PSPM 2 Q&A

PSPM I QS 025/1 Session 2016/2017

7. (a) Show that equation −4 2 + 5 + 7 = 0 has a root on the interval [−2, 0]. Use
the Newton-Raphson method to find the root of the equation correct to four
decimal places.

(b) Estimate the value of ∫−0 cos using trapezoidal rule with subinterval
. Give your answer correct to four decimal places.

4

SOLUTION CHOW
a) −4 2 + 5 + 7 = 0
( ) = −4 2 + 5 + 7
When x = -2
(−2) = −4(−2)2 + 5(−2) + 7
= −16 − 10 + 7
= −19 < 0
When x = 0
(0) = −4(0)2 + 5(0) + 7
= 7 >0
(−2) < 0 (0) > 0, ℎ
− 4 ^2 + 5 + 7 = 0 ℎ ℎ [−2, 0]

Page 15

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PSPM I QS 025/1 Session 2016/2017

Newton-Raphson Method

( ) = −4 2 + 5 + 7

′( ) = −8 + 5

+1 = − ( )
′( )

−4 2 + 5 + 7
+1 = − −8 + 5

4 2 − 5 − 7
+1 = − 8 − 5

0 = −1 CHOW

1 = −1 − 4(−1)2 − 5(−1) − 7 = −0.8462

8(−1) − 5

2 = −0.8462 − 4(−0.8462)2 − 5(−0.8462) − 7 = −0.8381

8(−0.8462) − 5

3 = −0.8381 − 4(−0.8381)2 − 5(−0.8381) − 7 = −0.8381

8(−0.8381) − 5

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PSPM I QS 025/1 Session 2016/2017

0

b) ∫ cos

−


ℎ=4

( ) =

x First and last ordinate Remaining ordinat

− 3.14159
3
1.66608
−4
2 0 CHOW

−4 -0.55536

0
−4
0 3.14159 1.11072
Total

∫ ℎ [( 0 ) 2( 1 + −1)]
( ) = + + + 2 + ⋯
2

0
∫ cos = 2(4) [(3.14159) + 2(1.11072)]

−


= 2(4) [(3.14159) + 2(1.11072)]

= 2.1061

Page 17

102

PSPM I QS 025/1 Session 2016/2017

8. Given the curve = 4 2 and the line = 6 .
a) Find the intersection points.
b) Sketch the region enclosed by the curve and the line.
c) Calculate the area of the region enclosed by the curve and the line.
d) Calculate the volume of the solid generated when the region is revolved
completely about the y-axis.

SOLUTION

a) = 4 2 ………………. (1)

= 6 ………………. (2)

(2) (1)

6 = 4 2 CHOW

4 2 − 6 = 0

2 (2 − 3) = 0

2 = 0 2 − 3 = 0

= 0 = 3

2

ℎ = 0  = 6(0) = 0  (0, 0)

ℎ = 3  = 6 (3) = 9  (3 , 9)

2 2 2

3
∴ : (0, 0) (2 , 9)

Page 18

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PSPM I QS 025/1 Session 2016/2017
b)
y
= 4 2
R 3
= 6 (0, 0) (2 , 9)

x

3 CHOW

c) , = ∫02 6 − 4 2

3
6 2 4 3 2
=[ 2 − ]
3
0

3
4 3 2
= [3 2 − ]
3
0

= [3 32 − 4 (323)3] − [3(0)2 − 4(0)3
(2) 3]

= [3 9 − 4 (3287)] − [0]
(4)

27 27
=[4 − 6]

= 9 2
4

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PSPM I QS 025/1 Session 2016/2017

d) = 4 2  2 =

4

= 6  =  2 = 2

6 36

, = ∫ 9 2
4 − 36
0

2 3 9
= [4(2) − 36(3)]0

2 3 9
= [ 8 − 108]

0

92 93 02 03
= [( 8 − 108) − ( 8 − 108)]
CHOW
81 729
= [( 8 − 108) − 0]

27
= ( 8 )

= 27 3
8

Page 20

105

PSPM I QS 025/1 Session 2016/2017

9. (a) If the line 1: 〈 , , 〉 = 〈1,1,2〉 + 〈2, −1,3〉 does not intersect with the plane
Π1: + + = 0, show that 2 − + 3 = 0. Hence, find the equation of
plane Π1 if the plane passes through the point (1, 0, 1).

