YISHUN JUNIOR COLLEGE Mathematics Department
Solution Examination : JC2 Preliminary Examination
: JC2 H2 Maths
Subject Qn 1(i)
Date
Paper No. : 1
Solution x
2 x
3
x 0 4
1 3 0 x 2 4
x
x 0,
2 x
x
Since x 0 , x (ii)
xn 1 xn
xn xn
4 3
lg
4
3 lg 2
4
3. lg 2
1
xn xn
4 3 xn xn 2 4 From graph, if 0 xn , xn
2 xn
lg
3
xn 0
2 xn 4 xn 1 xn 0 xn 1 xn
(iii)
Since the sequence converges, xn L as n . Therefore, xn 1 L . L
L L
: 17 Aug 11
1
L
2 4 From (i), L 0, .
From (ii), xn 1 xn if 0 x1 x2 .... , therefore sequence is increasing. Hence, L 4 lg Therefore, L 3 lg 2
Page 1 of 12
Qn 2(a) (i)
Solution
1 3 x
1 2
1 3 2 2 1 3x 2 1 3 x 2 2 1 3 5 2 2 2 3 x 3
1
1 3 x 2 1 3 x 1 x
3 2 1 3
6 27
x
8
x 2
135 16
1 1 3 9
4 3
1 3
.
1 2
31
27 1
2
135 1
3
1 29 8 9 16 9
1 2
3 2
373
373
3
(b)
3
.
Hence expansion valid for x (ii)
x
432 432 373
216 Let Pn be the statement “ un n 2 n ”, n 0. n 0 : LHS = u0 0 (Given)
RHS = 02 0 0 LHS RHS , P0 is true. Assume Pk is true for some k 0 . i.e. uk k 2 k To show Pk 1 is true. 2
i.e. uk 1 k 1 k 1 LHS uk 1 uk 2 k 1
k 2 k 2 k 1
Since P0 is true, Pk is true Pk 1 is also
k 2k 1 k 1 2
2
k 1 k 1 RHS (Shown) true, Hence by induction, Pn is true for all n 0.
Page 2 of 12
Qn 3(i)
Solution Let the volume of the water in the tank be V . 2
Sketch of x 2 y 5 52
y
5
5 x
Hence at height
V
h
0
h,
0
2
x d y
h
25 y 5 dy 0 2
h
10 y y 2 d y 0
h
1 5 y 2 y 3 3 0
(ii)
1 5h 2 h3 m 3 3 1 2 3 Initial V 5 10 10 3
500 3
Rate constant, dV dt
dV dt
500
1 500 3 3
dV d h dh d t
10h h 2
d h d t
1 2 d h When h 3 , 10 3 3 3 d t dh 1 Hence dt 63 1 Thus the rate of decrease is ms 1 63
Page 3 of 12
Solution
Qn 4(i)
2 4r 2 1
2
2r 1 2 r 1
(ii)
1 2r 1
1
2r 1 1 1
N 1 2 2 r 1 r 1 4 r 1 r 1 2 2 r 1 1 1 1 2 1 3 N
1
1
1
3 5
1
1
5 7
2 N 3
1 2 N 1
2 N 1 2 N 1 N 1 1 1 2 2 N 1 2 N 1 Let s r 1 . Then r 1 s 2 . Also r N s N 1
(iii)
1
N
1
1
N 1
1
1
2r 1 2r 3 2 s 1 2 s 1 r 1
s 2
N 1
s 2
N 1
s 1
1 4s 2 1 1 2 4s 1
N 1
1 2
4 1 1
1
2 N 1 1 3 3 N 1 2 N 3 3 2 N 3 N
3 2 N 3
Page 4 of 12
Qn 5
Solution x 3
3
x 2 x 3
x 2
3
x2 5
x 2 3 x x x 2
3
3 x 5 x x 2
3
x x 2 x2 5 x x 2
6(a)
(b) (i)
0
3
x2 5 x x 2
3
0
0 ve
ve
3
5
ve 0
3
ve
x
2
5 Hence, x , 0 x 2 3 3 x 4 10 3 y x 2 x2 3 x 4 10 Step 1: y y x 2 x2 Translation of 3 units downwards. 10 1 Step 2: y y x 2 x2 Scaling parallel to y-axis, scale factor 1 of 10 1 1 y Step 3: y x 2 x Translation of 2 units to the left. y f x y
A’(0, 0)
x = 2
B’(3, 0)
D’(6, 2)
x
C’(4, 3) y =
Page 5 of 12
Solution
Qn (ii)
y
B’(3, 3)
A(0, 2)
C’(4, 0)
D’(6, 0)
x
x = 2
7(i)
Let the length of a side of S 2 be a2 . 2
2
a1 a1 2 2 2 a2 a2
a1
2
Geometric sequence with first term = a 1 and common ratio =
1 2
31
Length of a side of S 3 = a1 (ii)
1 Length of a side of S n a1 2
1 2
.
a1
2
n 1
2
2 2 1 n1 a1 a1 Area of S n , T n a1 n1 2n2 2 2 2
a1
Area of S n 1 , T n 1
2
2n 11
2
a1
2n
.
a12 T n 1 T n
n 1 2 2 = a constant
a1
2
2n 1
Therefore, the areas of consecutive squares form a GP with common ratio Geometric progression with first term a1 and common ratio = 2
2
Sum to infinity =
a1
1
2a1
2
1 2
Page 6 of 12
1 2
.
