PSPM 2 Q&A

PSPM 2 PSPM 2 CHOW
2014 - 2019 (Q&A)

MATRICULATION

MATHEMATICS

CHOW CHOON WOOI
KEDAH
MATRICULATION
COLLEGE

TABLE OF CONTENTS Page CHOW
2
Year
2014/2015 39
2015/2016 86
2016/2017 136
2017/2018 190
2018/2019 222
2019/2020

1

PAPER 1 CHOW
SEMESTER 2
2014/2015

2

PSPM I QS 025/1 Session
2014/2015

1. Use the trapezoidal rule to estimate ∫01 ( ) dx from the data given below:
0.00 0.25 0.5 0.75 1.00

( ) 2.4 2.6 2.9 3.2 3.6

2. Given a parabola with vertex (−2, 1), opening to the right and passes through the point (3, 6).

Find the equation of the parabola and determine its focus.

3. Evaluate the following integrals:

a. ∫ sin 6 cos 4
b. ∫(3 tan + 4)5 2

4. Use the Newton-Raphson method with initial approximation 1 = 1 to find 6√2 on [0, 2] correct
to three decimal places.

5. Find the equation of a circle 2 + y2 + 2gx + 2fy + c = 0 which passes through the points

(0, 1), (3, −2) and (−1, −4). Hence, determine its center and radius. Find the points of CHOW

intersection of the circle with the y-axis.

6. Given that ( ) = 1 and ( ) = 4 .
+3

a. On the same axes, sketch the graphs of f and g for the values of x between x = 0 and x =

2. Shade the region R bounded by f, g, x = 0 and x = 2.

b. Find the area of region R.

c. Find the volume of the solid generated when the region R is rotated through 2 radian

about the x-axis.

7. (a) The amount Q(t) of radioactive substance present at time t in a reaction is given by the

differential equation


= −kQ
where k is a positive constant. If the initial amount of the substance is 100mg and is

decrease to 97 mg in 6 days, determine

i. The half-life of the substance

ii. The amount of radioactive substance present after 30 days.

b. Find the general solution to the differential equation

(1 + ) dy − y = 1 + x.
dx

8. Given two straight lines,

1: t = −1 = +2 = and 2: t = +2 = = −−74.
−3 8 −3 10 10

Page 2

3

PSPM I QS 025/1 Session
2014/2015

a. Show that L1 and L2 are not parallel and find the acute angle between the two straight
lines.

b. Determine intersection poin between L1 and plane Π: 2 − + 5 + 25 = 0.

c. Find an equation of the plane containing L1 and L2.

9. (a) Find the values of A, B, C and D if 2+9 = + + ( − 3).
2( −3) 2

(b) Hence, evaluate ∫−12 2+9 dx.
2( −3)

10. Given P, Q and R are three points in a space where

⃗⃗ ⃗⃗ ⃗ = = 3 − + , ⃗⃗ ⃗⃗ ⃗ = = 2 + − 3

and the coordinates of R is (3, 0, 1).

a. Hence, show that

i. a and b are not perpendicular. CHOW
ii. | |2 = | |2| |2 − ( . )2
b. Find the area of triangle PQR.

c. Find the Cartesian equation for the

i. Plane that passes through the points P, Q and R.

ii. Line that passes through the point R and perpendicular to the plane in part (i).

END OF QUESTION PAPER

Page 3

4

PSPM I QS 025/1 Session
2014/2015

1. Use the trapezoidal rule to estimate ∫01 ( ) dx from the data given below:
0.00 0.25 0.5 0.75 1.00

( ) 2.4 2.6 2.9 3.2 3.6

SOLUTION ( ) 2.4 2.6 CHOW
h  0.25  0.00  0.25 0 2.9
1 3.6 3.2
2 6.0
0 = 0.00 3 8.7
1 = 0.00 4
2 = 0.00
3 = 0.00 Total
4 = 0.00

b f (x) dx  h [( y0  yn )  2( y1  y2  yn1)]
a 2

 1 f (x) dx  0.25 [6.0  2(8.7)]

02
 2.925

Page 4

5

PSPM I QS 025/1 Session
2014/2015

2. Given a parabola with vertex (−2, 1), opening to the right and passes through the point (3, 6).
Find the equation of the parabola and determine its focus.

