PSPM 1 Q&A

PSPM 1 PSPM 1 CHOW CHOON WOOI
2012 - 2020 (Q&A)

MATRICULATION

MATHEMATICS

KEDAH
MATRICULATION
COLLEGE

1

TABLE OF CONTENTS Page CHOW CHOON WOOI
3
Year
2012/2013 49
2013/2014 92
2014/2015 137
2015/2016 194
2016/2017 247
2017/2018 289
2018/2019 330
2019/2020 368
2020/2021

2

2012/2013

3

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester 1
Session 2012/2013

4

PSPM 1 QS 015/1 Session
2012/2013

1. Find the value of x which satisfies the equation

2(5 − ) − 2( − 2) = 3 − 2(1 + )

2. Determine the solution set of the inequality

11 CHOW CHOON WOOI
2 − 1 < + 2

3. Given + 2, − 4, − 7 are the first three terms of a geometric series. Determine the
value of . Hence, find the sum to infinity of the series.

4. Given a complex number = 1 − √3 . Determine the value of if ̅ ̅ ̅2 = 1 .
̅

5. (a) Matrix M is given as [−34 −41]. Show that 2 = 7 − 8 , where I is the 2 x 2
identity matrix. Deduce that −1 = 7 − 1 .

88

+ 1 −1 1
(b) Given matrix = [ 3 2 4 ] and | | = 27. Find the value of , where

−1 0 + 2

is an integer.

6. The functions and are defined as ( ) = 3 +4, ≠ 2 and ( ) = 3 − .

−2

(a) Find −1( ) and −1( ).

(b) Evaluate ( ⃘ −1)(3).

(c) If ( ⃘ −1)( ) = 2, find the value of .

3

7. (a) Solve | 2 − − 3| = 3.

(b) Find the solution set of the inequality 2 2+9 −4 < 4.

+2

8. The first four terms of a binomial expansion (1 + ) is

1 + − 1 2 + 3 + ⋯
2

Find

(a) the values of a and n where ≠ 0.

Page 2

5

PSPM 1 QS 015/1 Session
2012/2013

(b) the value of p Hence, by substituting = 1, show that √3 is approximately

4 2

equal to 157.

128

9. Given ( ) = ln(2 + 3) and ( ) = −3.

2

(a) Show that ( ) is a one-to-one function algebraically. CHOW CHOON WOOI

(b) Find ( ⃘ )( ) and ( ⃘ )( ). Hence, state the conclusion about the results.

(c) Sketch the graphs of ( ) and ( ) on the same axes. Hence, state the
domain and range of ( ).

10. Given

223
= [1 5 4]

314

(a) Find the determinant of matrix .

(b) Find the minor, cofactor and adjoint of matrix .

(c) Given ( ( )) = | | where is 3 x 3 identity matrix, show that −1 =

1 ( ). Hence, find −1 .
| |

(d) By using −1 in part ( ), solve the following simultaneous equations.
2 + 2 + 3 = 49
+ 5 + 4 = 74
3 + + 4 = 49

END OF QUESTION PAPER

Page 3

6

PSPM 1 QS 015/1 Session
2012/2013

1. Find the value of x which satisfies the equation

2(5 − ) − 2( − 2) = 3 − 2(1 + )

SOLUTION

2(5 − ) − 2( − 2) = 3 − 2(1 + ) CHOW CHOON WOOI
(5 − )

2 ( − 2) = 3 − 2(1 + )

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

(5 − )(1 + ) = 23
( − 2)

(5 − )(1 + )
( − 2) = 8

(5 − )(1 + ) = 8( − 2)

5 + 5 − − 2 = 8 − 16

5 + 4 − 2 = 8 − 16

0 = 2 + 8 − 4 − 16 − 5

2 + 8 − 4 − 16 − 5 = 0

2 + 4 − 21 = 0

( + 7)( − 3) = 0

( + 7) = 0 or ( − 3) = 0

= −7 or = 3

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

When = −7
2 [5 − (−7)] − 2(−7 − 2) = 3 − 2(1 − 7)
≠ −7

When = 3
2 (5 − 3) − 2(3 − 2) = 3 − 2(1 + 3)
2 (2) − 2 (1) = 3 − 2(4)
1−0=3−2

1 = 1√

∴ = 3

Page 4

7

PSPM 1 QS 015/1 Session
2012/2013

2. Determine the solution set of the inequality CHOW CHOON WOOI
11

2 − 1 < + 2
SOLUTION

11
2 − 1 < + 2

11
2 − 1 − + 2 < 0
1( + 2) − 1(2 − 1)

(2 − 1)( + 2) < 0
+ 2 − 2 + 1
(2 − 1)( + 2) < 0

− + 3
(2 − 1)( + 2) < 0

− + 3 = 0 2 − 1 = 0 + 2 = 0
= 3
= 1 = −2
− + 3
2 − 1 2 1 1 (3, ∞)
+ 2 (−2, 2) (2 , 3) -
− + 3 (−∞, −2) +
(2 − 1)( + 2) + + + +

- -+ -

- ++

-+ +

1
∴ ℎ (−2, 2) ∪ (3, ∞)

Page 5

8

PSPM 1 QS 015/1 Session
2012/2013

3. Given + 2, − 4, − 7 are the first three terms of a geometric series. Determine the
value of . Hence, find the sum to infinity of the series.

