PSPM 1 Q&A

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

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. CHOW CHOON WOOI

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.

SOLUTION

r r
tan 60° = ℎ
h h
60° 60°
√3 = ℎ

= √3ℎ

a)

dv
dt = 25

ℎ ℎ
= .

= ℎ (25 )


∗∗ .

= 1 2ℎ
3

301 12

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

12 CHOW CHOON WOOI
= 3 (√3ℎ) ℎ

= 1 (3ℎ2)ℎ
3

= ℎ3

= 3 ℎ2
ℎ

ℎ ℎ
= ( ) .

1
= (3 ℎ2) . (25 )

25
= 3 ℎ2

25
= 3ℎ2

b) ℎ = 5, ℎ



dv
dt = 25


= .

ℎ3 = (25 ). (5)

ℎ3 = 125

ℎ=5

ℎ 25
= 3ℎ2

25
= 3(5)2

1
= 3 /

302 13

2018/2019 SM015/1
MATRICULATION PROGRAMME EXAMINATION

c) ℎ ℎ = 20

= ℎ3 CHOW CHOON WOOI

= (20)3

= 8000


= .

8000 = (25 ).

= 320

14

303

SM015/2 CHOW CHOON WOOI
MATEMATIK

2018/2019

Matriculation Programme Examination

304

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

1. Given 1 = 2 + 3 and 2 = 4 − 4 . Express ( 2) + [( 3 )] in Cartesian form.
(̅ ̅ 1̅)
− 2 CHOW CHOON WOOI

2. Solve

a. (12275)2 x (295)4 = (295) −3 x (68215)2

b. 1 ≥ 8
4−2

3. a) The first three terms of a geometric series are (3 − 27) , (3 − 2) 6.

Determine the value of . Hence, find the seventh term of this series.

b) Expand (3 2 − 3

2 1)

1 34

4. a) Given the matrix [ + 2 3 2] such that 11 = 7 and 12 = −1,
4 + 9

calculate the values of a and b.

134 7 13 −6 1

b) Let = [1 3 2] , = [−1 −7 2 ] = (1)

4 10 9 −2 2 0 2

i. Find determinant of by expanding first column.

ii. Evaluate ( 2 − ) .

5. a) Given ( ) = (5 4 + 1) and g( ) = √ +2−1−42. Find
i. the domain of g(x).

ii. ℎ( ), ( ℎ)( ) =

b) Given ( ) = (3 + 6) and ( ) = − 2. Show that ( ) ( ) are
3

inverses of each other.

2

305

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

6. The polynomial ( ) = 4 + 3 − 7 2 − 4 + has a factor ( + 3) and

remainder 60 when divided by ( − 3). Find the values of and . Hence, factorise CHOW CHOON WOOI

( ) completely.

7. a) Express 12 cos + 7 sin in the form of Rcos( − ), where > 0 and

0° ≤ ≤ 90°

b) Hence, show that the maximum value of 12 1 sin +15 is 1 (15 + √193).
cos +7 32

8. The function ( ) is defined by

2 , ≤ 2
− 2 , 2 < ≤ 8

( ) = √2 − 2 , > 8
|8 − |

{ − 8

Find

a) lim ( )

→2+

b) lim ( )

→8+

9. a) Find the derivative of ( ) = 6 using the first principle.
√

b) Find the value of when = 0 for each of the following:



i. = ln(9 − 2 )

ii. = −3

√3 +1

10. Given ( ) = 32 + 9, where > 0. Find the coordinates of the stationary point and

state it’s nature.

3

306

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

1. Given 1 = 2 + 3 and 2 = 4 − 4 . Express ( 2) + [( 3 )] in Cartesian form.
(̅ ̅ 1̅)
− 2 CHOW CHOON WOOI

SOLUTION

1 = 2 + 3

2 = 4 − 4

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

4 − 4 3
= 2 − 3 − (4 − 4 )

(4 − 4 )(4 − 4 ) − 3(2 − 3 ) = √−1
= (2 − 3 )(4 − 4 ) 2 = −1
3 = −
16 − 16 − 16 + 16 2 − 2 3 + 3 4 4 = 1
= 8 − 8 − 12 + 12 2

16 − 16 − 16 + 16(−1) − 2(− ) + 3(1)
= 8 − 20 + 12(−1)

