PSPM 1 Q&A

PSPM I QS 015/1
Session 2015/2016

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

SOLUTION

(a)

= 2 + bx + c

(2,11)➔ = 2, = 11

11 = 22 + b(2) + c

4 + 2b + c = 11 ………………. (1)

(−1, −16)➔ = −1, = −16

−16 = (−1)2 + b(−1) + c

− b + c = −16 ………………. (2)

(3,28)➔ = 3, = 28

28 = (3)2 + b(3) + c

9 + 3b + c = 28 ………………. (2)

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PSPM I QS 015/1
Session 2015/2016
(b)
………………. (1)
4 + 2b + c = 11 ………………. (2) CHOW CHOON WOOI
− b + c = −16 ………………. (2)
9 + 3b + c = 28

4 2 1 11

(1 −1 1) ( ) = (−16)

9 3 1 28

(c) 1
| |
−1 = ( )

421 ( ) = ( )
= (1 −1 1)

931

| | = (4) |−31 11| − (2) |91 11| + (1) |91 −31|

= 4[−1 − 3] − 2[1 − 9] + 1[3 + 9]

= −16 + 16 + 12

= 12



+ |−31 11| − |19 11| + |19 −31|
= − |32 11| + |94 11| − |49 23|

[+ |−21 11| − |41 11| + |41 −21|]

−1 − 3 −(1 − 9) 3 + 9
= [−(2 − 3) 4 − 9 −(12 − 18)]

2 + 1 −(4 − 1) −4 − 2

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PSPM I QS 015/1
Session 2015/2016
−4 8 12
= [ 1 −5 6 ] Page 16

3 −3 −6

CHOW CHOON WOOI

( ) =

−4 8 12
= [ 1 −5 6 ]

3 −3 −6

−4 1 3
= [ 8 −5 −3]

12 6 −6



− = ( )
| |

−1 = 1 −4 1 3
12 [8 −5 −3]
6
12 −6

4 1 3
− 12
=8 12 12
12 5 3
12
− 12 − 12
[ 12 6 6

12 − 12]

1 1 1
−3 12 4
=2 1
3 5 −4
[1 − 12 1
− 2]
1
2

153

PSPM I QS 015/1
Session 2015/2016

4 2 1 11

(1 −1 1) ( ) = (−16)

9 3 1 28

= −1

1 1 1 CHOW CHOON WOOI
−3 12
2 4 11
[ ] = 5 1 [−16]
3 − 12 −4
1 28
1
[1 2 − 2]

2
=[ 7 ]

−11

∴ = 2, = 7, = −11

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154

PSPM I QS 015/1
Session 2015/2016

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.

SOLUTION

( ) = √3 − 2

(a)

( 1) = √3 − 2 1

( 2) = √3 − 2 2
( 1) = ( 2)

√3 − 2 1 = √3 − 2 2
3 − 2 1 = 3 − 2 2
1 = 2
∴

(b)
( ) = √3 − 2

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155

PSPM I QS 015/1
Session 2015/2016
:
: 3 − 2 ≥ 0 Alternative CHOW CHOON WOOI

3 ≥ 2 ( ) = √3 − 2
2 ≤ 3 [ −1( )] =

3 √3 − 2 −1( ) =
≤ 2
3 − 2 −1( ) = 2
3
= (−∞, 2] 2 −1( ) = 3 − 2
= [ , ∞)
(c)

( ) = √3 − 2
= √3 − 2
2 = 3 − 2
2 = 3 − 2

3 − 2
=

2
−1( ) = 3 − 2

2

−1 ( ) = 3 − 2
2

Page 19

156

PSPM I QS 015/1
Session 2015/2016
(d)
y
( )
3 y=x
2

3 x CHOW CHOON WOOI
2

−1 ( )

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157

PSPM I QS 015/1
Session 2015/2016

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).