(b) Given the line 2: = 0 + 1, = 0 + 2, = 0 + 3, the plane Π2: −
+ 2 = 0, and a point ( 0, 0, 0) ≠ (0, 0, 0) is on the plane.

(i) If 2 is perpendicular to the plane Π2, show that
〈 1, 2, 3〉 = 2〈−1, 1, −2〉; 2 ≠ 0.

(ii) Give one example of the equation of straight line which satisfy part
9(b)(i)

SOLUTION CHOW

a) A line does not intersect with a plane  line is parallel to the plane 
 . = (See the diagram below)

: = +


Π1: . = .

1: 〈 , , 〉 = 〈1,1,2〉 + 〈2, −1,3〉  = 〈2, −1,3〉

Π1: + + = 0,  = 〈 , , 〉

. =

〈 , , 〉 . 〈2, −1,3〉 =

2 − + 3 = 0

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PSPM I QS 025/1 Session 2016/2017

Π1: + + = 0
2 − + 3 = 0 ……………(1)

Point (1, 0, 1) is on the plane + + = 0

(1) + (0) + (1) = 0

+ = 0

= − ……………(2)

(2) (1)

2 − + 3(− ) = 0

2 − − 3 = 0 CHOW

− − = 0

= − ……………(3)

∴ + + = 0  + (− ) + (− ) = 0

− − = 0

( − − ) = 0

− − = 0

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PSPM I QS 025/1 Session 2016/2017
b) line 2: = 0 + 1, = 0 + 2, = 0 + 3

Π2: − + 2 = 0
i)

Π2: − + 2 = 0

2: = 0 + 1, = 0 + 2, = 0 + 3 CHOW

2 ℎ Π2   = 0
= 〈 1, 2, 3〉
= 〈1, −1,2〉

= 0


| 1 2 3| = 0

1 −1 2

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

[2 2 + 3] − (2 1− 3) + (− 1 − 2) = 0

2 2 + 3 = 0  3 = −2 2
2 1− 3 = 0

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PSPM I QS 025/1 Session 2016/2017

1 + 2 = 0  1 = − 2

∴ 〈 1, 2, 3〉 = 〈− 2, 2, −2 2〉

= 2〈−1,1, −2〉; 2 ≠ 0

ii) = = 〈1, −1,2〉

At the point ( 0, 0, 0) = (−1, 1, 1)

Equation of straight line :

= −1 + (1)  = −1 +

= 1 + (−1)  = 1 − CHOW

= 1 + (2)  = 1 + 2

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PSPM I QS 025/1 Session 2016/2017

10. (a) Show that the expression 4 4+2 2−1 can be written as
(2 −3)2( +1)

+ 2 + + + .
2 −3 (2 −3)2
+1

(b) From part 10(a), determine the values of A, B and C. Hence, solve

∫ 4 4+2 2−1
(2 −3)2( +1)

SOLUTION

a) Improper Fraction  Use long division

4 4 + 2 2 − 1 4 4 + 2 2 − 1 CHOW
(2 − 3)2( + 1) = (4 2 − 12 + 9)( + 1)

4 4 + 2 2 − 1
= 4 3 + 4 2 − 12 2 − 12 + 9 + 9

4 4 + 2 2 − 1
= 4 3 − 8 2 − 3 + 9

x2
4x3  8x2  3x  9 4x4  0x3  2x2  0x 1

4x4  8x3  3x2  9x  0

8 3 + 5 2 − 9 − 1

8 3 − 16 2 − 6 + 18

21 2 − 3 − 19

4 4 + 2 2 − 1 21 2 − 3 − 19
(2 − 3)2( + 1) = + 2 + (2 − 3)2( + 1)

Page 25

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PSPM I QS 025/1 Session 2016/2017


= + 2 + 2 − 3 + (2 − 3)2 + ( + 1)

b)

21 2 − 3 − 19
(2 − 3)2( + 1) = 2 − 3 + (2 − 3)2 + ( + 1)