16 − 16 − 16 − 16 + 2 + 3
= 8 − 20 − 12

3 − 30
= −4 − 20

(3 − 30 ) (−4 + 20 )
= (−4 − 20 ) . (−4 + 20 )

−12 + 60 + 120 − 600 2
= 16 − 80 + 80 − 400 2

−12 + 60 + 120 − 600(−1)
= 16 − 80 + 80 − 400(−1)

588 + 180
= 416

4

307

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION
CHOW CHOON WOOI
588 180
= 416 + 416

147 45
= 104 + 104

5

308

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

2. Solve

a. (12275)2 x (295)4 = (295) −3 x (68215)2 CHOW CHOON WOOI

b. 1 ≥8

4−2

SOLUTION

a. (12275)2 x (295)4 = (295) −3 x (68215)2

(295)4 = (68215)2 −
(295) −3 (12275)2 ( ) = ( )

(295)4 [(35)4]2 9 −3 25 3−
(295)3− [(35)3]2 (25) = ( 9 )
=

[(53)2]4 = (35)8
[(35)2]3− (53)6

(53)8 = (53)8 = −
(53)6−2 (35)−6

5 8 −(6−2 ) 5 8−(−6)
(3) = (3)

5 10 −6 5 14
(3) = (3)

6

309

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

10 − 6 = 14 CHOW CHOON WOOI
= 2

b. 1 ≥8

4−2

18
4 − 2 − ≥ 0

− 8(4 − 2 )
(4 − 2 ) ≥ 0

− 32 + 16
(4 − 2 ) ≥ 0

17 − 32
(4 − 2 ) ≥ 0

:

= 32 = 0 = 2

17 (−∞, 0) 32 32 (2, ∞)
(0, 17) (17 , 2) +
-
+ +
17 − 32 - - +
+ -
(4 − 2 ) + +
○+
- +

17 − 32 ○+ -
(4 − 2 )

32
: { : < 0 ∪ 17 ≤ < 2}

7

310

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

3. a) The first three terms of a geometric series are (3 − 27) , (3 − 2) 6.

Determine the value of . Hence, find the seventh term of this series. CHOW CHOON WOOI

b) Expand (32 2 − 3

1)

SOLUTION

a) Geometric series

(3 − 7 , (3 − 2) 6
2)

2 = 3
1 2

3 − 2 6
=
7 3 − 2
3 − 2

(3 − 2)2 = 6 (3 − 7
2)

9 2 − 12 + 4 = 18 − 21

9 2 − 30 + 25 = 0

(3 − 5)(3 − 5) = 0

5
= 3

7
= 3 − 2

57
= 3 (3) − 2

8

311

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

3 CHOW CHOON WOOI
=2

6
= 3 − 2

6
= 3 (53) − 2
=2

7 = 6

= 3 (2)6
(2)

= 96

b) (3 2 − 3 = (03) (3 2 3 (−1)0 + (13) (3 2 2 (−1)1 + (32) (3 2 1 (−1)2 + (33) (3 2 0 (−1)3
2 2 2 2
2 1) ) ) ) )

27 6 9 4 3 2
= (1) ( 8 ) (1) + (3) ( 4 ) (−1) + (3) ( 2 ) (1) + (1)(1)(−1)

= 27 6 − 27 4 + 9 2 − 1
8 4 2

9

312

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

1 34 CHOW CHOON WOOI

i. a) Given the matrix [ + 2 3 2] such that 11 = 7 and 12 = −1,
4 + 9

calculate the values of a and b.

134 7 13 −6 1

b) Let = [1 3 2] , = [−1 −7 2 ] = (1)

4 10 9 −2 2 0 2

i. Find determinant of by expanding first column.

ii. Evaluate ( 2 − ) .

