − 2√

+ = 7 ………………… (1)
= 7 −

2√ = 3√5

√4 = √9 5 ………………… (2)
4 = 45

Substitute (1) into (2) = 5
4 (7 − ) = 45
28 − 4 2 = 45 2
4 2 − 28 + 45 = 0
(2 − 9)(2 − 5) = 0
= 9

2

Page 13

207

PSPM I QS 015/1 Session
2016/2017

= 7 − 9 = 7 − 5

2 2

= 5 = 9

2 2

95 59
∴ = 2 , = 2 = 2 , = 2
CHOW CHOON WOOI
b) 2 − 4 (3 + 4) = 0

=

2 − 2 (3 + 4) = 0
2 4

2 − 2 (3 + 4) = 0 log =
222

2 − 2 (3 + 4) = 0
2 2 2
= 1

2 − 2 (3 + 4) = 0
2(1)

1
2 − 2 2 (3 + 4) = 0

1
2 = 2 2(3 + 4)

1

2 = 2(3 + 4)2

1

= (3 + 4)2

2 = 3 + 4

2 − 3 − 4 = 0

( − 4)( + 1) = 0

= 4 = −1

> 0, ∴ = 4

Page 14

208

PSPM I QS 015/1 Session
2016/2017

8. (a) Solve the following equation | 3 | = 7, ≠ 4.

−4

(b) Find the solution set for the inequality, −4− ≥ + 4, ≠ 3.

−3

SOLUTION or 3 = −7 CHOW CHOON WOOI
(a) | 3 | = 7
−4
−4
3 = −7( − 4)
3 =7 3 = −7 + 28
7 = 28 − 3
−4 7 = 25
= 25
3 = 7( − 4)
3 = 7 − 28 7
7 = 3 + 28
7 = 31
= 31

7

(b) −4− ≥ + 4, ≠ 3

−3

−4 −
− 3 ≥ + 4

−4 −
− 3 − − 4 ≥ 0

−4 − − ( − 3) − 4( − 3) ≥0

− 3

−4 − − 2 + 3 − 4 + 12 ≥0

− 3

− 2 − 2 + 8
− 3 ≥ 0

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PSPM I QS 015/1 Session
2016/2017
−( 2 + 2 − 8) ≥0
( − 3) = 0
− 3

( 2 + 2 − 8)
≤0

− 3

( + 4)( − 2) ≤0 CHOW CHOON WOOI

− 3

( + 4) = 0 ( − 2) = 0

= −4 = 2 = 3

( + 4) (−∞, − ) (− , ) ( , ) ( , ∞)

( − 2) - + + +
- - + +
( − 3) - - - +
( + )( − ) - + - +

−

∴ ; { : − ∞ < ≤ −4 ∪ 2 ≤ < 3}

Page 16

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PSPM I QS 015/1 Session
2016/2017

CHOW CHOON WOOI

Page 17

211

PSPM I QS 015/1 Session
2016/2017
9. Given ( ) = and ( ) = | − 3|.

(a) Show that (  )( ) = { −( − −33) , ≥ 3
, < 3

(b) Sketch the graph of = (  )( ). Hence, state the interval in which

(  )−1( ) exists. CHOW CHOON WOOI

(c) Determine (  )−1( ),for ≥ 3.

(d) Find the function ℎ( ) for > 1, given that (ℎ  )( ) = 2 . Hence, show
1−3
3

that h(x) is a one to one function.

SOLUTION

( ) = and ( ) = | − 3|

a) (  )( ) = (| − 3|)

= | −3|

| − 3| = {−( −−33),, − 3 ≥ 0 | | = ቄ− ,, ≥ 0
− 3 < 0 < 0

= {−( −−33),, ≥ 3
< 3

(  )( ) = | −3|

= { −3, ≥ 3
−( −3), < 3

b)


3

1
3

Page 18

212

PSPM I QS 015/1 Session CHOW CHOON WOOI
(  )−1( ) exists for ≥ 3 < 3 2016/2017
c) (  )−1( ) ≥ 3
(  )( ) = −3 Page 19

