2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 1 of 16

Raffles Institution 2011 Year 6 Term 3 Common Test

H2 Mathematics 9740

SOLUTIONS 1

[3]

2

2 2 2

3

2 3 2 2

f ( ) e

f '( ) 3 e (2 e ) 1 e (3 2 ) 1

x

x x x

x x x

x x x x x x

= +

= + + = + +

For all values of x , we have 22 2 0, e 0, (3 2 ) 0xx x > + > , and so

22 2 e (3 2 ) 0xx x+ .

Hence f '( ) 1 0x > , and so 23f ( ) exx x x= + is strictly increasing for all values of x .

2

(i)

[1]

( )3 3 32

3

2

3

d d 1ln 3 ln(3 )d d 3

1 3 33 31 .3

x x x xx x

xx xx

x x

+ = + +

=+

+=

+

2(ii)

[3]

Using integration by parts, let

3 3 2dln 3 and 1 ,dvu x x xx

= + = +

giving

2

33

d 1 1and .d 3 3u x v x xx x x

+= = +

+

Hence

32 3

233 3 3

3

33 3 2

33 3 3

(1 ) ln 3 d

1 1 1ln 3 d3 3 3

1 1(3 ) ln 3 1 d3 31 1(3 ) ln 3 (3 ) .3 9

x x x x

xx x x x x x xx x

x x x x x x

x x x x x x C

+ +

+ = + + + +

= + + +

= + + + +

3(i)

[2]

2 2( 2) 1 ( 2) 1 2 1y x x y x y= = + = + .

Since 0 1x for Region R , 2 1x y= +

Hence volume of the solid formed when R is rotated through 360 about the -axisy

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 2 of 16

( )( )( )

3 2

023

0

3

0

3

0

d

2 1 d

4 4 1 1 d

5 4 1 d

x y

y y

y y y

y y y

=

= +

= + + +

= + +

3(ii) 2 2 2 55 2 2 52

yy x x y x = = =

Volume generated when S is rotated through 360 about the -axisy

( )5 30 0

3

5 d 5 4 1 d2

651217.0 units (3 s f)

y y y y y

= + +

=

=

4(i) [2]

0.5, 1.53, 0.5

p qr s= =

= =

4(ii)

[3]

( ) ( ) ( )hfg hf f 3x x x= = +

5 (i)

[1]

OF = Length of projection of b onto a

= |a.b|a

(OR: b a )

5(ii) By Cosine Rule, 2 2 2 2( )( ) cosAB OA OB OA OB AOB= +

( )0,0

hfg( )y x=

y

( )3,8

( )3,8

O

x

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 3 of 16

[4] 7 9 4 12cos AOB= + cos 0.5AOB =

So =a b a b cos AOB 3(2)(0.5)= 3=

From (i), =OFa ba

1=

Since is right-angled, 2 2BF OB OF= 22 1= 3=

Area of 1 ( )( )2

AOB BF OA = 1 3(3)2

= 3 3

2= units2

(i) [1]

Required amount = 24u = $415.95. (nearest cents) (from GC)

6(ii)

[5]

10.9 40n nu u = + = 20.9(0.9 40) 40nu + + = 2 20.9 40(0.9) 40nu + + = 2 30.9 (0.9 40) 40(0.9) 40nu + + + = 3 230.9 40(0.9 ) 40(0.9) 40nu + + + = 3 230.9 40(1 0.9 0.9 )nu + + + = ... = 2 10.9 40(1 0.9 0.9 ... 0.9 )n nn nu

+ + + + + (*)

= 01 0.90.9 401 0.9

nn u +

= 0.9 (600) 400(1 0.9 )n n+ = 200(0.9 ) 400n + , i.e. a = 200 and b = 400.

7

[6]

2

2

2

2

2

2 2

2 2

d 3d

d 13 d

d 1 d3

1 ln(3 )2ln(3 ) 23 e3 e

x C

x

yy yxy yy xy y xy

y x C

y x Byy A

+

= +

=+

=+

+ = +

+ = +

+ =

+ =

Given that 1 when 2y x= = ,

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 4 of 16

4

4

2 4 2

2 2 4

3 1 e4e

3 4e e4e 3

x

x

AA

yy

+ =

=

+ =

=

8(a) 5 3i 3 i 15 14i 3 12 14 6 7i= i3 i 3 i 9 1 10 10 5 5

w = = = + +

36 49 17| |25 25 5

w = + = or 3.4

8(b)

[5]

Let iz a b= + where ,a b .