SOLUTION

1 34
a) [ + 2 3 2]

4 + 9

11 = | 3 29|
+

= 27 − 2 − 2

27 − 2 − 2 = 7 ………………………….. (1)
2 + 2 = 20
+ = 10

12 = (−1)1+2 | + 2 92|
4

= (−1)[9 + 18 − 8]

10

313

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

= −9 − 18 + 8

−9 − 18 + 8 = −1 CHOW CHOON WOOI

9 + 18 = 9

+ 2 = 1 ………………………….. (2)

(2) − (1)

= 1 − 10 = −9

+ (−9) = 10

= 19

∴ = 19, = −9

bi)

134 7 13 −6 1

= [1 3 2] , = [−1 −7 2 ] = (1)

4 10 9 −2 2 0 2

| | = (1) |130 92| − (1) |130 94| + (4) |33 42|

= (1)[27 − 20] − (1)[27 − 40] + (4)[6 − 12]

= 7 + 13 − 24

= −4

314 11

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

134 134 7 −1 −2 1
bii) ( 2 − ) = [(1 3 2) (1 3 2) − ( 13 −7 2 )] (1)
CHOW CHOON WOOI
4 10 9 4 10 9 −6 2 0 2

20 52 46 7 −1 −2 1

= [(12 32 28 ) − ( 13 −7 2 )] (1)

50 132 117 −6 2 0 2

13 53 48 1
= [(−1 39 26 )] (1)

56 130 117 2

162
= ( 90 )

420

12

315

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

5. a) Given ( ) = (5 4 + 1) and g( ) = √ +2−1−42. Find CHOW CHOON WOOI

i. The domain of g(x).

ii. ℎ( ), ( ℎ)( ) =

b) Given ( ) = (3 + 6) and ( ) = − 2. Show that ( ) ( ) are
3

inverses of each other.

SOLUTION

( ) = 5 + 1
( 4 )

g( ) = √ +1−2
2−4

5ai) ( )

+ 1 ≥ 0 and 2 − 4 ≠ 0
≠ ±2
≥ −1 and

∴ : [−1,2) ∪ (2, ∞)
5 ) ( ℎ)( ) =

[ℎ( )] =
5ℎ( ) + 1
[ 4ℎ( ) ] =

5ℎ( ) + 1 = 4 ℎ( )

5ℎ( ) − 4 ℎ( ) = −1

ℎ( )[5 − 4 ] = −1

316 13

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

−1 CHOW CHOON WOOI
ℎ( ) = 5 − 4

5b) Given ( ) = (3 + 6) and ( ) = − 2

3


[ ( )] = [3 ( 3 − 2) + 6]

= [ − 6 + 6]
= [ ]
= [ ]
=

(3 +6)
[ ( )] = 3 − 2

3 + 6
= 3 −2

3 + 6 − 6
=3
=
[ ( )] = [ ( )] = , ℎ ( ) ( ) ℎ ℎ

14

317

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

6. The polynomial ( ) = 4 + 3 − 7 2 − 4 + has a factor ( + 3) and CHOW CHOON WOOI
remainder 60 when divided by ( − 3). Find the values of a and b. Hence, factorise
( ) completely.

SOLUTION

( ) = 4 + 3 − 7 2 − 4 +

(−3) = 0

(3) = 60

(−3) = (−3)4 + (−3)3 − 7(−3)2 − 4 (−3) + = 0

81 − 27 − 63 + 12 + = 0

15 − = 18 ……………………. (1)

(3) = (3)4 + (3)3 − 7(3)2 − 4 (3) + = 60

81 + 27 − 63 − 12 + = 60

15 + = 42 ……………………. (2)

(2) − (1)

2 = 24

= 12

15 − 12 = 18

= 2

∴ = 2, = 12

15

318

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

( ) = 4 + 2 3 − 7 2 − 8 + 12 CHOW CHOON WOOI

x3 − x2 − 4x + 4
x + 3 x4 + 2x3 − 7x2 − 8x + 12

x4 + 3x3
− x3 − 7x 2 − 8x + 12
− x3 − 3x2
− 4x2 − 8x + 12
− 4x2 −12x
4x + 12
4x + 12
0

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

= ( + 3)( − 1)( + 2)( − 2)

16

319

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

7. a) Express 12 cos + 7 sin in the form of Rcos( − ), where > 0 and

0° ≤ ≤ 90° CHOW CHOON WOOI

b) Hence, show that the maximum value of 12 1 sin +15 is 1 (15 + √193).
cos +7 32

SOLUTION

7a) 12 cos + 7 sin = Rcos( − )

12 cos + 7 sin = [cos cos + sin sin ]

12 cos + 7 sin = Rcos cos + Rsin sin

cos = 12 ……………………… (1)

sin = 7 ……………………… (2)

