1

2016 ACJC JC2 H2 Mathematics 9740

Preliminary Examination Paper 1 Markers Report

Qn Solution

1 2

2 50

3

w

w

OR

3 3w or 5

2w

2

2 5 sin0

3

y x

y

,

Since sin 0x for 3

2x

,

2

2 50

3

y

y

From above, 3 3y or 5

2y

0 3y or 5

2y

3 3y or 5

2y or

5

2y

2

Method 1 - Integration

Volume of solid S

2

2

0 1ln

e exdx x dx

e

22

2 0 1ln

e e

x dx x dxe

3

2

2 10

ln3

eex

x dxe

3

2

2 1ln

3

eex dx

e

Method 2 – Volume of Cone

Volume of solid S

2

1Vol of cone ln

e

x dx

22

1

1(1) ( ) ln

3

e

e x dx

2

1

1ln

3

e

e x dx

x

y

0 R

1

e 1

y = ln x

y = x/e

– + – +

33 5

233 5

2

For Int

ernal

Circula

tion

2

2

1

1ln

3

e

e x dx

2

1ln

e

x dx

2

11

ln 2 ln e e

x x x dx

2

1 11

ln 2 ln 1 e ee

x x x x dx

2

1 11

ln 2 ln 2 1 e ee

x x x x dx

2

1 11

ln 2 ln ln1 2e

e ex x e e x

2

1

ln 2 ln ln1 2 1e

x x e e e

2 2 2e e e

2e

Hence, volume of solid S

1( 2)

3e e

12

3

22

3

2(3 )

3

e e

e

e

3 (i)

2 500 000 000 = 1000 000(2)n-1

2n-1

= 2500

n-1 = 2ln

2500ln = 11.2877

n = 12. 2877

His net worth will first exceed 2.5 billion when n = 13

The year 1993 + (13 -1)(1) = 2005 or 1993 + 13 – 1 = 2005

3

(ii)

100 000 + 1000 + 1000(1.5) + 1000(1.52) + .......(15 terms) =

100 000 + 15.1

)15.1(1000 15

= $ 973 787.7808 = $ 973 788

4

(a)

xxxx

xg

sin2

1cos

2

1sin

2

1cos

2

1

1)( =

xx 22 sin

2

1cos

2

1

1=

)sin(cos

1

2221 xx

= x2cos

2

)2

1(

2

2

)2( x

= 221

2

x= 12 )21(2 x

)21(2 2x = 2 + 4x2

m must be sufficiently small for g(x) 2 + ax + bx2

2

1 lnu x

1 12 ln ( )

dux

dx x

1

1

1dv

dx

v x

2 lnu x

2 1du

dx x

2

2

1dv

dx

v x

For Int

ernal

Circula

tion

3

4

(b) (1 – x2)

-1/2 = 1 +

12 2

1x

+

312

2 2 2

2!x

+

3 513

2 2 2 2

3!x

= .....

16

51 64

832

21 xxx

1 2 4 63 512 8 162

3 5 7

1cos (1 ....)

1

1 3 5( )

6 40 112

x dx x x x dxx

x x x x C

When x = 0 , 2

0cos 1 = C

1 3 51 3cos ( ) ( )

6 40 2x x x x

5

1

11

1 1

2 12 1

3 23 2

11

1

1

1

1

1 21

12

33

2 3 3

3 3

3 3

...