= −3
ln = ln −3
ln = − 3
= ln + 3

∴ (  )−1( ) = ln + 3

d) ( ) = ,

2
(ℎ  )( ) = 1 − 3

2
ℎ[ ( )] = 1 − 3

ℎ[ ] = 1 2
− 3

=

2
ℎ[ ] = 1 − 3

2
ℎ[ ] = 1 − 3

Show that h(x) is a one to one function

ℎ( ) = 1 2
− 3

Let ℎ( 1) = ℎ( 2)

213

PSPM I QS 015/1 Session
2016/2017

1 2 1 = 1 2 2
− 3 1 − 3 2

2 1(1 − 3 2) = 2 2(1 − 3 1)

1(1 − 3 2) = 2(1 − 3 1)

1 − 3 1 2 = 2 − 3 1 2 CHOW CHOON WOOI

1 = 2

1 = 2, ℎ ℎ( )

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PSPM I QS 015/1 Session
2016/2017

10. (a) Given = [−21 0 −24] , = −1 31 2 1
6 [0 2] = [3 2 2].
4
−6 53 1

Find −1 by using the elementary row operations method.

Hence, if = 3 + , determine the matrix . CHOW CHOON WOOI

(b) Ahmad bought an examination pad, 2 pens and a tube of liquid paper for RM18.
Ali spent RM24 for 3 examination pads, 2 pens and 2 tubes of liquid paper. In
the meantime Abu spent RM36 at the same store for 3 examination pads, 4
pens and a tube of liquid paper. Let x, y and z represent the price per unit for
examination pad, pen and tube of liquid paper respectively.

i. Obatain the system of linear equations from the above information.
ii. Write the system in the form of matrix equation = .
iii. State the price of each unit of examination pad, pen and tube of liquid

paper.
iv. Aminah bought 4 examination pads, 5 pens and 1 tube of liquid paper.

What is the total price paid?

SOLUTION

a) = [−21 0 −24] , = −1 31 2 1
6 [0 2] = [3 2 2]
4
−6 53 1

2∗ = 3 1 − 2 1 2 11 0 0
3∗ = 3 1 − 3 (3 2 2|0 1 0)

3 4 10 0 1

1 2 11 0 0
(0 4 1|3 −1 0)

3 4 10 0 1

1 2 11 0 0
(0 4 1|3 −1 0 )

0 2 2 3 0 −1

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215

PSPM I QS 015/1 Session
2016/2017

2∗ 1 2 1 2 11 0 0
= 4 1 |34 1
(0 1 4 − 4 0)

0 2 −2 3 0 −1

3∗ = 2 2 − 3 12 1 1 0 0
13 1
01 4|4 − 4 0 CHOW CHOON WOOI

(0 0 − 3 − 3 − 1 1)
2 2 2

2 1 2 141|134 0 0
3 0 1 11 1
3∗ = − 3 (0 0 − 4 0

1 − 2
3)
3

1 10 0
4 1 2 11 1 1
2∗ = 2 − 3 0 1 0|2 − 3 6

(0 0 1 1 1 − 2
3 3)

0 − 1 2
001||12 3
1∗ = 1 − 3 1 2 1 3
0 1 1 3 1
0 0 −
( 6
1 2
− 3)
3

−1 1 1

1 0 010|| 3 3
0 1 1 1 1
1∗ = 1 − 2 2 0 0 2 − 3
( 6
1 2
1 − 3)
3

11
−1 3 3
−1 = 1 1 1
2 −3 6
12
[ 1 3 − 3]

= 3 +
−1 = −1(3 + )
= −1(3 + )

Page 22

216

PSPM I QS 015/1 Session
2016/2017

−1 11
=1
3 3 −1 3 + ([−21 0 −24]) ]
2 1 1 [3 [ 0 2] 6
−3
1 6 −6 5
2
[ 1 3 − 3]

−1 11 CHOW CHOON WOOI
=1
3 3 −3 92 −1
2 1 1 [[ 0 6]+[ 0 6 ]]
−3
1 6 −18 15 −4 2
2
[ 1 3 − 3]

−1 11
=1
3 3 −1 8
2 1 1 [0 12]
−3
1 6 −22 17
2
[ 1 3 − 3]

11 11
(−1)(−1) + (3) (0) + (3) (−22) (−1)(8) + (3) (12) + (3) (17)
1 11 1 11
= (2) (−1) + (− 3) (0) + (6) (−22) (2) (8) + (− 3) (12) + (6) (17)
1 2 1 2
[ (1)(−1) + (3) (0) + (− 3) (−22) (1)(8) + (3) (12) + (− 3) (17) ]

