Pioneer Junior College H2 Mathematics Prelim Exam Paper 1(Solution)
JC2-2015
Q1 dV
dt
5 kV , k 0
1
5 kV dv 1 dt 1
ln 5 kV t c k
ln 5 kV kt kc 5 kV e
kc kt
5 kV Ae
V
e
kt
A
,
1
e kc
5 Ae k kt
t = = 0, V = = 0. 1
Hence, 0 V
V
k
1
5 A A 5 .
5 5e k kt
5
1 k
As t ,
e
kt
V
5 k
Q2 (i)
Alternative solution y
esin
d y d x
1
esin
1 4 x 2
( 2 x )
1
ln y sin 1 ( 2 x)
( 2 x )
d y d x
2 1 ( 2 x) 2
2y
1 4 x d x d y
2
differentiate wrt x: 1 d y
2
y dx 1 (2 x) 2 1 4 x 2
2
4 y 2 [Shown]
d y d x
2y 2
1 4 x
2
d y d x 4 y 2 [Shown]
2 (ii) Differentiate implicitly, 2
dy d y d y dy 2 2 1 4 x 2 8 x 8 y dx d x dx dx 2
When x 0,
1,
y
x
y 1 2 x
2
2
d y d x
2 ,
d2 y dx 2
4
4 ...
1 2 x 2 x 2 ... (iii)
esin
1
(2 x )
(1 2 x 2 x ...)(1 2
cos x
(1 2 x 2 x x
1
2
2
x
2
2
...) 1
x 2 ...) 1 ... 2
2
2 x 2 x 2 ...
1 2 x
5
x
2
2
..
Q3 (i) y
4 x
2
4 x 4q
x q
4q
xq
The oblique asymptote is d y d x
4
When
4q
x
x
2
4,
y
4 x 4q .
2
q
2
d y d x
0
Therefore, 2 4q 4 0 2
x q 2
2q 4 4 q 2q 4q 2q 4q
2 2
or
2q 4q
2 (no solution)
2 r 4q 8 q
@PJC 2015
[Turn Over]
3
q 2, r 8
(ii)
y
(2,16)
0
One negative real root
(iii) y 0
or
x
y 32
(iv) 2
4 x 2 16 a x 2 x q 2
The equation given is actually
2 2 x 2 y 16
a
2
Therefore, the solution to the equation is given by the intersection between the
y
4 x
a
and
x
circle with radius
2
q
centred at 2,16 which is the point of intersection between the 2
asymptotes. Distance between origin and 2,16 = Distance between 2,16 and 4,32 22 162 260 For the equation to have 1 negative real root the graphs must i ntersect at least once for Hence, a 260
@PJC 2015
x
0.
[Turn Over]
4
Q4 (a) 1.
Translation by 1 unit in the direction of x-axis.
2.
Scaling parallel to the x-axis with a factor of
3.
Reflection about the y-axis.
(b)
y
1 2
Note: Can also do step 3 first followed by step 2. In fact there are many other possible ways.
.
y
x
0 y
x
0
y
a x
0 4a
0
f x a dx
1 2
0
x
1
3a 3a a a 5a 2 2
f x a d x 2 3 a a 2 a 2a 2 2a
1
8a 2 Therefore, 5a
2
k 8a 2
k
5 8
@PJC 2015
[Turn Over]
5
Q5 (i) a and
2a b are
perpendicular (2a b) 0
a
2a a a b 0
2a
2
a
b
2a
a b
0
2
8
3 8 4 1 2 b 8 2 a b cos
b
4
2
(ii) Length of projection of a onto b
= = =
Alternative method A
a b a
2
b b 8
Q
2
B
O
4 2
Length of projection of a onto b = OQ =
2 cos
2
4
(iii) PB
2
b
5
Area of triangle APB = = = = = =
@PJC 2015
1 2
AB PB
1 2
(b a)
1 2 2 5 1 5
1 5 1 5
bb
2
b
5 2 5
a b
ab
a b
3 4
sin
(2)(4 2)(
1 2
)
[Turn Over]
6
=
8 5
Q6 (i) Let the height of the shortest Singa sculpture be x. 14 Total height of all the Singa sculptures = 2 2 x (14 1)(10) x 15 110 3120 2 28 x 1820 x 140 3120 29 x x
1160
40
Height of the shortest Singa sculpture is 40 cm Height of the tallest Singa sculpture = 40 15 110 180 cm Alternative Let y be the height of the tallest Singa sculpture.
