Suggested Solutions to 2014 SH2 H2 Mathematics Preliminary Examination
Paper 1
1 (i) Since A(–1, 20) lies on the curve,
a(–1)2 + b(–1) + c = 20 ⇒a – b + c = 20 – (1)
Since B(3, 4) lies on the curve,
a(3)2 + b(3) + c = 4 ⇒9a + 3b + c = 4 – (2)
Since B(3, 4) is a stationary point,
2a(3) + b = 0 ⇒ 6a + b = 0 – (3)
a = 1, b = – 6, c = 13
1 (ii) 2 26 13 ( 3) 4= − + = − +y x x x
Hence 2
2
2 2
translate 3 units in the positive -direction
( 3)
translate 4 units in the positive -direction
( 3) 4 6 13
=
↓
= −
↓
= − + = − +
y x
x
y x
y
y x x x
2
( ) ( )2 22 2
2
1 8d d
81 4 1 4
1 1
8 1 4
x xx x
x x
cx
= ⋅+ +
= − + +
⌠⌡
∫
( ) ( )
( ) ( )
( ) ( )
( )
2
2 22 2
2 2
2 2
1
2
4d 4 d
1 4 1 4
1 14 4 d
8 1 4 8 1 4
1d
2 1 4 2 1 4
1tan 2
42 1 4
x xx x x
x x
x xx x
xx
x x
xx c
x
−
= ⋅+ +
= − − − + +
= − ++ +
= − + ++
∫ ∫
∫
∫
3 (i)
( ) ( )
( )
( )
2
2 2
2 22 2
22
2
2
e
e e
,
d (2 )
d
d(1 )
d
(1 )d
d
dAt 0,
d
e
.
e
t
t t t
t
tx y t
t a
x t a t t t a
t t a t a
yt t
t
t t ay
x t a
ayt a
x a
−
− − −
−
= =−
− − − −= =
− −
= − + = −
− −=
− −
−= = = −
−
Since the tangent to the curve C at t = 0 is perpendicular to the line 4y – x = 0,
( )11 4 (Shown)
4a a− = − ⇒ =
3 (ii)
For a = 4, ( )2
2
2
(1 ) 4d e.
d 4
tt ty
x t
− − −=
− −
As 2t → − , ( )22 (3) 4 4d
0.d 4 4
ey
x
−→ =
− −
The gradient of the curve approaches 0 as 2t → − .
3 (iii) Observe that
2
2as 2, , 2 .e e4
ttt x y t
t
−→ − = → ∞ = → −−
Thus, 22ey = − is a horizontal asymptote.
22ey = −
4 (i) Since A, B and P are collinear,
( )1OP λ λ+= −a b����
.
Since M, N and P are collinear,
( ) 12 1
3OP µ µ= + −a b����
.
Comparing, we have
( )
2
11 1
3
λ µ
λ µ
=
− = −
.
Solving, we have 2
5µ = and
4
5λ = .
Thus, ( )1 14
5
4
55OP + == +a b a b����
.
4 (ii) Method 1
Area of OMN1
3
1 12
2 3× ×= =ba a b
Area of APM1 1 1
2 5 1
1
5 0
× − × =
=
ba aa b
Area of the quadrilateral OAPN
1 1 7
3 10 30= × − × ×=a b a b a b
Method 2
Area of OAB1
2= ×a b
Area of BPN1 4 4 4
2 5 5
2
153
× − ×
= =b a b a b
Area of the quadrilateral OAPN
1 4 7
2 15 30= × − × ×=a b a b a b
Method 3
Area of OAP ( )1
1
2
1 14
5 0× + ×= =a ba a b
Area of ONP
( )1 14
3 5
1
2
2
15
= +
×=
×b a
a
b
b
Area of the quadrilateral OAPN
1 2 7
10 15 30= + =× × ×a b a b a b
5 (i) Method 1 2( )
f ( ) , x k
x x kx k
+= ≠
−
2
2
2
2
2( )( ) ( )f ( ) ,
( )
( )(2 2 ),
( 3)
( )( 3 ),
( )
x k x k x kx x k
x k
x k x k x kx k
x
x k x kx k
x k
+ − − +′⇒ = ≠
−
+ − − −= ≠
−
+ −= ≠
−
For f to be increasing, f ( ) 0′ >x
2
( )( 3 )0
( )
( )( 3 ) 0
or 3
x k x k
x k
x k x k
x k x k
+ −>
−
+ − >
< − >
Method 2 2
2 2
2 2 2
2
( )f ( ) ,
2
3 3 4
43
x kx x k
x k
x kx k
x k
x kx kx k k
x k
kx k
x k
+= ≠
−
+ +=
−
− + − +=
−
= + +−
( )
2
2
4f ( ) 1
kx
x k′⇒ = −
−
For f to be increasing, f ( ) 0′ >x
( )
( )
( )
2
2
2
2
22
41 0
41
4
2 or 2
or 3
k
x k
k
x k
k x k
x k k x k k
x k x k
− >−
<−
< −
− < − − >
< − >
5 (ii)
6 (i) y = cos
4x
π
+ 3.5
x = 4
π
cos −1
(y – 3.5)
f −1
(x) = 4
π
cos−1
(x – 3.5),
for x ∈ �, 5 7 2
2 2x
−< ≤
Range of f −1
is or [3, 4) or 3 < x < 4
6 (ii) As range of f −1
is [3, 4) ⊆ domain of g is (2, 4],
gf −1
exists.