22 12 17
1+0− 3 −8 + 3 + 3
1 22
= −2+0− 6 17
4−4+ 6
44
[−1 + 0 + 3 34
8+4− 3 ]

19 5
−3
25 3
= −6 17

41 6
2
[3 3]

bi) + 2 + = 18
3 + 2 + 2 = 24
3 + 4 + = 36

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PSPM I QS 015/1 Session
2016/2017
1 2 1 18
bii) [3 2 2] [ ] = [24]

3 4 1 36

biii) =

−1 = −1 CHOW CHOON WOOI

= −1

−1 11
=1
3 3 18
2 1 1 [24]
−3
1 6 36
2
[ 1 3 − 3]

11
(−1)(18) + (3) (24) + (3) (36)
1 1 1
= (2) (18) + (− 3) (24) + (6) (36)

[ (1)(18) + 1 (24) + (− 2 (36) ]
(3) 3)

−18 + 8 + 12
=[ 9−8+6 ]

18 + 8 − 24

2
= [7]

2

= , = , =

biv) − ; − ; −
= 4(2) + 5(7) + 1(2)
= 8 + 35 + 2
= 45.00

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218

QS 015/2 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2016/2017

219

PSPM I QS 015/2 Session
2016/2017

1. Express 2 in partial fractions form.
2 −2 −3

2. Evaluate the following limits, if exist.

a. lim −2
4 −16
→2

b. lim (2− )( −1) CHOW CHOON WOOI
( −3)2
→∞

3. Show that 2 = 1 + cos . Hence, solve 2 = cos 2 for 0° ≤ ≤ 360°.
1−cos 1−cos

4. Consider a function ( ) = 1 .

2−√

a. Find lim ( ) and state the equation of horizontal asymptote for .

→∞

b. By using the first principle of derivative, find ’( ).

5. (a) Use the derivative to find the maximum area of a rectangle that can be inscribed
in a semicircle of radius 10cm.

(b) A cone-shaped tank as shown below.

60°

A

Water flows through a hole A at rate of 6 3 per second. Find the rate of
change in height of the water when the volume of water in the cone is 24 3

6. (a) Polynomial ( ) has a raminder 3 when divided by ( + 3). Find the
remainder of ( ) + 2 when divided by ( + 3).

(b) Polynomial 1( ) = 3 + 2 − 5 − 7 has a factor ( − 1) and remainder 1
when divided by ( + 1), while a polynomial 2( ) = 3 − 2 + + 6 has a
remainder 2 when divided by ( – 1). Find the value of the constants and
if 1 + 2 = 5. Hence, obtain the zeroes for 1( ).

Page 2

220

PSPM I QS 015/2 Session
2016/2017

7. Consider a function ( ) = √3 cos 2 + 2 sin 2 .

a. Express in the form of cos(2 − ) for > 0, 0° ≤ ≤ 90° and to the
nearest minute. State the maximum and minimum values of .

b. Hence, solve √3 cos 2 + 2 sin 2 = −√2 for 0° ≤ ≤ 180°. Give your answer CHOW CHOON WOOI
to the nearest minute.

8. The parametric equations of a curve is given by

= 2 +1, = −(2 −1)

(a) Find and 2 when = 1.
2

(b) Given = 2 − . Express in terms of and find . Hence, duduce the set



value of t such that is positive.



9. (a) Given ( ) = 2| | + 5 . Compute lim ( ) and l→im0− ( ). Is continuous at
(b)
→0+

x=0? Give your reason.

The continuous function is defined by

( ) = {√5 − , <
3 − 1, ≥

Find the value of .

10. By writing in terms of and , show that

(tan ) = 2 .


a. If = , find 2 in terms of . Hence, determine the range of value of x
2

such that 2 > 0 for 0 < < .
2

b. If = ( + ), find in terms of and .



Hence, show that = − 22 when = = .