SOLUTION CHOW

Open to the right

( y  k)2  4 p(x  h)

V (h , k)  V (2 , 1)
h  2 , k  1
( y 1)2  4 p(x  (2))
( y 1)2  4 p(x  2)

At (3, 6) :
(6 1)2  4 p(3  2)
25  20 p
p5
4

( y 1)2  4 5 (x  2)
4

( y 1)2  5(x  2)

F (h  p , k)  F  2  5 , 1
 4

 F  3 , 1
4 

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6

PSPM I QS 025/1 Session
2014/2015
3. Evaluate the following integrals:
a. ∫ sin 6 cos 4 u  3 tan x  4
b. ∫(3 tan + 4)5 2 du  3sec2 x dx

SOLUTION

a)

 sin 6x cos 4x dx   1 [sin( 6x  4x)  sin( 6x  4x)] dx
2

 1  sin 10 x  sin 2x dx
2

 1   cos10 x  cos 2x   C
2  10 2

  1 cos10x  1 cos 2x  C
20 4

 b) (3 tan x  4)5 sec2 x dx  u5  du  CHOW
3

 1  u5 du
3

 1  u6   C
3 6

 1 (3tan x  4)6  C
18

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7

PSPM I QS 025/1 Session
2014/2015

4. Use the Newton-Raphson method with initial approximation 1 = 1 to find 6√2 on [0, 2] correct
to three decimal places.

SOLUTION

Let x  6 2
x6  2
x6  2  0
f (x)  x6  2

f (x)  6x5

xn1  xn  f (xn )
f (xn )

xn1  xn  (xn6  2) CHOW
6xn5

x1  1

x2  1  [(1)6  2]  1.1667
6(1)5

x3  1.1667  [(1.1667 )6  2]  1.1264
6(1.1667 )5

x4  1.1264  [(1.1264 )6  2]  1.1225
6(1.1264 )5

x5  1.1225  [(1.1225 )6  2]  1.1225
6(1.1225 )5

x  1.123

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8

PSPM I QS 025/1 Session
2014/2015

5. Find the equation of a circle 2 + y2 + 2gx + 2fy + c = 0 which passes through the points
(0, 1), (3, −2) and (−1, −4). Hence, determine its center and radius. Find the points of
intersection of the circle with the y-axis.

SOLUTION

x2  y 2  2gx  2 fy  c  0

(0, 1) : (0)2  (1)2  2g(0)  2 f (1)  c  0

1 2f c  0 ………………… (1)
2 f  c  1

(3, -2) : (3)2  (2)2  2g(3)  2 f (2)  c  0

13  6g  4 f  c  0
6g  4 f  c  13 ………………… (2)

(-1, -4) : (1)2  (4)2  2g(1)  2 f (4)  c  0

17  2g  8 f  c  0

 2g  8 f  c  17 ………………… (3) CHOW

(2) – (1) : 6g  6 f  12

g  f  2 ………………… (4)

(3) – (1) :  2g 10 f  16

 g  5 f  8 ………………… (5)

(4) + (5) :  6 f  10

f 5
3

(4) : g  5  2
3

g 1
3

(1) : 2 5   c  1
3

c  13
3

x 2  y 2  2  1 x  2 5  y    13   0
 3 3  3 

3x2  3y 2  2x 10 y 13  0

C(g ,  f )  C 1 ,  5 
3 3

r  g2  f 2 c

r    1 2   5 2    13   65  65
 3 3  3  9 3

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9

PSPM I QS 025/1 Session
2014/2015
At y-axis, x  0
3(0)2  3y2  2(0) 10 y 13  0
3y 2  10 y 13  0
(3y  13)( y  1)  0
y   13 or y  1
3

The points of intersection are  0 ,  13  and (0, 1)
 3

CHOW

Page 9

10

PSPM I QS 025/1 Session
2014/2015

6. Given that ( ) = 1 and ( ) = 4 .
+3

a. On the same axes, sketch the graphs of f and g for the values of x between x = 0 and x =

2. Shade the region R bounded by f, g, x = 0 and x = 2.

b. Find the area of region R.

c. Find the volume of the solid generated when the region R is rotated through 2 radian

about the x-axis.

SOLUTION

f (x)  1 , g(x)  x
x3 4

a)

( ) = CHOW


R

( ) =
+
R
2



0

b) x  1
4 x3
x2  3x  4
x2  3x  4  0

(x  4)(x 1)  0

x  4 or x  1
From graph, x  1

A  x2 y dx
x1

1  x dx  2 x  1
 1 1 4 x3 dx

Area 
0 x3 4

  x  3)  x2 1   x2 2
ln( 8   8  ln( x  3)
 0 
1

 (1) 2  ln 0  (2) 2   (1) 2 
ln 8   8 5  8 4
 4    3    ln    ln

 

 0.315 unit 2

Page 10

11

PSPM I QS 025/1 Session
2014/2015

c)

V   x2 y 2 dx
x1

 Volume   1  1 2   x 2 dx   2  x 2   1 2 dx   1 dx   (x  3)2 dx
0  x3 4 1 4  x3
(x  3)2