SOLUTION CHOW CHOON WOOI

+ 2 4
( ) = 3 − 4 , x ≠ 3
( )(x) = x

[ (x)] = x

+ 2
[3 − 4] = x

[ 3 + 24] + 2 = x
3 [ 3 − 24] − 4
+

−

+ 2 + 2
[3 − 4] + 2 = x [3 [3 − 4] − 4]

2 + 2 2(3 − 4) 3 2 + 6
[ 3 − 4 ] + [ 3 − 4 ] = [ 3 − 4 ] − 4

( 2 + 2 ) + 2(3 − 4) 3 2 + 6 4 (3 − 4)
[ 3 − 4 ] = [ 3 − 4 ] − [ 3 − 4 ]

2 + 2 + 6 − 8 3 2 + 6 − 12 2 + 16
[ 3 − 4 ] = [ ]
3 − 4

2 + 2 + 6 − 8 = 3 2 + 6 − 12 2 + 16

3 2 − 12 2 + 6 + 16 − 2 − 6 − 2 + 8 = 0

(3 − 12) 2 + (16− 2) + 8 − 2 = 0

2

3 − 12 = 0

= 4

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PSPM I QS 015/1 CHOW CHOON WOOI
Session 2015/2016
(b)
( ) = |3 + 2| Page 22
( ) = − + 2
(x) = [ ( )]
= [ |3 + 2|]
= − |3 +2| + 2
= |3 +2|−1 + 2
= |3 + 2|−1 + 2
1
= 3 + 2 + 2
1 + 2(3 + 2)
= 3 + 2
1 + 6 + 4
= 3 + 2
6 + 5
= 3 + 2

Let = 6 +5

3 +2

(3 + 2) = 6 + 5

3 + 2 = 6 + 5

3 − 6 = 5 − 2

(3 − 6) = 5 − 2

5 − 2
= 3 − 6

( )−1(y) = 5 − 2
3 − 6

( )−1(x) = 5 − 2
3 − 6

( )−1(3) = 5 − 2(3) = − 1
3(3) − 6 3

159

PSPM I QS 015/1
Session 2015/2016

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

+3

(b) Show that 2 x 42 = 22 .
8

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

SOLUTION CHOW CHOON WOOI

(a)

− 1
| | > 2
+ 3

− 1 Or − 1
+ 3 > 2 + 3 < −2

− 1 − 1
+ 3 − 2 > 0 + 3 + 2 < 0

− 1 2( + 3) − 1 2( + 3)
+ 3 − + 3 > 0 + 3 + + 3 < 0

− 1 − 2 − 6 >0 − 1 + 2( + 3) <0

+ 3 + 3

− − 7 − 1 + 2 + 6 <0
+ 3 > 0
+ 3

3 + 5
+ 3 < 0

Let − − 7=0 ➔ = −7 Let 3 + 5=0 ➔ = − 5
+ 3 = 0 ➔ = −3
3

+ 3 = 0 ➔ = −3

(−∞, −7) (−7, −3) (−3, ∞) (−∞, −3) 5 5
(−3, − ) (− , ∞)
33

− − 7 + - - 3 + 5 - -+

+ 3 - -+ + 3 - ++

- +- +-+

or

(−7, −3) 5
(−3, − 3)

-7 -3 5
−3

: (−7, −3) ∪ (−3, − 5)

3

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PSPM I QS 015/1
Session 2015/2016
(b)
Page 24
Show that 2 42 = 22
8

2 x 42 2 (22)2 CHOW CHOON WOOI
8 = (23)

2 24
= 23

25
= 23

= 22

2 x 42 − 13(2 ) + 36 ≥ 0
8

22 − 13(2 ) + 36 ≥ 0

(2 )2 − 13(2 ) + 36 ≥ 0

= 2

(2 )2 − 13(2 ) + 36 ≥ 0

2 − 13 + 36 ≥ 0

( − 9)( − 4) ≥ 0

Let ( − 9) = 0 ( − 4) = 0

u=9 u=4

161

PSPM I QS 015/1
Session 2015/2016
( − 9)
( − 4) (−∞, 4) (4, 9) (9, ∞)
( − 9)( − 4) −
− −+
+
++

−+ CHOW CHOON WOOI

≤ 4 or ≥ 9

2 ≤ 4 or 2 ≥ 9

ln 2 ≤ ln 4 or ln 2 ≥ ln 9

ln 2 ≤ ln 4 or ln 2 ≥ ln 9

≤ ln 4 or ≥ ln 9

ln 2 ln 2

≤ 2 or ≥ 3.170

Solution interval : (−∞, 2] ∪ [3.17, ∞)

