PSPM 1 Q&A

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
.

SOLUTION

= (−22 53) CHOW CHOON WOOI

2 + + = 0
(−22 53) (−22 35) + (−22 35) + (10 10) = (00 00)

(−44 − 6 −66++1255) + (−22 53 ) + ( 0 0 ) = (00 00)
− 10

(−−124 1219) + (−22 53 ) + ( 0 0 ) = (00 00)

(−−2 1+42− 2+ 21 + 3 ) = (00 00)
19 + 5 +

21 + 3 = 0
= −7

−2 + 2 + = 0
−2 + 2(−7) + = 0
= 16

∴ = −7, = 16

Page 4

251

PSPM I QS015/1 Session 2017/2018

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

SOLUTION

32 +1 − (16)3 + 5 = 0

32 31 − (16)3 + 5 = 0 CHOW CHOON WOOI

3. (3 )2 − (16)3 + 5 = 0
Let = 3
3 2 − 16 + 5 = 0

(3 − 1)( − 5) = 0

(3 − 1) = 0 ( − 5) = 0

= 1 = 5
3 = 5
3 ln 3 = ln 5

3 = 1

3

3 = 3−1

= −1 ln 3 = ln 5

= 1.465

∴ = −1 = 1.465

Page 5

252

PSPM I QS015/1 Session 2017/2018

3. The first and three more successive terms in a geometric progression are given as follows:
7, … , 189, , 1701, …

Obtain the common ratio r. Hence, find the smallest integer n such that the n-th term
exceeds 10,000.

SOLUTION CHOW CHOON WOOI

7, … , 189, , 1701, …
= 7

1701
189 =
2 = 321489
= 567

567
= 189
= 3
> 10000
−1 > 10000
(7)3 −1 > 10000
3 −1 > 1428.57
ln 3 −1 > ln 1428.57
( − 1) ln 3 > ln 1428.57
( − 1) > 6.61
> 6.61 + 1
> 7.61
= 8

Page 6

253

PSPM I QS015/1 Session 2017/2018

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. CHOW CHOON WOOI

SOLUTION

a. (1 − 1 = 1 + (1) (− 1 + (12)(−21) (− )2 + (12)(−12)(−23) (− )3

)2 2 )

3 1! 3 2! 3 3! 3

1 2 3 3
= 1 − 6 − 8 ( 9 ) − 48 (27)

2 3 3
= 1 − 6 − 72 − 48 (27)

= 1 − 1 − 1 2 − 1 3
6 72 432

The interval of for which the expansion is valid:

|3| < 1

−1 < 3 < 1

−3 < < 3

1

b. √9 − 3 = (9 − 3 )2

1

1 3 2
= 92 (1 − 9 )

= 3 (1 − 1

3 )2

c. √8.70 = √9 − 3(0.01) Page 7

254

PSPM I QS015/1 Session 2017/2018

= 0.01

√9 − 3 = 3 (1 − )12
3

√9 − 3 = 3 [1 − 1 − 1 2 − 1 3] CHOW CHOON WOOI
6 72 432

√9 − 3(0.01) = 3 [1 − 1 (0.01) − 1 (0.01)2 − 1 (0.01)3]
6 72 432

√8.7 = 2.95

Page 8

255

PSPM I QS015/1 Session 2017/2018

5. Solve the equation 3 9 = ( 3 )2.

SOLUTION

3 9 = ( 3 )2

3 3 = ( 3 )2 CHOW CHOON WOOI
3 9

3 3 = ( 3 )2
3 32

3 3 = ( 3 )2
2 3 3

3 3 = ( 3 )2
2

3 3 = 2( 3 )2

= 3

3 = 2 2

2 2 − 3 = 0

(2 − 3) = 0

= 0 2 − 3 = 0
= 0
3 = 0 = 3
= 30
2

3 = 3
2

3

= 32

= 1 = 5.196

Page 9

256

PSPM I QS015/1 Session 2017/2018

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.

̅ ̅

SOLUTION CHOW CHOON WOOI

(a) = 2 +

̅ − 1 = (2 − ) − 2 1
̅ −

1 (2 + )
= (2 − ) − (2 − ) (2 + )

(2 + )
= (2 − ) − 4 + 2 − 2 − 2

(2 + )
= (2 − ) − 5

21
= 2 − − 5 − 5

86
= 5 − 5

(b) | ̅ − 1| = |8 − 6 |

̅ 5 5

= √(85)2 + 62
(5)

= √64 + 36
25 25

= √100
25

= √4
=2

1 12
+ = | ̅ − ̅| ( ̅ − ̅)

Page 10

257

PSPM I QS015/1 Session 2017/2018

8 62
+ = 2 ( − )

55

82 8 6 62
+ = 2 [(5) − 2 (5) (5) + (5 ) ]

64 96 36
+ = 2 [25 − 25 − 25]