SOLUTION

+ 2, − 4, − 7 … CHOW CHOON WOOI
− 4 − 7
+ 2 = − 4
( − 4)( − 4) = ( − 7)( + 2)
2 − 4 − 4 + 16 = 2 + 2 − 7 − 14
2 − 8 + 16 = 2 − 5 − 14

16 + 14 = −5 + 8

3 = 30

= 10

When = 10 , the sequence is

10 + 2, 10 − 4, 10 − 7 …

12, 6, 3 …

= 12, = 6 = 1

12 2
∞ = 1 −

12
∞ = 1 − (21)

12
∞ = (21)

∞ = 24

Page 6

9

PSPM 1 QS 015/1 Session
2012/2013

4. Given a complex number = 1 − √3 . Determine the value of if ̅ ̅ 2̅ = 1 .
̅

SOLUTION

= 1 − √3 ̅ = 1 + √3

̅ ̅2̅ = 1 CHOW CHOON WOOI
̅

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

(̅̅1̅̅−̅̅̅√̅̅3̅̅ ̅)̅(̅1̅̅̅−̅̅̅√̅̅3̅ ̅ )̅ =
1 + √3

1̅̅̅−̅̅̅2̅√̅̅3̅̅ ̅ ̅+̅̅̅3̅ ̅ 2̅ =
1 + √3

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

̅1̅̅−̅̅̅2̅̅√̅̅3̅ ̅ ̅−̅̅̅3̅ =
1 + √3

−̅̅̅2̅̅−̅̅̅2̅̅√̅̅3̅̅ =
1 + √3


−2 + 2√3 =

1 + √3

= (−2 + 2√3 )(1 + √3 )

= −2 − 2√3 + 2√3 + 6 2
= −2 + 6(−1)

= −8

Page 8

10

PSPM 1 QS 015/1 Session
2012/2013
5. (a)
Matrix is given as [−34 −41]. Show that 2 = 7 − 8 , where is the 2 x 2
identity matrix. Deduce that −1 = 7 − 1 .

88

+ 1 −1 1 CHOW CHOON WOOI
(b) Given matrix = [ 3 2 4 ] and | | = 27. Find the value of , where

−1 0 + 2

is an integer.

SOLUTION

5(a)
= [−34 −41]
2 = [−34 −41] [−34 −41]
2 = [−192+−416 −4 3+−164]
2 = [−1238 −207]
7 − 8 = 7 [−34 −41] − 8 [10 01]
7 − 8 = [−2218 −287] − [80 08]
7 − 8 = [−1238 −207]
∴ 2 = 7 − 8

2 = 7 − 8

2 − 7 = −8

( − 7 ) = −8

−1 = − 1 ( − 7 )
8

−1 = − 1 + 7
8 8

Page 9

11

PSPM 1 QS 015/1 Session
2012/2013

−1 = 7 − 1
8 8

5(b)

+ 1 −1 1 CHOW CHOON WOOI

= [ 3 2 4 ]
−1 0 + 2

| | = 27

| | = +( + 1) |02 4 2| − (−1) |−31 4 2| + (1) |−31 02|
+ +

| | = +( + 1)[(2 + 4) − (0)] + [(3 + 6) − (−4)] + [(0) − (−2)]

| | = ( + 1)(2 + 4) + (3 + 6 + 4) + 2

| | = 2 2 + 4 + 2 + 4 + 3 + 6 + 4 + 2

| | = 2 2 + 9 + 16

2 2 + 9 + 16 = 27

2 2 + 9 − 11 = 0

(2 + 11)( − 1) = 0

= − 11 = 1

2

Since is an integer, ∴ = 1

Page 10

12

PSPM 1 QS 015/1 Session
2012/2013

6. The functions and are defined as ( ) = 3 +4, ≠ 2 and ( ) = 3 − .

−2

(a) Find −1( ) and −1( ).

(b) Evaluate ( ⃘ −1)(3).