END OF QUESTION PAPER

Page 3

221

PSPM I QS 015/2 Session
2016/2017

1. Express 2 in partial fractions form.
2 −2 −3

SOLUTION Improper Fraction

2
2 − 2 − 3

1 ( ) ( ) CHOW CHOON WOOI
x2 − 2x − 3 x2 ( ) = ( ) + ( )

x2 − 2x − 3
2x + 3

2 2 + 3
2 − 2 − 3 = 1 + 2 − 2 − 3

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

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

2 + 3 = ( + 1) + ( − 3)

ℎ = −1

2(−1) + 3 = ((−1) + 1) + ((−1) − 3)

1 = −4
1

= − 4
ℎ = 3

2(3) + 3 = ((3) + 1) + ((3) − 3)

9 = 4

9
= 4

2 + 3 9 1
2 − 2 − 3 = 4( − 3) − 4( + 1)

Page 4

222

PSPM I QS 015/2 Session
2016/2017
2 2 + 3
2 − 2 − 3 = 1 + 2 − 2 − 3

91
= 1 + 4( − 3) − 4( + 1)

CHOW CHOON WOOI

Page 5

223

PSPM I QS 015/2 Session
2016/2017

2. Evaluate the following limits, if exist.

a. lim −2
4 −16
→2

b. lim (2− )( −1)
( −3)2
→∞

SOLUTION CHOW CHOON WOOI

a. lim −2 = lim −2
4 −16 ( 2+4)( 2 −4)
→2 →2

= lim − 2

→2 ( 2 + 4)( + 2)( − 2)

= lim 1

→2 ( 2 + 4)( + 2)

1
= (22 + 4)(2 + 2)

1
= 32

b. lim (2− )( −1) = lim 2 −2− 2+
( −3)2 2 −6 +9
→∞ →∞

= lim − 2 + 3 − 2

→∞ 2 − 6 + 9

− 2 + 3 − 2

= lim 2
2 − 6 + 9
→∞

2

−1 + 3 − 2
2
= lim
1 − 6 + 9
→∞ 2

−1 + 0 + 0
=1

= −1

Page 6

224

PSPM I QS 015/2 Session
2016/2017

3. Show that 2 = 1 + cos . Hence, solve 2 = cos 2 for 0° ≤ ≤ 360°.
1−cos 1−cos

SOLUTION

2 1 − cos2 CHOW CHOON WOOI
1 − cos = 1 − cos

(1 − cos )(1 + cos )
= 1 − cos

= 1 + cos

2 2 + + = 0
1 − cos = cos 2 − ± √ 2 − 4
1 + cos = 2 2 − 1
2 2 − cos − 2 = 0 = 2
= cos

2 − − 2 = 0
−(−1) ± √(−1)2 − 4(2)(−2)

= 2(2)
1 ± √17

=4
= 1.2808, −0.7808
cos = 1.2808 , −0.7808
Since −1 ≤ cos ≤ 1
cos = − 0.7808




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PSPM I QS 015/2 Session
2016/2017
= −10.7808
= 38.67° CHOW CHOON WOOI

Given that 0° ≤ ≤ 360°
= 180° − 38.67, 180° + 38.67

= 141.33°, 218.67°

Page 8

226

PSPM I QS 015/2 Session
2016/2017

4. Consider a function ( ) = 1 .

2−√

a. Find lim ( ) and state the equation of horizontal asymptote for .

→∞

b. By using the first principle of derivative, find ’( ).

SOLUTION CHOW CHOON WOOI

( ) = 1
2 − √

a. lim ( ) = lim 1
→∞ →∞ 2−√

=0

∴ ( ) = 0 ℎ .

′ ( ) = lim ( + ℎ) − ( )
ℎ
ℎ→0

b. ′( ) = lim 1 [ ( + ℎ) − ( )]

ℎ→0 ℎ

11 1
= lim [ − ]
ℎ 2 − √ + ℎ 2 − √
ℎ→0

= lim 1 (2 − √ ) − (2 − √ + ℎ) ]
[
ℎ→0 ℎ (2 − √ + ℎ)(2 − √ )

= lim 1 2 − √ − 2 + √ + ℎ
[ ]
ℎ→0 ℎ (2 − √ + ℎ)(2 − √ )