(1)2 + (2)2

2 2 + 2 2 = 122 + 72

2( 2 + 2 ) = 193c

2(1) = 193

= √193
(2) + (1)

sin 7
cos = 12

7
tan = 12
= 30.3°

12 cos + 7 sin = √193 cos( − 30.3°)

320 17

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

7b) 1 +15 = 1
12 cos +7 sin (√193 cos( −30.3°))+15

−1 ≤ cos( − 30.3°) ≤ 1 CHOW CHOON WOOI

−√193 ≤ √193 cos( − 30.3°) ≤ √193

−√193 + 15 ≤ √193 cos( − 30.3°) + 15 ≤ √193 + 15

11 1
≤≤
√193 + 15 √193 cos( − 30.3°) + 15 −√193 + 15

11 1
√193 + 15 ≤ 12 cos + 7 sin + 15 ≤ −√193 + 15

1
ℎ 12 cos + 7 sin + 15

1 = 1 . (15 + √193)
−√193 + 15 (15 − √193) (15 + √193)

= 15 + √193
225 + 15√193 − 15√193 − 193

= 15 + √193
32

1
= 32 (15 + √193)

18

321

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

8. The function ( ) is defined by

2 , ≤ 2 CHOW CHOON WOOI
− 2 , 2 < ≤ 8

( ) = √2 − 2 , > 8
|8 − |

{ − 8

Find

a) lim ( )

→2+

b) lim ( )

→8+

SOLUTION

a) lim ( ) = lim ( −2 )
→2+ →2+ √2 −2

− 2 √2 + 2
= lim ( ) ( )
√2 − 2 √2 + 2
→2+

= lim ( ( − 2)(√2 + 2) )

→2+ 2 + 2√2 − 2√2 − 4

= lim ( ( − 2)(√2 + 2) )

→2+ 2 − 4

= lim ( ( − 2)(√2 + 2) )

→2+ 2( − 2)

= lim [(√2 + 2) ]

→2+ 2

√2(2) + 2
=2
=2

322 19

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

b) |8 − | = {−(88−− ) 8 − ≥ 0 CHOW CHOON WOOI
8 − < 0

= {−(88−− ) ≤ 8
> 8

lim ( ) = lim |8 − |

→8+ →8+ − 8

= lim −(8 − )

→8+ − 8

= lim − 8

→8+ − 8

= lim 1

→8+

=1

20

323

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

9. a) Find the derivative of ( ) = 6 using the first principle.
√

b) Find the value of when = 0 for each of the following: CHOW CHOON WOOI



i. = ln(9 − 2 )

ii. = −3

√3 +1

SOLUTION

a) ( ) = 6

√

6
( + ℎ) =

√ + ℎ

′( ) = lim ( + ℎ) − ( )
ℎ
ℎ→0

6 −6
√ + ℎ √
= lim ℎ

ℎ→0

= lim 6 61 .
( − ) (ℎ)
ℎ→0 √ + ℎ √

= lim (6√ − 6√ + ℎ . 1
√ √ + ℎ ) (ℎ)
ℎ→0

= lim 6(√ − √ + ℎ) . 1
[ ℎ ] (ℎ)
ℎ→0
√ √ +

= lim 6(√ − √ + ℎ)(√ + √ + ℎ) . 1
[ ℎ(√ + √ + ℎ) ] (ℎ)
ℎ→0
√ √ +

= lim [ 6 ( + √ √ + ℎ − √ √ + ℎ − ( + ℎ)) 1
] . (ℎ)
ℎ→0 √ √ + ℎ(√ + √ + ℎ)

324 21

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

= lim [ 6(−ℎ) . 1
√ √ ] (ℎ)
ℎ→0 + ℎ(√ + √ + ℎ) CHOW CHOON WOOI

= lim [ −6 ]

ℎ→0 √ √ + ℎ(√ + √ + ℎ)

−6
=

√ √ + 0(√ + √ + 0)

−6
=

√ √ (√ + √ )

−6
=

(2√ )

−3
=3

2

bi) = ln(9 − 2 )

= 9 1 (9 − 2 )
− 2

= 9 −12 (−2)