6 (iii)
g(3) = 1, g(4) = 1, g(3.5) = 0
Df
−1 → Rf
−1 = [3, 4) → Rgf
−1 = [0, 1]
Range of gf −1
is [0, 1]
7 (i)
7 (ii)
5sin
10
6
θ
πθ
=
=
Thus, ( )arg 6 3i6
zπ
+ − = .
From the right angled triangle,
5 35sin
3 2y
π= =
55cos
3 2x
π= =
Thus the complex number z representing P is 5 5 3
4 3 i2 2
− + +
, i.e.,
3 5 33 i
2 2
+ +
.
×
P
5
y
x
8 (a)
(i)
Total amount of prize fund needed
( ) ( ) ( )
( )
2 19
20
20
4 4 41000 1000 1000 1000
5 5 5
41000 1
5
41
5
45000 1 4942.35 nearest cents
5
= + + + +
− =−
= − =
�
8 (a)(ii) Assume that no two athletes will arrive at the same time.
8 (a) part
after
(ii)
( ) ( )( )
[ ]2
2 15 1 185 1000002
30 185 185 200000
185 155 200000 0
nn
n n
n n
+ − ≤
+ − ≤
− − ≤
By GC, 32.4 33.3n− ≤ ≤ .
Maximum n is 33.
8 (b) Given
1
1
210
5
n
n nS
+
−= − ,
( ) ( )
( )
1
1 1 2
1
2 1
2 210 10
5 5
2 2
5 5
25 2 5 2 .2
5 5
15 2, for 2
5
n n
n n n n n
n n
n n
n n
n n
n
n
T S S
n
+
− − −
+
− −
= − = − − −
= −
= −
= ≥
( )
1 1
1 1 1 1
1
1
210
5
15 26
5
T S+
−= = −
= =
( )15 2, for all
5
n
n nT n +∴ = ∈Z
( ) ( )
( )( )
1
1
1
1
1
15 2 15 2
5 5
15 2 5 2 2.5 2
5 5 5.2 515 2
−
−−
−
−
= ÷
= × = × =
n n
n
n n
n
n n n n
n n nn
T
T
Since 1
n
n
T
T −
gives a constant, the series is a geometric progression. Common ratio is 2/5.
9 (a) ( ) ( ) ( )( )
( )1
2 2 2
2
2
2
1
2
2
22
cos sin sin
3 4sin
5 5
61 36cos 48sin
61 36 1 482
25 48 18
25 48 18
18
25
18
2
5 6 2 5 6 cos
61 60 cos
61 60 cos
485 1
25
1 1
1 48 2 25 1
2 25 5
α θ
θ α θ α
θ θ
θ θ
θθ
θ θ
θ θ
θ θ
θ θ
−
+
= +
= − −
≈ − − −
= + −
= −
−
≈
= −
− ≈ + − +
= −
− +
+
+
+
AD
AD
2
2
2
48
2! 25
24 63
65 1
25
24
25
63
5 1255
θ
θ θ
θ θ
−
= −
−
−
= −
9 (b) ( ) ( )
( ) ( ) ( )
( )( )
( ) ( ) ( )
1
1
sin
sin
2 2 2
32
3/2 2 22
f e f 0 1
f e f f 01 ( ) 1
f f f f 011 ( )
−
−
= ⇒ =
′ ′= = ⇒ =− −
′′ ′ ′′= + ⇒ =−−
nx
nx
x
n nx x n
nx n x
n x nx x x n
n xnx
( )2
2f 12
= + +n
x nx x
Hence n = 2 and 22
2.2
= =b
10 (i) Let Pn be statement “un =
3
2
n
n +, for n
+∈� ”.
LHS of P1 = u1 = 1 and
RHS of P1 = 3(1)
1 2+ = 1 = LHS of P1
Hence, P1 is true.
Assume that Pk is true, i.e. uk = 3
2
k
k + for some .k +∈Z
Consider Pk+1 i.e. uk+1 = ( )
( )3 1
1 2
k
k
+
+ +.