= lim 1 −√ + √ + ℎ ]
[ √ )
ℎ→0 ℎ (2 − √ + ℎ)(2 −

1 √ + ℎ − √ (√ + ℎ + √ )
= lim [ ][ ]
ℎ→0 ℎ (2 − √ + ℎ)(2 − √ ) √ + ℎ + √

1 ( + ℎ) − (√ √ + ℎ) + (√ √ + ℎ) − ( ) ]
= lim [
ℎ→0 ℎ (2 − √ + ℎ)(2 − √ )(√ + ℎ + √ )

lim 1ℎ
= [ ]
ℎ→0 ℎ (2 − √ + ℎ)(2 − √ )(√ + ℎ + √ )

= lim [ 1 ]

ℎ→0 (2 − √ + ℎ)(2 − √ )(√ + ℎ + √ )

Page 9

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PSPM I QS 015/2 Session
2016/2017
1
=

(2 − √ )(2 − √ )(√ + √ )
1

= 2√ (2 − √ )2

CHOW CHOON WOOI

Page 10

228

PSPM I QS 015/2 Session
2016/2017

5. (a) Use the derivative to find the maximum area of a rectangle that can be inscribed
in a semicircle of radius 10cm.

(b) A cone-shaped tank as shown below.

60° CHOW CHOON WOOI
A
Water flows through a hole A at rate of 6 3 per second. Find the rate of
change in height of the water when the volume of water in the cone is 24 3

SOLUTION
(a)

10


2 + 2 = 100
2 = 100 − 2

1

= (100 − 2)2

Area of rectangle
=

1

= 2 (100 − 2)2

Page 11

229

PSPM I QS 015/2 Session
2016/2017

= 2 1
′ = 2
= (100 − 2)2

′ = 1 (100 − 2 )−12 (100 − 2)
2

1 CHOW CHOON WOOI
= 1 (−2 )

2(100 − 2)2

−2
=1

2(100 − 2)2
−

=1
(100 − 2)2

= ′ + ′


= (2 ) ( − 1

1) + ((100 − 2)2) (2)
(100 − 2)2

= −2 2 1

1 + 2(100 − 2)2
(100 − 2)2

11

−2 2 + 2(100 − 2)2(100 − 2)2
=1

(100 − 2)2

−2 2 + 2(100 − 2)
=1

(100 − 2)2
−2 2 + 200 − 2 2
=1

(100 − 2)2

200 − 4 2
=1

(100 − 2)2

Let = 0



Page 12

230

PSPM I QS 015/2 Session
2016/2017

200 − 4 2 CHOW CHOON WOOI
1=0

(100 − 2)2
200 − 4 2 = 0
4 2 = 200
2 = 50
= ±√50
Since ≥ 0

= √50

−4 2 + 200
=
1
(100 −
2 )2

= −4 2 + 200 1

′ = −8 = (100 − 2)2
′ = 1 (100 − 2)−12 (100 − 2)

2

= 1 (100 − 2)−12(−2 )
2
−

=1
(100 − 2)2

2 ′ − ′
2 = 2

(100 − 2)12(−8 ) − (−4 2 + 200) ( − 1)

= (100 − 2)2

12
((100 − 2)2)

−8 (100 − 2)12 + [ (−4 2 + 2010)]
(100 − 2)2
=
12
((100 − 2)2)

Page 13

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PSPM I QS 015/2 Session
2016/2017

11
[−8 (100 − 2)2(100 − 2)2 + (−4 3 + 200 ) ]

1
(100 − 2)2
= 100 − 2

=[ −8 (100 − 2) + (−4 3 + 200 ) ] [ 1 2]
100
1 − CHOW CHOON WOOI

(100 − 2)2

−800 + 8 3 − 4 3 + 200
=3

(100 − 2)2

4 3 − 600
=3

(100 − 2)2

ℎ = √50

2 4(√50)3 − 600√50
2 = = −8 < 0 (max)
3

[100 − (√50)2]2

1

The maximum area, = 2 (100 − 2)2

1

= 2√50 (100 − √502)2

1

= 2√50(50)2
= 2√50√50
= 2(50)
= 100 2

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PSPM I QS 015/2 Session
2016/2017
(b) r r
ℎ = tan 30°
h h 1 CHOW CHOON WOOI
ℎ = √3
30° ℎ
=
60° √3

A Page 15

= −6 3 −1


ℎ ℎ = 24 3


ℎ ℎ
= .