−2
= 9 − 2

ℎ = 0

−2
= 9 − 0

325 22

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

2 CHOW CHOON WOOI
= −9

bii) = −3 1

√3 +1 = √3 + 1 = (3 + 1)2

= −3 ′ = 1 (3 + 1)−21 (3 + 1)
′ = −3 −3 2

′ − ′ 3
= 2 =1

2(3 + 1)2

(3 + 1 (−3 −3 ) − −3 ( 3 1)

1)2

= 2(3 + 1)2

12
[(3 + 1)2]

ℎ = 0;

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

[(0 + 1)2]

= −3 − 3
1 2

9
= −2

326 23

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

10. Given ( ) = 32+ 9, where > 0. Find the coordinates of the stationary point and CHOW CHOON WOOI
state it’s nature.

SOLUTION

3 = 2 + 9
( ) = 2 + 9
= 3

′ = 3 ′ = 2

′( ) = ′ − ′
2

( 2 + 9)(3) − (3 )(2 )
= ( 2 + 9)2

3 2 + 27 − 6 2
= ( 2 + 9)2

−3 2 + 27
= ( 2 + 9)2

= −3 2 + 27 = ( 2 + 9)2

′ = −9 ′ = 2( 2 + 9) ( 2 + 9)


= 2( 2 + 9)(2 )

= 4 ( 2 + 9)

′ − ′
"( ) = 2

24

327

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

( 2 + 9)2(−9 ) − (−3 2 + 27)4 ( 2 + 9) CHOW CHOON WOOI
= ( 2 + 9)4

−9 ( 2 + 9)2 − 4 (−3 2 + 27)( 2 + 9)
= ( 2 + 9)4

′( ) = 0
−3 2 + 27
( 2 + 9)2 = 0
−3 2 + 27 = 0
3 2 = 27
2 = 9
= ±3
> 0 , = 3

=
3

( ) = 2 + 9
9

=9+9
1

=2
1

(3, 2 )

328 25

2018/2019 SM015/2
MATRICULATION PROGRAMME EXAMINATION

"( ) = −9 ( 2 + 9)2 − 4 (−3 2 + 27)( 2 + 9)

( 2 + 9)4 CHOW CHOON WOOI

−27(9 + 9)2 − 12(−27 + 27)(9 + 9)
= (9 + 9)4

−27(9 + 9)2 − 0
= (9 + 9)4
< 0 ( )

1
∴ (3, 2 )

26

329

2019/2020

330

CHOW CHOON WOOI

1 SESSION 2019/2020 SM015/1

PSPM 1

SM015/1 CHOW CHOON WOOI
PSPM 1

2019/2020

14 OCTOBER 2019

KOLEJ MATRIKULASI KEDAH

1

331

2 SESSION 2019/2020 SM015/1

PSPM 1 [2 marks]
[7 marks]
Questions [5 marks] CHOW CHOON WOOI
[5 marks]
SECTION A [45 marks] [3 marks]
[4 marks]
This section consists of 5 questions. Answer all questions. [2 marks]

1. Given the complex numbers 1 = − and 2 = 2 + √3. [2 marks]
[5 marks]
a. Express 12 and ̅ 2 in the form + , where , ∈ ℛ.
[2 marks]
b. From part 1(a), find = 12+ ̅ ̅2̅. Hence, find | | and argument . [4 marks]
[4 marks]
1

2. Solve the following:

a. 3(52 ) + 2512 +1 = 200
b. + 4 ≤ 2 + < 12

3. The sum of the first terms of a sequence is given by = 2 + 3−4 .

a. Find the value of constant c such that the -th term is 3−4 .
b. Show that the3 sequence is a geometric series.
c. Find the sum of the infinite series, ∞.

2 30
4. Given matrix = [−5 0 4].

0 21
a. Find the determinant of matrix by expanding the first row.
b. Calculate the adjoin of matrix . Hence, find −1.

1
c. Solve the equation = , where = [2], by using the answer obtained

2
in part 4(b).

5. Given ( ) = 1+11+1 1 .
a. Simplify ( ) and evaluate (1).

2

b. The domain of ( ) is a set of real number except three numbers.
Determine the numbers.

