PSPM 1 Q&A

PSPM I QS 015/1 Session 2013/2014

530 3 0

1. Given matrices = [3 −1 0] and = [3 0]. Find the values of c, d and e

002 0 0
such that AB = 14I, where I is the identity matrix. Hence, determine −1.

2. Consider the function ( ) = 1 + ln( ) , ≥ 1. Determine −1( ) and state its range. CHOW CHOON WOOI
Hence, evaluate −1(3).

3. Find the value of x which satisfies the equation
9 = ( 3 )2, > 1.

4. Solve the equation 22 −2 − 2 +1 = 2 − 23.

5. Given ( ) = +8, ≠ 5 where k is a constant.
4 −5 4

a. Find the value of k if ( ∘ )( ) = .

b. Find the value of k so that ( ) is not a one-to-one function.

6. Given ( ) = 3 + 4, ∈ ℜ.
a. Find −1( ).
b. On the same axes, sketch the graphs of ( ) and −1( ). State the domain of
( ) −1( ).

7. a. Find the values of p and q if + = 1 + 5 .
b. 4−2 4+2 2

Given 102 = and 10 7 = . Express x in terms of m and n if
(143 +1)(82 +3) = 7.

Page 2

51

PSPM I QS 015/1 Session 2013/2014

8. An osteoprorosis patient was advised by a doctor to take enough magnesium, CHOW CHOON WOOI
vitamin D and calcium to improve bone density. In a week, the patient has to take 8
units magnesium, 11 units vitamin D and 17 units calcium. The following are three
types of capsule that contains the three essential nutrients for the bone:

Capsule of type P: 2 units magnesium, 1 unit vitamin D and 1 unit calcium.

Capsule of type Q: 1 unit magnesium, 2 units vitamin D and 3 units calcium.

Capsule of type R: 4 units magnesium, 6 units vitamin D and 10 units calcium.

Let x, y and z represent the number of capsule of types P, Q and R respectively that
the patient has to take in a week.

(a) Obtain a system of linear equation to represent the given information and


write the system in the form of matrix equation AX = B, where = ( ).


(b) Find the inverse of matrix A from part (a) by using the adjoint method. Hence,
find the values of x, y and z.

(c) The cost for each capsule of type P, Q and R are RM10, RM15 and RM17

respectively. How much will the expenses be for 4 weeks if the patient follws
the doctor’s advice?

9. a. In an arithmetic progression, the sum of the first four terms is 46 and the
b. seventh term exceeds twice of the second term by 5. Obtain the first term and
the common difference for the progression. Hence, calculate the sum of the
first ten even terms of the progression.

A ball is dropped from a height of 2m. Each time the ball hits the floor, it
bounces vertically to a height that is 3 of its previous height.

4

i. Find the height of the ball at the tenth bounce.
ii. Find the total distance that the ball will travel before the eleventh

bounce.

10 a. Find the solution set of |2 − 3 | < | + 3|.
b.
If x + 1 < 0, show that

i. 2x – 1 < 0
ii. 2 −1 > 2

+1

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52

PSPM I QS 015/1 Session 2013/2014

530 3 0

1. Given matrices = [3 −1 0] and = [3 0]. Find the values of c, d and e

002 0 0
such that = 14 , where I is the identity matrix. Hence, determine −1.

SOLUTION

= 14 CHOW CHOON WOOI

5 3 0 3 0 14 0 0
[3 −1 0] [3 0] = [ 0 14 0 ]

0 0 2 0 0 0 0 14

5 + 9 15 + 3 0 14 0 0
[3 − 3 9 − 0 ] = [ 0 14 0 ]

0 0 2 0 0 14

5 + 9 = 14 ➔ = 1

15 + 3 = 0 ➔ = −5

2 = 14 ➔ = 7

∴ = 1, = −5, = 7

= 14

−1 = 1
14

−1 = 1 1 3 0
14 [3 −5 0]
0
0 7

13
14 14 0
3 −5
−1 = 14 14 0

7
[ 0 0 14]

13
14 14 0

−1 = 3 −5 0
14 14
1
[ 0 0 2]

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53

PSPM I QS 015/1 Session 2013/2014

2. Consider the function ( ) = 1 + ln( ) , ≥ 1. Determine −1( ) and state its range.
Hence, evaluate −1(3).

SOLUTION Method II CHOW CHOON WOOI

Method I ( ) = 1 + ln( )
( −1( )) =
= ln + 1, ≥ 1 1 + ln( −1( )) =
ln = − 1 ln( −1( )) = − 1
= −1 −1( ) = −1
−1( ) = −1 −1 = { : ≥ 1}
−1 = { : ≥ 1} −1(3) = 3−1 = 2
−1(3) = 3−1 = 2

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PSPM I QS 015/1 Session 2013/2014

3. Find the value of x which satisfies the equation
9 = ( 3 )2, > 1.

SOLUTION

9 = ( 3 )2 CHOW CHOON WOOI

3 = ( 3 )2
3 9

3 = ( 3 )2
332

3 = ( 3 )2
2 33

3 = ( 3 )2
2

3 = 2( 3 )2

= 3
= 2 2

2 2 − = 0

(2 − 1) = 0

= 0 or (2 − 1) = 0
= 0 or
= 1

2

3 = 0 or 3 = 1
2

= 30 or 1

= 32

= 1 or = √3

> 1, ℎ = √3

Page 6

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PSPM I QS 015/1 Session 2013/2014