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

21 2 − 3 − 19 = (2 − 3)( + 1) + ( + 1) + (2 − 3)2

= −

21(−1)2 − 3(−1) − 19 = [2(−1) − 3]2 CHOW

5 = 25

5
= 25

1
= 5


=

32 3 3
21 (2) − 3 (2) − 19 = [(2) + 1]

99 5
21 (4) − 2 − 19 = [2]

99 5
21 (4) − 2 − 19 = 2

95 5
4 = 2

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PSPM I QS 025/1 Session 2016/2017

95 2
= 4 5

19
= 2

=

−19 = [2(0) − 3][(0) + 1] + (0 + 1) + [2(0) − 3]2

−19 = −3 + + 9

19 1 CHOW
−19 = −3 + ( 2 ) + 9 (5)

19 9
−19 = −3 + ( 2 ) + (5)

113
−19 = −3 + 10

113
3 = 10 + 19

303
3 = 10

101
= 10

∴ = 101; = 19; = 1
10 2 5

4 4 + 2 2 − 1 101 19 1
(2 − 3)2( + 1) = + 2 + 10(2 − 3) + 2(2 − 3)2 + 5( + 1)

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PSPM I QS 025/1 Session 2016/2017

4 4 + 2 2 − 1 101 19 1
∫ (2 − 3)2( + 1) = ∫ + 2 + 10(2 − 3) + 2(2 − 3)2 + 5( + 1)

= ∫( + 2) + 101 1 3) + 19 − 3)−2 + 1 1 1)
10 ∫ (2 − 2 ∫(2 5 ∫ ( +

2 101 2 19 (2 − 3)−1 1
= 2 + 2 + (10)2 ∫ (2 − 3) + 2 [ (−1)(2) ] + 5 ln| + 1| +

2 101 19 1 1
= 2 + 2 + 20 ln|2 − 3| − 4 [(2 − 3)] + 5 ln⌈ + 1⌉ +

2 101 19 1
= 2 + 2 + 20 ln|2 − 3| − 4(2 − 3) + 5 ln⌈ + 1⌉ +

CHOW

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113

PAPER 2 CHOW
SEMESTER 2
2016/2017

114

PSPM II QS 025/2 Session 2016/2017
1. Given ( ) = 0.35 and ( ) = 0.45. Calculate

a. ( ∪ ) if events A and B are mutually exclusive.
b. ( ∩ ′) if events A and B are independent.

SOLUTION

a) ( ) = 0.35, ( ) = 0.45

Events A and B are mutually exclusive  ( ∩ ) = 0

( ∪ ) = ( ) + ( )

= 0.35 + 0.45

= 0.8 CHOW

b) ( ) = 0.35, ( ) = 0.45

Events A and B are independent  ( ∩ ) = ( ). ( )

( ∩ ) = ( ). ( )

= (0.35). (0.45) De Morgan Rule
= 0.1575 ( ′ ∪ ′) = ( ∩ )′
( ∩ ′) = ( ) − ( ∩ ) ( ′ ∩ ′) = ( ∪ )′
= 0.35 − 0.1575 ( ∩ ′) = ( ) − ( ∩ )

= 0.1925

Page 2

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PSPM II QS 025/2 Session 2016/2017

ALTERNATIVE 1(b)
( ) = 0.35, ( ) = 0.45
( ′) = 1 − ( )

= 1 − 0.45
= 0.55
( ∩ ′) = ( ). ( ′)

= (0.35). (0.55)
= 0.1925

CHOW

Page 3

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PSPM II QS 025/2 Session 2016/2017

2. The mean survival times (weeks), ̅, of a sample of 20 animals in a clinical trial is 28 with summary
statistics ∑ 2 = 18000.
a. Find the standard deviation correct to three decimal places.
b. It is known that the median is 26, compute Pearson’s Coefficient of Skewness. Comment on
your answer.