Total height of all the Singa sculptures = 2 28 y 280 1820 y
29 y
14 2
2 y 10 14 1 10 y 3120
3120
5220
y 180 The height of the tallest Singa sculpture is 180 cm.
Height of shortest Singa sculpture =
10 40 cm
180 15 1
or Height of shortest Singa sculpture = 170 14 1 10 40 cm Alternative Let the height of the shortest Singa sculpture be x. 14 15 Total height of all the Singa sculptures = 2 x (14 1)(10) 2x (15 1)(10) 3120 2 2 x
40
Height of the shortest Singa sculpture is 40 cm Height of the tallest Singa sculpture = 40 15 110 180 cm
@PJC 2015
[Turn Over]
7
(ii) T n
35
210 0.95
0.95
n 1
n 1
35
1 6
1
n 1 ln 0.95 ln 6 1 6
ln n 1
ln 0.95
n
35.93
n
36
The number of sports contested at the 2015 SEA Games is 36. (iii) T 36
210 0.95
36 1
34.88
The height of the shortest Nila sculpture is 34.88 cm. 210 1 0.95
S 36
36
1 0.95
3537.33
The total height of all the Nila sculptures is 3537.33 cm. Q7 (i) i
z
r e
is a root
z r e
A quadratic factor of
i
z re
z
2
zre
z
z z
2
2 2
i
zr
zr zr
z r e
i
i
zre
e
i
e
P
i
is another root since P( z ) has real coefficients.
z
2
r
i
r
2
cos i sin cos( ) isin( ) cos i sin cos i sin
2
r
2
r
z 2 2rz cos r 2 (shown) (ii) 4
z z
625 54 ei(
5e
n 4 2
i
@PJC 2015
2 n )
, n 0, 1, 2
[Turn Over]
8
so
z
5e
3 4
i
,5e
4
i
,5e
4
i
,5e
3 4
i
(iii) z
4
625 ( z 5e
i(
4
)
)( z 5e
i( ) 4
)( z 5e
i(
3 4
)
)( z 5e
i(
3 4
)
)
2 2 3 2 5 ][ z (2)(5) z cos 5 ] 4 4
= [ z 2 (2)(5) z cos
= ( z 2 5 2 z 25)( z 2 5 2 z 25) Q8 (i) Consider the diagonal length of the card in Fig 1. 2
2
Diagonal length a a 2a Let the height of the triangle side of the pyramid be b. 2b x b
2a
a
2a
x
b
2
a
x
Let the height of the pyramid be h. By Pythagoras Theorem, h
2
2 2
x
h V
1
2a x 2
2
b
2
x 2
2
2
3
x
2
x
x
2a x 2
2a 2 2
2
3
2
x h
3
2a x
2
a
2
3
2
x 2 2 2ax x 4
2
x2 4
2ax 2
(ii) Method 1 dV d x
2x
a
3
@PJC 2015
2
2ax 2
x
2
3
1a
2
2
2ax
2
1
2
2a 2
[Turn Over]
9
For stationary value of V , 2 x
a
3
2ax
2
2
2
2x
a
a
x
2
3
x
1
dV d x
0
2a
2
2
or
4
x
0 (N.A as
x
0)
5 2 4
x
Method 2 V
x
2
4
9
a 2 2ax 2
2
a x
2
4a x
2
4a x
dV
dx
5
3
3
5 2ax
4
18
When
0,
0
4
2ax
4a 5
3
5
dx
ax
2ax
18
dV
2V
4
2x
0
3
or
ax 4a 5 2 x 0
4a
3
ax
0 (NA
as
x
0)
5 2x
5 2
a
4
x
Q9 (a)(i) d
Consider
d x
e x
2
2
2 xe
Therefore, xe x dx
1 2
x
e
2
x2
C
(ii)
x
3
x
2
e dx
x2 = =
1 x
2
x
xe
x
e
2
2
1 2
x
2 x
e
2
2
d
x
xe 1
x
2
2
dx
e C 2
@PJC 2015
x
Let
u
du d x
x
2
dv d x
2 x
x
e
x
v
1 2
e
x
2
2
[Turn Over]
10
=
1
2
x
1
2
x e
2
e
x
2
C
2
(b) 1 u
=
u
d x
1 sin sin
2
sin
du
du
u
2
x
x
2 sin
x cos x
2
x
2sin x cos x
dx
cos x
sin x 2 sin x cos x dx = 2 cos d = cos 2 1 d =
2
x
x
x
=
sin2 x
x
x C
2
u
sin
2
x
x
= sin x cos x x C =
u 1 u
=
u u
2
sin
sin
1
1
sin x
u
u
C
sin
u 1
u
C
1
x
Q10 n
Let
P n be
the statement
r
n
sin 2r
cos cos 2n 1
1
2sin
,
n
1, 2,3,...