(c) If ( ⃘ −1)( ) = 2, find the value of . CHOW CHOON WOOI

3

SOLUTION

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

3 + 4
= − 2

( − 2) = 3 + 4

− 2 = 3 + 4

− 3 = 4 + 2

( − 3) = 4 + 2

4 + 2
= − 3

−1 ( ) = 4 + 2
− 3

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

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

Page 11

13

PSPM 1 QS 015/1 Session
2012/2013
9 − 3 + 4
= 3 − − 2 Page 12

13 − 3
= 1 −

( ⃘ −1)(3) = 13 − 3(3) CHOW CHOON WOOI
1 − (3)

( ⃘ −1)(3) = 13 − 9
−2

( ⃘ −1)(3) = −2

6(c) −1( ) = 4+2
( ) = 3 +4
−3
−2

( ) = 3 − −1( ) = 3 −

( ⃘ −1)( ) = 2
3

( ⃘ −1)( ) = [ −1( )]

4 + 2
= [ − 3 ]

4 + 2
= 3 − ( − 3 )

4 + 2 2
3 − ( − 3 ) = 3

4 + 2 2
− 3 = 3 − 3

4 + 2 7
− 3 = 3

3(4 + 2 ) = 7( − 3)

12 + 6 = 7 − 21

= 33

14

PSPM 1 QS 015/1 Session
2012/2013
7. (a)
(b) Solve | 2 − − 3| = 3.
Find the solution set of the inequality 2 2+9 −4 < 4.

+2

SOLUTION or 2 − − 3 = −3 CHOW CHOON WOOI
7(a) or 2 − = 0
or ( − 1) = 0
| 2 − − 3| = 3 or = 0 = 1
2 − − 3 = 3
2 − − 6 = 0
( − 3)( + 2) = 0
= 3 = −2

7(b)

2 2 + 9 − 4
<4

+ 2

2 2 + 9 − 4
+ 2 − 4 < 0

(2 2 + 9 − 4) − 4( + 2) <0

+ 2

2 2 + 9 − 4 − 4 − 8 <0

+ 2

2 2 + 5 − 12 <0

+ 2

(2 − 3)( + 4) <0

+ 2

2 − 3 = 0 + 4 = 0 + 2 = 0
= −2
= 3 = −4

2

Page 13

15

PSPM 1 QS 015/1 Session
2012/2013

2 − 3 (−∞, −4) (−4, −2) 3 3
- - (−2, 2) (2 , ∞)
+ 4 - +
- - - +
+ 2 - +
(2 − 3)( + 4) ++

+ 2 ++ CHOW CHOON WOOI

-+

3
∴ ℎ solution set { : < −4 ∪ −2 < < }
2

Page 14

16

PSPM 1 QS 015/1 Session
2012/2013

8. The first four terms of a binomial expansion (1 + ) is

1 + − 1 2 + 3 + ⋯
2

Find

(a) the values of a and n where ≠ 0. CHOW CHOON WOOI

(b) the value of p Hence, by substituting = 1, show that √3 is approximately

42

equal to 157.

128

SOLUTION

8(a)

Given that

(1 + ) = 1 + − 1 2 + 3 + ⋯
2

From binomial expansion

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

(1 + ) = 1 + + ( − 1) 2 2 + ( − 1)( − 2) 3 3 + ⋯
26

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

Coefficient of :
= 1 ...................... (1)

Coefficient of 2:

( − 1) 2 = − 1 = 1
2 2

( − 1) 2 = −1

( − 1) = −1

( − 1) = −1 ................(2)

(2) ( −1) = −1

(1) 1

( − 1)
= −1

− 1 = −

2 = 1

Page 15

17

PSPM 1 QS 015/1 Session
2012/2013
1 CHOW CHOON WOOI
= 2 Page 16
From (1)
= 1

1
(2) = 1

= 2

8(b)

Compare coefficient of 3:

( − 1)( − 2) 3 =
6

(12) [(12) − 1] [(12) − 2] (2)3
6
=

(12) (− 12) (− 32) (8)
6
=

24
8 = 6

6 = 3

1
= 2

(1 + 1 = 1 + − 1 2 + 1 3 + ⋯
2 2
2 )2

When = 1

4

1 1 1 12 1 13

12
[1 + 2 (4)] = 1 + 4 − 2 (4) + 2 (4) + ⋯

1

12 1 1 1 1 1
[1 + 2] = 1 + 4 − 2 (16) + 2 (64) + ⋯

1

32 1 1 1
[2] ≈ 1 + 4 − 32 + 128

18

PSPM 1 QS 015/1 Session
2012/2013

1
3 2 1(128) + 1(32) − 1(4) + 1
[2] ≈
128

1 CHOW CHOON WOOI

3 2 157
[2] ≈ 128

√3 157
2 ≈ 128

Page 17

19

PSPM 1 QS 015/1 Session
2012/2013

9. Given ( ) = ln(2 + 3) and ( ) = −3.

2

(a) Show that ( ) is a one-to-one function algebraically.

(b) Find ( ⃘ )( ) and ( ⃘ )( ). Hence, state the conclusion about the results.