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162

QS 015/2 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2015/2016

163

PSPM I QS 015/2 Session 2015/2016

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

2. Evaluate the following(if exist):

a) lim 2+4 −12 CHOW CHOON WOOI

→2 | −2|

b) lim 1−√

→1 1−

3. Find the derivative of the following functions:
a) ( ) = √4 2 + 1
b) ( ) = 2 ln(3 + 4)

4. Given 2 − cot = 3, show that 2 − cot − 2 = 0. Hence, solve the equation
2 − = 3 for 0 ≤ ≤ .

5. A polynomial ( ) = 2 4 + 3 + 2 − 17 + where , are constants, has
factors ( + 2) and ( − 1). When ( ) is divided by ( + 1), the remainder is 8. Find the
values of , . Hence, factorize ( ) completely and state its zeroes.

6. (a) Evaluate lim √2 2+3 .

→∞ 5 +1

5 − , −2 < ≤ −1
(b) Given ( ) = { 2 + + , −1 < ≤ 2

2−4 , > 2

−2

(i) Find the values of and if ( ) is continuous for all real values of .

(ii) Sketch the graph of ( ) using the values p and q obtained in part (i).

Page 2

164

PSPM I QS 015/2 Session 2015/2016

7. A curve is given by the parametric equations = − 1, = + 1.



a) Find and 2 in terns of .
2

b) Obtain the coordinates of the stationary points of the curve and determine the

nature of the points. CHOW CHOON WOOI

8. (a) If 2 − 2 √(1 + 2) + 2 = 0, show that = .
√(1+ 2
)

(b) Water is running at a steady rate of 36 3 −1 into a right inverted circular cone
with a semi-vertical angle of 45°.

(i) Find the rate of increasing in water depth when the water level is 3 cm.

(ii) Find the time taken when the depth of the water is 18cm.

9. (a) Determine the values of and , where > 0 and 0° < < 90° so that 3 sin −
4 cos = ( − ).

(b) Hence, solve the equation 3 sin − 4 cos = 2 for 0° ≤ ≤ 360°.

(c) From the answer obtained in part (b), find the value of for 0° ≤ ≤ 360° so that
( ) = 1 is minimum. Hence, find the minimum value of .

3 −4 +15

10. (a) Find the value of if the slope of the curve 3 + 2 − 2 2 = 0 at the point (-1,1)
is -3.

(b) Given = sin .
1+cos

(i) Find and 2 in terms of .
2

(ii) Hence, show that 3 − 2 − ( 2 = 0.
3 2
)

Page 3

165

PSPM I QS 015/2 Session 2015/2016

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

SOLUTION

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

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


= ( − 2) + ( + 2) + ( + 2)2

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

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

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

ℎ = 2:

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

32 = 16

= 2

ℎ = −2:

5(−2)2 + 4(−2) + 4 = [(−2) − 2]

16 = −4

= −4

ℎ = 0, = 2, = −4:

5(0)2 + 4(0) + 4 = 2[(0) + 2]2 + (0 + 2)(0 − 2) + (−4)(0 − 2)

4 = 2(4) − 4 + 8

Page 4

166

PSPM I QS 015/2 Session 2015/2016

4 = 12
= 3

5 2 + 4 + 4 2 3 4 CHOW CHOON WOOI
( 2 − 4)( + 2) = ( − 2) + ( + 2) − ( + 2)2

Page 5

167

PSPM I QS 015/2 Session 2015/2016

2. Evaluate the following(if exist):

a) lim 2+4 −12

→2 | −2|

b) lim 1−√

→1 1−

SOLUTION CHOW CHOON WOOI

(a)

lim 2 + 4 − 12 = lim ( + 6)( − 2)

→2 | − 2| →2 | − 2|

| − 2| = {−(( −−22)), , ≥ 2
< −2

( + 6)( − 2) , ≥ 2
, < −2
( + 6)( − 2) = − 2
| − 2| + 6)( −
( −( − 2) 2)
{

= {−( ++66, ), ≥ 2
< −2

lim ( + 6)( − 2) = lim −( + 6)

→2− | − 2| →2−

= −(2 + 6)

= −8

lim ( + 6)( − 2) = lim ( + 6)

→2+ | − 2| →2+

= (2 + 6)

=8

( + 6)( − 2)
l→im2− | − 2|

≠ lim ( + 6)( − 2) , ℎ lim 2 + 4 − 12 .

