Aim for A : PYQ Mathematics SM025 /2

PYQ Mathematics
Paper 2
SM025/2
2018/19

REVISION 3

Prepared by : Lim Hwee Cheng

QUESTION 1
(a) Evaluate  x(1 x)8 dx by using a suitable

substitution.

(b) Find the area of the region boundedby the

curve y  x cos x and x  axis between x  0

and x   . Give your answer in term of 

2

Solution 1(a)

u 1 x du  dx x 1 u

 x(1 x)8 dx  (1 u)u8(du)
 u9  u8 du

 u10  u9  C
10 9

 1 x10  1 x9  C

10 9

Solution 1(b)



2 u  x dv  cosx dx

A  x cos x dx du  dx v  sin x

0


2

   x cos x dx  x sin x  sin x dx 2
0

0

 x sin x  cos x02

  sin    cos    0  cos 0
  
 2 2 2

   1 unit2

2 

QUESTION 2

Solve dy  2(x 1) y2  0,
dx

Given that y  1 when x  1.

Express y in term of x.

Solution 2

dy  2(x 1) y2  0  1  x2  2x  C
dx y

dy  2(x 1) y2 when x 1, y 1
dx
 1  12  2  C
dy 1
y2
 2( x 1) dx C2

 1  2( x2  x)  C  1  x2  2x  2
y2 y

y  x2  1  2
2x

QUESTION 3

Use Newton - Raphson method to solvethe
equation ex  x  2  0 correct to four decimal
places by taking x  1as the first approximat ion.

Solution 3

f (x)  ex  x  2 , f '(x)  ex 1

x1  1

x2 1  e1 1  2   1.16395
 
 e1 1 

x3  1.16395   e1.16395 1.16395  2  1.14642
 e1.16395 1 
 

x4  1.14642   e1.14642 1.14642  2  1.14619
 
 e1.14642 1 

x5  1.14619

The solution is x  1.1462

QUESTION 4

An ellipse Ax2  y2  Bx  Cy 1  0 passes

through points 0,1,1,1and 2,1

(a) Find the equation of the ellipse in the standard
form. Hence, statethe centre and vertices of the
ellipse.

(b) Find the foci of the ellipse

(c) Sketch the graph of the ellipse

Solution 4 (a)

At0,1 0 12  0  C(1) 1  0

C  2

At1,1 A  (1)2  B  C(1) 1  0

A  B  4 - - - -(1)

At2,1 4A 12  B(2)  C(1) 1  0

4A  2B  0
2A  B  0 - - - -(2)

(2)  (1) A  4

from (1) B  8

4x2  y2 8x  2y 1 0

4x2 8x  y2  2y 1 0

 4 x 12 1  (y 1)2 11  0

4x 12  ( y 1)2  4
x 12  ( y 1)2 1

4

Center 1,1 Vertices 1,1 2  1,1,1,3

Solution 4 (b)

c2  4 1

c 3

     Foci 1,1 3  1,1 3 , 1,1 3

Solution 4 (c) V 1,3

y  F 1,1 3

C 1,1 x

 F 1,1 3

V 1,1

QUESTION 5

The line L1 and L2 passesthrough the point

R2,4,3and S8,5,9in the direction of

2 i  3 j 4k and i  2 j 3k, respectively

~~ ~ ~~~

(a) State the equations for line L1 and L2

in the vector form. Hence, calculate the acute

angle between the line L1 and L2 .

QUESTION 5

(b) Find the equation of plane containing the line

L1 and the point 7,3,5in the Cartesian form.

(c) Determine whether the line L2 is parallel to the
plane x  5y  3z  5

Solution 5 (a)