(c) Sketch the graphs of ( ) and ( ) on the same axes. Hence, state the CHOW CHOON WOOI
domain and range of ( ).

SOLUTION

9(a)

( ) = ln(2 + 3)

( 1) = ln(2 1 + 3)

( 2) = ln(2 2 + 3)

Let ( 1) = ( 2)

ln(2 1 + 3) = ln(2 2 + 3)

2 1 + 3 = 2 2 + 3

2 1 = 2 2

1 = 2

Since 1 = 2, therefore ( ) is a one-to-one function.

9(b)

( ) = ln(2 + 3), ( ) = −3

2

( ⃘ )( ) = [ ( )]

− 3
= [ 2 ]

− 3
= ln [2 ( 2 ) + 3]

= ln[ − 3 + 3]

Page 18

20

PSPM 1 QS 015/1 Session
2012/2013

= ln[ ]

=

= ln = CHOW CHOON WOOI
( ⃘ )( ) = [ ( )] ln(2 +3) = 2 + 3

= [ (2 + 3)]

ln(2 +3) − 3
=2

(2 + 3) − 3
=2
=

Conclusion
Since  ( ) =  ( ) = , therefore −1( ) = ( ) and −1( ) = ( ).

9(c)

3 y
= − 2
( )

( )


3

= − 2

Page 19

21

PSPM 1 QS 015/1 Session
2012/2013

= (− , ∞) =(−∞, ∞)


CHOW CHOON WOOI

Page 20

22

PSPM 1 QS 015/1 Session
2012/2013

10. Given
223

= [1 5 4]
314

(a) Find the determinant of matrix .

(b) Find the minor, cofactor and adjoint of matrix . CHOW CHOON WOOI

(c) Given ( ( )) = | | where is 3 x 3 identity matrix, show that −1 =

1 ( ). Hence, find −1 .
| |

(d) By using −1 in part ( ), solve the following simultaneous equations.
2 + 2 + 3 = 49
+ 5 + 4 = 74
3 + + 4 = 49

SOLUTION

10(a)

223
= [1 5 4]

314

| | = (2) |15 44| − (2) |31 44| + (3) |31 15|

= (2)(20 − 4) − (2)(4 − 12) + (3)(1 − 15)

= (2)(16) − (2)(−8) + (3)(−14)

= 32 + 16 − 42

=6

10(b)

|15 44| |13 44| |13 15|
, = |21 34| |32 34| |23 21|

[|25 43| |12 34| |12 25|]
(20 − 4) (4 − 12) (1 − 15)

= [ (8 − 3) (8 − 9) (2 − 6) ]

(8 − 15) (8 − 3) (10 − 2)

Page 21

23

PSPM 1 QS 015/1 Session
2012/2013
16 −8 −14
= [ 5 −1 −4 ] −1 =

−7 5 8

+ 11 − 12 + 13 CHOW CHOON WOOI
, = [− 21 + 22 − 23]
− 32 + 33
+ 31

+(16) −(−8) +(−14)
= [ −(5) +(−1) −(−4) ]

+(−7) −(5) +(8)

16 8 −14
= [−5 −1 4 ]

−7 −5 8

, =

16 8 −14
= [−5 −1 4 ]

−7 −5 8

16 −5 −7
= [ 8 −1 −5]

−14 4 8

10(c)

( ( )) = | |

−1 ( ( )) = −1| |

( ( )) = −1| |

( ( )) = −1| |

−1| | = ( ( ))

−1 = 1 ( ( ))
| |

−1 = 1 16 −5 −7
6 [8 −1 −5]
4
−14 8

Page 22

24

PSPM 1 QS 015/1 Session
2012/2013

16 −5 −7 CHOW CHOON WOOI
6 66
= 8 −1 −5
6 66
−14 4 8
[ 6 6 6]

8 −5 −7
366
= 4 −1 −5
366
−7 2 4
[3 3 3]

10(d) =
−1 = −1
2 + 2 + 3 = 49
+ 5 + 4 = 74 = −1
3 + + 4 = 49 = −1

2 2 3 49
[1 5 4] [ ] = [74]
3 1 4 49

8 −5 −7 8 −5 −7

3 6 6 2 2 3 3 6 6 49
4 −1 −5 [1 5 4] [ ] = 4 −1 −5 [74]

36 6 3 1 4 36 6 49
−7 2 4 −7 2 4

[3 3 3] [3 3 3]

8 −5 −7

3 6 6 49
[ ] = 4 −1 −5 [74]

36 6 49
−7 2 4

[3 3 3]

 71
 
=  6 
 73 

6
1
 3 


∴ = , = , =

Page 23

25

QS 015/2 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester 1
Session 2012/2013