28 96 CHOW CHOON WOOI
+ = 2 [25 − 25 ]

56 192
+ = 25 − 25

= 56 = − 192

25 25

Page 11

258

PSPM I QS015/1 Session 2017/2018

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

+3

b. |3 −2| > 2

2 +3

SOLUTION

a) 5 − 1 ≤ 0 + 3 = 0 CHOW CHOON WOOI
= −3
+3

5 − ( + 3)
+ 3 ≤ 0

5 − − 3
+ 3 ≤ 0

2 −
+ 3 ≤ 0
2 − = 0

= 2

2 − (−∞, −3) (−3, 2) (2, ∞)
+ 3 + + -
2 − - + +
+ 3
- + -
{ : < −3 ∪ ≥ 2}

b) |3 −2| > 2 | | > ⟺ > < −

2 +3

3 − 2 3 − 2
2 + 3 > 2 2 + 3 < −2

3 − 2 3 − 2
2 + 3 − 2 > 0 2 + 3 + 2 < 0

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

2 + 3 2 + 3

3 − 2 − 4 − 6 >0 3 − 2 + 4 + 6 <0

2 + 3 2 + 3

− − 8 7 + 4
2 + 3 > 0 2 + 3 < 0

Page 12

259

PSPM I QS015/1 Session 2017/2018

− − 8 = 0 2 + 3 = 0 7 + 4 = 0 2 + 3 = 0
= −8 = − 3
= − 4 = − 3
2
7 2
33
(−∞, −8) (−8, − 2) (− 2 , ∞) 3 34 4
(−∞, − 2) (− 2 , − 7) (− 7 , ∞)
--
− − 8 + -+ 7 + 4 - -+
-
2 + 3 +- 2 + 3 - ++
− − 8 -
2 + 3 7 + 4 + -+ CHOW CHOON WOOI
2 + 3

3 or 34
(−8, − 2) (− 2 , − 7)

−8 34
−2 −7

3 34
(−8, − 2) ∪ (− 2 , − 7)

Page 13

260

PSPM I QS015/1 Session 2017/2018

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).

SOLUTION CHOW CHOON WOOI

( ) = ( − 2)2 + 1, > 2
( ) = ln( + 1), > 0.

(a) Let = −1( )
( ) =
( − 2)2 + 1 =
( − 2)2 = − 1

− 2 = √ − 1

= √ − 1 + 2
−1( ) = √ − 1 + 2
Let = −1( )
( ) =
ln( + 1) =
+ 1 =
= − 1
−1( ) = − 1
−1: (1, ∞)
−1: (2, ∞)
−1: (0, ∞)
−1: (0, ∞)

(b) ( )( )
[ ( )] = [( − 2)2 + 1]
= ln[[( − 2)2 + 1] + 1]
= ln[( − 2)2 + 2]

( )(2) = ln[(2 − 2)2 + 2] Page 14
= ln[2]

261

PSPM I QS015/1 Session 2017/2018

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( ) = .

SOLUTION CHOW CHOON WOOI

( ) = 1 5
2 −

(a) : 2 − 5 ≠ 0

5
≠ 2

55
: (−∞, 2) ∪ (2 , ∞)
: (−∞, 0) ∪ (0, ∞)

(b) ( ) = 1

2 −5

1
( 1) = 2 1 − 5

1
( 2) = 2 2 − 5
Let ( 1) = ( 2)

11
2 1 − 5 = 2 2 − 5
2 1 − 5 = 2 2 − 5
2 1 = 2 2
1 = 2
1 = 2 ℎ ( 1) = ( 2), ℎ ( ) .

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

Page 15

262

PSPM I QS015/1 Session 2017/2018

1 + 5 =
2 = ( )

1 + 5
= 2

−1( ) = 1 + 5
2

(c) CHOW CHOON WOOI

−1( )

(d) −1( ) = [ −1( )]

1 + 5
= [ 2 ]

= 1

2 [1 + 5 ] − 5
2

1
= [1 + 5 ] − 5

= 1

[1 + 5 − 5 ]


1
= [1 ]

=

Page 16

263

PSPM I QS015/1 Session 2017/2018

10. Given the system of linear equations as follow: CHOW CHOON WOOI

2 + 4 + = 77

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 , .