SOLUTION CHOW
= 20, ̅ = 28, ∑ 2 = 18000
∑
̅ =
∑
28 = 20
∑ = 20(28)

= 560
a) Standard deviation

= √∑ 2 − 1 (∑ )2
1

−

= √18000 − 1 (560)2
20 20

−1

= √18000 − 1 (560)2
20 20

−1

= √2320
19

= 11.050

Page 4

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PSPM II QS 025/2 Session 2016/2017

b) Median = 26

Pearson’s Coefficient of Skewness

3( − )
=

= 3( ̅ − )



3(28 − 26)
= 11.050

= 0.543

:

CHOW

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PSPM II QS 025/2 Session 2016/2017

3. The table below shows the classification of 200 shirts based on sizes and colours.

Small Medium Large

White 40 35 5

Blue 10 30 15

Black 25 20 20

A shirt is selected randomly. Find the probability that the shirt is

a. Small ini size. CHOW
b. Either blue or white.
c. Medium size given that it is blue.

SOLUTION

a)

White Small Medium Large Total
Blue 40 35 5 80
Black 10 30 15 55
Total 25 20 20 65
65 85 40 200

( ) = 40 + 10 + 25
200

3
=8

= 0.375

Page 6

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PSPM II QS 025/2 Session 2016/2017

White Small Medium Large Total
Blue 40 35 5 80
Black 10 30 15 55
Total 25 20 20 65
65 85 40 200

b) ( ∪ ℎ ) = 10+30+15 + 40+35+5

200 200

135
= 200

27
= 40

= 0.675

White Small Medium Large Total CHOW
Blue 40 35 5 80
Black 10 30 15 55
Total 25 20 20 65
65 85 40 200

c) ( | ) = ( ∩ )

( )

30

= 200
55

200

30
= 55

6
= 11

= 0.545

Page 7

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4. For every class of 40 students, on average there are 4 of them are left-handed. Find the
probability that
a. Exactly 5 students are left-handed in any class.
b. Between 4 and 17 students are left-handed in any two classes.

SOLUTION

a) = 4 ( 40 )

~ (4) − .
−4. 45 ( = ) = !

( = 5) = 5! CHOW
= 0.1563

b) = 8 ( 80 )

~ (8)
(4 < < 17) = ( ≥ 5) − ( ≥ 17)

= 0.9004 − 0.0037

= 0.8967

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PSPM II QS 025/2 Session 2016/2017

5. The following list is the number of car thefts during the year 2013 in 11 particular cities.

110 340 210 300 660 115 135 400 180 145 265

a. Find the median
b. Draw a box-and-whisker plot to represent the data. Hence, state the shape of the distribution

of the data and give your reason.

SOLUTION

110 115 135 145 180 210 265 300 340 400 660

a) = 11, (112+1) = 6

Median = 210 CHOW

b) 1 = 135
Median = 210

3 = 340
= 3 − 1 = 340 − 135 = 205
Lower fence = 1 − 1.5 = 135 − 1.5(205) = −172.5
Upper fence = 3 + 1.5 = 340 + 1.5(205) = 647.5

x

110 135 210 340 400 660

:



Page 9

122

PSPM II QS 025/2 Session 2016/2017

6. (a) A total of 6 students can sit on 10 chairs which are arranged in a row.
i. Find the number of different ways that all the 6 students can sit.
ii. If both seats at the ends are to be seated, find the number of different ways
this can be done.
iii. If 2 particular students do not sit next to each other, find the number of
different ways that all 6 students can sit.

(b) A committee consisting of 2 males and 3 females is to be formed from 5 males and 7
females. Find the number of different ways if

i. A particular female must be in the committee.
ii. 2 particular males cannot be in the committee.

SOLUTION CHOW

ai) 10P6 = 151200
aii) 6P2 x 8P4 = 50400
aiii) 10P6 - (9P5)(2!)=120960
bi) The total number of possible selection = 5C2 x 6P2 = 150
bii) The total number of possible selection = 3C2 x 7P3 = 105

Page 10

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PSPM II QS 025/2 Session 2016/2017

7. The number of times, X, a certain statistics book is borrowed from a library per semester is
modeled as probability distribution function below

( = ) = { (7 − 2 ), = 0, 1, 2, 3
0, ℎ

With k as a constant. Find k.