26

PSPM 1 QS 015/2 Session 2012/2013

1 + , < 1
1. Given that ( ) = { 1, = 1

2 − , > 1

Find lim f (x) and lim f (x) . Does the lim f (x) exist? State your reason.
x →1− x →1+ x→1

2. Prove that 1 + tan 2 tan = sec 2 .

3. Find the following limits: CHOW CHOON WOOI

2x2 + x − 4
(a) lim

x→ 1− x2
(b) lim 3 − x + 7

x→2 x2 − 4

4. Express 2 3 −7 2 +17 −19 in the form of partial fractions.
2 2 −7 +6

5. (a) Given that ( ) = | 2− −2| , ≠ 0, 2

{ 2−2

0, = 2

Find the lim f (x) . Is ( ) continuous at = 2?
x→2

+ 6 , < 4
(b) A function ( ) is defined by ( ) = { 2 + 2 , 4 ≤ < 6

2 − , ≥ 6

Determine the values of the constants and if ( ) is continuous.

6. The polynomial ( ) = 2 3 + 2 + − 24 has a factor ( – 2) and a remainder
15 when divided by ( + 3).

(a) Find the values of a and b.
(b) Factorise ( ) completely and find all zeroes of ( ).

7. Given ( ) = 3 sin − 2 cos .

(a) Express ( ) in the form of ( − ), where > 0, 0 ≤ ≤ .

2

Hence, find the maximum and minimum values of ( ).

(b) Solve ( ) = √3 0° ≤ ≤ 360°.

2

Page 2

27

PSPM 1 QS 015/2 Session 2012/2013

8. (a) Given that = 1 .

√2 +1

i. By using the first principle of derivative, find .



ii. Find 2 2 .

(b) Find of the following: CHOW CHOON WOOI



i. = 2 tan
ii. = sec

9. (a) A conical tank is of height 12m and surface diameter 8m. Water is pumped
into the tank at the rate of 50 3/ . How fast is the water level increasing

when the depth of the water is 6m?

(b) A cylindrical container of radius and height h has a constant volume . The
cost of the materials for the surface of both of its ends is twice the cost of its
sides. State ℎ in terms of and . Hence, find h and in terms of such that
the cost is minimum.

10. (a) Given 3 2 − + 2 = 3. By using implicit differentiation,

i. Find the values of at = 1.


ii. Show that (6 − ) 2 + 6 ( 2 − 2 + 2 = 0.
2
)

(b) Consider the parametric equations

= 3 − 2, = 3 + 2 where ≠ 0.



i. Show that = 1 − 3 24+2.


ii. Find 2 when = 1.
2

END OF QUESTION PAPER Page 3

28

PSPM 1 QS 015/2 Session 2012/2013

1 + , < 1
1. Given that ( ) = { 1, = 1

2 − , > 1

Find lim f (x) and lim f (x) . Does the lim f (x) exist? State your reason.
x →1− x →1+ x→1

SOLUTION

1 + , < 1 CHOW CHOON WOOI
( ) = { 1, = 1

2 − , > 1

lim f (x) = lim (1+ ex )
x →1− x→1−

= (1+ e1)

= 1+ e

lim f (x) = lim (2 − x)
x →1+ x →1+

=2−1

=1

Since lim f (x)  lim f (x) . Therefore lim f (x) does not exist.
x →1− x →1+ x→1

Page 4

29

PSPM 1 QS 015/2 Session 2012/2013

2. Prove that 1 + tan 2 tan = sec 2 .
SOLUTION

sin 2 sin cos( − ) = cos cos + sin sin CHOW CHOON WOOI
1 + tan 2 tan = 1 + (cos2 ) (cos ) cos(2 − ) = cos 2 cos + sin 2 sin

sin 2 sin
= 1 + (cos2 cos )

cos2 cos + sin 2 sin
= cos2 cos

cos(2 − )
= cos2 cos

cos( )
= cos2 cos

1
= cos2
= sec 2 .

Page 5

30

PSPM 1 QS 015/2 Session 2012/2013
Page 6
3. Find the following limits:

(a) lim 2x2 + x − 4
x→ 1− x2

(b) lim 3− x+7
x→2
x2 − 4

SOLUTION CHOW CHOON WOOI

3(a)

2x2 + x − 4

lim 2x2 + x − 4 = lim x2
1− x2 1− x2
x→ x→

x2

2 + 1 − 4
x x2
= lim
x→ 1
x2 −1

= 2+0−0
0 −1

= −2

3(b)

lim 3 − x + 7 = lim 3 − x + 7 . 3 + x + 7
x→2 x2 − 4 x→2 x2 − 4 3 + x + 7

= lim 9 − (x + 7)

x→2 (x + 2)(x − 2)(3 + x + 7)

= lim 2−x

x→2 (x + 2)(x − 2)(3 + x + 7)

= lim − (x − 2)

x→2 (x + 2)(x − 2)(3 + x + 7)