1: 1
sin 2 sin 2
LHS
r
1
r
RHS
cos cos 3
2sin 1 1 2sin 3 sin 3 2 2 2sin 2sin 2 sin( )
2sin
sin 2 LHS RHS ,
Assume
P k is
P 1
true for some
k
i.e.
sin 2r
r 1
@PJC 2015
is true. k
1, 2,3...... .
cos cos 2k 1
2sin
[Turn Over]
11 To show
P k 1 is
true.
k 1
i.e.
sin 2r
Note: 2 sin s in(2 k 2)
cos cos 2k 3
2sin
r 1
cos 2k 3 cos 2k 1
k 1
LHS
sin 2r
cos 2k 3 cos[ (2 k 1) ]
r 1 k
cos 2k 3 cos[(2 k 1) ]
sin 2r sin 2k 2 r 1
cos cos 2k 1
2sin cos cos 2k
Factor formula should be applied on 2sin sin(2k 2) to avoid the negative angle.
sin 2 k 2
1 2 sin 2 k 2 sin 2sin
cos cos 2k
1 cos 2 k 3 cos 2k 1 2sin
cos cos 2k 3
2 sin
RHS
P k 1 is true Since P 1 is true and P k is true P k+ 1 is also true, by the principle of Mathematical Induction, P n is true for all n 1, 2,3...... .
n
sin(2 ) r
r
cos cos(2 n 1) 2sin
1
n
2 sin( ) cos( r
r
1
)
r
cos cos(2 n 1) 2sin
2(sin cos sin 2 cos 2 ... sin n cos n)
cos cos(2 n 1)
2sin cos cos(2 n 1) sin cos sin 2 cos 2 ... sin n cos n 4sin
@PJC 2015
[Turn Over]
12
Q11 (i)
Using
ur 1 ur
n 1
1 (ln k )r n 1
u 1 u
r
r
,
r 1
r 1
r 1
1 (ln k ) r
u2 u1 u u n 1 3 2 1 1 1 u4 u3 ln ln k k 1 u u 1 ln k n 2 n3 un 1 un 2 un un1 1 ln k ln k un 1 1 1 1
n
ln k
n
1 ln k ln k 1 1 1 1
un
ln k
n
1 1 1 ln k ln k ln k 1 1 1
ln k
1 1 ln k 1
n
1
ln k
(ii) (a) 1
un 1
lim un
n
1
ln k
ln k
ln k ln k 1
n
1 Note: As n , 0, ln k therefore sequence converges and limit exists.
ln k 1
@PJC 2015
[Turn Over]
13
(b) Since
u
1
n
converges, y
1
ln k
1 ln k ln k
1
0k
or
1
ln k 1 k
e
k e
Q12 (i)
Subst coordinates of A into the Cartesian equations of
p1 and p2
….. (1) 3 3 ….(2) From (2), 0 3 2
3
(ii)
2 1 3 3 Direction vector of l is given by 0 3 1 1 1 0 6 6 Point A 0,1, 3 lies on both p and p and hence on l . 1
2
0 3 Vector equation of line l is r 1 t 1 , 3 6
t
.
(iii) Since the 3 planes have no common point, l must be parallel to p3 but not contained in p3 .
1 0
3 1 0 3 1 0 6
1 3
0 1 1 3 0 1 @PJC 2015
[Turn Over]
14
(ii)
Equation of line through B and perpendicular to p1 is
1 2 2 s 0 , r 4 1
s
This line intersects p1 at N . Hence
1 2 s 2 4 s
2 0 3 1
2 4 s 4 s 3 s 1
1 2 3 ON 2 0 2 4 1 3 Let B’ be the image of B upon reflection in p1 By ratio theorem ON
OB OB ' 2
5
OB 2 2 5 AB ' OB ' OA 1 5 OB ' 2ON
0 5 Equation of reflected line is r 1 1 , 3 5
@PJC 2015
Alternative method By ratio theorem’ AN
AB AB ' 2
5 AB ' 2 AN AB 1 5 Equation of reflected line is
0 5 1 1 , r 3 5
[Turn Over]
15
@PJC 2015
[Turn Over]