→2+ | − 2| →2 | − 2|

Page 6

168

PSPM I QS 015/2 Session 2015/2016
2 + 2 = ( + )( − )
(b)

lim 1 − √ = lim 1 − √

→1 1 − →1 12 − (√ )2

1 − √ CHOW CHOON WOOI
= lim

→1 (1 + √ )(1 − √ )
1

= lim
→1 (1 + √ )
1

=
(1 + √1)
1

=2

Page 7

169

PSPM I QS 015/2 Session 2015/2016

3. Find the derivative of the following functions:
a) ( ) = √4 2 + 1
b) ( ) = 2 ln(3 + 4)

SOLUTION CHOW CHOON WOOI
(a)

( ) = √4 2 + 1

1

( ) = (4 2 + 1)2

′( ) = − 2 (4 2 1 (4 2 + 1

+ 1)2 1)2

= − 2 (4 2 1 1 (4 2 + 1)−21 (4 2 + 1)
(2)
+ 1)2

= − 2 (4 2 1 1 (4 2 + 1)−21(8 )
(2)
+ 1)2

= − 1 (8 )(4 2 + 1)−12 2 (4 2 + 1
(2)
1)2

=− (4 ) 1

1 2 (4 2 + 1)2
(4 2 + 1)2

4 2√(4 2 + 1)
=−

√(4 2 + 1)

Page 8

170

PSPM I QS 015/2 Session 2015/2016

(b)

( ) = 2 ln(3 + 4)

= 2 = ln(3 + 4)

′ = 2 (2 ) ′ = 1 (3 + 4) CHOW CHOON WOOI

3 +4

= 2 2 =3

3 +4

′( ) = ( 2 ) 3 4) + [ln(3 + 4)](2 2 )
(3 +

= 3 2 + 2 2 ln(3 + 4)
3 + 4

= 2 3 4 + 2 ln(3 + 4)]
[3 +

Page 9

171

PSPM I QS 015/2 Session 2015/2016

4. Given 2 − cot = 3, show that 2 − cot − 2 = 0. Hence, solve the equation
2 − = 3 for 0 ≤ ≤ .

SOLUTION CHOW CHOON WOOI
2 − cot = 3
(1 + 2 ) − cot = 3
2 − cot + 1 = 3
2 − cot + 1 − 3 = 0
2 − cot − 2 = 0

2 − = 3 , 0 ≤ ≤

2 − cot − 2 = 0

Let =

2 − − 2 = 0

( − 2)( + 1) = 0

= 2 = −1

= 2 or = −1

1 =2 or 1 = −1

tan tan

tan = 1 or tan = −1

2

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172

PSPM I QS 015/2 Session 2015/2016 CHOW CHOON WOOI
0.464
For tan = 1

2 4

0.464

= 0.464

For tan = −1

4
3

= − 4 = 4


∴ = . ,

Page 11

173

PSPM I QS 015/2 Session 2015/2016

5. A polynomial ( ) = 2 4 + 3 + 2 − 17 + where , are constants, has
factors ( + 2) and ( − 1). When ( ) is divided by ( + 1), the remainder is 8. Find the
values of , . Hence, factorize ( ) completely and state its zeroes.

SOLUTION CHOW CHOON WOOI

(−2) = 0

(1) = 0

(−1) = 8

For (− ) =

( ) = 2 4 + 3 + 2 − 17 +

(−2) = 2(−2)4 + (−2)3 + (−2)2 − 17(−2) +

0 = 32 − 8 + 4 + 34 +

8 − 4 − = 66 ………………………………….. (1)

For ( ) =

(1) = 2(1)4 + (1)3 + (1)2 − 17(1) +

0 = 2 + + − 17 +

+ + = 15 ………………………………….. (2)

For (− ) = Page 12
(−1) = 2(−1)4 + (−1)3 + (−1)2 − 17(−1) +
8 = 2 − + + 17 +

174

PSPM I QS 015/2 Session 2015/2016 CHOW CHOON WOOI
………………………………….. (3)
− − = 11
(2) + (3) ………………………………….. (4)