   1 1  x2 dx   2 x2  1 dx  ( x  3) 1 
 1


0 (x  3)2 16 1 16 (x  3)2 1
x3
   1  x3  1    x3  1  2
   0   3 1
x 3 48   48 x

    1  1     1  0    (2)3  1    1  1 
4 48   3 48 5  48 4

 19  unit3
120

CHOW

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12

PSPM I QS 025/1 Session
2014/2015

7. (a) The amount Q(t) of radioactive substance present at time t in a reaction is given by the

differential equation


= −kQ
where k is a positive constant. If the initial amount of the substance is 100mg and is

decrease to 97 mg in 6 days, determine

i. The half-life of the substance

ii. The amount of radioactive substance present after 30 days.

b. Find the general solution to the differential equation

(1 + ) dy − y = 1 + x.
dx

SOLUTION

a) dQ  kQ CHOW
dt

i)  1 dQ    k dt
Q

ln Q  k t  C

Q  ek tC

Q ekteC

Q  Aek t where A  eC

t  0 , Q  100

100  Aek (0)

A  100
Q  100 ek t

t  6, Q  97

97  100 ek (6)

0.97  e6k

ln( 0.97)   6k

k  0.00508
Q  100 e0.00508t

50  100 e0.00508t

0.5  e 0.00508t
ln( 0.5)  0.00508 t

t  136 days
ii) Q  100 e0.00508(30)

Q  85.86 mg

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PSPM I QS 025/1 Session
2014/2015

b) (1 x) dy  y  1 x
dx

dy  1 y  1
dx 1 x

dy  P(x) y  Q(x)
dx
P(x)   1 , Q(x)  1

1 x

V (x)  e P(x)dx

V (x)  e  1 dx
1 x

 e  ln(1 x)
 eln(1x)1
 (1  x)1

1 CHOW
1 x

V (x) y   V (x)Q(x) dx

 1 x  y    1 x (1) dx
1   1  

  1 x dx
1

y  ln(1 x)  C
1 x

y  (1  x)(ln(1  x)  C)

Page 13

14

PSPM I QS 025/1 Session
2014/2015

8. Given two straight lines,

1: t = −1 = +2 = and 2: t = +2 = = −−74.
−3 8 −3 10 10

a. Show that L1 and L2 are not parallel and find the acute angle between the two straight

lines.

b. Determine intersection poin between L1 and plane Π: 2 − + 5 + 25 = 0.

c. Find an equation of the plane containing L1 and L2.

SOLUTION

L1 : t  x 1  y2  z and L2 : t  x2  y  z4
3 8 3 10 10 7

a) v1  3i  8 j  3k a and b are parallel CHOW
v2  10 i  10 j  7 k  ab  0
i jk
v1  v2   3 8  3
10 10  7
 (56  30)i  (21  30) j  (30  80)k
 26i  51 j 110k
0
L1 and L2 are not parallel

v1  v2  (3i  8 j  3k)  (10 i  10 j  7k)
 30  80  21
 71

v1  (3)2  82  (3)2  82
v 2  10 2  10 2  (7)2  249

v1  v 2  v1 v 2 cos

71  ( 82 )( 249 ) cos
  60.20

b)  : 2x  y  5z  25  0
Parametric equation of line L1
x  1  3t , y  2  8t , z  3t
substitute equation of line into equation of plane
2(1  3t)  (2  8t)  5(3t)  25  0
2  6t  2  8t 15t  25  0

Page 14

15

PSPM I QS 025/1 Session
2014/2015
29  29t  0
 29t  29 CHOW

t 1
x  1  3(1)  2
y  2  8(1)  6
z  3(1)  3
The intersection point is (-2, 6, -3)

c) n  v1  v2
 26i  51 j 110k

a i2j
Equation of plane

rnan

 26x  51y 110 z  26(1)  51(2) 110(0)
 26 x  51y 110 z  76
26 x  51y  110 z  76

Page 15

16

PSPM I QS 025/1 Session
2014/2015

9. (a) Find the values of A, B, and C if 2+9 = + + ( − 3).
2( −3) 2

(b) Hence, evaluate ∫−12 2+9 dx.
2( −3)

SOLUTION

a) x2 9  A  B  C
x2 (x  3) x x2 (x  3)

x2  9  Ax(x  3)  B(x  3)  Cx2

x  0 : 9  B(3)

B  3

x  3 : (3)2  9  C(3)2 CHOW

18  9C
C2

x2 : 1 AC

1 A2
A  1

x2 9   1  3  2
x2 (x  3) x x2 x  3

1 x2  9 dx  1  1  3  2 dx
2 x2 (x  3) 2 x x2 x  3
 b)

  ln x  3  2ln x  3  1
x  2

  ln 1  3  2ln 2    ln 2  3  2ln  5 
1  (2) 

 3.36

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17

PSPM I QS 025/1 Session
2014/2015

10. Given P, Q and R are three points in a space where

⃗ ⃗ ⃗⃗ ⃗ = = 3 − + , ⃗⃗ ⃗⃗ ⃗ = = 2 + − 3

and the coordinates of R is (3, 0, 1).

a. Hence, show that

i. a and b are not perpendicular.

ii. | |2 = | |2| |2 − ( . )2

b. Find the area of triangle PQR.

c. Find the Cartesian equation for the

i. Plane that passes through the points P, Q and R.

ii. Line that passes through the point R and perpendicular to the plane in part (i).

