AJC H2 Math 2013 Prelim P1 Solutions

8/11/2019 AJC H2 Math 2013 Prelim P1 Solutions

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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


2/15

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)


3/15

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


4/15

(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


5/15

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.


6/15

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


7/15

7

From graph, there are 2 points of intersection. Thus, there are 2 solutions.


8/15

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


9/15

=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


10/15

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

)


11/15

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)


12/15

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


13/15

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,


14/15

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


15/15

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.

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AJC H2 Math 2013 Prelim P1 Solutions