BT2 Revision Package Solutions (2008 Prelims)

7/30/2019 BT2 Revision Package Solutions (2008 Prelims)

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Anglo-Chinese Junior CollegeH2 Mathematics 9740

2008 JC 2 PRELIM PAPER 1 Solutions

1

11 22

2 12 3 1

31 3

xx x

xx

11 1

22 2

1 12 1

33

x x

x

1 1

2 22

1 1 12 1

6 243x x

x x

1 1 3

2 2 21 13 3

6 83x x x

range of validity:1

3x or

1

3x

but since1

3x will result in the sq rt of a negative

number, reject1

3x .

range of validity:1

3x .

2 no. of terms in the sum: 2n

sum = 2 1 1

12 2 2

nm d m n d

= 2

1 2 12

nm n d

3 2 cos 2 1cos 1 121 cos22

x x

x x

xe x dx e dx

e x e dx

11

11

22

cos2 , cos2

1 1 1sin 2 sin 2 , sin 22 2 2

1 1 1 1sin 2 cos 2 cos 2 ,2 2 2 2

x x

x x x

x x x x

dvI e x dx u e x

dx

due x e x dx e v x

dx

dve x e x e x dx u e

2

2

sin2

1 1 1 1sin 2 cos 2 , cos 22 4 4 2

5 1 1sin 2 cos 24 2 4

2 1sin 2 cos 25 5

x x x

x x

x x

xdx

due x e x I e v x

dx

I e x e x

I e x e x

Thus,

2 1 2 1 1cos 1 sin 2 cos 22 5 5 21 1 1sin 2 cos 25 10 2

x x x x

x x x

e x dx e x e x e C

e x e x e C


2/111

4 5 5 5: 10, , , ,

2 8 32x

5 5 5: 5, , , ,

4 16 64y

x: GP with 10a and1

4r

y: GP with 5a and

1

4r 10

: 81

14

x S

,5

: 41

14

y S

the ant will eventually end up at 8,4 .

5 x > 2 or x 2When x > 2, |x 2| = x 2

2

2

( 2) 1

2 1 0

( 1) 2 0

( 1 2)( 1 2) 0

1 2 (N.A.) or 1 2

x x

x x

x

x x

x x

Hence, 1 2, 1x x

( 1) 1 1x x

Let u =x + 1, 2 1u u

1 1 2 2,

1 1 0

u x x

u x x

6

22

2

ln(cos ), ln(sin )

sin cos

cos sin

cos cos cos

cotsin sin sin

x y

dx dy

d d

dy

dx

when , 14

dy

dx

1ln

2 11

ln2

y

x

The equation of tangent is ln 2y x .If this tangent meets the curve again,ln(sin ) ln(cos ) ln 2

ln(sin 2 ) 0 sin 2 1

2 02 4 2

The tangent will not meet the curve again as there is only one solution in

range 02

i.e.

4

.

Whenx 2, |x 2| = 2 x

No real solution exceptx = 1

OTHERWISE

sketch of

hence,


3/111


4/111

7 By Newtons Law of Cooling,

20

20

d k dt

d k dt

ln 20 kt C

Given: 0, 60t A

10, 45t BFrom: ln40 C , ln 25 10k C

40ln 1025

k

1 8ln10 5

k

Also given: 0, ?t

20, 45t C

From &:

2 1

1 8ln 20 ln10 5

t C ,

11 8ln 25 20 ln10 5 C 20 8ln 2ln

25 5 [Modulus can be removed as initial

temperature is higher than 45C, hence20 0 .]

2

20 825 5

2

825 20 845 C

8 6, 1a b

Asymptotes1

2,3

y x

Axial intersection : (0,1) , (1/6,0)

Asymptotes

1

2, 3y x Axial intersection : (0,1) , (0,-1), (1/6,0)

6 1 12

3 1 3 1

xy

x x


5/111

I : A translation of 1 unit in direction of the positivexaxis

II : A scaling parallel to thexaxis with scale factor 13

units

III: A translation of 2 unit in direction of the positive yaxis

OR

I : A scaling parallel to thexaxis with scale factor 13

units

II : A translation of 13

unit in direction of the positivexaxis

III: A translation of 2 unit in direction of the positive yaxis

9 1siny x 2

1

1

dy

dx x

.