SOLUTION

2 + 4 + = 77

4 + 3 + 7 = 114

2 + + 3 = 48

(a)

2 4 1 77

(4 3 7) ( ) = (114)

2 1 3 48

= 77
= (114)
241
= (4 3 7); 48

213

(b)
| | = +(2) |31 73| − (4) |42 73| + (1) |42 13|
= 2(9 − 7) − 4(12 − 14) + (4 − 6)
= 2(2) − 4(−2) + (−2)
= 10

Page 17

264

PSPM I QS015/1 Session 2017/2018

(i) = = ℎ = −1
= −1
| | = | −1| | −1| = 1
1 | |
= | |
1 ( ) = CHOW CHOON WOOI
= 10 , ℎ | | = | |

(ii) = (2 ) | | = | |
| | = |(2 ) |
= 23| |
= 8| |
= 8(10)
= 80

(iii) ( ) =

241
= (4 3 7)

213

+ |13 37| − |42 73| + |42 31|
, = − |41 13| + |22 31| − |22 14|

(+ |43 71| − |42 71| + |24 43|)

2 2 −2
= (−11 4 6 )

25 −10 −10

( ) =

2 2 −2
= (−11 4 6 )

25 −10 −10

2 −11 25
= ( 2 4 −10)

−2 6 −10

−1 = 1 ( )
| |

1 2 −11 25
= (2 4 −10)
10 −2 6 −10

Page 18

265

PSPM I QS015/1 Session 2017/2018

2 −11 25

10 10 10
= 2 4 −10

10 10 10
−2 6 −10
( 10 10 10 )

1 −11 5 CHOW CHOON WOOI

5 10 2
=1 2
5 5 −1
−1 3
( 5 5 −1)

=
−1 = −1
= −1
= −1

1 −11 5

5 10 2 77
1 2 −1 (114)
( ) = 5 5
−1 3 48
5
(5 −1)

10
= (13)

5

∴ = 10, = 13, = 5

Page 19

266

QS 015/2 CHOW CHOON WOOI
Matriculation Programme Examination

Semester I
Session 2017/2018

267

PSPM I QS015/2 Session 2017/2018

1. Express 3 2 −5 in partial fractions.
( −3)( 2+2)

2. Solve the equation cos + cos 5 = 2 cos 3 0 ≤ ≤ . Give your answers in
terms of .

3. Evaluate the following limits:

a. lim 3 −8 CHOW CHOON WOOI
2 −2
→2

b. lim √5 +7

→∞ 6 −5

4. Given = −2 sin 3 . Find and 2 2 .


Hence, show that 2 + 4 + 13 = 0.
2

5. Given the polynomial ( ) = 2 − 4 and ( ) = 4 + 3 + 2 2 + + 28.

a. Find all zeroes of ( ).

b. When ( ) is divided by ( ), the remainder is 14 + 52. Use the remainder
theorem to find the values of and .

c. Using the values of and obtained from part 5( ), find the remainder when
2 ( ) + is divided by ( ).

6. Express cos + √2 sin in the form sin( + ), where > 0 and is an acute
angle.

Hence,

a. Solve the equation cos + √2 sin = √3 by giving all solutions between
2

0° and 360°.

b. Show the greatest value of 1 is 5+√3.

cos +√2 sin +5 22

7. State the conditions for continuity of ( ) at = .

a. By using the conditions for continuity of ( ) at = , find the values of
and such that

− 2 cos , < 0

( ) = {2 + 2, 0 ≤ < 2

− , ≥ 2

is continuous on the interval (−∞, ∞).

Page 2

268

PSPM I QS015/2 Session 2017/2018

b. If = −2 and = 4, determine whether ( ) is differentiable at = 2 or
not.

8. A curve with equation 2 − 3 2 = −2 + − 6, where and are constants,
passes through the point (1, 2).

a. Given = 1 at (1, 2), determine the values and . CHOW CHOON WOOI



b. Evaluate 2 at (1, 2).
2

9. The function is defined by ( ) = ln( −1) for > 1.

−1

a. By considering the first and second derivatives of ( ), show that there is
only one maximum point on the graph = ( ).

b. Use the result obtained in part 9(a) to state the exact coordinates of the
maximum point.

c. Find the -coordinate of the function f when 2 = 0.
2

10. A curve is defined by the parametric equations = 3 − 1 and = + 3, where ≠ 0.



a. Show that = 3 22−+31. Hence, find 2 2 .


b. Show that can be expressed as = 1 − 3 (31 20+1). Hence, deduce that
3

−3 < < 1.

3

END OF QUESTION PAPER

Page 3

269

PSPM I QS015/2 Session 2017/2018

1. Express 3 2 −5 in partial fractions.
( −3)( 2+2)

SOLUTION

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

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

3 2 − 5 = ( 2 + 2) + ( + )( − 3)

=

3(3)2 − 5 = [(3)2 + 2] + [ (3) + ][(3) − 3]

22 = 11

= 2

=
3(0)2 − 5 = (2)[(0)2 + 2] + [ (0) + ][(0) − 3]
−5 = 4 − 3
3 = 9
= 3

=

3(1)2 − 5 = (2)[(1)2 + 2] + [ (1) + (3)][(1) − 3]

−2 = 6 − 2[ + 3]

2[ + 3] = 8

+ 3 = 4

= 1

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

Page 4

270

PSPM I QS015/2 Session 2017/2018

2. Solve the equation cos + cos 5 = 2 cos 3 0 ≤ ≤ . Give your answers in
terms of .

SOLUTION



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

a) 22

cos + cos 5 = 2 cos 3 b) sin − sin = 2 cos ( + ) sin ( − )