4. Solve the equation 22 −2 − 2 +1 = 2 − 23.

SOLUTION

22 −2 − 2 +1 = 2 − 23

22 2−2 − 2 21 = 2 − 23 CHOW CHOON WOOI

(2 )2 2(2 ) = 2 − 8
4−

= 2

( )2
4 − 2( ) = − 8

2 − 8 = 4 − 32

2 − 8 − 4 + 32 = 0

2 − 12 + 32 = 0

( − 8)( − 4) = 0

− 8 = 0 or − 4 = 0

= 8 or = 4

2 = 8 or 2 = 4

= 3 or = 2

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PSPM I QS 015/1 Session 2013/2014

5. Given ( ) = 4 +−58, ≠ 5 where k is a constant.
4

a. Find the value of k if ( ∘ )( ) = .

b. Find the value of k so that ( ) is not a one-to-one function.

SOLUTION

5a. CHOW CHOON WOOI

( ) = + 8
4 − 5

( ∘ )( ) =

[ ( )] =

(4 + 58) + 8 =
4 (4 − 85) − 5
+

−

( 4 + 85) + 8(4 − 5) =
4 ( 4 − 58) − (4 − 5)

+ 5(4 − 5)
− (4 − 5)

( + 8) + 8(4 − 5)

4( + 4 − 5 − 5) =
8) − 5 (4

4 − 5

( + 8) + 8(4 − 5)
4( + 8) − 5 (4 − 5) =

( 2 + 8 ) + (32 − 40) = [(4 + 32) − (20 − 25)]

2 + 8 + 32 − 40 = [4 + 32 − 20 + 25]

( 2 + 32) + (8 − 40) = 4 2 + 32 − 20 2 + 25

( 2 + 32) + (8 − 40) = (4 − 20) 2 + 57

ℎ 2

4 − 20 = 0

= 5

Page 8

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PSPM I QS 015/1 Session 2013/2014

5b.

( ) = + 8
4 − 5

Let 2 ≠ 1, but ( 2) = ( 1)

2 + 8 = 1 + 8 CHOW CHOON WOOI
4 2 − 5 4 1 − 5

( 2 + 8)(4 1 − 5) = ( 1 + 8)(4 2 − 5)

4 1 2 − 5 2 + 32 1 − 40 = 4 1 2 − 5 1 + 32 2 − 40

32 1 − 5 2 − 32 2 + 5 1 = 0

32 1 − 32 2 + 5 1 − 5 2 = 0

32( 1 − 2) + 5 ( 1 − 2) = 0

( 1 − 2)(32 + 5 ) = 0

1 − 2 ≠ 0 ➔ 32 + 5 = 0
−32

= 5

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PSPM I QS 015/1 Session 2013/2014

6. Given ( ) = 3 + 4, ∈ ℜ.
a. Find −1( ).
b. On the same axes, sketch the graphs of ( ) and −1( ). State the domain of
( ) −1( ).

SOLUTION

6a CHOW CHOON WOOI
( ) = 3 + 4

Method I Method II

Let = 3 + 4 [ −1( )] =

3 = − 4 3 −1( ) + 4 =

ln( 3 ) = ln ( − 4) 3 −1( ) = − 4

3 ln = ln ( − 4) ln 3 −1( ) = ( − 4)

3 = ln ( − 4) 3 −1( ) ln = ( − 4)

1 3 −1( ) = ( − 4)
= 3 ln ( − 4)

∴ −1 ( ) = 1 ln ( − 4) −1 ( ) = 1 ln ( − 4)
3 3

6b
y f(x)
y=x

5 −1( )
4 x

45

Domain ( ): {−∞, ∞} Domain −1( ): {4, ∞}

Page 10

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PSPM I QS 015/1 Session 2013/2014

7. a. Find the values of p and q if + = 1 + 5 .
4−2 4+2 2

b. Given 102 = and 107 = . Express x in terms of m and n if
(143 +1)(82 +3) = 7.

SOLUTION CHOW CHOON WOOI

7a

5
4 − 2 + 4 + 2 = 1 + 2

(4 + 2 ) + (4 − 2 ) 2 + 5
(4 − 2 )(4 + 2 ) = 2

4 + 2 + 4 − 2 2 + 5
16 + 8 − 8 − 4 2 = 2

4 + 4 + 2 − 2 2 + 5
16 − 4(−1) = 2

(4 + 4 ) + (2 − 2 ) 2 + 5
20 = 2

2[(4 + 4 ) + (2 − 2 ) ] = 20[2 + 5 ]

[(4 + 4 ) + (2 − 2 ) ] = 20 [2 + 5 ]
2

[(4 + 4 ) + (2 − 2 ) ] = 10[2 + 5 ]

[(4 + 4 ) + (2 − 2 ) ] = [20 + 50 ]

(4 + 4 ) = 20

+ = 5 ......................... (1)

2 − 2 = 50

− = 25........................ (2)

(1) + (2)

2 = 30

= 15

= −10

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PSPM I QS 015/1 Session 2013/2014

7b

10 2 = , 107 =
(143 +1)(82 +3) = 7

[(2 7)3 +1][(23)2 +3] = 7

[23 +1 73 +1][(2)6 +9] = 7 CHOW CHOON WOOI

[23 +1 (2)6 +9 ][73 +1] = 7

[23 +1+6 +9 ][73 +1] = 7

[29 +10 ][73 +1] = 7

{[29 +10 ][73 +1]} = log 7

29 +10 + 73 +1 = log 7

(9 + 10) log 2 + (3 + 1) log 7 = log 7

(9 + 10) + (3 + 1) =

9 + 10 + 3 + =

9 + 3 = − − 10

(9 + 3 ) = −10

−10
= 9 + 3

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61

PSPM I QS 015/1 Session 2013/2014

8. An osteoprorosis patient was advised by a doctor to take enough magnesium, CHOW CHOON WOOI
vitamin D and calcium to improve bone density. In a week, the patient has to take 8
units magnesium, 11 units vitamin D and 17 units calcium. The following are three
types of capsule that contains the three essential nutrients for the bone:

Capsule of type P: 2 units magnesium, 1 unit vitamin D and 1 unit calcium.

