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ANDERSON JUNIOR COLLEGE2013 Preliminary Examinations
H2 MATHEMATICS (JC2)
PAPER 1 (Solutions)
1 21For small , cos 12
12 2
2 2
22
1 1 2 1 21 1
7 7 7 72(7) 1
7
a ay x x
ax
2
2
2
1 27 49
1 18
7 49
a x
x
2 2 18 3a a
3 d sin2 cos
d 2
u xu x
x
1
2 22
0
0 2 2
2
02 2
2
22
0
2
0
4 1 4 d
sincos 1 cos d 2
1cos sin d
2
1sin 2 d
8
11 cos 4 d
16
u u u
xx x x
x x x
x x
x x
or
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2
0
1 sin 4
16 4
32
xx
1 322
0
113 12
2 222 2
00
1 12 22 2
0
1 4 d
31 4 1 4 8 d
2
3 4 1 4 d
332
3
32
u u
u u u u u u
u u u
Alternatively,
1
2 22
0
1 1
2 2 22 2
0 0
02
2
2
0
2
0
1 4 1 4 d
1 4 d 4 1 4 d
1sin d
2 32
11 cos 2 d
4 32
1 sin 2
4 2 32
3
32
u u u
u u u u u
x x
x x
xx
3 (i)
From the tables, ff (6) = f(14) = 261f (8) = 4
(ii) Since Rg= [0, 1] Df= [0, ), fg exists.
(iii) Dfg= Dg= 0,8
.
Taking Rgas new Df , Rfg = [2, 0.5] (f is increasing)
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fg 1x f (tan 2 ) 1x
But Rfg = [2, 0.5], 1< f (tan 2 )x 0.5
1 tan 2 13
x
26 4 12 8
x x
4 1sin (2 )y x diff. w.r.t.x
2
d 2
d1 2
y
xx
2 d1 4 2d
yx
x
2
2 d(1 4 ) 4d
yx
x
(i)
22
2
2
22
2
d d d2(1 4 ) 8 0
d d d
d d d
2(1 4 ) 8 0d d d
y y yx x
x x x
y y y
x xx x x
Sinced
0d
y
x ,
22
2
3 2 22
3 2 2
d d2(1 4 ) 8 0
d d
d d d d 2(1 4 ) 16 8 8 0
d d d d
y yx x
x x
y y y yx x x
x x x x
Subx= 0
1sin 0y = 0 ,2
d 22
d 1 0
y
x
,
2
2
d0
d
y
x ,
3
3
d8
d
y
x
f 0 0 f ' 0 2 f '' 0 0 f ''' 0 8
2 3f '' (0) f ''' (0)f ( ) f (0) f '(0)2! 3!
x x x x
34
f 2 ...3
x x x
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(ii)
Volume
21
321
2
42 d3
x x x
The approximated volume is an under-estimation of the actual volume. This can be seen
from the above diagram, the region under the graph of 1sin (2 )y x is larger than the
region under the graph of 34
23
y x x .
5 (i) Using similar triangles ABC and DEC,
5
5 2 2
w hw h
Volume of the water in the tank = Base area length
V=1
82wh
V= 4hw= 4h5
2h =10h
2
(ii)d d d
d d d
V V h
t h t
d d
20d d
V h
ht t ---- (1)
At t= 2 seconds, V= 2(5) = 10 m3
To find h, 10 = 10 h2
h = 1 m
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d20 5
d
d 1 /d 4
h
t
h m st
(iii) from (1),
d2 20
d
d 1m / s
d 10
hh h
t
h
t
Since the rate of change for his a constant,time taken for hfrom 1 to 2 m = 10 seconds
Therefore the time taken for the trough to be completely filled is 12 seconds.
6(a)
First term =1, common difference = d
5 10 20, ,S S S form a GP
10 20
5 10
S S
S S
2
10 5 202 9 2 4 . 2 19
2 2 2d d d
2
2 9 2 4 . 2 19d d d 25 10 0d d
2d or d= 0 (rejected as AP is increasing)
1
2
2
2 2
2
2
2
21
100
21
1 100
2 ( 1) 1000
1 100
2 990
1 100
n
n
S
S
n
n
n n
n
n n
n
Since 2 2 99 0n n as discriminant < 0 and coefficient of n2is +ve.
2
1 100 0
( 1 10)( 1 10) 0
( 11)( 9) 0
9
n
n n
n n
n
least nis 10.
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6
(b) nth
month
Outstanding amt owed at
the start of the month (inhundreds)
Outstanding amt owed at the end of the nth
month (in hundreds)
1 34 34
2 (34)2 (34)2
3 (34)22
34(2)2-70
4 22 34(2) 70 334(2) 70(2) 70
5 3
2 34(2) 70(2) 70 4 2
34(2) 70(2) 70(2) 70
n 1 334(2) 70(2) .......... 70(2) 70n n
Total amount of money owed at the end of nth month
= 100( 1 334(2) 70(2) .......... 70(2) 70n n )2
1 70((2) 1)100 34(2)2 1
nn
1
100 70 2
n
To be free from debt,170 2 0n
2 140
ln1407.129
ln 2
n
n
Least n= 8
Earliest month to be free from debt = Jan 2014
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7
From graph, there are 2 points of intersection. Thus, there are 2 solutions.
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8 (i) The points Q (0,5,0) is on plane p
0
5 15 5 15 30 4
a
b b b
(ii)2
1
3 0
4 0sin sin
1 25
3 0
4 0
a
a
a a
i.e.