1 2
.
Qn 8(a) (i)
Solution Let AB y
40 x 2
x h y 2
2
2
A
2
40 x x 2 2
y
2
B
1
40 40 2 x
2
y h
x/2
x/2
C
2 100 5 x z
1 2
xh
1 2
x 2 100 5 x
x 100 5 x (ii)
d z d x
1
100 5 x x 5 100 5 x
1 2
2
100 5 x
5
2 100 5 x
x
15 x 2 100 5 x 100
z d x d x
0 100 40 3
x z d x d
x
15 x 0 2
40 3 +ve
3
40 3
0
ve
40
1 40 40 y 40 x 3 2 3 3
40
Hence area is maximised when triangle is equilateral
Page 7 of 12
Qn (b) (i)
Solution x3 2 y 3 3 xy k 2 2 3 x 6 y
d y d x
3x
d y d x
3y 0
3 y 3 x 2 d x 6 y 2 3x y y x 2 d d x 2 y 2 x d y
(ii)
Tangent parallel to x-axis
y d x d
0
y x 2 0 2 y 2 x y x 2 Subst y x 2 into x 3 2 y 3 3 xy k
x 2 x 3
(iii)
2
3
3x x 2 k
x3 2 x 6 3x3 k 2 x 6 2 x 3 k 0 the line y = 1 is a tangent to the curve C 1 x 2 x 1 When x 1, 2 2 k 0 k 4
When x 1, 2 2 k 0 k 0 9 (i)
y e 2cos
x
y d x d y d
2 sin x cos x e 2 cos x sin x 2 y cos x sin x
x d d 2 y
d x 2
2 2
d y d x y d x d
cos x sin x 2 y sin x cos x cos x sin x 2 y sin x cos x
Page 8 of 12
Qn (ii)
Solution
d2 y d y x x x x 2 cos sin sin cos 2 d x3 d x dx y d 2 cos x sin x y sin x cos x x d d3 y
d3 y d x3
2
d2 y
cos x sin x 4 2
dx 2 y cos x sin x
d y d x
cos x sin x
When x 0, y e , 2
d y d x
2e , 2
y e 2e x 2
2
d 2 y d x
x 2
2
2e , 2
d 3 y d x
x3
3
6e2 .
2e 3! 6e 2! 2
2
e 2 1 2 x x 2 x3 (iii)
2cos xsin x
e x2 e2cosx e2ex e
1 1 e2 1 x x2 x3 e2 1 2x x2 x3 2 6
10 (i)
Cartesian equations: 2 x z 3
x 3 y 2 x 1 z 1, y 1 2 1 3 0 3 1 1 0 6 1 3 Hence l : r 1 t 1 , t R 1 6
Page 9 of 12
1 7 xe2 x 2 x 1 2 6
Qn (ii)
Solution Let A and B be 2 points that lie on the line l. Let t 1 , 2
0 7
OA
Since the line l will lie on 3 ,
2 0 0 7 14 98 Let t 0 , 1 OB 1 1 1 0 1 98 1 14 14 98 84 (iii)
Let Q be the reflection of P in 2
5 1 Then OQ 2 3 for some R 7 0 5 1 and 2 3 is on 2 . 7 2 0 5 1 1 2 2 3 3 2 0 0 7
5 6
2
1 9 2
5 13
13 5
5 1 12 5 13 OQ 2 3 29 5 7 5 0 7 12 29 Hence Q , , 7 5 5
Page 10 of 12
Qn (iv)
Solution
12 7 5 1 5 29 1 24 5 5 7 1 8 Hence reflection of plane 1 in 2 ,
7 1 3 5 : r 1 s 1 t 24 , s, t R 5 1 6 8 11
12 (i)
Let a, b, c be the quantity of commodities Towns A, B, and C sold respectively. b c Town A: 0.4 a 5 2 2770 2 2 0.4a 2.5b c 2770 a c Town B: 5b 0.4 2 3260 2 2 0.2a 5b c 3260 a b Town C: 1.75c 0.4 5 740 2 2 0.2a 2.5b 1.75c 740 Using G.C. Thus a = 1200, b = 900, c = 1000. Thus Town A sold 1200 kWh of electricity, Town B sold 900 fishes, Town C sold 1000 kg of vegetables. y
23
y = 20
y From the graph of y f x , the line y = 20 cuts the graph of f twice, therefore the graph of f is not one to one. Hence, f (ii)
Minimum point of f x at x Hence largest integer a 1
(iii)
-1
3 2
does not exist.
1.5
2
3 f x 4 x 14, x 1 2
Page 11 of 12
Solution
Qn 2
3 Let y 4 x 14 2 2
3 4 x y 14 2 x
3
x
1
2 2
1
y 14
2
3
y 14
1 1 1 x f 3 x 14 y x rej. 14 a s 1 2 2
Rf 1 Df , 1 Df 1 Rf 15,
(iv)
hf x h 4 x 2 12 x 23
e4 x
2
12 x 23
Dhf D f ,1 2
hf : x e 4 x 12 x 23 , x 1 Since Rf 15, , from the graph of h, Rhf , e15
Page 12 of 12