( )( ) ( )

( ) ( )

*

*

(2 i) 3 i

2 2 i i i 3 i

2 2 1 i 3 i

z z

a b a b

a b b a

+ = +

+ + = + +

+ = + +

Compare real and imaginary parts to get

( )

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

solving (1) and (2):4 8 and 3 3

a b b ab a a b

a b

= =

+ = + + =

= =

4 8 i3 3

z = .

x

y

0

2 2 44e 3xy =

32 ln2

+

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 5 of 16

9(i)

[1] ( )

2

2

4 5

2 1 0

x x

x

+

= < for all real values of x.

Explain (for soln)

9(ii)

[3]

( )( )

2

2

2

2

2

2

2

3 7 14 53 7 1 0

4 53 7 4 5 0

4 57 12 04 5

7 12 04 3 0

xx x

xx xx x x

x xx xx x

x xx x

+

+ +

+

+ +

+

+

+

3x or 4x .

Alt solution:

Since ( )22 4 5 2 1 0x x x + = < for all real values of x.

( )

( )( )

2

2

2

3 7 14 5

3 7 1 4 5

7 12 04 3 0

xx x

x x x

x xx x

+ +

+

+

3x or 4x .

9

(iii)

[2]

From (ii), replace x with e x .

0 e 3x < or e 4x

ln 4x or ln 3x

4 3 x

4 3 x

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 6 of 16

10(i)

[1] ( )22h( ) 2 2 1 1x x x x= + = +

Since the horizontal line 5y = , does not cut the graph of h at one and only one point, h is not a one-one function.

10(ii) [1]

1a =

10(iii)

[2]

Let ( )gy x= . ( )

( )

2

2

1 1

1 1

1 1

y x

y x

x y

= +

=

=

Since ( )2,1x , positive is rejected. Now ( )1g 1,10gD R = = ,

( )1g 1 1, 1 10.x x x = < + +

=+ +

=+ +

Thus 2 6,s = giving 6s = as 0.s >

2

2 2 2 2 2 2 2 2

2 9 4 2 5( 30)f '( ) .5 ( 9) ( 4) 5 ( 9) ( 4)s s ss

s s s s

= = + + + + This implies that

s 6 6 6

+

f '( )s 0< 0 0>

f ( )s __

Since 2cos is decreasing for 0,

2

, when f is minimum, 2cos is minimum and

this implies is maximum. Hence the soccer player should shoot when 6.s = Alt soln : graph

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 11 of 16

14(i) [2]

3 0 6 24 1 3 3 12 1 3 1

= =

2 2 21 0 1 41 0 1

= =

r

9 22 1 18 42 1

=

. Therefore A does not lie in the plane 1 .

14(ii) [4]

Let the foot of the perpendicular from A to plane 2 be N

Vector equation of the line AN is r = 9 12 5 ,2 1

t t +

9 1 12 5 5 52 1 1

t + =

(9 + t) + 5(2 + 5t) + (2 + t) = 5 t = 49

The position vector of N is 771 2

9 22

14(ii) [1] The shortest distance is

9 7712 292 22

=4 3

3

14(iii) [2] The position vector of the reflection of A in the plane 2 is

9 77 912 2 2 292 22 2

+

= 731 22

9 26

(using ratio theorem)

14(iv) [2]

21 4 2 41

x y z = + + =

r and 15 5 5 51

x y z = + + =

r

Using GC , r

5 43 92 13 90 1

= +

14(v) [3]

Since 1 , 2 and 3 meets in a line, then we have

0 4 2 42 1 1 0 0 1 02 0 9 9

m

m

= =

Solving, m = 8 9

2011 Year 6 Term 3 H2 Mathematics Common Test Solution Page 12 of 16

OR 4 49 90 21 12 1 0 0 09 92 0 1 1

m

m

= =

Solving, m = 8 9