= lim −1

x→2 (x + 2)(3 + x + 7 )

31

PSPM 1 QS 015/2 Session 2012/2013

= −1 CHOW CHOON WOOI
(2 + 2)(3 + 2 + 7)

= −1
(4)(3 + 9)

= −1
24

Page 7

32

PSPM 1 QS 015/2 Session 2012/2013

4. Express 2 3 −7 2 +17 −19 in the form of partial fractions.
2 3 −7 +6

SOLUTION

2 3 −7 2+17 −19 ➔ Improper fraction
2 2 −7 +6

x CHOW CHOON WOOI
2x2 − 7x + 6 2x3 − 7x2 +17x −19

2x3 − 7x2 + 6x

11x −19

2 3 − 7 2 + 17 − 19 11x −19
= +
2 2 − 7 + 6 2x2 − 7x + 6

2 3 − 7 2 + 17 − 19 11x − 19
= + (2 − 3)( − 2)
2 2 − 7 + 6

11x − 19
(2 − 3)( − 2) = (2 − 3) + ( − 2)

11x − 19 ( − 2) + (2 − 3)
(2 − 3)( − 2) = (2 − 3)( − 2)

11x − 19 = ( − 2) + (2 − 3)

ℎ = 2

11(2) − 19 = [(2) − 2] + [(2(2) − 3)]

3 = [0] + [(1)]

= 3

3 3
ℎ = 2
33
11 (2) − 19 = [(2) − 2] + [(2 (2) − 3)]

51
− 2 = [− 2] + [(0)]

= 5

11x − 19 53
(2 − 3)( − 2) = (2 − 3) + ( − 2)

2 3 − 7 2 + 17 − 19 11x − 19
= + (2 − 3)( − 2)
2 2 − 7 + 6

2 3 − 7 2 + 17 − 19 53
= + (2 − 3) + ( − 2)
2 2 − 7 + 6

Page 8

33

PSPM 1 QS 015/2 Session 2012/2013

5. (a) Given that ( ) = | 2− −2| , ≠ 0, 2

{ 2−2

0, = 2

Find the lim f (x) . Is ( ) continuous at = 2?
x→2

+ 6 , < 4 CHOW CHOON WOOI
(b) A function ( ) is defined by ( ) = { 2 + 2 , 4 ≤ < 6

2 − , ≥ 6

Determine the values of the constants and if ( ) is continuous.

SOLUTION

5(a)

| 2 − − 2| ≠ 0, 2
( ) = { 2 − 2 , = 2

0,

lim f (x) = lim x2 − x − 2

x→2− x→2− x2 − 2x

(x +1)(x − 2)
= lim

x→2− x(x − 2)

= lim (x + 1)[−(x − 2)]
x→2− x(x − 2)

= lim − (x +1)
x→2− x

= − (2 +1)
2

=−3
2

lim f (x) = lim x2 − x − 2

x→2+ x→2+ x2 − 2x

(x +1)(x − 2)
= lim

x→2+ x(x − 2)

Page 9

34

PSPM 1 QS 015/2 Session 2012/2013
Page 10
= lim (x + 1)(x − 2)
x→2+ x(x − 2)

= lim x +1
xx→2+

= 2+1 CHOW CHOON WOOI
2

=3
2

lim f (x)  lim f (x)
x→2− x→2+

lim f (x) does not exist

x→2

f (x) discontinuous at x = 2

5(b)

+ 6 , < 4
( ) = { 2 + 2 , 4 ≤ < 6

2 − , ≥ 6

f is continuous at x = 4

lim f (x) exists.

x→4

lim f (x) = lim f (x)
x→4− x→4+

lim x + 6 = lim x2 + 2
x→4− x→4+

 (4) + 6 = (4)2 + 2

4 = 12
 =3
f is continuous at x = 6

lim f (x) exists.

x→6

lim f (x) = lim f (x)
x→6− x→6+

lim x2 + 2 = lim 2 − x
x→6− x→6+

35

PSPM 1 QS 015/2 Session 2012/2013
(6)2 + 2 = 2 −  (6)
38 = 2 − 6 CHOW CHOON WOOI
6 = −36
 = −6

∴ = 3, = −6

Page 11

36

PSPM 1 QS 015/2 Session 2012/2013

6. The polynomial ( ) = 2 3 + 2 + − 24 has a factor ( – 2) and a remainder 15
when divided by ( + 3).

a. Find the values of a and b.
b. Factorise ( ) completely and find all zeroes of ( ).