2

332

3 SESSION 2019/2020 SM015/1

PSPM 1 [6 marks]
[7 marks]
SECTION B [25 marks] [2 marks] CHOW CHOON WOOI
[7 marks]
1. Solve the following: [3 marks]
a. 22 = 2 4( + 4)

b. 2 | −3 | ≥ 1

2 −1

2. Given a function ( ) = ln (2 + 1)
a. State the domain and range of ( ).
b. Find the inverse function of ( ) and state its domain and range. Hence, find
the value of for which −1( ) = 0.
c. Sketch the graph of ( ) and −1( ) on the same coordinate axes.

END OF QUESTION PAPER

3

333

4 SESSION 2019/2020 SM015/1

PSPM 1

Question A1 CHOW CHOON WOOI

1. Given the complex numbers 1 = − and 2 = 2 + √3.
a. Express 12 and ̅ 2 in the form + , where , ∈ ℛ.
b. From part 1(a), find = 12+̅ ̅2̅. Hence, find | | and argument .

1

SOLUTION

a) = −

= + √
= (− )

= − +
̅ ̅ ̅ = − √

b) = +̅ ̅ ̅



= (− )+( − √ )

(− )

= − √
(− )

= ( − √ )( )

(− )( )

= − √
−

= +√



= √ +

| | = √√ +

4

334

5 SESSION 2019/2020 SM015/1

PSPM 1

= √ + CHOW CHOON WOOI

=
= − ( )

√

=



5

335

6 SESSION 2019/2020 SM015/1

PSPM 1

Question A2 CHOW CHOON WOOI

2. Solve the following:
a. 3(52 ) + 2521 +1 = 200
b. + 4 ≤ 2 + < 12

SOLUTION

a) 3(52 ) + 2521 +1 = 200

( ) + ( ) + =

( ) + ( ) + =

( ) + ( ) ( ) =

( ) + ( ) =

=

+ − =

( + )( − ) =

= − =



= − (ignored)



=

=

b) + 4 ≤ 2 + < 12 And + <
+ ≥ + + − <
+ − − ≥ ( + )( − ) <
− ≥ = − =
( + )( − ) ≥
= − = 6

336

7 SESSION 2019/2020 SM015/1

PSPM 1

(−∞, − ) (− , ) ( , ∞) (−∞, − ) (− , ) ( , ∞)
+
+ - + + + - + +
+
− - - ○+ − - - +

○+ - + ○- CHOW CHOON WOOI

(−∞, − ] ∪ [ , ∞) And (− , )

− −

(− , − ] ∪ [ , )

7

337

8 SESSION 2019/2020 SM015/1

PSPM 1

Question A3 CHOW CHOON WOOI

3. The sum of the first terms of a sequence is given by = 2 + 3−4 .
a. Find the value of constant c such that the -th term is 3−4 .
b. Show that the sequence is a geometric series.
c. Find the sum of the infinite series, ∞.

SOLUTION

a) = 2 + 3−4

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

b) = − ( − )

= − ( − )
− [ − ( − )]
−

= ( − )
[ − + ]
−

= − + −

−

= −

−

8

338

9 SESSION 2019/2020 SM015/1

PSPM 1

=

−

,

−

c) = − ( − ) CHOW CHOON WOOI

= = − ( − ( )) = −


=



∞ =
−

(− )

=

−( )



(− )

=

( )



= −

9

339

10 SESSION 2019/2020 SM015/1

PSPM 1

Question A4 CHOW CHOON WOOI

2 30
4. Given matrix = [−5 0 4].

0 21

a. Find the determinant of matrix by expanding the first row.

b. Calculate the adjoin of matrix . Hence, find −1.

1
c. Solve the equation = , where = [2], by using the answer obtained in

2

part 4(b).

SOLUTION

2 30
a) = [−5 0 4]

0 21

| | = ( ) | | − ( ) |− | + ( ) |− |

= ( )( − ) − ( )(− − ) + ( )|− − |
= − + +

= −

b) Adjoin of matrix

+ | | − |− | + |− |
, = − | | + | | − | |

(+ | | − |− | + |− |)

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

+( − ) −( − ) +( + )

10

340

11 SESSION 2019/2020 SM015/1

PSPM 1

− − CHOW CHOON WOOI
= (− − )

−

=

− −
= ( − )

− −

− =
| |

− −
−
= ( − )
− −

−
= (− − )

−

c) =

= −


[− ] [ ] = [ ]

−

[ ] = (− − ) [ ]
−

+ −
= (− − + )

+ −

11

341

12 SESSION 2019/2020 SM015/1

PSPM 1

− CHOW CHOON WOOI
= ( )

−

∴ = − , = , = −

12

342

13 SESSION 2019/2020 SM015/1

PSPM 1

Question A5 CHOW CHOON WOOI

5. Given ( ) = 1+11+11 .
a. Simplify ( ) and evaluate (1).

2

b. The domain of ( ) is a set of real number except three numbers. Determine

the numbers.