3 3

2 123 3

11

33

13 1 3 3 3 1

3

rrr

n nr rr r

r r

n nn n

n nn n

nn

n nn

uu r

u ur

u u

u u

u u

u u nn

un n

u n n n n

6 Rate of change = rate of growth – rate of decrease

Rate of change = rate of earning interest – rate of withdrawal

400005.0 Mdt

dM

11

0.05 4000dM dt

M

1

0.05ln 0.05 4000M t C

ln 0.05 400020 20

t CM

For Int

ernal

Circula

tion

4

20 200.05 4000t C

M e

0.05M – 4000= 20t

Ae where 20C

A e

2080000 20t

M Ae

When t = 0 , M = K

K = 80000 + 20A

20A = K – 80000

Hence 2080000 ( 80000)t

M K e

Money is completely withdrawn if K < 80000

7(i) Let 5sinx

5cosdx

d

225 x dx

225 (5sin ) (5cos )d

225 25sin (5cos )d

225(1 sin )(5cos )d

225(cos )(5cos )d

2(5cos ) d

225 cos d

251 cos 2

2d

25 1sin 2

2 2c

25

sin cos2

c

2125 25

sin2 5 5 5

x x xc

1 225 1sin 25

2 5 2

xx x c

7(ii) 2 2( ) 25x y b

2 2( ) 25y b x

Since 0y , 0y b ,

225y b x

225y b x

θ

x

Let 5sinx

sin5

x

1sin

5

x

21cos 25

5x

80000

K > 80000

K < 80000

M

t

For Int

ernal

Circula

tion

5

Area of region R

2

025

a

b x dx

2

0 0 25

a a

b dx x dx

1 2

00

25 1sin 25

2 5 2

aa x

bx x x

1 225 1sin 25

2 5 2

aab a a

2 2 11 2525 25 sin

2 2 5

aa a a a

2 1 1 21 25 25 125 sin sin 25 .

2 2 5 2 5 2

a aa a a a

8 (i)

8

(ii)

z k i 2 2 2 2 2( ) 2( )( ) ( ) ( 1) (2 )z k i k k i i k k i 3 3 3 2 2 3( ) 3( ) ( ) 3( )( ) ( )z k i k k i k i i

3 2( 3 ) (3 1)k k k i

3 2 2 4 0z iz z i

3 2 2[( 3 ) (3 1) ] [( 1) (2 ) ] 2[ ] 4 0k k k i i k k i k i i 3 2 2[( 3 ) 2 2 ] [(3 1) ( 1) 2 4] 0k k k k i k k

3 2( 3 ) (2 6) 0k k i k 2( 3) 0k k and 22 6 0k

( 0 or 3)k k and 3k

Hence, 3k

3 0z i k

1 3 2z

arg( )6

z

Method 1: By Polar Form & Trigonometry

62 2 cos sin6 6

iz e i

62 2 cos sin6 6

n n in n n nz e i

nz is real sin 06

n

, where 6

nk k

6 , where n k k

Hence, 0, 6, 12, 18, ...n

Substituting x = a

and y = 0 into the

equation

2 2( ) 25x y b ,

we have

2 2

2

(0 ) 25

25

a b

b a

x

y

0

1

For Int

ernal

Circula

tion

6

Method 2: By Properties of arg(z)

arg( ) arg( )6

n nz n z

nz is real, the point representing nz on the Argand diagram is on

the x-axis.

Thus, arg( ) , where 6

n nz k k

6 , where n k k

i.e. 0, 6, 12, 18, ...n

Given 100nz .

2nn nz z

Hence, 2 100n

But n is a multiple of 6. We then have 6

12

2 64 100

2 4096 100

The least value of n is then 12.

9

Let be the statement

1

7 93 1

1 2 4 1 2

n

r

n nr

r r r n n

for

all integers n ≥ 1.

When n = 1, LHS = 1

1

3 1 3 1 2

( 1)( 2) 1(2)(3) 3r

r

r r r

RHS = 7 9 2

4(2)(3) 3

= LHS is true.

Assume that is true for some positive integer k, k ≥ 1,

i.e.

1

7 93 1

1 2 4 1 2

k

r

k kr

r r r k k

Need to prove is true,

i.e.

1

1

( 1) 7 163 1

1 2 4( 2) 3

k

r

k kr

r r r k k

.

LHS of =

1

1

3 1

1 2

k

r

r

r r r

=

1

3 1 13 1

1 2 ( 1)( 2)( 3)

k

r

kr

r r r k k k

= (7 9) 3 4

4( 1)( 2) ( 1)( 2)( 3)

k k k

k k k k k

=3 27 30 39 16

4( 1)( 2)( 3)

k k k

k k k

Pn

1P

Pk

1Pk

1Pk

For Int

ernal

Circula

tion

7

= 2(7 23 16)( 1)

4( 1)( 2)( 3)

k k k

k k k

=

( 1) 7 16

4( 2) 3

k k

k k

= RHS

is true is true.