2 = 26
= 13

Substitute (4) into (1) ………………………………….. (5)
8(13) − 4 − = 66
4 + = 38

Substitute (4) into (2) ………………………………….. (6)
13 + + = 15
+ = 2

(5) – (6) ………………………………….. (7)
3 = 36
= 12

Substitute (7) into (6) Page 13
12 + = 2
= −10
∴ = 13, = 12, = −10

175

PSPM I QS 015/2 Session 2015/2016

( ) = 2 4 + 13 3 + 12 2 − 17 − 10 Factor Theorem
(−2) = 0 ➔ ( + 2) ( ) ( ) = 0 ℎ
(1) = 0 ➔ ( − 1) ( ) ( − ) ( )
( ) = ( + 2)( − 1) ( )
CHOW CHOON WOOI
= ( 2 + − 2) ( )

2 x2 +11x + 5
x2 + x − 2 2x4 +13x3 +12x2 −17x −10

2x4 + 2x3 − 4x2
11x3 +16x2 −17x −10
11x3 +11x2 − 22x
5x2 + 5x −10
5x2 + 5x −10
0

( ) = ( + 2)( − 1)(2 2 + 11 + 5)
= ( + 2)( − 1)(2 + 1)( + 5)

When ( ) = 0
( + 2)( − 1)(2 + 1)( + 5) = 0
1
= −2, = 1, = − 2 , = −5

Page 14

176

PSPM I QS 015/2 Session 2015/2016

6. (a) Evaluate lim √2 2+3 . CHOW CHOON WOOI

→−∞ 5 +1

5 − , −2 < ≤ −1
(b) Given ( ) = { 2 + + , −1 < ≤ 2

2−4 , > 2

−2

(i) Find the values of and if ( ) is continuous for all real values of .

(ii) Sketch the graph of ( ) using the values p and q obtained in part (i).

SOLUTION

(a)

√2 2 + 3 √2 22 + 3
5 + 1 2
lim = lim
5 1
→−∞ →−∞ +

√2 + 3

= lim
1
→−∞ 5 +

= −√2 + 0
5+0

√2
=− 5
( )

5 − , −2 < ≤ −1

( ) = 2 + + , −1 < ≤ 2
2 − 4 , > 2

{ − 2

(i) ( ) is continuous at x = -1.

lim ( ) = lim ( )
→−1− →−1+

Page 15

177

PSPM I QS 015/2 Session 2015/2016

lim 5 − = lim 2 + +
→−1− →−1+

5 − (−1) = (−1)2 + (−1) +

5 + = 1 − +

2 − = −4 …………………….. (1) CHOW CHOON WOOI

( ) is continuous at x = 2.

lim ( ) = lim ( )
→2− →2+

lim 2 + + = 2 − 4
lim
→2− →2+ − 2

22 + 2 + = lim ( + 2)( − 2)

→2+ − 2

4 + 2 + = lim ( + 2)

→2+

4 + 2 + = 2 + 2

2 + = 0 …………………….. (2)

(2) - (1)
2 = 4
= 2
= −1

Page 16

178

PSPM I QS 015/2 Session 2015/2016

5 + , −2 < ≤ −1 Sketch = − +
1. a = 1, b = -1, c = 2
( ) = { 2− + 2 , −1 < ≤ 2 2. a > 0 (open upwards)
2 − 4 , > 2 3. Minimum point

− 2 − 1
= 2 = 2
5 + , −2 < ≤ −1 CHOW CHOON WOOI
2 − 4 1 − 8 7
= {( +2 − + 2 , −1 < ≤ 2 = −4 = −4 = 4
2)( − 2) > 2 4. Intercept
, When = 0, = 2
− 2

5 + , −2 < ≤ −1

= { 2 − + 2 , −1 < ≤ 2

+ 2 , > 2

(ii) Sketch the graph of ( )
( )

4

-2 -1 2 x

Page 17

179

PSPM I QS 015/2 Session 2015/2016

7. A curve is given by the parametric equations = − 1, = + 1.



a) Find and 2 in terns of .
2

b) Obtain the coordinates of the stationary points of the curve and determine the nature

of the points. CHOW CHOON WOOI

SOLUTION

(a)

= − 1 = + 1



= − −1 = + −1

= 1 + −2 = 1 − −2



= 1 + 1 = 1 − 1
2 2

= 2+1 = 2−1
2 2


= .