LHS of Pk+1= 1ku + ku= + 6
( 2)( 3)k k+ +
= 3
2
k
k + +
6
( 2)( 3)k k+ +
= 3 ( 3) 6
( 2)( 3) ( 2)( 3)
k k
k k k k
++
+ + + +
=
23 9 6
( 2)( 3)
k k
k k
+ +
+ +=
( )23 3 2
( 2)( 3)
k k
k k
+ +
+ +
= ( )( )3 1 2
( 2)( 3)
k k
k k
+ +
+ + =
( )3 1
( 1) 2
k
k
+
+ += (RHS)
Thus, Pk is true ⇒ Pk+1 is true.
Since P1 is true and Pk is true ⇒ Pk+1 is true, by mathematical induction, Pn is true for
all .n +∈Z
10 (ii) =
1
6
( 2)( 3)
N
n n n= + +∑
= u2 – u1
+ u3 – u2
+ …
+ uN – uN − 1
+ uN + 1 – uN
= uN+1 − u1
=( )3 1
13
N
N
+−
+
=2
3
N
N + or
62
3N−
+
10 (iii)
10
3
( 2)( 3)n n n
∞
= + +∑
= 9
1 1
1 6 6
2 ( 2)( 3) ( 2)( 3)n nn n n n
∞
= =
− + + + +
∑ ∑
= 3
lim 13N N→∞
− +
−9
9 3+= 1 −
9
12 =
1
4
10 (iv) Let j − 3 = n + 2
⇒ j = n + 5
6
6
( 2)( 3)
N
j j j= − −∑
=6
6
( 2)( 3)
j N
j j j
=
= − −∑
= 5
5 6
6
( 5 2)( 5 3)
n N
n n n
+ =
+ = + − + −∑
= 5
1
6
( 2)( 3)
N
n n n
−
= + +∑ =
2 10
2
N
N
−
−
11 (a) Method 1 2
2
2 2
e
e
d de 2 e
d d
x
x
x x
z y
y z
y zz
x x
−
− −
=
=
= −
( )
( )
( )
2
2
2
2 2 2
2
d2 1 e
d
de 2 e 2 e 1 e
d
d1 e
d
x
x x x x
x x
yy x
x
zz z x
x
zx
x
− − −
+
+ = +
− + = +
= +
Method 2 2
2 2 2
e
d d de 2e e 2
d d d
x
x x x
z y
z y yy y
x x x
=
= + = +
( )
( )
( )
2
2
2
2 2
2
d2 1 e
d
de 2 1 e
d
d1 e
d
x
x x x
x x
yy x
x
yy x
x
zx
x
+
+
+ = +
+ = +
= +
( )
( )
( )
2
2
2
2
2
2
2
2
2
2
2 2
2
d1 e
d
d 12 2 e
d 2
12 2 e d
2
1e , where is an arbitrary constant
2
1e e
2
1e e
2
x x
x x
x x
x x
x x x
x x
zx
x
zx
x
z x x
c c
y c
y c
+
+
+
+
+
−
= +
= +
= +
= +
= +
= +
⌠⌡
11 (b)
(i) ( )d20
dk
t
θθ= − −
As d
1dt
θ= − when 70θ = , ( )1 70 20 0.02− = − − ⇒ =k k
( )d0.02 20
dt
θθ= − −
When 40θ = , ( )d0.02 40 20 0.4
dt
θ= − − = −
It is cooling at a rate of 0.4°C per minute.
( )
0.02
0.02
0.02
0.02
d0.02 20
d
1 d0.02
20 d
1 d d 0.02 d
20 d
ln 20 0.02
20 e
20 e
e , where e
20 e
t c
t c
t c
t
t
t
t tt
t c
A A
A
θθ
θθ
θθ
θ
θ
θ
θ
− +
− +
−
−
= − −
= −−
= −−
− = − +
− =
− = ±
= = ±
= +
⌠⌡ ∫
Given 95θ = when 0t = ,
( )95 20 1A= +
75A = 0.0220 75e tθ −= +
11(b) (ii) It is a good model because the equation reflects the decrease of the temperature to a
steady state temperature, which is what would happen in real life.
12 (i)
12 (ii)
2 2
2 2
1 : 1 14 4
x xC y y+ = ⇒ = −
2 22 2
2 : 1 14 4
x xC y y− = ⇒ = −
Volume generated
( )( )2 2
2 82
0 2π 1 8 1 d π 1 d
4 4
3.4701 (to 4 dec places)
π
= − − − −
=
∫ ∫x x
x x
12 (iii) 22 1
4
xy+ =
2
14
xy⇒ = −
Using the substitution 2 cosx θ= , d
2sind
xθ
θ= − .
When 1,3
πθ= =x ; when 2, 0x θ= = .
Area of region ( )2 2
2 0
13
4cos1 d 1 2sin d
4 4π
θθ θ= − = − −∫ ∫
xx
( )
( )
π
23 3
0 0
3
0
2sin d 1 cos 2 d
1sin 2
2
1 2 1sin 0 sin 0
3 2 3 2
3
3 4
π
π
θ θ θ θ
θ θ
π π
π
= = −
= −
= − − −
= −
∫ ∫