21 1dy

xdx

22

2 2

22

2

3 2 22

3 2 2

3 22

3 2

diff. w.r.t ,

21 0

2 1

1 0

diff. w.r.t ,

1 2 0

1 3 0 ( )

x

d y dy xx

dxdx x

d y dyx x

dxdx

x

d y d y d y dyx x x

dxdx dx dx

d y d y dyx x Shown

dxdx dx

4 3 22

4 3 2

5 4 32

5 4 3

diff. w.r.t ,

1 5 4 0

diff. w.r.t ,

1 7 9 0

x

d y d y d yx x

dx dx dx

x

d y d y d yx x

dx dx dx

(0) 0, '(0) 1, ''(0) 0, '''(0) 1, ''''(0) 0, '''''(0) 9f f f f f f 3 53

...

6 40

x xy x

1

2

3 5

2 4

1sin

1

3...

6 40

31 ...

2 8

dx

dxx

d x xx

dx

x x

Using1

2x ,

2 4

2

1 13

1 2 21 ...

2 811

2


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2 4

2

1 13

1 2 21 ...

2 811

2

147

128

256

3147

10(a) Area

12 24

01 4x x dx

Let 1 sin2

x 1 cos2

dx d

When 0, 0x ; 1 ,4 6

x .

Substituting,

2 26

0

2 26

0

26

0

6

0

6

0

1 1Area sin 1 sin cos4 2

1sin cos8

1 sin 232

1 1 cos 464

1 1 sin464 4

1 364 6 8

d

d

d

d

10

(b)

From G.C., approximate coordinates of the points of intersection of the

two curves are 1.36759, 0.65435 and 1.87156, 1.08307 .

21.87156

1.36759

1Volume ln 0.0766 (3 s.f.)6

x x x e dx

A


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11 2

323

23

23 2 6

arg ( 2)

arg[ ( 2 )]

arg arg( 2 )

arg ( 2 )

iz

i z i

i z i

z i

54 6 12

2 2

512

3 3 3 2

sin 4.10 (3 sig fig)3 2

NPQ

PQ

NQNQ

Least value of |z 3 + i | is 4.10

Range of values ofm for which there is exactly one complex numberw

is m = 4.10, 3 2m

12

(i)

(ii)

(iii)

Asymptotes , 2y x a x

2

2

41

2

dy a

dx x

For stationary points, 0dy

dx

2

2

40 1

2

2 2

a

x

x a

when 2 2 , 2 5x a y a when 2 2 , 2 3x a y a

6

4

P(0, 2)

Q(3, 1)

N

X


8/111

13 2

33 1 , 1, 0, 1k i

w w e k

2

3

3 3

3 3 3 3

3

( ) ( ) 0

( ) ( ) ( 1) ( )

1

k i

z i z i

z i z i z i

z i

z iz i

ez i

23

23

2 23 3

2 23 3

23 3 3

23 3 3

3 3 3

3 3 3

3

( 1 )

1( 1 cos sin )

1 cos sin

1 1 2sin 2sin cos

1 2cos 1 2sin cos

2sin cos sin

2cos cos sin

tan

k

k

i

i

k k

k k

k k k

k k k

k k k

k k k

k

i ez

ei i

i

i i

i

i

i

ALTERNATIVE:

23

23

23 3

23 3

3 3

3 3

3

3

3

( 1 )

1( 1 )

.1

( )

(2 sin )

2cos

tan

k

k

k k

k k

k k

k k

i

i

i i

i i

i i

i i

k

k

k

i ez

ei e e

e ei e e

e ei i

3 3

3 2 3 2

3

2

( ) ( ) 0

3 3 3 3 0

2 3 0

3 0

0, 3

z i z i

z z i z i z z i z i

z z

z z

z

Since tan3

z

is negative (when k= 1) ,

tan 3 tan 33 3 .