SOLUTION

a) ( ) = 1
1+1+1 1

=
+ +



=

+ +

=
+ +
+

= +

+

( ) = ( )+



( )+

= ( )



=



b) ( ) = 1
1+1+1 1

13

343

14 SESSION 2019/2020 SM015/1

PSPM 1

1 ≠ 0 ➔ ≠ 0



1 + 1 ≠ 0 ➔ ≠ −1



1 + 1 ≠ 0 ➔ 2 +1 ≠ 0 ➔ ≠ − 1 CHOW CHOON WOOI
1+1
2

∴ ≠ 0; ≠ −1; ≠ − 1

2

14

344

15 SESSION 2019/2020 SM015/1

PSPM 1

Question B1 CHOW CHOON WOOI

1. Solve the following:
a. 22 = 2 4( + 4)

b. 2 | −3 | ≥ 1

2 −1

SOLUTION

a) 22 = 2 4( + 4)

= ( + )


= ( + )


= ( + )


= ( + )
( )

= ( + )
= +

=

b) 2 | −3 | ≥ 1

2 −1

| − | ≥

−

− ≥ OR − ≤ −
−
−

− − ≥ − + ≤

− −

( − )−( − ) ≥ ( − )+( − ) ≤
( − ) ( − )

− − + ≥ − + − ≤

− −

15

345

16 SESSION 2019/2020 SM015/1

PSPM 1 − ≤

− ≥ −

− = ; =

− <
<
CHOW CHOON WOOI
(−∞, ) ( , ) ( , ∞)

< - +
- +

− +

+
-

−

+ ○-

OR ( , ]
(−∞, )






(−∞, ) ∪ ( , ]


16

346

17 SESSION 2019/2020 SM015/1

PSPM 1

Question B2 CHOW CHOON WOOI

2. Given a function ( ) = ln (2 + 1)
a. State the domain and range of ( ).
b. Find the inverse function of ( ) and state its domain and range. Hence, find
the value of for which −1( ) = 0.
c. Sketch the graph of ( ) and −1( ) on the same coordinate axes.

SOLUTION

a) ( ) = ln (2 + 1)

:

: + >

> −



: (− , ∞)

:

= (−∞, ∞)

b) ( ) = ln (2 + 1)

[ − ( )] =
[ − ( ) + ] =
− ( ) + =
− ( ) = −
− ( ) = −



:
− = = (−∞, ∞)

:

− = = (− , ∞)



17

347

18 SESSION 2019/2020 SM015/1

PSPM 1 CHOW CHOON WOOI

− ( ) =
− =



− =
=
=
=

c)

18

348

1 SESSION 2019/2020 SM015/2

PSPM 1

SM015/2 CHOW CHOON WOOI
PSPM 1

2019/2020

14 OCTOBER 2019

KOLEJ MATRIKULASI KEDAH

1

349

2 SESSION 2019/2020 SM015/2

PSPM 1 [8 marks]
[5 marks]
Questions [5 marks]

SECTION A [45 marks] [7 mark] CHOW CHOON WOOI
This section consists of 5 questions. Answer all questions. [3 mark]
[5 mark]
1. Polynomial ( ) = 12 3 − 2 − + 8 is divisible by 3 2 − 7 + 4. Find the [7 marks]
values of and . Hence, factorise ( ) completely.

2. a) Show that − cot can be simplified as tan .

2

b) Hence, find the values of ° and °. Give your answer in the form

+ √ , where a, b and c are integers.

3. a) A function ( ) is defined as

− <
− , =
( ) = − ,
− >
{ − ,

Determine whether ( ) is continuous at = .

b) Find −+ − .

→∞

4. a) Given = 2 −3 , find 2 2 . Give your answer in the simplest form.
b) Given = ( + ).

Show that ( + ) + + = ( + ).


2

350