SOLUTION CHOW CHOON WOOI

6(a)

(2) = 0
(−3) = 15
( ) = 2 3 + 2 + − 24
(2) = 2(2)3 + (2)2 + (2) − 24 = 0

16 + 4 + 2 − 24 = 0
4 + 2 = 8 .................(1)
(−3) = 2(−3)3 + (−3)2 + (−3) − 24 = 15
−54 + 9 − 3 − 24 = 15
9 − 3 = 93
3 − = 31.................(2)
(2) 2
6 − 2 = 62.................(3)
(1)+(2)
10 = 70
= 7
= −10

Page 12

37

PSPM 1 QS 015/2 Session 2012/2013

6(b) CHOW CHOON WOOI

P(x) = 2x3 + 7x2 −10x − 24
= (x − 2)Q(x)

2x 2 + 11x + 12
x − 2 2x3 + 7x 2 − 10x − 24

2x3 − 4x2
11x 2 − 10x − 24
11x 2 − 22x
12x − 24
12x − 24
0

P(x) = (x − 2)(2x2 + 11x + 12)

= (x − 2)(2x + 3)(x + 4)

when P(x) = 0

(x − 2)(2x + 3)(x + 4) = 0
x = 2 , x = − 3 , x = −4

2
The zeroes are 2 , − 3 and − 4

2

Page 13

38

PSPM 1 QS 015/2 Session 2012/2013

7. Given ( ) = 3 sin − 2 cos .

a. Express ( ) in the form of ( − ), where > 0, 0 ≤ ≤ .

2

Hence, find the maximum and minimum values of ( ).

b. Solve ( ) = √3 0° ≤ ≤ 360°. CHOW CHOON WOOI

2

SOLUTION

a) 3sin  − 2 cos = R sin( −  )
= R sin  cos  − R cos sin 

R cos  = 3 (1)

R sin  = 2 (2)

(1)2 + (2)2 :
R 2 cos 2  + R 2 sin 2  = 32 + 22
R2 (1) = 13

R = 13

(2)  (1) :
R sin  = 2
R cos 3

tan = 2
3

 = 0.588

3sin − 2 cos = 13 sin( − 0.588)
−1  sin( − 0.588)  1

− 13  13 sin( − 0.588)  13

− 13  3sin − 2 cos  13

− 13  f ( )  13

Minimum value of f ( ) = − 13
Maximum value of f ( ) = 13

Page 14

39

PSPM 1 QS 015/2 Session 2012/2013

b) f ( ) = 13 13
2 2

13 sin( − 0.588) =
sin( − 0.588) = 1

2

 − 0.588 = 0.785 ,  − 0.785 CHOW CHOON WOOI

 = 1.373 , 2.945

 = 78.70 , 168.70

Page 15

40

PSPM 1 QS 015/2 Session 2012/2013

8. (a) Given that = 1 . CHOW CHOON WOOI

√2 +1

i. By using the first principle of derivative, find .



ii. Find 2 2 .
(b) Find of the following:



iii. = 2 tan

iv. = sec

SOLUTION

(a) y = 1
2x +1

i) Let f (x) = 1
2x +1

f (x + h) = 1

2(x + h) +1

f (x) = lim f (x + h) − f (x)
h→0 h

dy = lim 1− 1
dx h→0 2(x + h) + 1 2x +1

h

= lim 2x +1 − 2(x + h) +1
h→0 2(x + h) +1 2x + 1

h

= lim 2x + 1 − 2x + 2h + 1 2x + 1 + 2x + 2h + 1
h→0 h 2(x + h) + 1 2x + 1 2x + 1 + 2x + 2h + 1

= lim (2x +1) − (2x + 2h +1)

h→0 h 2(x + h) +1 2x +1( 2x +1 + 2x + 2h +1)

= lim − 2h

h→0 h 2(x + h) +1 2x +1( 2x +1 + 2x + 2h +1)

= lim −2

h→0 2(x + h) + 1 2x +1( 2x +1 + 2x + 2h +1)

Page 16

41

PSPM 1 QS 015/2 Session 2012/2013
Page 17
= −2
2x +1 2x +1( 2x +1 + 2x +1)

= −2
(2x +1)(2 2x +1)

=− 1 CHOW CHOON WOOI
3
(2x + 1) 2

ii) dy −3

dx = −(2x + 1) 2

d2y = 3 (2x −5

dx2 2 + 1) 2 (2)

=3
5
(2x + 1) 2

b i) y = e2x tan x

dy = e2x (sec2 x) + tan x(2e2x )
dx

= e2x (sec 2 x + 2 tan x)

ii) y = xsec x

ln y = ln xsec x

ln y = sec x ln x

1 dy = sec x 1  + (ln x)(sec x tan x)
y dx  x 

dy = y  sec x + x(ln x) sec x tan x 
dx x 

= x sec x sec x1 + x(ln x) tan x
x 

42

PSPM 1 QS 015/2 Session 2012/2013

9. (a) A conical tank is of height 12m and surface diameter 8m. Water is pumped
into the tank at the rate of 50 3/ . How fast is the water level increasing

when the depth of the water is 6m?