(iii) The angle between
4
ab
and10
0
is obtuse 10 0 0
4 0
ab a
2
22
145
2 25
25 2
a
a
a a
Since2 2a a ,
2 25a
Since a< 0, a= 5
(iv) Equation of plane p : 5 3 4 15x y z
Equation ofx-zplane : 0y
Equation ofx-yplane : 0z
Solving simultaneously
3 , 0 , 0x y z i.e.
3
0
0
OW
(v)M (1,0,5) is on planep (given) and
is also on the x-zplane (y-coordinates of 0M )
M is on their line of intersection l .
PointsM and W are on l , and the shortest distance of Qto the line l
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=2 2 2 2
3 4 25
5 0 15
0 5 20 1250units
414 5 4 5
WQ WM
WM
orMQ WM
WM
9
(a) (i) Let( 1)
be the proposition that for .( 1)!
n n
a n nP u n
n
For n = 1, LHS of 1P = 2a
RHS of 1P =(1 1)1 2
= 2(1 1)! 1
a aa
= LHS of 1P
Thus,1
P is true.
Assume thatk
P is true for some k .
i.e.( 1)
( 1)!k
a k ku
k
To prove that1k
P is true:
1 2
2
2
( 2) ( 1)
( 1)!
( 2)( 1) =
( 1)!
( 2)( 1) =
!
k k
ku u
k
k a k k
k k
a k k
k k
a k k
k
Thus 1kP is true.
Since1
P is true, andk
P true 1kP is true. By Mathematical Induction,
is true for all .nP n
(ii) Consider
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1 2
2
2
2
2
2
2 ( 1)
2 ( )
( 1)( 2) 0
( 1)( )since for 3 and 0, ( 2) 0 and 0
( 1)!
n n n n
n
n
n
n
nu u u u
n
nun
n nu
n
n nu
n
a n nn a n u
n
1 for 3n nu u n
Alternatively,
1
2
2
( 2)( 1) ( 1)( )
! ( 1)!
( 1)2
!
( 1) ( 2)
!
n nu u
a n n a n n
n n
a nn n
n
a n n
n
Since 3n and 0a , 2( 1) 0n and ( 2) 0n
10
n nu u for 3n
16
3
3 4 5 16
3 3 3 3 1
3
3
16
3
...
... since for 3
14
84 since 6
84
r
r
r r
r
r
u
u u u u
u u u u u u r
u
a u a
u a
9(b
)
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222
2 2
2
2 2
2
2d
1
1ln
1
1ln ln 3
1
1ln ln 3
1
ln 1 ln 1 ( 1) ln 3
ln(1) ln(3)ln(2) ln(4)
ln(3) ln(5)
ln( 2) ln( )
ln( 1) ln( 1)
nr
r
rn
r
n
r
n n
r r
n
r
xx
x
x
r
r
r
r
r r n
n n
n n
1
1
( 1) ln(3)
ln 2 ln( ) ln( 1) ln 3
2(3)ln
( 1)
n
n
n
n n
n n
12 2
226
2 12
222 6
14
228
14 7
2 22 22 2
13 6
6
2d
1
2d
1
2d
1
2 2d d
1 1
2(3) 2(3)ln ln
14(15) 7(8)
4(3)ln
5
r
r
mm
m
m
m
m m
m m
xx
xx
xx
x xx x
10(a)
(i)
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2
2
4
3
d 2 4
d 4 4
2( 2)
( 2)
2
( 2)
y x
x x x
x
x
x
Since3
20
( 2)x
,
d0
d
y
x , there are no stationary point when C= 4.
(ii)
WhenC=4 WhenC>4 WhenC
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2
2 2
1 1d dt
8 2
1 1d24 4
1 4 4 1ln
8 4 4 2
1 8 1ln
8 2
rr r
r t Cr
rt C
r
rt C
r
[M1]
4 8 4 8
4
8ln 4 8
8
where =
8
1
t C t C
t
rt C
r
r
e Be B er
rBe
When t= 0, r= 4
0
84
1
1
Be
B
4
8
1 tr
e
As 4, 0, 8tt e r The radius of the circular shaped leaf will grow to a radius of 8 cm for large values of t.
11
(a)
6s w i The real part of s and w are the same.
Let s a bi and w a ci
6b c --- (1)
2
10
( ) 10
a bi a ci
a bc a b c i
2 10a bc and 0b c 1, 3a b and 3c
Since a> 0,
1 3 , 1 3s i w i
Alternatively,
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Subst10
ws
into 6s w i ,
2
2
10 6
10 6
6 10 0
6 36 4( 10)1 3
2
s is
s i s
s i s
is i
Since Re(s) > 0, 1 3s i and 1 3w i
Let u is and v iw and we would arrive at the original pair of given equations.
3u i and 3v i
Alternatively,,s iv w iu 3v i and 3u i
11
(b
)
3 3 12i and5
arg( 3 3 )6
i
5 23 612i k i
z e
1 5 2
6 18 312 , 0, 1k i
i
z e k
1 7 1 5 1 17
6 18 6 18 6 181 2 312 , 12 and 12
i i i
z e z e z e
1 2 2 3 1 3Z , Z Z and ZOZ O OZ arecongruenttriangleswith
1
61 2 3 12z z z and
1 2 2 3 1 3
2
3Z OZ Z OZ Z OZ
Areaoftriangle 1 1
6 61 2 3
1 23 12 12 sin
2 3Z Z Z
1
33 3
124
Leti
c e
55
18182
ii
icz e e e
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Since2
cz isapositiverealnumber,2
arg( ) 0cz
5 5018 18
5
18
i
c e
Simliarly, we can also consider 2cz or 3cz
The corresponding values for c would be17
18
i
e
and7
18
i
e
respectively.
Any one of the above 3 values of c is acceptable.