= 1 2ℎ
3

ℎ
=

√3

1 ℎ2
= () ℎ
3 √3

1 ℎ3
= 3 3

= ℎ3
9

ℎ2
ℎ = 3

ℎ 3
= ℎ2

ℎ 3
= ℎ2 . (−6)

ℎ 18
= − ℎ2

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PSPM I QS 015/2 Session
2016/2017

ℎ = 24

ℎ3 = 24
9

ℎ3 = 24 9


= 216 CHOW CHOON WOOI

ℎ=6

ℎ 18
= − 62
ℎ 18
= − (6)2

= − 1 −1

2

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PSPM I QS 015/2 Session
2016/2017

6. (a) Polynomial ( ) has a raminder 3 when divided by ( + 3). Find the CHOW CHOON WOOI
remainder of ( ) + 2 when divided by ( + 3).

(b) Polynomial 1( ) = 3 + 2 − 5 − 7 has a factor ( − 1) and remainder
1 when divided by ( + 1), while a polynomial 2( ) = 3 − 2 + + 6
has a remainder 2 when divided by ( – 1). Find the value of the constants
and if 1 + 2 = 5. Hence, obtain the zeroes for 1( ).

SOLUTION
(a) (−3) = 3
( ) = ( ) ( ) + ( )
( ) = ( )( + 3) + 3
( ) + 2 = ( )( + 3) + 3 + 2
( ) + 2 = ( )( + 3) + 5
∴ ( ) = 5 ℎ ( ) + 2 ( + 3)

(b) 1( ) = 3 + 2 − 5 − 7
1(1) = 0
(1)3 + (1)2 − 5 (1) − 7 = 0

1 + − 5 − 7 = 0

− 5 = 6 ……………………… (1)

1(−1) = 1

(−1)3 + (−1)2 − 5 (−1) − 7 = 1

−1 + + 5 − 7 = 1

+ 5 = 1 + 8

1 = + 5 − 8 ……………………… (2)

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PSPM I QS 015/2 Session CHOW CHOON WOOI
2016/2017
2( ) = 3 − 2 + + 6
2(1) = 2 ……………………… (3)

(1)3 − (1)2 + (1) + 6 = 2
1 − + + 6 = 2
− + = 2 − 7
2 = − + + 7

(2) + (3)- ……………………… (4)
1 + 2 = 6 − 1

ℎ 1 + 2 = 5
5 = 6 − 1
6 = 6
= 1

(4) (1)
− 5(1) = 6
= 11

∴ = 11, = 1 1( ) = 3 + 2 − 5 − 7 has a
factor of ( − 1)
1( ) = 3 + 11 2 − 5 − 7
= ( − 1) ( )

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PSPM I QS 015/2 Session
2016/2017

x2 + 12x + 7 CHOW CHOON WOOI
x −1 x3 + 11x2 − 5x − 7

x3 − x2

12x2 − 5x − 7
12x2 −12x

7x − 7
7x − 7

0

1( ) = 3 + 11 2 − 5 − 7
= ( − 1)( 2 + 12 + 7)

ℎ 1( )
1( ) = 0
( − 1)( 2 + 12 + 7) = 0

− 1 = 0 ( 2 + 12 + 7) = 0
= 1 − ± √ 2 − 4

= 2
−12 ± √144 − 28

=2
−12 ± √116

=2
= −6 ± √29

, ℎ 1( ) 1, −6 ± √29

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PSPM I QS 015/2 Session
2016/2017

7. Consider a function ( ) = √3 cos 2 + 2 sin 2 . CHOW CHOON WOOI

a. Express in the form of cos(2 − ) for > 0, 0° ≤ ≤ 90° and to the
nearest minute. State the maximum and minimum values of .

b. Hence, solve √3 cos 2 + 2 sin 2 = −√2 for 0° ≤ ≤ 180°. Give your anser
to the nearest minute.