Since is true, and is true is true, by the Principle of

Mathematical Induction,

1

7 93 1

1 2 4 1 2

n

r

n nr

r r r n n

is

true for all integers n ≥ 1

9 (i)

7 9 7 6 7

4 1 2 4 2 1 2

6 7Since is a positive integer, 0

2 1 2

n n n

n n n n

nn

n n

1

7 93 1 7 6 7 7

1 2 4 1 2 4 2 1 2 4

n

r

n nr n

r r r n n n n

9

(ii)

3 3 2

3 2 3

3

31 1 1

( 1) 3 3 1

( 1)( 2) 3 2 ( 1) ( 1)( 2)

1 1

( 1)( 2)( 1)

3 3 3 1 7< .

( 1)( 2) ( 1)( 2) 4( 1)

n n n

r r r

r r r r

r r r r r r r r r r

Sor r rr

r r r

r r r r r rr

10

2 2

2 2

2 2

2 2

2 2

2 2

4where is a constant such that 0

4

( 4 ) 0

is real discriminant 0

4 4 0

4 16 0

2 12 0

This inequality is true for all values of .

Therefore can take

x ky k k

x k

xy ky x k

x xy ky k

x

y ky k

y ky k

y k k

y

y

the set of all real numbers.

Pk

1Pk

1P Pk 1Pk

For Int

ernal

Circula

tion

8

Alternative Method:

2 2

2

2 2

2

0

2 40

2 4 0

discriminant = 12 0 no real roots. Hence, no turning pts.

.

dy

dx

x xk k

x k

x xk k

k

y

10

(i) 2 2 24 3x k k

y x kx k x k

Asymptotes: y x k and x k

Points of intercept with axes: (0,4 ), ( 2 ,0), (2 ,0)k k k

10

(ii) 2

1 1 1

1 0 0

12

2

0

2 2

2

3f d 2 f 2 d

2 3 ln2

12 3 ln 1 3 ln

2

1 2 6 ln .1

kx x x dx x k x

x k

xkx k x k

k k k k k

kk k

k

y

x O

For Int

ernal

Circula

tion

9

10

(iii)

2

2

2 2

2

22 2

44 2

(scaling)

2

42 162

translation

2

2

2 2

kx k

y yx k

k

kx k k

y

x

x

x

xk

k

x k

11

(i)

4

4

4

4

a b a c

a b a c 0

a b a c 0

a b c 0

These are the three possible conclusion that can be drawn from

the above equation.

1) a = 0

2) 4b c = 0

3) a is parallel to b – 4c , hence b – 4c = a

We will present no. 3) as we are expected to show the given

expression in (i).

(ii)

1126

2

14 126

2

126

2

4 126

23

3 1264

2

378

2

a b

a c

a c

b cc

b c c c

b c

(iii) Area of parallelogram with adjacent sides OB and OC.

(iv) (b – 4c) . (b – 4c) = 3|a|2

|b|2 – 8 b . c + 16|c|

2 = 3|a|

2

b . c = 10

8

0

10

. c 8cos| || | 1(2)

128.7 and not 51.3

b

b c

For Int

ernal

Circula

tion

10

12

(i)

5

5

(2 )5 2 2

0

, where i k

i

z i

z i

z e e k

2( )

5 10

ki

z e

Putting 2, 1,0,1,2n , 7 3 9

10 10 10 102, , , ,i i i ii

z e e e e e

Given 1 2 3 4 5arg( ) arg( ) arg( ) arg( ) arg( )z z z z z .

i.e. 7 3 9

10 10 10 1021 2 3 4 5, , , ,

i i i ii

z e z e z e z e z e

Let the points P1, P2, P3, P4 and P5 on the Argand diagram

represent the complex numbers 1 2 3 4 5, , , , and z z z z z .

.