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14(i)

(ii)

(iii)

(iv)

(v)

1

1 1

3 . 2

1 2cos 107.5 (1 d.p)

11 9BOC

Area of OBC 1

sin 107.52

OB OC

1

11 81sin 107.52 214.2 units

2 2 3 1 2

1 4 6 7 4

3 4 6 3 4

t t

CP t t

t t

,

2

9

5

CB

Since,

. 110

1 2 2

7 4 . 9

3 4 5110

110

CBCP

CB

t

t

t

| 60t+ 80 | = 110

60 80 110 or 60 80 110

1 1or 3 ( 3.17)

2 6

t t

t

1

2 2

2

PA t

1

2

2

AB

Since 2PA t AB , hence,P,A andB are collinear for all values oft.

[OR 1 2PB t AB , OR 1 12

PB PAt

]

OBC and OPC share the same base ofOC. Since //OC BP , OBC and OPC also has the same perpendicular height. Area of OPC = Area of 214.2 units .OBC


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Anglo-Chinese Junior CollegeH2 Mathematics 9740

2008 JC 2 PRELIM PAPER 2 Solutions

1Let

3 2

2 23 12 1

1( 1)(3 1) 3 1

x x B Cx DAxx x x

Using any appropriate & reasonable method,

1, 4, 3, 2A B C D

3 2

2 23 12 1 4 3 21

1( 1)(3 1) 3 1

x x xxx x x

3 2

2 2

2 2

2 1

2 1

3 12 1 4 3 211( 1)(3 1) 3 1

4 1 6 2 111 2 3 13 1

3

1 24ln 1 ln 3 1 3 tan2 3 1 3

1 2 34ln 1 ln 3 1 tan 32 3

x x xdx dxxx x x

xdx dx dx dxx x x

xx x x C

x x x x C

2 2 2 0x Ay By C

Sub (2.15,0) , 22.15 4.6225C

Sub (1,6.8) ,21 6.8 6.8 0

46.24 6.8 3.6225

A B C

A B

Sub (0,7) , 49 7 0

49 7 4.6225

A B C

A B

Solving,A = 0.6381827731 ,B = 3.806922269

2 2

2 2

2 2

2 2

0.6382 3.8069 4.6225 0

0.6382( 5.9651 ) 4.6225 0

0.6382[( 2.9826) 8.8959] 4.6225 0

0.6382( 2.9826) 10.2999

x y y

x y y

x y

x y

Wheny = 2.9826, 2 10.2999 3.2093x x Widest part of tunnel is 6.4 cm (to 1 d.p.)

31 2 3 4

1 1 3 1, , ,2 2 8 4

u u u u

1

1

1 32

2 2r

r

u

2

1

1 1 42

2 2 4r

r

u

3

1

11 52

8 8r

r

u

4

1

13 628 16

r

r

u

conjecture for 2

f2n

nn


11/111

1 1

2 22 , i.e., 2 2

2 2

n nr

r n nr r

n nu r

Proof:

Let P n be the statement:1

22 2

2

nr

nr

nr

To show P 1 is true: LHS = 11

1 22

RHS = 3 12 2 2

P 1 is true.

Assume P k is true, i.e.1

22 2

2

kr

kr

kr

.

To show P 1k is true:

LHS = 1

1

1 1

2 2 1 2k k

kr r

r r

r r k

= 122 1 22

k

kk k

= 11

2 2 2 12k

k k

=1

32

2kk

= RHS

Since P(1) is true and P(k) is true P(k+1) is true, hence by induction, P n is true for 1n .

4 Graphical Method:

Since the horizontal line y k , f,k R cuts the graph of fy x more than once, f is a not a

one-one function. Hence, f-1 does not exist.

(Analytical Method):

f 2 f 4 3 , f is a not a one-one function. Hence, f-1 does not exist.

Largest f 3,5D .