Capsule of type Q: 1 unit magnesium, 2 units vitamin D and 3 units calcium.

Capsule of type R: 4 units magnesium, 6 units vitamin D and 10 units calcium.

Let x, y and z represent the number of capsule of types P, Q and R respectively that
the patient has to take in a week.

(a) Obtain a system of linear equation to represent the given information and


write the system in the form of matrix equation AX = B, where = ( ).


(b) Find the inverse of matrix A from part (a) by using the adjoint method. Hence,
find the values of x, y and z.

(c) The cost for each capsule of type P, Q and R are RM10, RM15 and RM17

respectively. How much will the expenses be for 4 weeks if the patient follws
the doctor’s advice?

SOLUTION

8a

2 + + 4 = 8

+ 2 + 6 = 11

+ 3 + 10 = 17
2 1 4 8
[1 2 6 ] [ ] = [11]
1 3 10 17

8b

21 4
= [1 2 6 ]

1 3 10
| | = (2) |23 160| − (1) |11 160| + (4) |11 23|
| | = (2)[20 − 18] − (1)[10 − 6] + (4)[3 − 2]

| | = (2)[2] − (1)[4] + (4)[1]

| | = 4

Page 13

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PSPM I QS 015/1 Session 2013/2014

+ |32 160| − |11 160| + |11 23|
= − |13 140| + |12 140| − |12 13|

[ + |12 64| − |12 46| + |12 12|]

+(20 − 18) −(10 − 6) +(3 − 2) CHOW CHOON WOOI
= [−(10 − 12) +(20 − 4) −(6 − 1)]

+(6 − 8) −(12 − 4) +(4 − 1)

2 −4 1
= [ 2 16 −5]

−2 −8 3

=

2 2 −2
= [−4 16 −8]

1 −5 3

−1 = 1
| |

−1 = 1 2 2 −2
4 [−4 16 −8]
−5
1 3

=

−1 = −1

= −1

1 2 2 −2 8
= [−4 16 −8] [11]
4 1 −5 3 17

1
= [2]

1

∴ = 1, = 2, = 1

8c Page 14
= 10, = 15, = 17
4 = 4[10(1) + 15(2) + 17(1)]
4 = 4[57]
4 = 228

63

PSPM I QS 015/1 Session 2013/2014

9. a. In an arithmetic progression, the sum of the first four terms is 46 and the
seventh term exceeds twice of the second term by 5. Obtain the first term and
the common difference for the progression. Hence, calculate the sum of the
first ten even terms of the progression.

b. A ball is dropped from a height of 2m. Each time the ball hits the floor, it CHOW CHOON WOOI
bounces vertically to a height that is 3 of its previous height.

4

i. Find the height of the ball at the tenth bounce.
ii. Find the total distance that the ball will travel before the eleventh

bounce.

SOLUTION



Arithmetic progression = [2 + ( − 1) ], = + ( − 1)

2

4 = 46

4
2 [2 + (4 − 1) ] = 46

2[2 + 3 ] = 46

4 + 6 = 46

2 + 3 = 23 .................................. (1)

7 = 2 2 + 5
+ (7 − 1) = 2[ + (2 − 1) ] + 5
+ 6 = 2[ + ] + 5
+ 6 = 2 + 2 + 5
− 4 = −5
2 − 8 = −10 .........................(2)
(1) − (2)
11 = 33
= 3
= 7

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PSPM I QS 015/1 Session 2013/2014

( )

1st bounce ➔ 1 = 2 (3)

4

2nd bounce ➔ 2 = 2 (3) (3) = 2 (3)2

44 4

3rd bounce ➔ 3 = 2 (3)3 CHOW CHOON WOOI

4

3 3
= 2 (4) , = (4)

= −1

3 3 10−1
10 = [2 (4)] [(4) ]

3 39
10 = [2 (4)] [(4) ]

( ) 3 10
10 = 2 (4)
32
3 2 [2 (4) ]
2 [2 (4)]

2m ……………… 3 10
2 [2 (4) ]

3 32 3 10
= 2 + 2 [2 (4)] + 2 [2 (4) ] + … + 2 [2 (4) ]

3 32 3 10
= 2 [1 + 2 (4) + 2 (4) + ⋯ + 2 (4) ]

= 2 {1 + 2 [(3) + (3)2 + ⋯ + (3)10]}
44 4

=2 1+2 (34) [1 − (43)10]
{[ = ( ) , = ( )

1 − (34) ]} ( − )
= −

= 13.324

Page 16

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PSPM I QS 015/1 Session 2013/2014

10 a. Find the solution set of |2 − 3 | < | + 3|.