2 − 1 2
= 2 . 2 + 1

2 − 1
= 2 + 1

2
2 = ( ) .

2 − 1
( ) = ( 2 )
+ 1

= 2 − 1 = 2 + 1

′ = 2 ’ = 2

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180

PSPM I QS 015/2 Session 2015/2016

′ − ′ CHOW CHOON WOOI
( ) = 2

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

(2 3 + 2 ) − (2 3 − 2 )
= ( 2 + 1)2

2 3 + 2 − 2 3 + 2
= ( 2 + 1)2

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

(b)

To obtain the coordinates of the stationary points,

Let = 0



2 − 1
2 + 1 = 0

2 − 1 = 0

2 = 1

= ±1

Page 19

181

PSPM I QS 015/2 Session 2015/2016

ℎ = 1 = + 1 CHOW CHOON WOOI
= − 1


= 1 + 1
= 1 − 1
1
1
= 2
= 0

ℎ = −1 = + 1
= − 1


= −1 + 1
= −1 − 1
−1
−1
= −2
= 0

ℎ (0, −2) (0, 2)

ℎ = 1, ℎ (0,2)
2 4 3
2 = ( 2 + 1)3
4(1)3
= ((1)2 + 1)3

> 0 (Min)

ℎ = −1, ℎ (0, −2)
2 4 3
2 = ( 2 + 1)3
4(−1)3
= ((−1)2 + 1)3

< 0 (Max)

∴ (0, −2) , (0,2)

Page 20

182

PSPM I QS 015/2 Session 2015/2016

8. (a) If 2 − 2 √(1 + 2) + 2 = 0, show that = .
√(1+ 2
)

(b) Water is running at a steady rate of 36 3 −1 into a right inverted circular cone
with a semi-vertical angle of 45°.

(i) Find the rate of increasing in water depth when the water level is 3 cm. CHOW CHOON WOOI

(ii) Find the time taken when the depth of the water is 18cm.

SOLUTION

(a) 2 − 2 √(1 + 2) + 2 = 0

1

2 − 2 (1 + 2)2 + 2 = 0

2 − [(2 ) 1 (1 + 2)−21(2 ) + (1 + 1 (2 + 2 = 0
(2) )]
2 )2

2 − 2 + 2√1 + 2 + 2 = 0
[ ]
√1 + 2


2 − 2 − 2√1 + 2 + 2 = 0
√1 + 2

2 − 2√1 + 2 = 2 − 2
√1 + 2

(2 − 2√1 + 2) = 2 − 2 √1 + 2
√1 + 2

2 − 2 √1 + 2
= (2 − 2√1 + 2)√1 + 2

2 ( − √1 + 2)
= 2( − √1 + 2)√1 + 2


=

√1 + 2

Page 21

183

PSPM I QS 015/2 Session 2015/2016

( ) = 36 3 −1 r
h tan 45° = ℎ

45°
(i) Find ℎ when h = 3 cm 1=ℎ
ℎ =


ℎ ℎ CHOW CHOON WOOI
= .

= 1 2ℎ
3

= ℎ

= 1 ℎ2ℎ
3

= 1 ℎ3
3

= ℎ2
ℎ

ℎ ℎ
= .

1
= ( ℎ2) . (36 )

36
= ℎ2

ℎ ℎ = 3

ℎ 36
= 32

= 4 −1

Page 22

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PSPM I QS 015/2 Session 2015/2016

(ii) ℎ ℎ = 18 .

= 1 ℎ3
3

= 1 (18)3
3
CHOW CHOON WOOI
= 1944


= 36
1944

= 36
= 54

Page 23

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PSPM I QS 015/2 Session 2015/2016

9. (a) Determine the values of and , where > 0 and 0° < < 90° so that
3 sin − 4 cos = ( − ).

(b) Hence, solve the equation 3 sin − 4 cos = 2 for 0° ≤ ≤ 360°.

(c) From the answer obtained in part (b), find the value of for 0° ≤ ≤ 360° so that CHOW CHOON WOOI
( ) = 1 is minimum. Hence, find the minimum value of .