Finding f-1, let 2( 6 5)y x x

-1

3 4

3 4 , 3 4 (N.A since 3 5)

f 3 4 , 0 4

x y

x y x y x

x x x

g ,R and h , 2D . Since Rg Dh , hgdoes not exist.

hr exists if r hR D .

h , 2D and maximum r , 2R .

x

y

2 5 6


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Finding k,2ln 2x

x e

Maximum value of .k e

hr 0,R

5(i)

(ii)

(iii)

(iv)

(v)

8 0 8

0 0 0

10 6 4

EF

,

2 0 2

4 0 4

4 6 2

EG

Let n be the normal of PlaneEFG.

8 2 2

0 4 8 3

4 2 4

n

PlaneEFG:

2 0 2

. 3 0 . 3 24

4 6 42 3 4 24 (Shown).x y z

r

Let be the angle between PlaneEFG and the horizontal ground.

Horizontal plane:

0

r. 0 0.

1

1

0 2

0 . 3

1 4cos

1 4 9 16

4cos 42.0 . (1 d.p)

29

Let h be the vertical distance from the drilled hole to the ground.

3 2

2 . 3 24

4

6 units.

h

h

Point of intersection of line:

3 0

2 0

0 1

r and planeEFG:

2

. 3 24.

4

r

Vertical wall partition: 2 8x y PlaneEFG: 2 3 4 24x y z Using GC,

2 1 0 8 1 0 0.5 0Rref

2 3 4 24 0 1 1 8

Line GM:

0 0.5

8 1

0 1

r .


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Vertical wall partition:

2

. 1 8.

0

r

2 8 6

4 0 4

4 0 4

AG

or

6

4

0

AB

Let the shortest distance be d.

2 6 2

1 4 1

0 4 0 83.58 units.

4 1 4 1 5

AG

d

6a)i)

ii)

b)

It is possible to use systematic sampling.Random and stratified sampling would be awkward because the sampling frame is not fixedORThe exact number of customers is not known in advance.

The first 100 customers will not be representative of all supermarket customers midday andlate night shoppers will be unrepresented.

No.Each household has different number of people, so the probability of being chosen is not equallylikely.OR (give example)A person in a household of one will stand more chance of being chosen than a person on ahousehold of four. Hence the method does not produce a random person.

7 To test H0 : = 9.5 against

H1 : 9.5 at 5% level of significance

Given : Reject H0 at 5% level of significance

Reject H0 at 2.5% level of significanceNot necessarily true nor necessarily false

The critical region at 2.5% level of significance is smaller than that at 5% level so the teststatistic may lie inside or outside the critical region at 2.5% level of significance

OR

Reject H0 at 5% level of significance

p-value < 0.05

Ifp-value = 0.01 < 0.025 < 0.05 , reject H0 at 2.5% level

Ifp-value = 0.03 < 0.05 but 0.03 > 0.025 , do not reject H0 at 2.5% level8 LetXbe the number of boys out of 500 babies.

X~ B(500, 0.005)Since n is large and p is small and np = 2.5 < 5therefore X~ Po( 2.5 ) approximately

P ( at most 495 of the babies are girls)

= 5P X

= 1 4P X = 0.109

9i) No of ways = 4! 5!


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ii)

= 2880

No. of ways = 6 3 3! 5!C = 14400

10i)

10ii)

10iii)

LetA be the event that the orange cat is in the right cage.LetB be the event that the black cat is in the right cage.

Required probability =

PP( | )

P

C

C

C

A BA B

B

=

1 3.

5 4

115

=

3

16or 0.188

Required probability = P 1 PC CA B A B

=1 1 1 1

1 .5 5 5 4

= 0.65

LetXbe the number of times out of 100 such that the black kitten and the orange kitten areplaced in the wrong cages.X ~ B( 100, 0.65 )But n is large andp is not too small and np = 65 > 5 and nq = 22.75 > 5therefore X~ N (65, 22.75) approximately

P(X = 60) = P(59.5 < X < 60.5) ( applying continuity correction)= 0.0483

11i)

ii)

LetR be the mass of a randomly chosen red apple. R ~ N (75, 212.5 )

Let G be the mass of a randomly chosen green apple. G ~ N (55, 10.5 2 )

1 2 3

64692 ~ 15,

12G G G R N

Required probability = 1 2 3P 2 0G G G R = 0.741

P( 70)R = 0.344578 = 0.345LetXbe the number of red apples out of 10 which are underweight.