b. If x + 1 < 0, show that

i. 2x – 1 < 0
ii. 2 −1 > 2

+1

SOLUTION CHOW CHOON WOOI



|2 − 3 | < | + 3|

(2 − 3 )2 < ( + 3)2

4 − 12 + 9 2 < 2 + 6 + 9

9 2 − 2 − 12 − 6 + 4 − 9 < 0

8 2 − 18 − 5 < 0

(2 − 5)(4 + 1) < 0

51
= 2 , − 4

15
−4 2

15
∴ { : − 4 < < 2}
( )
2 − 1 = 2 + 2 − 3 = 2( + 1) − 3
∵ + 1 < 0 ➔ 2( + 1) − 3 < 0
∴ 2 − 1 < 0

Page 17

66

PSPM I QS 015/1 Session 2013/2014

10b(ii)

2 − 1
+ 1 > 2

2 − 1
+ 1 − 2 > 0

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

2 − 1 − 2 − 2
= + 1

−3
= + 1

∵ + 1 < 0

−3
+ 1 > 0
2 − 1
+ 1 − 2 > 0

2 − 1
∴ + 1 > 2

Page 18

67

PSPM I QS 015/2 Session
2013/2014

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

2. State the values of R and α such that 3 + 6 = ( + ) where R > 0 and 0° ≤
< 90°. Hence, solve 3 + 6 = √5 for 0° ≤ < 180°.

3. (a) Find the value of if lim mx + 3x2 = 3. CHOW CHOON WOOI
4x −8x2
x→0

(b) Evaluate lim 3−x − 3

x→0 x .

4. (a) Find if = { [ ( + 1)]}.


(b) Obtain the second derivative of = cos 3 and express your answer in the simplest
2

form.

5. A cubic polynomial ( ) has remainders 3 and 1 when divided by ( − 1) and ( − 2) ,
respectively.

(a) Let ( ) be a linear factor such that ( ) = ( − 1)( − 2) ( ) + + , where
are constants. Find the remainder when ( ) is divided by ( − 1)( − 2).

(b) Use the values of from part (a) to determine ( ) if the coefficient of 3 for
( ) is 1 and (3) = 7. Hence, solve for x if ( ) = 7 – 3 .

6. (a) State the definition of the continuity of a function at a point. Hence, find the value of
d such that

( ) = {3 3 + + 5 ,, ≤ 0
> 0

( )Is continuous at = 0.

(b) A function f is defined by

( ) = { ( 2 − 1, ≤ 1
− 1), > 1

Determine the value(s) of k if is:

(i) Continuous for all ∈ ℝ.

Page 1

68

PSPM I QS 015/2 Session
2013/2014

(ii) Differentiable for all ∈ ℝ.

7. (a) Find the derivative of ( ) = 1 by using the first principle.

+1

(b) Use implicit differentiation to find:

(i) if = − . CHOW CHOON WOOI



(ii) the value of if 1 − 1 = 3 when = 1.
x 2

8. A curve is defined by parametric equations

= ln(1 + ), = 2 for > −1.

(a) Find and 2 in terms of t.
2

(b) Show that the curve has only one relative extremum at (0,1) and determine the nature

of the point.

9. (a) A cylindrical container of volume 128 3 is to be constructed with the same
material for the top, bottom and lateral side. Find the dimensions of the container
that will minimise the amount of the material needed.

(b) Gravel is poured onto a flat ground at the rate of 3 3 per minute to form a conical-

20

shaped pile with vertex angle 60° as shown in the diagram below.

60°
h

r

Compute the rate of change of the height of the conical pile at the instant t = 10
minutes.

Page 2

69

PSPM I QS 015/2 Session
2013/2014

10. (a) Show that sin +sin = cot [ −2 ].
cos −cos

(b) Use trigonometric identities to verify that

(i) sin = 2 tan
2

1+tan2
2

(ii) cos = 1−tan2 CHOW CHOON WOOI
2

1+tan2
2

Hence, solve the equation 3 sin + cos = 2 for 0° ≤ ≤ 180°. Give your answers
correct to three decimal places.

END OF QUESTION PAPER

Page 3

70

PSPM I QS 015/2 Session
2013/2014
1. Express 2 in partial fractions form.
2 +3 +2 Page 4

SOLUTION

1
2 + 3 + 2 x2 + 0x + 0

2 3 + 2 −3 − 2 CHOW CHOON WOOI
2 + 3 + 2 = 1 − 2 + 3 + 2

3 + 2
= 1 − ( + 1)( + 2)

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

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

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

When = −1

3(−1) + 2 = (−1 + 2) + (−1 + 1)

−1 = (1)

= −1

When = −2

3(−2) + 2 = (−2 + 2) + (−2 + 1)

−4 = (−1)

= 4

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

2 3 + 2
2 + 3 + 2 = 1 − ( + 1)( + 2)

−1 4
= 1 − [( + 1) + ( + 2)]

14
= 1 + ( + 1) − ( + 2)

71

PSPM I QS 015/2 Session
2013/2014

2. State the values of R and α such that 3 + 6 = ( + ) where R > 0 and 0° ≤ <
90°. Hence, solve 3 + 6 = √5 for 0° ≤ < 180°.