(ii) The locus 2 3z z z z is a perpendicular bisector of the line

segment P2P3 where 2 2P z and

3 3P z .

Since OP2P3 is an isosceles triangle, the perpendicular cuts

through the origin and bisect the angle P2OP3.

Thus the perpendicular bisector is inclined at angle 10

radian

below the positive x-axis.

Cartesian equation of locus 2 3z z z z is tan( )10

y x

.

x

y

0

P1 P2

P3

P4(0, 1)

P5

1

-1

-1

For Int

ernal

Circula

tion

11

(ii) z OP

1 1z OP

1 1z z OP OP i.e. 1 1z z PP

The least value of 1z z is the shortest distance from P1 to the

perpendicular bisector (line OP5).

1

1

1 1

4sin

10

4 2sin sin 0.951

10 5

PN

OP

PN OP

(iii) The locus 1 4arg( ) arg( )z z z is a half-line with its initial point

at P1 and above and excluding P1, parallel to the y-axis. (Refer to

diagram.)

12

(iv)

The intersection point between the 2 loci has the same x-

coordinates as the point P1, i.e. 7 3

cos cos10 10

x

.

Substituting 3

cos10

x

into equation of the perpendicular

bisector tan( )10

y x

, we have

3 3( cos ) tan cos tan 0.191

10 10 10 10y

3cos 0.588

10x

Coordinates of the intersection point are ( 0.588,0.191) .

x

y

0

P1 P2

P3

P4(0, 1)

P5

1

-1

-1

0

P1

P5 N

For Int

ernal

Circula

tion

12

13

(i)

Let E be the centre of the circle and F be the foot of

perpendicular from the point E to BC.

Since EC and ED are both radii of the circle, 4ED EC k .

Let H be the length of EF, which is the height of the trapezium.

By Pythagora’s theorem,

2 2 2

2 2 2

2 2

(4 )

16

16

H x k

H k x

H k x

2 2

1 1( ) 2 8

2 2

( 4 ) 16 ( )

S H BC AD H x k

x k k x shown

(ii)

12 2 2 2 2

2 2

2 2

2 2

16 4 16

2 4 160

16

2 8 0

2 4 0

2 4 (N.A, 0)

dSk x x k x k x

dx

x kx k

k x

x kx k

x k x k

x k or x k x

2 2 2 2

2 2 2 2

12 2 2 2 2 2 2

2

2 2 2

2 2 3 2 2 2 2 2

2 2 3 2 3 2 2 3

2 4 16 2( 2 8 )

16 16

116 2 2 ( 2 8 ) 16 ( 2 )

2216

2(16 ) 2( )(16 ) ( 2 8 )

2(16 ) 2( 16 16 )

dS x kx k x kx k

dx k x k x

k x x k x kx k k x xd S

dx k x

k x x k k x x x kx k

k x x kx k x k

3 2 2

2 2 3 2 3 2 3

2 32 2 3 2 3 3 3

2 3

( 2 8 )

2(16 ) 24 32

2

2(72 )2(16 4 ) 8 48 32 12 0

12 12

x kx k x

k x x k x k

when x k

d S kk k k k k

dx k

Area of trapezium is maximum when x=2k.

A D

B

E

F C

H 4k

x

For Int

ernal

Circula

tion

13

(iii)

Given 2x k and from (i) we have 2 216H k x .

2 2 2 2

2 2 2

16 16 (2 )

16 4 12 2 3

H k x k k

k k k k

Using similar triangles:

2

22 3 3

y h

k H

h hy k

k

area of trapezium length of container

1 = 4 (4 2 ) (3)

2

3 = 8 2

2

=3 4

=3 4 ( )3

V

h k k y

h k y

h k y

hh k shown

(iv) 1

4 3 33 3

6 12

3

3 , 18

0.2 18

1/

90

dV hk h

dh

hk

dVWhen h k k

dh

dV dV dh

dt dh dt

dhk

dt

dhm s

dt k

2k 2k 2k 2k

2k 2k

2k 2k y y

h

H

h

H y

2k

For Int

ernal

Circula

tion