SOLUTION CHOW
a) PQ  a  3i  j  k , PR  b  2i  j  3k , R(3, 0, 1)

i) a  b  (3i  j  k)  (2i  j  3k) a and b are perpendicular
 ab 0
 613
2
0
a and b are not perpendicular

ijk
ii) a  b  3 1 1

2 1 3
 (3 1)i  (9  2) j  (3  2)k
 2i 11 j  5k

a  b  22  112  52  150
a  b 2  150

a  32  (1)2  12  11
a 2  11

b  22  12  (3)2  14
b 2  14
ab  2
(a b)2  22  4
a 2 b 2  ( a b )2  (11)(14)  (4)  150
ab 2  a 2 b 2  (ab)2

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PSPM I QS 025/1 Session
2014/2015
b) PQ  PR  a  b  2i 11 j  5k
PQ  PR  a  b  150 CHOW
Area of triangle PQR  1 PQ  PR
2
 1 150
2
 6.124 unit 2

c i) n  PQ  PR  2i 11 j  5k
R(3, 0, 1)
Cartesian equation of plane
2x  11y  5z  2(3)  11(0)  5(1)
2x  11y  5z  11

ii) v  n  2i 11 j  5k
a1  3i  k
Cartesian equation of line
x 3  y  z 1
2 11 5

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19

PAPER 2 CHOW
SEMESTER 2
2014/2015

20

PSPM I QS 025/2 Session
2014/2015

1. A survey found that 32% of teenage consumers earned their spending money from working
part-time. If five teenagers are selected at random, find the probability that at least two of
them are working part-time.

2. Number of accidents at a particular location of a highway occurs at the rate of 1.6 per week.
Find the probability
a. There will be two accidents in a week
b. There are more than 10 accidents in a five weeks period.

3. Given ( ∩ ′) = 0.25, ( ) = 0.48 ( ) = 0.42. Find ( ∩ ). Is A and B mutually
exclusive events? Hence, determine whether A and B are independent events.

4. The following table shows the frequency distribution of the total time (hours) spent by 60

students in a week for revision: CHOW

Total time (Hours) Number of students

0 to less than 5 7

5 to less than 10 12

10 to less than 15 15

15 to less than 20 13

20 to less than 25 8

25 to less than 30 5

Find the mean, mode and standard deviation.

5. The following data are collected from a number of patients X in a clinic and is represented by the

stem-and-leaf diagram as below:

28 8 9

31 2 3 6 6

40 1 5 7 9

52 3 3 6 6 6 8

60 2 3 5

72 4

80

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PSPM I QS 025/2 Session
2014/2015

Based on the given diagram,
a. Find the mode, median, first and third quartiles.
b. Find the mean and standard deviation given that ∑ = 1335 and ∑ 2 = 71783.
c. Calculate Pearson’s coefficient of skewness and state the skewness of the data
distribution.

6. Seven identical boxes are labelled with numbers 1, 2, 3, 4, 5, 6 and 7. If five boxes are chosen at CHOW
random,
a. Find the number of different ways to arrange the boxes in a row such that
i. There are two odd and three even numbered boxes
ii. There are only one even numbered box.
b. Find the probability that there are only two odd numbered boxes next to each
other.

7. In a college there are 150 students taking courses in Chemistry, Physics and Biology. Among the
students, 92 are females. There are 48 students taking Chemistry which 28 are females. Half of
the 68 students taking Physics are females.
a. Construct the contingency table for the given data.
b. A student is chosen at random. Find the probability that the student
i. Takes Biology
ii. Is a male, given that he takes Biology
iii. Takes Biology or a female.
c. Two students are chosen at random, find the probability at least one student is a
female and takes Biology.

8. An egg is classified as grade A if it weights at least 100 grams. Suppose eggs lay at a particular
farm has the probability of 0.4 being classified as grade A eggs.
a. If 15 eggs are selected at random from the farm, calculate the probability that
more than 20% of them are not grad A eggs.
b. A retailer bought 500 eggs from the farm.
i. Approximate the percentage that the retailer would have bought from 220
to 230 grade A eggs.

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PSPM I QS 025/2 Session
2014/2015

ii. If the probability not more than m of the eggs bought are of grade A is
0.9956, determine the value of m.