22

cos 5 + cos = 2 cos 3 c) + = ( + ) ( − )



5 + 5 − d) cos − cos = −2 sin ( + ) sin ( − )
2 cos ( 2 ) cos ( 2 ) = 2 cos 3
22

2 cos 3 cos 2 = 2 cos 3

2 cos 3 cos 2 − 2 cos 3 = 0

cos 3 [2 cos 2 − 2] = 0

cos 3 = 0 2 cos 2 − 2 = 0
0 ≤ ≤ cos 2 = 1
0 ≤ 3 ≤ 3
0 ≤ 2 ≤ 2

3 = , 2 − , 2 + 2 = 0,
5
222
= 0, 6 , 2 , 6 ,
5
= 6 , 2 , 6

Page 5

271

PSPM I QS015/2 Session 2017/2018

3. Evaluate the following limits:

a. lim 3 −8
2 −2
→2

b. lim √5 +7

→∞ 6 −5

SOLUTION CHOW CHOON WOOI

a. lim 3 −8 = lim ( −2)( 2 +2 +4)
2 −2 ( −2)
→2 →2

= lim ( 2 + 2 + 4)

→2

22 + 2(2) + 4
=2

=6

b. lim √5 +7 = lim √5 +7
− 5
→∞ 6 −5 →∞ 6


= lim 5 + 7
√ −
→∞ 5
6


= √5 + 0
6 − 0

= √5
6

Page 6

272

PSPM I QS015/2 Session 2017/2018

4. Given = −2 sin 3 . Find and 2 2 .


Hence, show that 2 + 4 + 13 = 0.
2

SOLUTION

= −2 sin 3 CHOW CHOON WOOI

= −2 = sin 3

′ = −2 −2 ′ = 3 cos 3

= ′ + ′


= ( −2 )(3 cos 3 ) + (sin 3 )(−2 −2 )

= 3 −2 cos 3 − 2 −2 sin 3

= −2 (3 cos 3 − 2 sin 3 )


= −2 = 3 cos 3 − 2 sin 3

′ = −2 −2 ′ = −9 sin 3 − 6 cos 3

= ′ + ′


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

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

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

= ( −2 )[−5 sin 3 − 12 cos 3 ]

2
2 + 4 + 13
= ( −2 )[−5 sin 3 − 12 cos 3 ] + 4[3 −2 cos 3 − 2 −2 sin 3 ] + 13[ −2 sin 3 ]
= −5 −2 sin 3 − 12 cos −2 3 + 12 −2 cos 3 − 8 −2 sin 3 + 13 −2 sin 3
=0

Page 7

273

PSPM I QS015/2 Session 2017/2018

5. Given the polynomial ( ) = 2 − 4 and ( ) = 4 + 3 + 2 2 + + 28. CHOW CHOON WOOI

a. Find all zeroes of ( ).

b. When ( ) is divided by ( ), the remainder is 14 + 52. Use the remainder
theorem to find the values of and .

c. Using the values of and obtained from part 5( ), find the remainder when
2 ( ) + is divided by ( ).

SOLUTION
( ) = 2 − 4
( ) = 4 + 3 + 2 2 + + 28
(a) When ( ) = 0
2 − 4 = 0
2 = 4
= ±√4
= ±2
ℎ , − 2, 2

(b) ( ) = ( ) = 2 − 4 = ( + 2)( − 2)
( ) = 4 + 3 + 2 2 + + 28
( ) = 14 + 52 When a polynomial P(x) is divided by (x – a), then the
remainder is P(a)

4 + 3 + 2 2 + + 28 = ( )( 2 − 4) + (14 + 52) Page 8
4 + 3 + 2 2 + + 28 = ( )( + 2)( − 2) + (14 + 52)
ℎ = 2

(2) = (2)
(2)4 + (2)3 + 2(2)2 + (2) + 28 = [14(2) + 52]

274

PSPM I QS015/2 Session 2017/2018

16 + 16 + 2 + 28 = [28 + 52]

16 + 2 = 36

8 + = 18 ………………… (1)

ℎ = −2

(−2) = (−2) CHOW CHOON WOOI

(−2)4 + (−2)3 + 2(−2)2 + (−2) + 28 = [14(−2) + 52]

16 − 8 + 8 − 2 + 28 = [−28 + 52]

16 − 2 + 28 = 24

16 − 2 = −4

8 − = −2 ………………… (2)

(1) + (2)

16 = 16

= 1

8 − = −2

= 10

∴ = 1, = 10

(c) ( ) = 4 + 3 + 2 2 + 10 + 28

( ) = 2 ( ) +

( ) = 2 − 4

( 4 + 3 + 2 2 + 10 + 28) = [ ( )( + 2)( − 2) + (14 + 52)]