(b) A cylindrical container of radius and height h has a constant volume . The CHOW CHOON WOOI
cost of the materials for the surface of both of its ends is twice the cost of its
sides. State ℎ in terms of and . Hence, find h and in terms of such that
the cost is minimum.

SOLUTION

9(a) 4
8m

4m

12 m r r ℎ
h 12 4 = 12

h ℎ
= 3

= 50 3/ [ ℎ =? ℎ ℎ = 6]



ℎ ℎ
= .

= 1 2ℎ
3

1 ℎ2
= 3 (3) ℎ

1 ℎ3
= 3 9

= 1 ℎ3
27

= 1 ℎ2 = ℎ2
ℎ 9 9

Page 18

43

PSPM 1 QS 015/2 Session 2012/2013

ℎ 9 r CHOW CHOON WOOI
= ℎ2
ℎ ℎ = 2 ℎ
= .
ℎ 9 h
= ( ℎ2) . (50)
ℎ 450
= ℎ2
ℎ ℎ = 6
ℎ 450
= (6)2
ℎ 450
= 36
ℎ 25
= 2 /

9(b)

r

h

2

2

V =  r2h
h= V

 r2

Page 19

44

PSPM 1 QS 015/2 Session 2012/2013

The cost,
C = 2 r2 (2) + 2 rh(1)

= 4 r2 + 2 r V 2 
r

C = 4 r 2 + 2V
r

dC = 8 r − 2V CHOW CHOON WOOI
dr r 2

when dC = 0
dr

8 r − 2V =0
r2

8 r = 2V
r2

r3 = V
4

1

r =  V  3
 4 

d 2C = 8 + 4V
dr 2 r3

1

when r =  V  3 ,
 4 

d 2C = 8 + 4V = 24  0 (min)
dr 2 V

4

1

 r =  V  3
 4 

1

 h = V 2 =V =  16V  3
r 2   

  V  3
 4 

Page 20

45

PSPM 1 QS 015/2 Session 2012/2013

10 (a) Given 3 2 − + 2 = 3. By using implicit differentiation,

i. Find the values of at = 1.



ii. Show that (6 − ) 2 + 6 ( 2 − 2 + 2 = 0.
2
)

(b) Consider the parametric equations CHOW CHOON WOOI

= 3 − 2, = 3 + 2 where ≠ 0.



iii. Show that = 1 − 3 24+2.


iv. Find 2 when = 1.
2

SOLUTION
10[a(i))]

3 2 − + 2 = 3
6 − [ + ] + 2 = 0




6 − − + 2 = 0


(6 − ) = − 2
− 2
= 6 −
ℎ = 1

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

3

Page 21

46

PSPM 1 QS 015/2 Session 2012/2013
Page 22
6 y dy − x dy  + y(−1) + 2x = 0
dx  dx 

dy (6 y − x) = y − 2x
dx

dy = y − 2x
dx 6 y − x

For x = 1, y = − 2 : CHOW CHOON WOOI
3

 − 2  − 2(1) =8
dy =  3 
dx 6 − 2  −1 15
 3

For x = 1, y = 1 :

dy = 1 − 2(1) = − 1
dx 5
6(1) −1

10[a(ii))]

dy (6 y − x) = y − 2x
dx

dy  6 dy − 1 + (6 y − x) d2y  = dy − 2
dx  dx  dx 2 dx

6 dy  2 − dy + (6 y − x) d2y  − dy + 2 = 0
 dx  dx dx 2 dx

(6 y − x) d2y + 6 dy  2 − 2 dy + 2 = 0
dx 2  dx  dx

10(b)

x = 3t − 2 , y = 3t + 2
tt

i) dx = 3+ 2 = 3t 2 + 2
dt t2 t2

dy = 3 − 2 = 3t 2 − 2
dt t 2 t2

dy

dy = dt
dx dx

dt

47

PSPM 1 QS 015/2 Session 2012/2013

3t 2 − 2
= t2

3t 2 + 2
t2

= 3t 2 − 2
3t 2 + 2

1 CHOW CHOON WOOI
3t 2 + 2 3t 2 − 2

3t 2 + 2
−4

dy = 1 − 4
dx 3t 2 + 2

ii) d  dy  = d 1 − 3t 4 2 
dt  dx  dt  2+ 

= d [1− 4(3t2 + 2)−1]
dt

= 4(3t 2 + 2)−2 (6t)

= 24t
(3t 2 + 2)2

d2y = d  dy  24t = 24t 3
dt  dx  = (3t 2 + 2)2

dx2 dx 3t 2 + 2 (3t 2 + 2)3

dt t 2

when t = 1,

d 2 y = 24(1)3 = 24
dx2 (3(1)2 + 2)3 125

Page 23

48

2013/2014

49

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2013/2014

50