SOLUTION = √ 2 + 2 CHOW CHOON WOOI
3 + 6 = ( + )

= √32 + 62 = √45 = 3√5 = tan−1
( )
6
= tan−1 3 = tan−1 2 = 63.4°

3 + 6 = √5

3√5 ( + 63.4°) = √5

( + 63.4°) = √5

3√5

( + 63.4°) = 1 + 63.4° + 63.4°
3

+ 63.4° = sin−1 (1) = 19.5°, 160.5°

3

= 19.5° − 63.4°, 160.5° − 63.4°

= −43.9°, 97.1°

Given that 0° ≤ < 180°.

∴ = 97.1°

Page 5

72

PSPM I QS 015/2 Session
2013/2014
3. (a) Find the value of if lim mx + 3x2 = 3.
x→0 4x − 8x2 Page 6

(b) Evaluate lim 3 − x − 3 .
x→0 x
SOLUTION
3(a) CHOW CHOON WOOI

lim mx + 3x2 = 3
4x −8x2
x→0

lim x(m + 3x) = 3
x→0 x(4 − 8x)

lim (m + 3x) = 3
x→0 (4 − 8x)

m + 3(0) = 3
4 − 8(0)

m =3
4
= 12
3(b)

lim 3 − x − 3 = lim 3 − x − 3 X 3 − x + 3
x→0 x x→0 x 3−x + 3

(3 − x) − 3
= lim

x→0 x( 3 − x + 3)

−x
= lim

x→0 x( 3 − x + 3)

−1
= lim

x→0 3 − x + 3

= −1
3−0 + 3

= −1
23

73

PSPM I QS 015/2 Session
2013/2014

4. (a) Find if = { [ ( + 1)]}.


(b) Obtain the second derivative of = cos 3 and express your answer in the simplest
2

form.

SOLUTION

4(a) CHOW CHOON WOOI

= { [ ( + 1)]}


= − { [ ( + 1)]} { [ ( + 1)]} { [ ( + 1)]}

= − { [ ( + 1)]} { [ ( + 1)]} cos[ ( + 1)] [ ( + 1)]
1
= − { [ ( + 1)]} { [ ( + 1)]} cos[ ( + 1)] ( + 1) ( + 1)

= − { [ ( + 1)]} { [ ( + 1)]} cos[ ( + 1)] ( 1 1) (1)
+

− { [ ( + 1)]} { [ ( + 1)]} cos[ ( + 1)]
=
( + 1)

4(b)

= cos 3
2

= cos 3 = 2 ′ − ′
′ = 2 2 = 2
′ = −3sin 3

( 2 )(−3sin 3 ) − (cos 3 )(2 2 )
=
( 2 )2

( 2 )[−3sin 3 − 2cos 3 ]
=
( 2 )2

−3sin 3 − 2cos 3
=
2

= −3sin 3 − 2cos 3 = 2 ′ − ′
′ = 2 2 2

′ = −9cos 3 + 6 3

2 ( 2 )(−9cos 3 + 6 3 ) − (−3sin 3 − 2cos 3 )(2 2 )
2 =
( 2 )2

Page 7

74

PSPM I QS 015/2 Session
2013/2014

2 ( 2 )[(−9cos 3 + 6 3 ) − (2)(−3sin 3 − 2cos 3 )]
2 =
( 2 )2

2 ( 2 )[−9cos 3 + 6 3 + 6sin 3 + 4cos 3 ]
2 =
( 2 )2

2 −9cos 3 + 6 3 + 6sin 3 + 4cos 3
2 = CHOW CHOON WOOI
2

2 12 3 − 5cos 3
2 =
2

Page 8

75

PSPM I QS 015/2 Session
2013/2014

5. A cubic polynomial ( ) has remainders 3 and 1 when divided by ( − 1) and ( − 2) , CHOW CHOON WOOI
respectively.

(a) Let ( ) be a linear factor such that ( ) = ( − 1)( − 2) ( ) + + , where
are constants. Find the remainder when ( ) is divided by ( − 1)( − 2).

(b) Use the values of from part (a) to determine ( ) if the coefficient of 3 for
( ) is 1 and (3) = 7. Hence, solve for x if ( ) = 7 – 3 .

SOLUTION

5(a)

(1) = 3

(2) = 1

( ) = ( − 1)( − 2) ( ) + +

(1) = (1 − 1)(1 − 2) (1) + (1) + = 3

+ = 3 ............................... (1)

(2) = (2 − 1)(2 − 2) (2) + (2) + = 1

2 + = 1............................... (2)

(1) − (2)

− 2 = 3 − 1 ( ) = ( ) ( ) + ( )

− = 2 ( ) = ( − 1)( − 2) ( ) − 2 + 5

= −2 ( ) ( )

= 5

( ) = ( − 1)( − 2) ( ) − 2 + 5

∴ ℎ ℎ ( ) ( − 1)( − 2) − 2 + 5

Page 9

76

PSPM I QS 015/2 Session
2013/2014
5(b)
( ) = ( − 1)( − 2) ( ) − 2 + 5 CHOW CHOON WOOI
Since the coefficient of 3 for ( ) is 1 ➔ ( ) = ( + )
( ) = ( − 1)( − 2)( + ) − 2 + 5
ℎ (3) = 7
(3) = (3 − 1)(3 − 2)(3 + ) − 2(3) + 5 = 7
6 + 2 = 8
= 1
( ) = ( + 1)
( ) = ( − 1)( − 2)( + 1) − 2 + 5
( ) = 7 – 3
( − 1)( − 2)( + 1) − 2 + 5 = 7 − 3
( − 1)( − 2)( + 1) − 2 + 3 + 5 − 7 = 0
( − 1)( − 2)( + 1) + − 2 = 0
( − 1)( − 2)( + 1) + ( − 2) = 0
( − 2)[( − 1)( + 1) + 1] = 0
( − 2)[ 2 + − − 1 + 1] = 0
( − 2)[ 2] = 0
= 2 = 0

Page 10

77

PSPM I QS 015/2 Session
2013/2014

6. (a) State the definition of the continuity of a function at a point. Hence, find the value of
d such that

( ) = {3 3 + + 5 ,, ≤ 0
> 0

Is continuous at = 0.