2( 4 + 3 + 2 2 + 10 + 28) = 2[ ( )( + 2)( − 2) + (14 + 52)]

2[ 4 + 3 + 2 2 + 10 + 28] + = 2[ ( )( + 2)( − 2) + (14 + 52)] +

2[ 4 + 3 + 2 2 + 10 + 28] + = 2 ( )( + 2)( − 2) + 2(14 + 52) +

( ) = 2(14 + 52) +

= 28 + 104 +

= 29 + 104

Page 9

275

PSPM I QS015/2 Session 2017/2018

6. Express cos + √2 sin in the form sin( + ), where > 0 and is an acute CHOW CHOON WOOI
angle.

Hence,

a. Solve the equation cos + √2 sin = √3 by giving all solutions between

2

0° and 360°.

b. Show the greatest value of 1 is 5+√3.

cos +√2 sin +5 22

SOLUTION

cos + √2 sin = sin( + )

cos + √2 sin = (sin cos + cos sin )

cos + √2 sin = sin cos + cos sin

cos + √2 sin = cos sin + sin cos

cos sin = cos

sin = 1 …………. (1)

sin cos = √2 sin

cos = √2 …………. (2)

(1)2 + (2)2

( 2 sin2 ) + ( 2 cos2 ) = 12 + 2

(√2)

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

2 = 3

= √3

Page 10

276

PSPM I QS015/2 Session 2017/2018

(1) ÷ (2) CHOW CHOON WOOI
sin 1
cos = √2

1
tan =

√2
= 35.26°

cos + √2 sin = √3 sin( + 35.26°)

(a) cos + √2 sin = √3 0° < < 360°

2

√3 sin( + 35.26°) = √3
2

sin( + 35.26°) = 1
2

0° + 35.26° < + 35.26° < 360° + 35.26°

35.26° < + 35.26° < 395.26°

−1 (12) = 30° 30°

+ 35.26° = 180° − 30°; 360° + 30°
+ 35.26° = 150°; 390°
= 150° − 35.26° = 150°; 390° − 35.26°
= 114.7°, 354.74°

Page 11

277

PSPM I QS015/2 Session 2017/2018

(b) Show the greatest value of 1 is 5+√3

cos +√2 sin +5 22

cos + √2 sin = √3 sin( + 35.26°)
−1 ≤ sin( + 35.26°) ≤ 1

−√3 ≤ √3 sin( + 35.26°) ≤ √3 CHOW CHOON WOOI

−√3 ≤ cos + √2 sin ≤ √3

−√3 + 5 ≤ cos + √2 sin + 5 ≤ √3 + 5

11 1
≤≤
√3 + 5 cos + √2 sin + 5 −√3 + 5

111
≤≤

√3 + 5 cos + √2 sin + 5 5 − √3

The greatest value of 1:
cos +√2 sin +5

1 1 (5 + √3)
=

5 − √3 (5 − √3)(5 + √3)

5 + √3
= 25 − 3

5 + √3
= 22

Page 12

278

PSPM I QS015/2 Session 2017/2018

7. State ℎ ( ) = .

a. By using the conditions for continuity of ( ) at = , find the values of
and such that

− 2 cos , < 0

( ) = {2 + 2, 0 ≤ < 2

− , ≥ 2 CHOW CHOON WOOI

is continuous on the interval (−∞, ∞).

b. = −2 = 4, determine whether ( ) is differentiable at = 2 or
not.

SOLUTION

ℎ ( ) =

i. ( )
ii. lim ( )

→

iii. lim ( ) = ( )

→

− 2 cos , < 0

(a) ( ) = {2 + 2, 0 ≤ < 2

− , ≥ 2

( ) = 0 = 2 .

ℎ = 0

lim ( − 2 cos ) = lim (2 + 2)
→0− →0+

− 2 cos(0) = 2 + (0)2

− 2 = 2

= 4

ℎ = 2

lim (2 + 2) = lim ( − )
→2− →2+

2 + (22) = − 2

3 + 4 = 0

Page 13

279

PSPM I QS015/2 Session 2017/2018

4
= − 3

4
∴ = − 3 , = 4

(b) = −2 = 4 CHOW CHOON WOOI

4 − 2 cos , < 0

( ) = { 2 − 2 2, 0 ≤ < 2

−2 − , ≥ 2

= 2 ′ ( ) = lim ( ) − ( )
−
→

′ (2− ) = l→im2− ( ) − (2) ′ ( − ) = lim ( ) − ( )
− 2 −
→ −

= l→im2− (2 − 2 2) − (−2 − ) ′ ( + ) = lim ( ) − ( )
−
− 2 → +

= lim 2 − 2 2 + 2 +

→2− − 2

= lim −2 2 + + 4

→2− − 2

−2(2)2 + (2) + 4
= 2−2

= +∞ ((Undefined)

′(2−) , ℎ ( ) = 2.