Hence,

a. Construct a probability distribution table for X.
b. Find ( ≤ 2)
c. Calculate (2 + 3)
d. Find (2 + 3)

SOLUTION CHOW

( = ) = { (7 − 2 ), = 0, 1, 2, 3
0, ℎ

0 1 2 3
( = ) 7 5 3

Probability Distribution Function  ∑ ( = ) = 1
∑ ( = ) = 1
7 + 5 + 3 + = 1
16 = 1
1
= 16

Page 11

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PSPM II QS 025/2 Session 2016/2017
a)

0 1 2 3

( = ) 7 5 3 1
16 16 16 16

b) ( ≤ 2) = ( = 2) + ( = 1) + ( = 0)
357

= 16 + 16 + 16
15

= 16

c) (2 + 3) = 2 ( ) + 3 CHOW

7531
( ) = 0 (16) + 1 (16) + 2 (16) + 3 (16)

14 Properties of expectation
= 16
a) ( ) =
7 b) ( ) = ( )
=8 c) ( + ) = ( ) +
(2 + 3) = 2 ( ) + 3

7
= 2 (8) + 3

19
=4

d) (2 + 3) = 4 ( )
( ) = ( 2) − [ ( )]2
7
( ) = 8

Page 12

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PSPM II QS 025/2 Session 2016/2017

( 2) = 02 7 + 12 5 + 22 3 + 32 1
(16) (16) (16) (16)

26
= 16

13
=8

( ) = ( 2) − [ ( )]2

13 7 2
= 8 − [8]

13 7 2
= 8 − [8]

55 CHOW
= 64

(2 + 3) = 4 ( )

55
= 4 (64)

55
= 16

= 3.4375

Page 13

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PSPM II QS 025/2 Session 2016/2017

8. Let the probability density function of a continuous random variable X be defined by

2
( ) = {18 , − < <

0, ℎ

a. Show that c = 3.
b. Find the cumulative distribution function of X.
c. Hence, find

i. (0 ≤ ≤ 2).
ii. ℎ .

SOLUTION

2 CHOW
( ) = {18 , − < <

0, ℎ

a) ∫−∞∞ ( ) = 1

− 2 ∞
∫ 0 + ∫ +∫ 0 = 1
18
−∞ −

3
[54]− = 1

3 − 3
[54] − [ 54 ] = 1

3 3
54 + 54 = 1
2 3
54 = 1
2 3 = 54
3 = 27

= 3

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PSPM II QS 025/2 Session 2016/2017

b) Cumulative Distribution Function Of X



( ) = ∫ ( )

−∞

2
( ) = {18 , −3 < < 3

0, ℎ

<-3
−3 ≤ < 3
( ) = ∫ 0
≥ 3
−∞

=0

( ) −3 2
= ∫ 0 + ∫
18 CHOW
−∞ −3

3
= 0 + [54]−3

3 −33
= [54] − [ 54 ]

3 27
= 54 + 54

3 1
= 54 + 2

( ) = 1

0, < −3

( ) = { 3 + 1 , −3 ≤ < 3
54 2 ≥ 3
1,

Page 15

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PSPM II QS 025/2 Session 2016/2017
c) i)
(0 ≤ ≤ 2) = (2) − (0)
23 1 03 1

= [54 + 2] − [54 + 2]
4

= 27

c) ii) ( ) = 0.5
3 1 1
54 + 2 = 2 CHOW
3
54 = 0
3 = 0
= 0

Page 16

129

PSPM II QS 025/2 Session 2016/2017

9. The amount of grains packed in a sack is normally distributed with mean weight and standard
deviation 6 kg. Given ( < 24) = 0.1587. The sack is seperated from the others if it weighs less
that 25kg.
a. Find the value of .
b. Hence,
i. Find the probability that a randomly chosen sack has weights of more than 33 kg.
ii. Find the probability that a randomly chosen sack will be separated.
c. A total of 5 sacks are chosen at random, find the probability that
i. All the sacks are to be separated.
ii. At least 4 of the sacks are to be separated.