L1 : r1  2 i 4 j 3 k  t 2 i 3 j  4 k 
 
~ ~ ~ ~ ~ ~

L2 : r2  8i 5 j 9k s i  2 j 3k 
~ 
~ ~ ~ ~ ~

 2  6 12 

  cos1
  22  (3)2  42  12  (2)2  32  

    cos1 20 14   6.98
29

Solution 5 (b) n  v  AP

P 7,3,5 ~~

 7   2  5  i jk
AP    3   4    7 n 2 3 4

 5   3  8  ~

 5i7 j8k 5 7 8

~~~ n  4i 4 j k

~ ~ ~~

Solution 5 (b)

r  (4 i  4 j k)  (7 i  3 j 5k)  (4 i  4 j k)

~ ~~ ~ ~~ ~ ~~

4x  4y  z  28 12  5

4x  4y  z  21

Solution 5 (c)

n  i 5 j 3k v  i 2 j3k

~~ ~ ~ ~~ ~ ~

n v  (i  5 j 3k)  (i  2 j 3k)

~~ ~ ~ ~ ~ ~ ~

n v  110  9

~~

n v  0

~~

thus,the line L2 is parallel to the plane.

QUESTION 6

TABLE 1shows the frequency distribution of the
diameter of 100 pebbles which are measured to
nearest millimeter (mm)

Diameter (mm) Frequency
10-14 22
15-19 20
20-24 25
25-29 15
30-34 18

QUESTION 6

Calculate the
(a) mean
(b) mode
(c) median

Solution 6

Diameter Frequency x fx C.F
(mm)
22 12 264 22
10-14 20 17 340 42
15-19 25 22 550 67
20-24 15 27 405 82
25-29 18 32 576 100
30-34

 fx  2135

Solution 6

(a) mean  2135  21.35
100

(b) mode 19.5   5 (5)  21.17
 5 10 

(c) median 19.5   50  42 (5)  21.1
 25 

QUESTION 7

A committee of 5 is to be selected from a group
of 7 men and 6 women. How many different
committeescouldbe formedif
(a) there is no women in the committee?
(b) a particular man must be in the committee and

the remaining has equal number of men
and women?
(c) at least 3 men are in the committee?

Solution 7

(a) 7C56C0  21
(b) 1C1 6C2  6C2  225

(c) 7C3 6C2 7C4 6C17C5 6C0
 525 210 21
 756

QUESTION 8

The probabilities of events X and Y are given

as P(X )  3 , PX ' Y   31 and PX Y   2
5 45 25

(a) Show that P(Y )  9
35

(b) Find PX  Y '

Solution 8(a)

P(X )  3 , PX ' Y   31 and PX Y   2
5 45 25

PX ' Y  PX 'Y 
P(Y )
31 P(Y )  P(Y )  P( X  Y )
45

14 P(Y )  2
45 25
P(Y )  2  45  9 ( Shown)

25 14 35

Solution 8(b)

PX  Y '  P( X )  P(Y ')  P( X  Y ')
PX Y '  3 1 9  3  2

5 35 5 25

PX Y '  144

175

QUESTION 9

A discrete random variable X has the probability
distribution function

f (x)  x 1, x  2,3,4
16
kx, x  6,8
0, otherwise

QUESTION 9

(a) Show that k  1
56

(b) Hence, calculate P3  X  8.

(c) Determine the values of E(X ) andVar(X )
Thus,evaluate Var( 3X 1).

Solution 9

(a)  f (x)  1 (b) P(3  X  8)

3  4  5  6k  8k  1  P(X  3)  P(X  4)  P(X  6)
16 16 16 456

14k  1 16 16 56
4  75  0.6696

k  1 (shown) 112
56

Solution 9(c)

E( X )  xf (x)

E(X )  2 3   3 4   4 5   6 6   8 8   233
16  16  16   56   56  56

E(X 2 )  22 3   32 4   42 5   62 6   82 8   21
16  16  16   56   56 

Var( X )  21   233 2  3.69
 56 

Var( 3X 1)  3Var(x)
 3(3.69)  11.07

QUESTION 10

Thenumberof batteriessold at a servicecenter
on any particular day follow a Poisson distribution

with mean

(a) If the probability of selling exactly 4 batteries
divided by theprobability of selling exactly

2 batteries is 225 ,show that   15

12

QUESTION 10

(b) On any particular day,calculate the probability
that theservicecentersells between5 and14
batteries.

(c) Given that the probability of selling less than k

batteries on any particular day is 0.917, find the
valueof k
(d) Find the probability that exactly 40 batteries
are sold in 2 working days.Give your
answer in four decimal places.

Solution 10(b) & (c)

(b) X ~ P (15) (c) P(X  k)  0.917
P(5  X  14) P(X  k) 1 P(X  k)

 P(X  6)  P(X  14) P(X  k)  1 0.917

 0.9972 0.6368 P(X  k)  0.083
 0.3604 fromtable
P(X  21)  0.083

k  21

Solution 10(a)

P(X  4)  225 2  225
P(X  2) 12   15
 15
e 4

4!  225
12
e 2

2!

2  225

12 12

Solution 10(d)

X ~ P (30)

P(X  40)  e30(30)40  0.0139
40 !

Or Refer PoissonTable
P(X  40)  P(X  40)  P(X  41)
P(X  40)  0.0463  0.0323
P(X  40)  0.0140