Page 14

280

PSPM I QS015/2 Session 2017/2018

8. A curve with equation 2 − 3 2 = −2 + − 6, where and are constants,
passes through the point (1, 2).

a. Given = 1 at (1, 2), determine the values and .



b. Evaluate 2 at (1, 2).
2

SOLUTION CHOW CHOON WOOI

2 − 3 2 = −2 + − 6

( , )

12 − 3(2)2 = 2−2(1) + (2) − 6

1 − 12 = + 2 − 6

+ 2 = −5 ………………. (1)


( , ), =

2 − 3 2 = −2 + − 6

2 − 6 = −2 ( − 2 ) +


2 − 6 = ( −2 ) − 2] +
[

2(1) − 6(2)(1) = ( (2)−2(1))[1 − 2] + (1)

−10 = − +

− = 10 ………………. (2)

(1) – (2)

3 = −15

= −5

+ 5 = 10

= 5

∴ = 5, = −5

Page 15

281

PSPM I QS015/2 Session 2017/2018

( ) Evaluate 2 at (1, 2)
2

2 − 3 2 = 5 −2 − 5 − 6

2 − 6 = (5 −2 ) − 2] − 5
[


2 − 6 = 5 −2 − 10 −2 − 5
CHOW CHOON WOOI

2 − [6 2 + 6 2 = [5 −2 2 + 5 −2 ( − 2 )] − 10 −2 ( − 2 ) − 5 2
2 ( ) ] 2 2

2 − [6 2 + 6 2 = [5 −2 2 + 5 −2 − 2)] − 10 −2 − 2) − 5 2
2 ( )] 2 ( ( 2




( , ), =

2 − [6 2 + 6 2 = [5 −2 2 + 5 −2 − 2)] − 10 −2 − 2) − 5 2
2 ( )] 2 ( ( 2


2 − [6(2) 2 + 6(1)2] = [5 2 −2(1) 2 + 5 2−2(1)(1)((1) − 2)] − 10 2−2(1)((1) − 2) − 5 2
2 2 2

2 2 2
2 − [12 2 + 6] = [5 2 − 5] + 10 − 5 2

2 2 2
2 − 12 2 − 6 = 5 2 − 5 + 10 − 5 2

2 2 2
5 2 + 12 2 − 5 2 = 2 + 5 − 10 − 6

2
12 2 = −9

2 9
2 = − 12

2 3
2 = − 4

Page 16

282

PSPM I QS015/2 Session 2017/2018

9. The function is defined by ( ) = ln( −1) for > 1.

−1

a. By considering the first and second derivatives of ( ), show that there is
only one maximum point on the graph = ( ).

b. Use the result obtained in part 9(a) to state the exact coordinates of the
maximum point.

c. Find the -coordinate of the function f when 2 = 0. CHOW CHOON WOOI
2

SOLUTION

(a) ( ) = ln( −1) , > 1

−1

= ln( − 1) = − 1

′ = 1 ( − 1) ′ = 1

−1

1
= − 1

′ − ′
′( ) = 2

( − 1) ( 1 1) − [ln( − 1)](1)
−
=
( − 1)2

1 − ln( − 1)
= ( − 1)2

′( ) = 0

1 − ln( − 1)
( − 1)2 = 0

1 − ln( − 1) = 0

ln( − 1) = 1

− 1 = 1 = ⟺ =
= + 1 = ⟺ =

Page 17

283

PSPM I QS015/2 Session 2017/2018

1 − ln( − 1)
′( ) = ( − 1)2

= 1 − ln( − 1) = ( − 1)2
′ = 2( − 1)
′ = − 1
− 1
CHOW CHOON WOOI
′′( ) ( − 1)2 (− 1 1) − [1 − ln( − 1)]2( − 1)
− [( − 1)2]2
=

[−( − 1)] − [1 − ln( − 1)](2 − 2)
= ( − 1)4

ℎ = + 1

"( ) = (−( + 1 − 1)) − [1 − ln( + 1 − 1)][(2( + 1) − 2)]

( + 1 − 1)4

(− ) − [1 − ln( )](2 )
= ( )4

(− )
= ( )4

1
= − 3 < 0 ( )

(b) Use the result obtained in part 9(a) to state the exact coordinates of the maximum
point.