(b) A function is defined by CHOW CHOON WOOI

( ) = { ( 2 − 1, ≤ 1
− 1), > 1

Determine the value(s) of k if is:

(iii) Continuous for all ∈ ℝ.
(iv) Differentiable for all ∈ ℝ.

SOLUTION

6(a)

Definition of continuity:

A function is continuous at a point = in the domain of f if the following three
conditions are satisfied:

i. f(c) is defined

ii. lim f (x) exists or finite
x→c

iii. lim f (x) = f (c)
x→c

For ( ) to be continuous at x = 0,

(i) (0) = 3(0)+ =

(ii) lim f (x) = lim 3x + 5 = 5
x→0+ x→0+

(iii) = 5

= 5

6(bi)

( ) = { ( 2 − 1, ≤ 1
− 1), > 1

lim f (x) = lim f (x)
x →1− x →1+

lim x2 -1 = lim k(x −1)
x →1− x →1+

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PSPM I QS 015/2 Session
2013/2014

12 − 1 = (1 − 1)

0=0

Thus, k can take any real value for the continuity: ∈ ℝ

6(bii)

( ) = { ( 2 − 1, ≤ 1 f '(a) = lim f (x) − f (a) CHOW CHOON WOOI
− 1), > 1 x→1− x − a

f '(1− ) = lim f (x) − f (1)
x→1− x −1

f '(1− ) = lim (x2 −1) − (12 −1)
x→1− x −1

f '(1− ) = lim x2 −1
x→1− x −1

f '(1− ) = lim (x +1)(x −1)
x→1− x −1

f '(1− ) = lim (x +1)
x→1−

f '(1− ) = 1+1

f '(1− ) = 2

f '(1+ ) = lim f (x) − f (1)
x→1+ x −1

f '(1+ ) = lim (k(x -1)) − (12 −1)
x→1+ x − 1

f '(1+ ) = lim k(x −1)
x→1+ x −1

f '(1+ ) = lim k
x →1+

f '(1+ ) = k

For to be differentiable at = 1,
′(1+) = ′(1−)
= 2

Page 12

79

PSPM I QS 015/2 Session
2013/2014
7. (a)
(b) Find the derivative of ( ) = 1 by using the first principle.
+1

Use implicit differentiation to find:

(i) if = − .


(ii) the value of if 1 − 1 = 3 when = 1. CHOW CHOON WOOI
x
2

SOLUTION

7(a)

( ) = 1 First principle

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

1 1 1
= lim (( − 1)
ℎ + ℎ) + 1 +
ℎ→0

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

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

1 (( −ℎ
= lim 1))
ℎ + ℎ + 1)( +
ℎ→0

−1
= lim (( 1))
+ ℎ + 1)( +
ℎ→0

−1
= ( + 0 + 1)( + 1)

−1
= ( + 1)2

7(bi)

= −

1 + = − ( − )
( ) ( )

+ = − [1 −
( ) ]

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PSPM I QS 015/2 Session
2013/2014
+ = − − −
( ) Page 14

+ − = − −
( )

[ + − ] = − −
( )
CHOW CHOON WOOI
= − −
( ) +

−

7(bii)

11
− x = 3

1
ℎ = 2

11
− (12) = 3
1
− 2 = 3

1
= 3 + 2

1
= 5

1
= 5
11
− x = 3

−1 − −1 = 3

(−1) −2 − (−1) −2 = 0


−1 1
2 + 2 = 0

−1 1
2 = − 2

81

PSPM I QS 015/2 Session
2013/2014

1 2
= − 2 . −1

2
= 2

1 1 CHOW CHOON WOOI
ℎ = 2 , = 5

= (15)2
(21)2

1

= 25
1

4

1 4
= 25 . 1

4
= 25

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82

PSPM I QS 015/2 Session
2013/2014

8. A curve is defined by parametric equations

= ln(1 + ), = 2 for > −1.

(c) Find and 2 in terms of t.
2

(d) Show that the curve has only one relative extremum at (0,1) and determine the nature

of the point. CHOW CHOON WOOI

SOLUTION

8(a)

= ln(1 + ), = 2

= 1 (1 + ) = 2 ( 2)

1+

= 1 (1) = 2(2 )

1+

= 1 = 2 2

1+

= 1 +




= .

= (2 2). (1 + )


= 2 2 (1 + )


2
2 = [ ] . [ ]

2
2 = [ ] . [ ]

2 = [2 2 (1 + )]} . [1 + ]
2 {

= 2 2 = 1 +
′ = (2 )(2 2) + ( 2)(2) ′ = 1
′ = 4 2 2 + 2 2

2 = {(2 2 )(1) + (1 + )(4 2 2 + 2 2 )}. [1 + ]
2

Page 16

83

PSPM I QS 015/2 Session
2013/2014

2 = {2 2 + 4 2 2 + 2 2 + 4 3 2 + 2 2 }. [1 + ]
2

2 = {2 2 + 4 2 + 4 2 2 + 4 3 2 }. [1 + ]
2

2 = {2 2 (1 + 2 + 2 2 + 2 3)}. [1 + ] CHOW CHOON WOOI
2

8(b)

For extremum point(s), let = 0


= 2 2 (1 + )


2 2(1 + ) = 0

= 0 or = −1

Since given that > −1, the curve has only one extremum point when = 0.