SOLUTION cos( − ) = cos cos + sin sin
(a) cos(2 − ) = cos 2 cos + sin 2 sin

( ) = √3 cos 2 + 2 sin 2

√3 cos 2 + 2 sin 2 = cos(2 − )

√3 cos 2 + 2 sin 2 = [cos 2 cos + sin 2 sin ]

√3 cos 2 + 2 sin 2 = cos 2 cos + Rsin 2 sin

cos = √3 ………………………. (1)

sin = 2 ………………………. (2)

(1)2 + (2)2

cos2 + sin2 = 1

2 cos2 + 2 sin2 = 2 + (2)2

(√3)

2(cos2 + sin2 ) = 3 + 4

2 = 7

= √7
(2) ÷ (1)

sin = 2

cos √3

2
tan =

√3
= 49.11°

∴ ( ) = √7 cos(2 − 49.11°)
− 1 ≤ cos(2 − 49.11°) ≤ 1

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PSPM I QS 015/2 Session
2016/2017
−√7 ≤ √7 cos(2 − 49.11°) ≤ √7
∴ = √7
∴ = −√7

(b) CHOW CHOON WOOI
√3 cos 2 + 2 sin 2 = −√2 for 0° ≤ ≤ 180°
√7 cos(2 − 49.11°) = −√2
cos(2 − 49.11°) = − √2
√7
2 − 49.11° = 180° − 57.69°, 180° + 57.69°
2 − 49.11° = 122.31°, 237.69°
2 = 171.12°, 286.80°
= 85.56°, 143.4°

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PSPM I QS 015/2 Session
2016/2017

8. The parametric equations of a curve is given by

= 2 +1, = −(2 −1)

(c) Find and 2 when = 1.
2

(d) Given = 2 − . Express in terms of and find . Hence, duduce the set CHOW CHOON WOOI



value of t such that iz positive.



SOLUTION

= 2 +1, = −(2 −1)

= 2 2 +1 = −2 −(2 −1)




= .

= −2 −(2 −1). 1
2 2 +1

−2 −(2 −1)
= 2 2 +1

− −(2 −1) = −
= 2 +1

1
= − 4

ℎ = 1
1
= − 4(1)
1
= − 4

2
2 = [ ] .

1 )] 1 +1 ]
= [ (− . [
4 2 2

= (− −4 )] . 1
[ [2 2 +1 ]

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PSPM I QS 015/2 Session
2016/2017
4 −4 . [ 1
= 2 2 +1 ] ( ) =
+ = +
4 −4
= 2 2 +1

2 −4 CHOW CHOON WOOI
= 2 +1

2
= 6 +1

ℎ = 1
2 2
2 = 6 +1
2
= 6(1)+1
2
= 7

= 2 −

= ( 2 +1)2 − ( 2 +1)( −(2 −1))

= 2(2 +1) − 2 +1−2 +1

= 4 +2 − 2

= 4 4 +2



> 0

4 4 +2 > 0

4 +2 > 0

: { : ∈ ℛ}

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PSPM I QS 015/2 Session
2016/2017
9. (a)
Given ( ) = 2| | + 5 . Compute lim ( ) and lim ( ). Is continuous at
(b) →0+ →0−

x=0? Give your reason.

The continuous function is defined by

( ) = {√5 − , < CHOW CHOON WOOI
3 − 1, ≥

Find the value of .

SOLUTION

(a) ( ) = 2| | + 5



| | = {− ,, ≥ 0
< 0

2 > 0
( ) = {2(− ) + 5 , < 0

+ 5 ,

( ) = {−22 + 5 , > 0
+ 5 , < 0

lim ( ) = lim 2 + 5
→0+ →0+

= 2 + 5(0)

=2

lim ( ) = lim −2 + 5
→0− →0−

= −2 + 5(0)

= −2

lim ( ) ≠ lim ( )
→0+ →0−

lim ( )

→0

lim ( ) , ℎ = 0

→0

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PSPM I QS 015/2 Session
2016/2017

(b) ( ) = {√5 − , <
3 − 1, ≥

( ) = 3( ) − 1 = 3 − 1

lim ( ) = lim 3 − 1 = 3( ) − 1 = 3 − 1
→ + → +

lim ( ) = lim √5 − ) = √5 − CHOW CHOON WOOI
→ − → −

= ➔ lim ( ) = l→im − ( )

→ +

3 − 1 = √5 −

(3 − 1)2 = (√5 − 2

)

9 2 − 6 + 1 = 5 −

9 2 − 5 − 4 = 0

(9 + 4)( − 1) = 0

(9 + 4) = 0 ( − 1) = 0

= − 4 = 1

9

ℎ = − 4,

9

4 − 1 ≠ √5 − 4
3 (− 9) (− 9)
=

4 lim ( ) = lim ( )
∴ ≠ − 9 → + → −
ℎ = 1,
3 − 1 = √5 −

3(1) − 1 = √5 − (1)

∴ = 1

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PSPM I QS 015/2 Session
2016/2017

10. By writing in terms of and , show that

(tan ) = 2 .


a. If = , find 2 in terms of . Hence, determine the range of value of x
2

such that 2 > 0 for 0 < < . CHOW CHOON WOOI
2

b. If = ( + ), find in terms of and .