9. Let X be the random variable representing the number obtained when a biased dice is rolled.
The probability of the biased dice to give odd numbers is three times higher than even numbers
when it is rolled.
a. If the dice is rolled once,
i. Construct a probability distribution table for X.
ii. Find the probability of getting a number less than 2.
iii. Find the mean and variance of X.
b. If the dice is rolled 100 times, find the expected value of getting the number “6”.

10. The cumulative distribution function of a contimuous random variable, X is given as follows:

0, < 0 CHOW
0 ≤ ≤ 4
( ) = {1 ( + 4),
32 ≥ 4
1,

a. Calculate P(| − 1| < 1).
b. Find the median.
c. Determine the probability density function of X. Hence, evaluate (3 2 − 1).

END OF QUESTIONS PAPER

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PSPM I QS 025/2 Session
2014/2015

1. A survey found that 32% of teenage consumers earned their spending money from working part-
time. If five teenagers are selected at random, find the probability that at least two of them are
working part-time.

SOLUTION

Let X be the number of teenage consumers working part-time

X ~ B(5, 0.32)

P( X  2)  1  [P( X  0)  P( X  1)]

  1  5C0 (0.32)0 (0.68)5 5C1(0.32)1(0.68)4

 0.5125

CHOW

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PSPM I QS 025/2 Session
2014/2015

2. Number of accidents at a particular location of a highway occurs at the rate of 1.6 per week.
Find the probability
a. There will be two accidents in a week
b. There are more than 10 accidents in a five weeks period.

SOLUTION CHOW

a) Let X be the number of accidents in one week

X ~ P0 (1.6)

P( X  2)  P( X  2)  P( X  3)
 0.4751  0.2166
 0.2585

b) Let Y be the number of accidents in five weeks

  51.6  8

Y ~ P0 (8)

P( X  10)  P( X  11)
 0.1841

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PSPM I QS 025/2 Session
2014/2015

3. Given ( ∩ ′) = 0.25, ( ) = 0.48 ( ) = 0.42. Find ( ∩ ). Is A and B mutually
exclusive events? Hence, determine whether A and B are independent events.

SOLUTION

P( A  B)  0.25 , P( A)  0.48 and P(B)  0.42
P( A  B)  P( A)  P( A  B)
0.25  0.48  P( A  B)
P( A  B)  0.23
P(A  B)  0
A and B are not mutually exclusive events

P( A)  P(B)  0.48  0.42 CHOW
 0.2016

P( A  B)  P( A)  P(B)
A and B are not independent events

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PSPM I QS 025/2 Session
2014/2015

4. The following table shows the frequency distribution of the total time (hours) spent by 60

students in a week for revision:

Total time (Hours) Number of students

0 to less than 5 7

5 to less than 10 12

10 to less than 15 15

15 to less than 20 13

20 to less than 25 8

25 to less than 30 5

Find the mean, mode and standard deviation.

SOLUTION CHOW
7
Total Time 12 2.5 17.50 43.75
0 ≤ < 5 15 7.5 90.00 675.00
5 ≤ < 10 13 12.5 187.50 2343.75
10 ≤ < 15 8 17.5 227.50 3981.25
15 ≤ < 20 5 22.5 180.00 4050.00
20 ≤ < 25 27.5 137.50 3781.25
25 ≤ < 30 60
90 840 14875
Total

n  60 ,  fx  840 ,  fx2  14875

Mean, x   fx  840  14
n 60

d1  15 12  3 , d2  15 13  2 , Lk  10 , C  5

Mode  Lk   d1 d1 d C  10   3 3 2 (5)  13
   
2

 fx2   fx 2 14875  (840)2
60  7.266
 Standard deviation, s  n 
60 1
n 1

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PSPM I QS 025/2 Session
2014/2015

CHOW

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PSPM I QS 025/2 Session
2014/2015

5. The following data are collected from a number of patients X in a clinic and is represented by the
stem-and-leaf diagram as below:
Based on the given diagram,
a. Find the mode, median, first and third quartiles.
b. Find the mean and standard deviation given that ∑ = 1335 and ∑ 2 = 71783.
c. Calculate Pearson’s coefficient of skewness and state the skewness of the data
distribution.

SOLUTION

a) Mode  56 CHOW
Median  52
 Q1  36
 Q3  60

b)  x  1335 ,  x2  71783 , n  27

Mean, x   x  1335  49.44
n 27

 x2   x2 71783  (1335 )2
 Standard deviation, s  n 27  14.90
n 1
27 1

c) Pearson coefficient of skewness,

 sk  3(mean  median)  3(49.44  52)  0.515
s tan dard deviation 14.90

Data is skewed to the left

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PSPM I QS 025/2 Session
2014/2015

6. Seven identical boxes are labeled with numbers 1, 2, 3, 4, 5, 6 and 7. If five boxes are chosen at
random,
a. Find the number of different ways to arrange the boxes in a row such that
i. There are two odd and three even numbered boxes
ii. There are only one even numbered box.
b. Find the probability that there are only two odd numbered boxes next to each
other.