SOLUTION CHOW

a) = 6

( < 24) = 0.1587 −
=
24 −
( < 6 ) = 0.1587

24− = −1 (From Statiscal table)
6

24 − = −6

= 30

b) i) ( > 33) = ( > 33−630)
= ( > 0.5)

= 0.3085

b) ii) ( < 25) = ( < 25−30)

6

= ( < −0.83)

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PSPM II QS 025/2 Session 2016/2017

= 0.2033

c) ~ (5, 0.2033) ~ ( , )
i. ( = 5) = 5 5(0.2033)5(0.7967)0 ( = ) = ⬚ ( ) ( ) −

= 0.0003473

ii. ( ≥ 4) = 5 4(0.2033)4(0.7967)1 + 5 5(0.2033)5(0.7967)0

= 0.00715

CHOW

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PSPM II QS 025/2 Session 2016/2017

10. A games is conducted by tossing a biased coin 3 times. The coin has probability ( ) = 1 and
3

( ) = 23, where the event in obtaining head is H and the event in obtaining tail is T.

a. Construct a tree diagram and hence, show that the probability of getting one head is 2172.

b. Let X be the number of heads that appears, find the probability distribution of X.

c. Suppose a player wins RM2 each time a tail appears. If Y is the profit,

i. Find the probability distribution of Y.

ii. Calculate ( ) and ( ).

SOLUTION Outcomes Probabilities CHOW
a)
H H,H,H
H
H H,H,T
H,T,H
T H,T,T
T,H,H
H
T,H,T
T,T,H
T

T



H

T T

H

T

H

T T,T,T



(1 ) = ( ∩ ∩ ) + ( ∩ ∩ ) + ( ∩ ∩ )

444

=++

27 27 27
12

=

27

Page 19

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PSPM II QS 025/2 Session 2016/2017
b) − ℎ ℎ .

Outcomes Probabilities X
3
H H,H,H 2
2
H H,H,T 1
H,T,H 2
T H,T,T 1
T,H,H 1
H
T,H,T 0
H T,T,H
T

T



H

T T

H

T

H

T T,T,T CHOW



0 1 2 3

( = ) 8 12 6 1

27 27 27 27

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PSPM II QS 025/2 Session 2016/2017
c) i) −

Outcomes Probabilities Y
0T= 0
H H,H,H 1T= 2
1T= 2
H H,H,T 2T= 4
H,T,H 1T= 2
T H,T,T 2T= 4
T,H,H 2T= 4
H
T,H,T
H T,T,H
T

T



H

T T

H

T

H

T T,T,T CHOW

3T= 6



( = )



c) ii) ( ) = 0 ( 1 ) + 2 ( 6 ) + 4 (12) + 6 ( 8 )

27 27 27 27

12 48 48
= 0 + (27) + 4 (27) + (27)

=4

Page 21

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PSPM II QS 025/2 Session 2016/2017

( 2) = 02 1 + 22 6 + 42 12 + 62 8
(27) (27) (27) (27)

24 192 288
= 0 + (27) + ( 27 ) + ( 27 )

504
= 27

( ) = ( 2) − [ ( )]2

= 504 − [4]2
27

= 2.667

CHOW

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135

PAPER 1 CHOW
SEMESTER 2
2017/2018

136

PSPM 2 QS 025/1 Session 2017/2018



1. Evaluate ∫06 tan 2 22 .
2. Given vectors = 3 − 6 + and = − 4 + 5 where and are contants.

a) Find the values of and if and are parellel.

b) Given = 1, find if and are perpendicular.

3. Given three points P(-3, 2, -1), Q(-2, 4, 5) and R(1, -2, 4). Calculate the area of triangle

PQR.

4. Determine the vertices and foci of the ellipse 25 2 + 4 2 − 250 − 16 + 541 = 0.

Sketch the ellipse and lable the focy, center and vertices.

5. Show that the equation ln + − 4 = 0 has a root between 1 and 3. From the Newton-

Raphson formula, show that iterative equation of the root is +1 = (5−ln ). Hence, if

1+

the initial value is 1 = 2, calculate the root correct to three decimal places.

6. Show that the line 2 − 5 + 4 = 0 does not intersect the circle 2 + 2 + 3 − 2 +

2 = 0. Find centre and radius of the circle. Hence, determine the shortest distance CHOW

between the line and the circle.

7. Given the line : −1 = −3 = −2 and the planes 1: 2 − − 2 = 17 and 2: −4 −
2 −1 −3

3 + 5 = 10.

Find

a) The intersection point between and 1.
b) The acute angle between 1 and 2.
c) The parametric equations of the line that passes through the point (2, −1, 3) and

perpendicular to the plance 2.