Hence, show that = − 22 when = = .



SOLUTION

sin
(tan ) = (cos )

= sin = cos

′ = cos ′ = −sin

=

(cos )(cos ) − (sin )(−sin )
= (cos )2

cos2 + sin2
= cos2

1
= cos2
= 2

(a) = 1 + 2 = 2
= 2



= 1 + 2
= 1 + 2
2
2 = 2

= 2 (1 + 2)

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PSPM I QS 015/2 Session
2016/2017
2 >0 0 < <
2 tan

2 (1 + 2) > 0

1 + 2 > 0

2 > 0 CHOW CHOON WOOI

> 0

tan > 0

2

∴ (0, )
2

(b) = ( + )

= 2( + ) [ + ]


= 2( + ) [1 +
]


= 2( + ) + 2 ( + )


− 2( + ) = 2 ( + )


[1 − 2 ( + )] = 2 ( + )


2( + )
= 1 − 2( + )

when = =

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

22
= 1 − 22

2 2
= − 22

1

= 22

− 22
22

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PSPM I QS 015/2 Session
2016/2017

1 22 CHOW CHOON WOOI
= 22 (− 22 )

1
= − 22
= − 22

Page 28

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2017/2018

247

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2017/2018

248

PSPM I QS015/1 Session 2017/2018

1. Given matrix = (−22 35) such that 2 + + = 0, are constants, where
and are identity matrix and zero matrix of 2 2 respectively. Determine the value of

.
2. Solve the equation 32 +1 − (16)3 + 5 = 0.

3. The first and three more successive terms in a geometric progression are given as follows:

7, …, 189, y, 1701, … CHOW CHOON WOOI

Obtain the common ratio r. Hence, find the smallest integer n such that the n-th term

exceeds 10,000.

1

4. a) Expand (1 − )2 in ascending power of up to the term in 3 and state the interval

3

of for which the expansion is valid.

1

b) From part 4(a), express √9 − 3 in the form of (1 − )2 , where is an integer.

3

c) Hence, by substituting the suitable value of , approximate √8.70 correct to two

decimal places.

5. Solve the equation 3 9 = ( 3 )2.
6. Given a complex number = 2 + .

a. Express ̅ − 1 in the form + , where and are real numbers.

̅

b. Obtain | ̅ − 1|. Hence, determine the values of real numbers and if +

̅

= | ̅ − 1| ( ̅ − 1)2.

̅ ̅

7. Find the interval of for which the following inequalities are true.
a. 5 − 1 ≤ 0.

+3

b. |3 −2| > 2.

2 +3

8. Consider functions of ( ) = ( − 2)2 + 1, > 2 and ( ) = ln( + 1), > 0.
a. Find −1( ) and −1( ) , and state the domain and range for each of the

inverse function.
b. Obtain ( )( ). Hence, evaluate ( )(2).
9. Given the function ( ) = 1 .

2 −5

a. Find the domain and range of ( ).
b. Show that ( ) is a one-to-one function. Hence, find −1( ).
c. On the same axis, sketch the graph of ( ) and −1( ).
d. Show that −1( ) = .

10. Given the system of linear equations as follow:

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PSPM I QS015/1 Session 2017/2018

2 + 4 + = 77 CHOW CHOON WOOI

4 + 3 + 7 = 114

2 + + 3 = 48

a. Express the system of equations in the form of matrix equation = where


= ( ). Hence, determine matrix and matrix .


b. Based on part 10(a), obtain | |.

Hence, find
i. | | if = , where is an identity matrix 3 3.
ii. | | if = (2 ) .

iii. .
Hence, obtain −1 and find the values of , .

END OF QUESTION PAPER

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