SOLUTION

Odd : 1, 3, 5, 7 CHOW
Even : 2, 4, 6

a) i) The different ways 4C23C3  5!  720
ii) The different ways 3C14C4  5!  360

b) Let A be only two odd numbered boxes next to each other

P(A)  4C2 3C3  4! 2! 4
7 P5 35

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PSPM I QS 025/2 Session
2014/2015

7. In a college there are 150 students taking courses in Chemistry, Physics and Biology. Among the
students, 92 are females. There are 48 students taking Chemistry which 28 are females. Half of
the 68 students taking Physics are females.
a. Construct the contingency table for the given data.
b. A student is chosen at random. Find the probability that the student
i. Takes Biology
ii. Is a male, given that he takes Biology
iii. Takes Biology or a female.
c. Two students are chosen at random, find the probability at least one student is a
female and takes Biology.

SOLUTION

a) CHOW
C PBT

M 20 34 4 58
F 28 34 30 92
T 48 68 34 150

b i) P(B)  34  17
150 75

ii) P(M B)  4  2
34 17

iii) P(B  F )  P(B)  P(F )  P(B  F )

 34  92  30
150 150 150

 16
25

c) Let A be at least one student is a female and takes Biology

P( A)  30C1120C1  30C2 120C0  269
150C2 150C2 745

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PSPM I QS 025/2 Session
2014/2015

8. An egg is classified as grade A if it weights at least 100 grams. Suppose eggs lay at a particular
farm has the probability of 0.4 being classified as grade A eggs.
a. If 15 eggs are selected at random from the farm, calculate the probability that
more than 20% of them are not grade A eggs.
b. A retailer bought 500 eggs from the farm.
i. Approximate the percentage that the retailer would have bought from 220
to 230 grade A eggs.
ii. If the probability not more than m of the eggs bought are of grade A is
0.9956, determine the value of m.

SOLUTION

a) Let X be the number of grade A eggs

X ~ B(15 , 0.4) CHOW

20% of 15 eggs = 0.20 15  3 More than 20% are not grade A eggs =
Less than 80% are grade A eggs
80% of 15 eggs = 0.80 x 15 = 12

P( X  12)  1  P( X  12)
 1 0.0019
 0.9981

b) Let Y be the number of grade A eggs in 500

Y ~ B(500 , 0.4)
  np  (500)(0.4)  200

 2  npq  (500)(0.4)(0.6)  120

Y ~ N (200 , 120)

i) P(220  Y  230)  P(219.5  Y  230.5)

 P 219 .5  200  Z  230.5  200 
120 120

 P(1.78  Z  2.78)

 P(Z  1.78)  P(Z  2.78)

 0.0375  0.00272

 0.03478

Percentage  0.03478 100%  3.478%

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PSPM I QS 025/2 Session
2014/2015

ii) P(Y  m)  0.9956

P(Y  m  0.5)  0.9956

P Z  m 0.5  200   0.9956
120

P Z  m  199 .5   0.9956
120

P Z  m  199 .5   0.0044
120

m 199.5  2.62
120

m  228.2

m  228

CHOW

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PSPM I QS 025/2 Session
2014/2015

9. Let X be the random variable representing the number obtained when a biased dice is rolled.
The probability of the biased dice to give odd numbers is three times higher than even numbers
when it is rolled.
a. If the dice is rolled once,
i. Construct a probability distribution table for X.
ii. Find the probability of getting a number less than 2.
iii. Find the mean and variance of X.
b. If the dice is rolled 100 times, find the expected value of getting the number “6”.

SOLUTION 1 23 45 6
3a a 3a a 3a a
a i)

X
P(X  x)

 P(X  x)  1 CHOW

P( X  1)  P( X  2)  P( X  3)  P( X  4)  P( X  5)  P( X  6)  1

3a  a  3a  a  3a  a  1
12a  1
a 1

12

X 1 23456
P(X  x) 1 1 1 1 1 1

4 12 4 12 4 12

ii) P( X  3)  P( X  1)  P( X  2)
1 1
4 12
1
3

iii) E( X )   x P( X  x)

E( X )  1 1   2 1   3 1   4 1   5 1   6 1 
 4  12   4  12   4  12 

 13
4

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PSPM I QS 025/2 Session
2014/2015

E( X 2 )  x 2 P( X  x) CHOW

E( X 2 )  12  1   22  1   32  1   42  1   52  1   62  1 
 4  12   4  12   4  12 

 161
12

Var(X )  E(X 2 )  [E(X )]2
Var( X )  161  13 2

12  4 
 137

48
b) E(X  6)  100  1  25

12 3

Page 16

35

PSPM I QS 025/2 Session
2014/2015

10. The cumulative distribution function of a contimuous random variable, X is given as follows:

0, < 0
0 ≤ ≤ 4
( ) = {1 ( + 4),
32 ≥ 4
1,

a. Calculate P(| − 1| < 1).
b. Find the median.
c. Determine the probability density function of X. Hence, evaluate (3 2 − 1).