Whent = 0,

= ln(1 + ), = 2
= ln(1 + 0), = 02

= ln(1), = 0

= 0, = 1

 (0, 1) is the extremum point.

At = 0, the second derivative test gives

2 = {2 02 (1 + 2(0) + 2(02) + 2(03))}. [1 + (0)]
2

2
2 = {2(1)(1 + 0 + 0 + 0)}. [1 + (0)]

2
2 = 2 > 0

Thus, the extremum point (0,1) is a relative minimum point.

Page 17

84

PSPM I QS 015/2 Session
2013/2014
9. (a)
A cylindrical container of volume 128 3 is to be constructed with the same
(b) material for the top, bottom and lateral side. Find the dimensions of the container
that will minimise the amount of the material needed.

Gravel is poured onto a flat ground at the rate of 3 3 per minute to form a conical-
20

shaped pile with vertex angle 60° as shown in the diagram below.

60° CHOW CHOON WOOI
h

r

Compute the rate of change of the height of the conical pile at the instant t = 10
minutes.

SOLUTION

= 2ℎ = 128 ➔ To minimise the surface area.

128
ℎ = 2

128
ℎ = 2

To minimise the surface area

= 2 2 + 2 ℎ

= 2 2 + 2 128
( 2 )

= 2 2 + 256


= 2 2 + 256 −1

= 4 − 256 −2


256
= 4 − 2

Page 18

85

PSPM I QS 015/2 Session
2013/2014

CHOW CHOON WOOI
= 0

256
4 − 2 = 0
256

2 = 4
4 3 = 256

3 = 256
4

3 = 256
4

3 = 64

= 4

When = 4,

128
ℎ = 2

128
ℎ = 42
ℎ=8

256
= 4 − 2
2 512
2 = 4 + 3
ℎ = 4

2 512
2 = 4 + 43 = 12 > 0
ℎ = 8 = 4 ℎ

Page 19

86

PSPM I QS 015/2 Session
2013/2014
9(b)
Page 20
60°
h

r CHOW CHOON WOOI

= 3 3 Find ℎ when = 10

20


tan 30° = ℎ
1
√3 = ℎ

√3 = ℎ

ℎ
=

√3

ℎ ℎ
= .

= 1 2ℎ
3

= 1 ℎ2 ℎ
()
3 √3

1 ℎ2
= 3 ( 3 ) ℎ

= ℎ3
9

= ℎ2
ℎ 3

ℎ2
ℎ = 3

ℎ 3
= ℎ2

ℎ ℎ
= .

87

PSPM I QS 015/2 Session
2013/2014

ℎ 3 3
= ℎ2 . 20

ℎ 9
= 20 ℎ2

ℎ = 10,

= 3 ➔ = 3 CHOW CHOON WOOI
20 20

= 3 (10) = 3
20 2

3
ℎ = 2

3 = ℎ3
2 9

ℎ3 = 9 . 3
2

ℎ3 = 27
2

1

27 3
ℎ = (2 )

ℎ 9
=
2

20 (22 7 )3

2

ℎ 9 2 3
= 20 (27) = 0.0542

Page 21

88

PSPM I QS 015/2 Session
2013/2014

10. (a) Show that sin +sin = cot [ −2 ].
cos −cos

(b) Use trigonometric identities to verify that

(i) sin = 2 tan
2

1+tan2
2

(iii) cos = 1−tan2 CHOW CHOON WOOI
2

1+tan2
2

Hence, solve the equation 3 sin + cos = 2 for 0° ≤ ≤ 180°. Give your answers
correct to three decimal places.

SOLUTION

10(a)

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

cos ( − )
2
=
( − )
− sin 2

−
= − ( 2 )

−
= ( 2 )

( )


sin = 2 2 2

cos
2 2 [
= 2 cos 2 ]


2

2 cos2
2 2
=

cos 2

= 2 cos2
2 2

Page 22

89

PSPM I QS 015/2 Session
2013/2014
1
= 2 2 1 Page 23

cos2 2

= 2 1

2 sec2
2
CHOW CHOON WOOI
= 2
sec2 2

2

= 2
2

1 + tan2 2

( )

cos = 1 − 2 sin2
2

= (1 − 2 sin2 . cos2
2) [ 2 ]
cos2 2

(1 − 2 sin2 2 ) cos2
2
=

cos2 2

= 1 − 2cosisn22 2 2 ) . cos2
( 2 2

cos2

= (sec2 − 2 tan2 . 1
2 2) 1

cos2
2

= (sec2 − 2 tan2 . 1
2 2) sec2 2

= sec2 − 2 tan2
2 2

sec2 2

(1 + tan2 2 ) − 2 tan2
1 + tan2 2
=
2

90

PSPM I QS 015/2 Session
2013/2014

= 1 − tan2
1 + tan2 2

2

( ) CHOW CHOON WOOI

3 sin + cos = 2

3 ( 2 ) + 1 − tan2 = 2
1 2 2 1 + tan2 2

+ tan2 2

Let =
2

2 1 − 2
3 (1 + 2) + 1 + 2 = 2

6 + 1 − 2
1 + 2 = 2

6 + 1 − 2 = 2(1 + 2)