( ) = ln( − 1)
− 1

ℎ = + 1

( ) = ln( + 1 − 1)
+ 1 − 1

ln
=

1
=

1
∴ ℎ ℎ : ( + 1, )

Page 18

284

PSPM I QS015/2 Session 2017/2018

(c) Find the -coordinate of the function f when 2 = 0
2

2
ℎ 2 = 0

[−( − 1)] − [1 − ln( − 1)](2 − 2) =0 CHOW CHOON WOOI

( − 1)4

[−( − 1)] − [1 − ln( − 1)](2 − 2) = 0

[−( − 1)] − 2( − 1)[1 − ln( − 1)] = 0

( − 1)[−1 − 2(1 − ln( − 1))] = 0

( − 1)[−1 − 2 + 2ln ( − 1)] = 0

( − 1)[−3 + 2ln ( − 1)] = 0

( − 1) = 0 −3 + 2 ln( − 1) = 0

= 1 2 ln( − 1) = 3

ln( − 1) = 3

2

3

− 1 = 2

3

= 2 + 1

3

≠ 1, ℎ = 2 + 1

Page 19

285

PSPM I QS015/2 Session 2017/2018

10. A curve is defined by the parametric equations = 3 − 1 and = + 3, where ≠ 0.



a. Show that = 3 22−+31. Hence, find 2 2 .


b. Show that can be expressed as = 1 − 3 (31 20+1). Hence, deduce that
3

−3 < < 1. CHOW CHOON WOOI

3

SOLUTION

= 3 − 1 and = + 3



(a) Show that = 3 22−+31. Hence, find 2
2

1 3
= 3 − = +
1 3
= 3 + 2 = 1 − 2

3 2 + 1 2 − 3
= 2 = 2


= .

2 − 3 2
= 2 . 3 2 + 1

2 − 3
= 3 2 + 1

Page 20

286

PSPM I QS015/2 Session 2017/2018

2
2 = ( ) .

2 − 3 = 3 2 + 1 CHOW CHOON WOOI
= 3 2 + 1 ′ = 6

= 2 − 3
′ = 2

′ − ′
( ) = 2

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

6 3 + 2 − 6 3 + 18
= (3 2 + 1)2

20
= (3 2 + 1)2

2
2 = ( ) .

20 2
= (3 2 + 1)2 . 3 2 + 1

20 3
= (3 2 + 1)3

(b) Show that can be expressed as = 1 − 3(31 20+1). Hence, deduce that
3

Page 21

287

PSPM I QS015/2 Session 2017/2018

−3 < < 13.


2 − 3
= 3 2 + 1

1 CHOW CHOON WOOI
3

3 2 + 1 2 − 3

2 + 1
3

10
−3

= 1 + − 10
3 3 2 3
+1

1 10
3 − 3(3 2 + 1)

2 > 0

3 2 > 0

3 2 + 1 > 0 + 1

3 2 + 1 > 1

3(3 2 + 1) > 3

3 < 3(3 2 + 1) < ∞

11
0 < 3(3 2 + 1) < 3

10 10
0 < 3(3 2 + 1) < 3

10 10
− 3 < − 3(3 2 + 1) < 0

1 10 1 10 1
3 − 3 < 3 − 3(3 2 + 1) < 3 + 0

1 10 1
−3 < 3 − 3(3 2 + 1) < 3

1
−3 < < 3

Page 22

288

2018/2019

289

CHOW CHOON WOOI

SM015/1 CHOW CHOON WOOI
MATHEMATICS

2018/2019

Matriculation Programme Examination

290

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

1. a) Solve √6 + 1 − √ = 3.

b) Determine the solution set of x which satisfies the inequality. CHOW CHOON WOOI
2
2. a)
b) + 1 < + 3
Find the sum of all integers from 5 to 950 which are divisible by 3.

1

Expand 3(1 + )4 in ascending powers of up to the fourth term. Hence,

approximate 4√80 correct to four decimal places.

3. Find the gradient of the curve cos(4 ) = tan( 2) − 3 at the point where = 0.

4. The parametric equations of a curve are = + 2 and = 2 − 4 , where ≠ 0. Show that


= 2 + 28−2. Hence, find 2 in term of .
2

5. Water is poured into a right inverted cone of height ℎ with a semi-vertical angle of 60° at a

constant rate of 25 3 per second.

a. Show that the rate of change of the height of water is ℎ = 32ℎ52.


b. Find the rate of change of the height of water after 5 seconds.

c. Given the height of the cone is 20cm, find the time taken to fill the cone

completely with water.

END OF QUESTION PAPER

2

291

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

1. a) Solve √6 + 1 − √ = 3. CHOW CHOON WOOI
Determine the solution set of x which satisfies the inequality.
b)
2
+ 1 < + 3

SOLUTION
1a)

√6 + 1 − √ = 3

(√6 + 1 − 2 = 32

√ )

(6 + 1) + ( ) − 2(√6 + 1)(√ ) = 9

7 + 1 − 2(√6 + 1)(√ ) = 9

7 − 8 = 2(√6 + 1)(√ )

(7 − 8)2 = [2(√6 + 2

1)(√ )]

49 2 + 64 − 112 = 4(6 + 1)( )

49 2 + 64 − 112 = 24 2 + 4

25 2 − 116 + 64 = 0

(25 − 16)( − 4) = 0

= 16 , = 4

25

Check:

√6 + 1 − √ = 3 √6 + 1 − √ = 3
When = 16 When = 4

25

3

292

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

√6 + 1 − √ = √6 16 + 1 − √16 √6 + 1 − √ = √6(4) + 1 − √4 CHOW CHOON WOOI
(25) 25 = √25 − 2
=3
= √121 − √16
25 25

11 4
= 5 −5

7
=5≠3

∴ = 4 4

1b)
2

+ 1 < + 3
2

+ 1 − + 3 < 0
2( + 3) − ( + 1)

( + 1)( + 3) < 0
2 + 6 − 2 −
( + 1)( + 3) < 0
− 2 + + 6
( + 1)( + 3) < 0

2 − − 6
( + 1)( + 3) > 0
( + 2)( − 3)
( + 1)( + 3) > 0

293

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

Critical value: (−2, −1) (−1,3) (3, ∞) CHOW CHOON WOOI
= −3, −2, −1, 3
+ + +
(−∞, −3) (−3, −2)

+ 2 - -

− 3 - - - - +

+ 1 - - - + +

+ 3 - + + + +

○ ○ ○( + 2)( − 3)
+ - + -+

( + 1)( + 3)

: { : < −3 ∪ −2 < < −1 ∪ > 3}

5

294

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

2. a) Find the sum of all integers from 5 to 950 which are divisible by 3. CHOW CHOON WOOI
b)
1

Expand 3(1 + )4 in ascending powers of up to the fourth term. Hence,
approximate 4√80 correct to four decimal places.

SOLUTION
2a)

Integers:
5, 6, 7, 8, 9, 10,...950

Integers which are devisible by 3:
6, 9, 12, 15, ... 945, 948

= 6, = 3
= 948
+ ( − 1) = 948
6 + ( − 1)3 = 948
6 + 3 − 3 = 948
3 = 945
= 315


= 2 [2 + ( − 1) ]

315
315 = 2 [2(6) + (315 − 1)3]

= 150255

6

295

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

2b)

3(1 1 [1 (14) ( ) (14) (− 34) ( )2 (14) (− 34) (− 47) ( )3] CHOW CHOON WOOI
1! 2! 3!
+ )4 = 3 + + +

1 3 2 + 7 3 ]
= 3 [1 + −
4 32 128

= 3 + 3 − 9 2 + 21 3
4 32 128

| | < 1
−1 < < 1

11

3(1 + )4 = [34(1 + )]4

1

= (81 + 81 )4

4√80 = 1

804

81 + 81 = 80

1
= − 81

4√80 = 3 + 3 (− 1 − 9 (− 12 + 21 (− 13
4 81) 32 81) 128 81)

11 7
= 3 − 108 − 23326 − 22674816

= 2.9907

7

296

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

3. Find the gradient of the curve cos(4 ) = tan( 2) − 3 at the point where = 0.

SOLUTION CHOW CHOON WOOI

cos(4 ) = tan( 2) − 3

− sin(4 ) (4 ) = 2( 2) ( 2) − 3


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


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


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


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


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

ℎ = 0

cos(4 ) = tan( 2) − 3

cos[4(0) ] = tan[(0) 2] − 3

cos 0 = tan 0 − 3

1 = 0 − 3

1
= − 3

1
ℎ = 0; = − 3

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

8

297

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

4 (− 13) sin [4(0) (− 31)] + (− 13)2 2 [(0) (− 31)2] CHOW CHOON WOOI
= 3 − 4(0) sin [4(0) (− 31)] − 2(0) (− 31) 2 [(0) (− 13)2]

4 (− 31) sin 0 + (− 13)2 2 0
3
=

4 (− 13) sin 0 + (− 31)2 ( 1 20)
3
=

1
= 27

9

298

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

4. The parametric equations of a curve are = + 2 and = 2 − 4, where ≠ 0. Show that



= 2 + 28−2. Hence, find 2 in term of . CHOW CHOON WOOI
2

SOLUTION

= + 2 = + 2 −1 = 2 − 4 = 2 − 4 −1


= 1 − 2 −2 = 2 + 4 −2


2 4
= 1 − 2 = 2 + 2

2 − 2 2 2 + 4
= 2 = 2

dy dy
dx = dt .

2 2 + 4 1
= ( 2 ) . ( 2 −2 2)

2 2 + 4 2
= ( 2 ) . 2 − 2

2 2 + 4
= 2 − 2

2
t 2 − 2 2t 2 + 4

2t 2 − 4

8

299 10

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

8 CHOW CHOON WOOI
= 2 + 2 − 2

2
2 = [ ] .

2 = [2 + 8( 2 − 2)−1] . 2 2)
2 ( 2 −

= [0 − 8( 2 − 2)−2 ( 2 − 2)] . ( 2 )
2
2 −

8 2
= [( 2 2)2 (2 )] . ( )
− 2 −2

16 2
= [( 2 − 2)2] . ( 2 − 2)

16 3
= ( 2 − 2)3

11

300