3 −4 +15

SOLUTION
(a)

3 sin − 4 cos = ( − ).

3 sin − 4 cos = ( cos − cos sin )

3 sin − 4 cos = cos − Rcos sin

sin : cos = 3 ………………………….. (1)

cos : sin = 4 ………………………….. (2)

(1)2 + (2)2 cos2 + sin2 = 1
2 cos2 + 2 sin2 = 32 + 42
2(cos2 + sin2 ) = 25
2 = 25

= 5

(2) ÷ (1)
sin 3
cos = 4
sin 3
cos = 4

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PSPM I QS 015/2 Session 2015/2016

3
tan = 4
= 53.1°
∴ 3 sin − 4 cos = 5 ( − 53.1°)

(b) CHOW CHOON WOOI

3 sin − 4 cos = 2 0° ≤ ≤ 360°

5 ( − 53.1°) = 2
2

( − 53.1°) = 5
( − 53.1°) = 0.4

0° ≤ ≤ 360°

0° − 53.1° ≤ − 53.1° ≤ 360° − 53.1°

−53.1° ≤ − 53.1° ≤ 306.9°

( − 53.1°) = 0.4

23.6° 23.6°

− 53.1° = 23.6°, 180° − 23.6° Page 25
− 53.1° = 23.6°, 156.4°
= 23.6° + 53.1°, 156.4° + 53.1°
= 76.7°, 209.5°

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PSPM I QS 015/2 Session 2015/2016

(c)

( ) = 3 − 1 + 15
4

( ) = 5 ( − 1 + 15 CHOW CHOON WOOI
53.1°)

Since −1 ≤ ( − 53.1°) ≤ 1

For ( ) minimum

( − 53.1°) = 1

− 53.1° = 90°

= 143.1°

Therfore, the minimum value of ( )

( ) = 5 1 53.1°) + 15
(143.1° −

1
= 5 (90°) + 15

1
= 5 (1) + 15

1
= 20

Page 26

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PSPM I QS 015/2 Session 2015/2016

10. (a) Find the value of if the slope of the curve 3 + 2 − 2 2 = 0 at the point (-1,1)
is -3.

(b) Given = sin .

1+cos

(i) Find and 2 in terms of . CHOW CHOON WOOI
2

(ii) Hence, show that 3 − 2 − ( 2 = 0.
3 2
)

SOLUTION

(a)

3 + 2 − 2 2 = 0

3 2 + [ 2 + 2 ] − 4 = 0


3 2 + 2 + 2 − 4 = 0


2 − 4 = −2 − 3 2


[ 2 − 4 ] = −2 − 3 2


−2 − 3 2
= 2 − 4

At the point (-1,1) ➔ x = -1, y =1 = −3



−2 − 3 2
= 2 − 4

−2 (−1)(1) − 3(−1)2
−3 = (−1)2 − 4(1)

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PSPM I QS 015/2 Session 2015/2016
Page 28
2 − 3 CHOW CHOON WOOI
−3 = − 4
−3( − 4) = 2 − 3
−3 + 12 = 2 − 3
12 + 3 = 2 + 3
5 = 15
= 3

(b)

(i) = sin = 1 + cos

1+cos

= sin

′ = cos ′ = − sin

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

cos + cos2 + sin2
= (1 + cos )2

cos + 1
= (1 + cos )2

1
= 1 + cos
= (1 + cos )−1

2 = −(1 + cos )−2 (1 + cos )
2

= −(1 + cos )−2(− sin )

sin
= (1 + cos )2

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PSPM I QS 015/2 Session 2015/2016

(ii) show that 3 − 2 − ( )2 = 0
3 2


2 sin
2 = (1 + cos )2

= sin = (1 + cos )2

′ = cos ′ = 2(1 + cos ) (1 + cos ) CHOW CHOON WOOI



= 2(1 + cos )(− sin )

= −2 sin (1 + cos )