8. Express 6 2− +7 in partial fractions. Hence, show that ∫01 6 2− +7 = 1 + ln 2.
(4−3 )(1+ )2 (4−3 )(1+ )2

9. a) Find the general solution of the differential equation = 2 −2 . Give your



answer in the form = ( ).

b) Find the particular solution of the differential equation + = √1 + 2, given
1+ 2

that y = 1 when x = 0.

10. Given the curve 2 = and the line = −2 + 1.
a) Determine the points of intersection between the curve and the line.

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b) Sketch the curve and the line on the same axes. Shade the region bounded by
the curve and the line. Label the points of intersection.

c) Find the area of the region .
d) Calculate the volume of the solid generated when the region is rotated 2

radian about the y-axis.

END OF QUESTION PAPER

CHOW

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1. Evaluate ∫06 tan 2 22 .

SOLUTION


6 sin 2
6 22 = ∫ 22
cos 2
∫ tan 2 0

0


6

= ∫ sin 2 cos 2

0

sin 2 = 2 sin cos
61 1
=∫ 2 sin 2(2 )
sin cos = 2 sin 2
0 1

sin 2 cos 2 = 2 sin 2(2 )
16
= ∫ sin 4
2
0 CHOW

= − 1 [cos
8
4 ]06

1
= − 8 {[cos 4 (6)] − [cos 4(0)]}

1 2
= − 8 [cos 3 − 1]

11
= − 8 [− 2 − 1]

13
= − 8 [− 2]

3
= 16

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2. Given vectors = 3 − 6 + and = − 4 + 5 where and are contants.
a) Find the values of and if and are parellel.
b) Given = 1, find if and are perpendicular.

SOLUTION

= 3 − 6 +

= − 4 + 5

a) and are parellel  = .

=

CHOW
|3 −6 | = 0
−4 5

(−30 + 4 ) − (15 − ) + (−12 + 6 ) = 0 + 0 + 0

−30 + 4 = 0  = 30 = 15

42

−12 + 6 = 0  = 12 = 2

6

b) Given = 1. and are perpendicular . =

. =

(3 − 6 + ) . ( − 4 + 5 ) =

(3)( ) + (−6)(−4) + (1)(5) = 0

3 + 24 + 5 = 0

3 = −29

−29
= 3

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3. Given three points P(-3, 2, -1), Q(-2, 4, 5) and R(1, -2, 4). Calculate the area of triangle
PQR.

SOLUTION

P(-3, 2, -1)  ⃗⃗ ⃗⃗ ⃗ = −3 + 2 −

Q(-2, 4, 5)  ⃗⃗⃗ ⃗⃗ ⃗ = −2 + 4 + 5

R(1, -2, 4)  ⃗⃗ ⃗⃗ ⃗ = − 2 + 4

R

PQ CHOW

= 1 |⃗ ⃗ ⃗⃗ ⃗ ⃗ ⃗ ⃗⃗ ⃗ |
2

⃗⃗ ⃗⃗ ⃗ = ⃗⃗⃗ ⃗⃗ ⃗ − ⃗ ⃗ ⃗⃗ ⃗

= (−2 + 4 + 5 ) − (−3 + 2 − )

= + 2 + 6

⃗ ⃗ ⃗⃗ ⃗ = ⃗⃗ ⃗⃗ ⃗ − ⃗ ⃗ ⃗⃗ ⃗

= ( − 2 + 4 ) − (−3 + 2 − )

= 4 − 4 + 5


⃗⃗ ⃗⃗ ⃗ ⃗ ⃗ ⃗⃗ ⃗ = |1 2 6|

4 −4 5

= (10 + 24) − (5 − 24) + (−4 − 8)

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= 34 + 19 − 12

| ⃗⃗ ⃗⃗ ⃗ ⃗ ⃗ ⃗⃗ ⃗ | = √(34)2 + (19)2 + (−12)2

= √1661

= 1 |⃗ ⃗ ⃗⃗ ⃗ ⃗ ⃗ ⃗⃗ ⃗ |
2

1
= 2 √1661
= 20.38 2

CHOW

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4. Determine the vertices and foci of the ellipse 25 2 + 4 2 − 250 − 16 + 541 = 0.
Sketch the ellipse and lable the focy, center and vertices.

