-
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