3 (1 + cos )2(cos ) − [−2 sin (1 + cos )](sin )
3 =
[(1 + cos )2]2

(1 + cos )2(cos ) + 2 sin2 (1 + cos )
= (1 + cos )4

(1 + cos )[cos (1 + cos ) + 2 sin2 ]
= (1 + cos )4

cos + cos2 + 2 sin2
= (1 + cos )3

3 2 2
3 − 2 − ( )

cos + cos2 + 2 sin2 sin 12
= ( (1 + cos )3 ) − ((1 + cos )2) − (1 + cos )

cos + cos2 + 2 sin2 sin 1
= (1 + cos )3 − (1 + cos )2 − (1 + cos )2

cos + cos2 + 2 sin2 (1 sin ) sin 1
(1 + cos )3 + cos cos
= − −
(1 + cos )2 (1 + )2

cos + cos2 + 2 sin2 sin2 1 + cos
= (1 + cos )3 − (1 + cos )3 − (1 + cos )3

cos + cos2 + 2 sin2 − sin2 − (1 + cos )
= (1 + cos )3

Page 29

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PSPM I QS 015/2 Session 2015/2016

cos + cos2 + 2 sin2 − sin2 − 1 − cos CHOW CHOON WOOI
= (1 + cos )3

cos2 + sin2 − 1
= (1 + cos )3

1−1
= (1 + cos )3
=0

Page 30

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PSPM I QS 015/2 Session 2015/2016

CHOW CHOON WOOI

Page 31

193

2016/2017

194

CHOW CHOON WOOI

QS 015/1 CHOW CHOON WOOI
Matriculation Programme

Examination
Semester I
Session 2016/2017

195

PSPM I QS 015/1 Session
2016/2017

1. Solve for and where ≠ , such that ( + ) = (3 + √−16 − 3).
3

SOLUTION

( + ) = (3 + √−16 − 3)
3
CHOW CHOON WOOI
+ = (3 + √−16 − 3)(3 )

+ = (3 + √16√−1 − 3)(3 )

+ = (3 + 4 − 3)(3 )

+ = 9 + 12 2 − 3 4

+ = 9 + 12(−1) − 3(1)

+ = 9 − 15

+ = −15 + 9

= − , =

Page 2

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PSPM I QS 015/1 Session
2016/2017

2. Determine the values of x which satisfy the equation 32 −1 = 4(3 ) − 9.

SOLUTION

32 −1 = 4(3 ) − 9

32 3−1 = 4(3 ) − 9 CHOW CHOON WOOI

(3 )2 1 = 4(3 ) − 9
()
3

1 (3 )2 = 4(3 ) − 9
(3)

(3 )2 = 12(3 ) − 27

= 3

( )2 = 12( ) − 27
2 − 12 + 27 = 0

( − 3)( − 9) = 0

= 3 = 9
3 = 3 3 = 9

= 1 = 2

Page 3

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PSPM I QS 015/1 Session
2016/2017

3. The seventh term of a geometric series is 16, the fifth term is 8 and the sum of
the first ten terms is positive. Find the first term and the common ratio. Hence,
show that 12 = 126(√2 + 1).

SOLUTION

Geometric series CHOW CHOON WOOI

➔ = −

7 = 6 = 16 …… (1)
5 = 4 = 8 …… (2)

(1) ÷ (2)

6 16
4 = 8

2 = 2

= ±√2

ℎ = ±√2
(±√2)4 = 8

4 = 8

= 2

Page 4

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PSPM I QS 015/1 Session
2016/2017
➔ = ( − )
− Page 5

Given that > CHOW CHOON WOOI
( − )

= −

ℎ = 2; = √2

2 [(√2)10 − 1]
10 = √2 − 1

2[32 − 1]
=

√2 − 1

62
=

√2 − 1

= 149.68

ℎ = 2; = −√2
2 [(−√2)10 − 1]

10 = −√2 − 1
2[32 − 1]

=
−√2 − 1
62

=
−√2 − 1

= −25.68

ℎ > 0, ➔ = 2 = √2
2 [(√2)12 − 1]

12 = √2 − 1

199

PSPM I QS 015/1 Session
2016/2017

2[64 − 1] CHOW CHOON WOOI
=

√2 − 1
126
=
√2 − 1
126 √2 + 1
=
√2 − 1 √2 + 1
126(√2 + 1)
=
2 + √2 − √2 − 1
126(√2 + 1)
=1
= 126(√2 + 1)

Page 6

200