SOLUTION

25 2 + 4 2 − 250 − 16 + 541 = 0

25 2 − 250 + 4 2 − 16 = −541

25( 2 − 10 ) + 4( 2 − 4 ) = −541

25 [ 2 − 10 + −10 2 − −10 2 + 4 [ 2 − 4 + −4 2 − −4 2 = −541
(2) ( 2 )] (2) (2) ]

25[( − 5)2 − (−5)2] + 4[( − 2)2 − (−2)2] = −541 CHOW
25[( − 5)2 − 25] + 4[( − 2)2 − 4] = −541
25( − 5)2 − 625 + 4( − 2)2 − 16 = −541
25( − 5)2 + 4( − 2)2 = −541 + 625 + 16
25( − 5)2 + 4( − 2)2 = 100
25( − 5)2 4( − 2)2 100

100 + 100 = 100
( − 5)2 ( − 2)2

4 + 25 = 1
( − ℎ)2 ( − )2

2 + 2 = 1
ℎ = 5, = 2, = 2, = 5
2 = 2 − 2

= 52 − 22

= 21

= √21

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(ℎ, ) = (5, 2)

, (ℎ, ± ) = (5, 2 ± 5) = (5,7) (5, −3)

, (ℎ, ± ) = (5, 2 ± √21) = (5, 2 + √21) (5, 2 − √21)

( , ) CHOW
( , 2 + √21)
C(5, 2)

( , 2 − √21)
( , − )

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5. Show that the equation ln + − 4 = 0 has a root between 1 and 3. From the Newton-

Raphson formula, show that iterative equation of the root is +1 = (5−ln ). Hence, if

1+

the initial value is 1 = 2, calculate the root correct to three decimal places.

SOLUTION

ln + − 4 = 0

Let

( ) = ln + − 4

(1) = ln(1) + (1) − 4 = −3 < 0

(3) = ln(3) + (3) − 4 = 0.099 > 0

(1) < 0 (3) > 0, ℎ ln + − 4 = 0 ℎ 1 3. CHOW

( ) = ln + − 4

′( ) = 1 + 1


1 +
=

+1 = − ( )
′( )

= − ln + − 4
1 +


= − (ln + − 4 )
1 +

= (1 + ) − (ln + − 4 )
1 +

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= [1 + − (ln + − 4 )]
1+

= [1 + − ln − + 4 ]
1 +

= [5 − ln ]
1 +

1 = 2

2[5 − ln 2]
2 = 1 + 2 = 2.8712

3 = 2.8712[5 − ln 2.8712] = 2.9261 CHOW

1 + 2.8712

4 = 2.9261[5 − ln 2.9261] = 2.9263

1 + 2.9261

∴ = 2.926

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6. Show that the line 2 − 5 + 4 = 0 does not intersect the circle 2 + 2 + 3 − 2 +
2 = 0. Find centre and radius of the circle. Hence, determine the shortest distance
between the line and the circle.

SOLUTION

2 − 5 + 4 = 0

= 5 −4 …………………………… (1)
…………………………… (2)
2

2 + 2 + 3 − 2 + 2 = 0

Substitute (1) into (2)

2 + (5 −4)2 + 3 − 2 (5 −4) + 2 = 0 CHOW

22

2 + 25 2 − 40 + 16 + 3 − (5 − 4) + 2 = 0
4

4 2 + 25 2 − 40 + 16 + 12 − 20 + 16 + 8 = 0

29 2 − 48 + 40 = 0

2 − 4 = 482 − 4(29)(40)

= −2336 < 0

2 − 4 < 0,

ℎ 2 − 5 + 4 = 0 does not intersect the circle 2 + 2 + 3 − 2 + 2 = 0

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

= 3 = −1

2

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: (− , − ) = (− 3 , 1)
2

: = √ 2 + 2 −

= √(32)2 + (−1)2 − 2

= √5
4

= √5
2

2 − 5 + 4 = 0 CHOW

3
(− 2 , 1)

r
dD

= −
| ℎ + + |

=
√( )2 + ( )2
|(−5) (− 3) + (2)(1) + (4)|

=2
√(2)2 + (−5)2

= |227|
√4 + 25

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27
=

2√29

= −

= 27 − √5
2√29 2

= 1.39

CHOW

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