SOLUTION

0 , x0

F x   1 x( x  4) , 0 x4
 32
 CHOW
 1 , x4

a) P( X 1  1)  P(1  X 1  1)

 P(0  X  2)
 F (2)  F (0)

 1 (2)(2  4)  1 (0)(0  4)
32 32

3
8

b) F (m)  0.5
1 m(m  4)  0.5
32
m 2  4m  16
m2  4m  16  0
m  2.47 or m  6.47

Since 0  m  4

m  2.47

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PSPM I QS 025/2 Session
2014/2015
c) f (x)  d [F(x)]
dx Page 18

x  0, f (x)  d [0]  0 37
0 x4, dx

x  4, f (x)  d 1 (x2  4x)
dx  32

 1 (2x  4)
32

 1 (x  2)
16

f (x)  d [1]  0
dx

( ) = 1 ( + 2), 0 ≤ < 4
{16 ℎ
CHOW
0,

E(X 2 )   x2 f (x) dx


E( X 2 )  4 x 2 1 (x  2) dx
0 16

 1 4 x3  2x2 dx

16 0

 1 x4  2x3 4
16  
 4 3 0

1  (4) 4 2  0 0
16 4 3
  (4) 3   

 20
3

E(aX  b)  aE(X )  b

E(3X 2 1)  3E(X 2 ) 1
 3 20  1
3

PSPM I QS 025/2 Session
2014/2015

 19

CHOW

Page 19

38

PAPER 1 CHOW
SEMESTER 2
2015/2016

39

PSPM I QS 025/1 Session 2015/2016

1. Find the equation of a circle that is passing through points (1, 2), (-1, 2) and (0, -1). Hence,
determine its center.

2. Show that ∫1 ln = 1 (1 + ).
4

3. Find y in terms of x given that = (1 − 2 2)y where > 0 and y = 1 when x = 1.


4. Find the general solution of the differential equation + y cot = 2 sin .


5. Express 1−4 in partial fractions and hence, find the exact value of ∫01 1−4 . CHOW
3+ −2 2 3+ −2 2

6. (a) Given 1( ) = 2x and 2( ) = − ln .
i. Without using curve sketching, show that = 1( ) and = 2( ) intersect on
the interval of [0.1, 1].
ii. Use Newton-Raphson’s method to estimate the intersection point of = 1( )
and = 2( ), with the initial value 1 = 1. Iterate until | ( )| < 0.005. Give
your answer correct to three decimal places.

b) By using the tropezoidal rule, find the approximate valur for ∫01 √ + 1 dx when n = 4,
correct to four decimal places.

7. (a) If = 3 − + 2 and = 2 + 2 − , show that

| x |2 = | |2| |2 − ( . )2

(b) Given a triangle ABC with ⃗ ⃗ ⃗⃗ ⃗ = 2 and ⃗⃗ ⃗⃗ ⃗ = 3 . Use the result in part (a), show that the
area of the triangle is 3√| |2| |2 − ( . )2. Hence, deduce the area of the triangle if =
and =
Page 2

40

PSPM I QS 025/1 Session 2015/2016

8. Given a line : = 2 − , = −3 + 4 , = −5 − 3 , and two planes 1: 2x − y + 7z = 53
and 2: 3x + y + z = 1. Find
a) The point of intersection between the line and the plane 1.
b) The acute angle between the line and the plane 1.
c) The acute angle between planes 1 and 2.

9. (a) Find the equation in standard form of an ellipse which passes through the point (-1, 6)
and having foci at (-5, 2) and (3, 2).

(b) From the result obtained in part (a), sketch the graph of the ellipse.

10. (a) Sketch and shade the region R bounded by the curve = √ , line = 2 − and y – CHOW
axis. Hence, find the area of the region R.

(b) If 1 is a region bounded by the curve = √ , line = 2 − and -axis, deduce the
ratio of : 1.

(c) Find the volume of the solid generated when the region R is rotated through 3600
about the -axis.

END OF QUESTION PAPER

Page 3

41

PSPM I QS 025/1 Session 2015/2016

1. Find the equation of a circle that is passing through points (1, 2), (-1, 2) and (0, -1). Hence,
determine its center.