6 + 1 − 2 = 2 + 2 2

6 + 1 − 2 = 2 + 2 2

2 + 2 2 + 2 − 6 − 1 = 0

3 2 − 6 + 1 = 0

− ± √ 2 − 4
= 2

−(−6) ± √(−6)2 − 4(3)(1)
= 2(3)

6 ± √36 − 12 or = 1.8165
= 6 = 1.8165
= 0.1835
2
= 0.1835
= tan−1 1.8165
2
2
= tan−1 0.1835
= 61.167°
2
2
= 10.398°
= 122.334°
2

= 20.796°

Page 24

91

2014/2015

92

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2014/2015

93

PSPM I QS 015/1 Session
2014/2015

1. Solve the equation 3 + 3(3−x) = 12.

2. Solve the inequality 1 < x−11.
6−

100 100

3. Given matrices = [4 1 0] and = [ 1 0] where is the inverse of . Find , and CHOW CHOON WOOI

1 1

in terms of and .

4. Using algebraic method, find the least value of n for which the sum of the first n terms of a
geometric series

0.88 + (0.88)2 + (0.88)3 + (0.88)4 + ⋯

is greater than half of its sum to infinity.

3

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.

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,

2

sketch the graph of ( ) and −1(x).

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 − ) = (z̅ + 1)(1 + i).

Page 2

94

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 CHOW CHOON WOOI

+ 2
2 − ( − 4) < 5

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 (x). 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.

(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.

10. (a) Determine whether ( ) = 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 ( ).
(ii) Find −1(x), if it exists. Hence, state the domain and range of −1(x).

END OF QUESTION PAPER

Page 3

95

PSPM I QS 015/1 Session
2014/2015

1. Solve the equation 3 + 3(3−x) = 12.

SOLUTION

3 + 3(3−x) = 12

3 + 33. 3−x = 12

3 + 33. 1 = 12 CHOW CHOON WOOI
3x
27
3 + 3x = 12

Let = 3 ( − 3) = 0
27
= 3
+ y = 12 3 = 3
2 + 27 = 1

y = 12
2 + 27 = 12y
2 − 12y + 27 = 0
(y − 9)( − 3) = 0
(y − 9) = 0

= 9
3 = 9

= 2

Page 4

96

PSPM I QS 015/1 Session
2014/2015

2. Solve the inequality 1 < x−11.
6−

SOLUTION CHOW CHOON WOOI
11

6 − < x − 1

11
6 − − x − 1 < 0

(x − 1) − (6 − )
(6 − )(x − 1) < 0

− 1 − 6 +
(6 − )(x − 1) < 0

2 − 7
(6 − )(x − 1) < 0

2 − 7 = 0 6 − = 0 − 1 = 0
= 7 = 6 = 1

2 (−∞, 1) 7 7 (6, ∞)
(1, 2) (2 , 6)
2 − 7 - +
6 − + - + -
x−1 - + + +
2 − 7 + +
(6 − )(x − 1) + -
- +

7
∴ (1, 2) ∪ (6, ∞)

Page 5

97

PSPM I QS 015/1 Session
2014/2015

1 00 100

3. Given matrices = [4 1 0] and = [ 1 0] where is the inverse of . Find , and

1 1

in terms of and .

SOLUTION 100 CHOW CHOON WOOI
= [ 1 0]
100
= [4 1 0] 1

1

is the inverse of ➔ =

=

100100 100
[4 1 0] [ 1 0] = [0 1 0]
1 1 0 0 1

1+0+0 0+0+0 0+0+0 1 0 0
[ 4 + + 0 0 + 1 + 0 0 + 0 + 0] = [0 1 0]

+ + 0 + + 0 + 0 + 1 0 0 1

1 0 0 100

[ 4 + 1 0] = [0 1 0]

+ + + 1 0 0 1

4 + = 0 ➔ = −4

+ + = 0 ➔ = − + 4

+ = 0 ➔ = −

Page 6

98

PSPM I QS 015/1 Session
2014/2015

4. Using algebraic method, find the least value of n for which the sum of the first n terms of a
geometric series

0.88 + (0.88)2 + (0.88)3 + (0.88)4 + ⋯

is greater than half of its sum to infinity.

SOLUTION CHOW CHOON WOOI

1
> 2 ∞ (1 − )

= 0.88, = 0.88 = 1 −
∞=(1− )
(1 − ) 1
1 − > 2 (1 − )

(0.88)[1 − (0.88) ] 1 0.88
> 2 (1 − 0.88)
1 − 0.88

[1 − (0.88) ] > 1 (1 0.88 1 − 0.88
2 − 0.88) ( 0.88 )

1 − (0.88) > 1
2

1 − 1 > (0.88)
2

1 > (0.88)
2

(0.88) < 1
2

(0.88) < 1
(2)

log 0.88 < −0.3010

(−0.0555) < −0.3010

−0.0555 < −0.3010

0.3010 < 0.0555

0.0555 > 0.3010

Page 7

99

PSPM I QS 015/1 Session
2014/2015
0.3010
> 0.0555
> 5.422
= 6

CHOW CHOON WOOI

Page 8

100