SOLUTION

General Equation of circle:

2 + 2 + 2 + 2 + = 0

At (1, 2)

12 + 22 + 2 (1) + 2 (2) + = 0

2 + 4 + = −5 …………………………………… (1)

At (-1, 2)

(−1)2 + 22 + 2 (−1) + 2 (2) + = 0 CHOW

−2 + 4 + = −5 …………………………………… (2)

At (0, -1)

02 + (−1)2 + 2 (0) + 2 (−1) + = 0

0 − 2 + = −1 …………………………………… (3)

(1) + (2)

8 + 2 = −10 …………………………………… (4)

(3) x 2

−4 + 2 = −2 …………………………………… (5)

(4) – (5)

12 = −8

−8
= 12

= − 2 …………………………………… (6)
3

Page 4

42

PSPM I QS 025/1 Session 2015/2016

(6) into (5)

2
−4 (− 3) + 2 = −2

8
3 + 2 = −2

8
2 = −2 − 3

14
2 = − 3

= − 7 …………………………………… (7)

3

Substitute (6) & (7) into (2) CHOW

27
−2 + 4 (− 3) − 3 = −5

87
−2 − 3 − 3 = −5

15
−2 − 3 = −5

15
−2 = −5 + 3
−2 = 0

= 0

General Equation of circle:

2 + 2 + 2 + 2 + = 0

2 + 2 + 2(0) + 2 (− 2 − 7 = 0
3) 3

2 + 2 − 4 − 7 = 0
3 3

Page 5

43

PSPM I QS 025/1 Session 2015/2016
Center of circle,

= (− , − )
2

= (−0, − (− 3))
2

= (0, 3)

CHOW

Page 6

44

PSPM I QS 025/1 Session 2015/2016

2. Show that ∫1 ln = 1 (1 + ).
4

SOLUTION

∫ ln =

= ln =
1 ∫ = ∫
=
2
1 = 2
=

∫ = − ∫

2 2 1 CHOW
∫ ln = (ln ) ( 2 ) − ∫ ( 2 ) ( )

2 1
= 2 ln − 2 ∫

2 2
= 2 ln − 4

∫ 2 2
ln = [2 ln − ]
1 4
1

2 2 12 12
= [ 2 (ln ) − 4 ] − [ 2 (ln 1) − 4 ]

2 2 1
= [ 2 − 4 ] − [0 − 4]

2 2 2 1
= [ 4 − 4]+4

2 + 1
=4

= 1 (1 + 2)
4

Page 7

45

PSPM I QS 025/1 Session 2015/2016

3. Find y in terms of x given that = (1 − 2 2)y where > 0 and y = 1 when x = 1.


SOLUTION

= (1 − 2 2)y


(1 − 2 2)
= dx

1 2 2
= ( − ) dx

11
dy = ( − 2 ) dx

11 CHOW
∫ dy = ∫ ( − 2 ) dx

ln = ln − 2 +

ℎ = 1, = 1

ln 1 = ln 1 − 12 +

0 = 0 − 1 +

= 1

Particular Solution

ln = ln − 2 + 1

ln − ln = 1 − 2

= ↔ =

ln = 1 − 2
( )

= 1− 2


= 1− 2

Page 8

46

PSPM I QS 025/1 Session 2015/2016

4. Find the general solution of the differential equation + y cot = 2 sin .


SOLUTION

+ ( ) = ( )
+ y cot = 2 sin

( ) = = cos
sin

( ) = 2 sin

,

( ) = ∫ ( ) ′( )
= ∫(csoins ) ∫ ( ) = ln| ( )|
= ln(sin )
CHOW
= sin

( ) = ∫ ( ) ( )

(sin ) = ∫(sin )(2 sin ) sin2 = 1 − cos 2
sin = 2 ∫ sin2 2

1 − cos 2 cos2 = 1 + cos 2
y sin = 2 ∫ 2 2
y sin = ∫ 1 − cos 2

sin 2
y sin = − 2 +

Page 9

47

PSPM I QS 025/1 Session 2015/2016

5. Express 1−4 in partial fractions and hence, find the exact value of ∫01 1−4 .
3+ −2 2 3+ −2 2

SOLUTION

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

1 − 4
3 + − 2 2 = (3 − 2 ) + (1 + )

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

1 − 4 = (1 + ) + (3 − 2 )

ℎ = −1 CHOW

1 − 4(−1) = (1 − 1) + [3 − 2(−1)]

5 = 5

= 1

3
ℎ = 2

33 3
1 − 4 (2) = (1 + 2) + [3 − 2 (2)]

5
−5 = 2

= −2

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

1 − 4 −2 1 ′( )
∫ 3 + − 2 2 = ∫ (3 − 2 ) + (1 + ) ∫ ( ) = ln| ( )|

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PSPM I QS 025/1 Session 2015/2016

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

= ln(3 − 2 ) (1 + ) ln + ln = ln( )


ln − ln = ln ( )

∫ 1 1 − 4 = [ln(3 − 2 ) (1 + )]10
3 + − 2 2
0

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

= [ln 2] − [ln 3]
2

= ln (3)

CHOW

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