X~ B(10, 0.344578)P( X < 2 ) = 0.0915378 = 0.0915

Let Ybe the number of bags out of 20 which pass the quality test.Y~ B ( 20 , 0.0915378 )

Required probability = P(Y= 0) = 0.14712a)i)

n

xx

_

=50

1230= 24.6

( 20) (20 x 50)x x = 1230 1000 = 230

2

2

2

Unbiased estimate of population variance

( 20)1( 20)

49

1 230112949 50

711.448979592 1.45 (3 )

49

xx

n

sf


15/111

12a)

ii)

(b)

Test H0 : 25 against

H1 : 25

Under H0, since n = 50 is large,

X ~71

N 25,(49)50

approx. by CLT.

Using two-tailedZtest at 5% level,p-value = 0.0188 < 0.05

Reject H0 and conclude that there is sufficient evidence at 5% level of significance that there has

been a change in the mean of the distribution.

No, Since sample size is large, Central Limit Theorem applies.

A t-test should be used because the population variance is unknown and the sample size is small

13i)

ii)

iii)

Let Tbe the total number of deaths in the country in 5 days.T~ Po(15)P(T= 9) = 0.0324

LetD be the number of deaths in the northern parts in 10 days.D ~ Po(20)

Let D be the mean number of deaths over 50 periods of 10 days.

D ~2

N 20,5

approx by Central Limit Theorem since n = 50 is large

Required probability = P 21 30D = 0.0569

Let Sbe the number of deaths in the southern parts in 2n days.S~ Po(2n)Since 2n > 10, S~ N(2n, 2n) approximately

25 0.01P S

P 24.5 0.01 (applying continuity correction)

24.5 2P 0.01

2

24.5 2P 0.99

2

S

nZ

n

nZ

n

24.5 22.326348

2

n

n

2 2.326348 2 24.5 0n n

4.417831n or 2.772854n 7.688720n

least n = 8

14a)

(i)

(ii)

FalseMethod 1: The linear relation holds only for the sample. The population may not have a linearrelationship.

Method 2: Use a diagram

False

sample


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(b)(i)

(ii)

(iii)

(iv)

r= 0 an absence of linear correlation betweenXand Ybut there might be some curvilinearrelationship. ThusXan Ymay be correlated.

Least square line of regression ofy on tis y= 2.88112t + 1.65431i.e y= 2.88 t + 1.65 ( to 3 sf)

lnx= 2.88112t + 1.65431x = e1.65431e-2.88112t

Hence x0 = e1.65431 = 5.23 (to 3 sf)k = 2.88 (to 3 sf)

When tincreases by 1, y decreases by 2.88.

x = 1.5,y = ln1.5, t=88112.2

65431.15.1ln

= 0.433

Usey on tbecause(1) tis the independent variable andy is the dependent variable; or(2) tis exact

r2 = 0.92 = bd where b = 2.4

Hence d=4.2

9.0 2= 0.3375

Hence y = 0.3375u + k

Since_

y = 0.074364,_

u = 2.4_

y + 10 = 2.4(-0.074364) + 10 = 9.82153

Hence k=_

y 0.3375_

u = 0.074364 0.3375(9.82153) = 3.39 (to 3 sf)

Hence the least squares line of regression ofy on u is

y = 0.338u 3.39 (to 3sf)


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2008 CJC H2 Maths Prelim Paper 1

1. Let x be the no. of toy cars of type ALet y be the no. of toy cars of type BLet z be the no. of toy cars of type C

2x + 3y + z = 100 (1)x + 4y +3z = 150 (2)x + 5y + 2z = 130 (3)

Using GC, x = 20, y = 10, z = 30

No. of toy cars of type A is 20No. of toy cars of type B is 10No. of toy cars of type C is 30

Total profit = 20($1) +10($2)+30($3) = $130

2.0)5x(

2x

10x

02x

)5x)(2x(10x

02x

x4x 2

02x

)4x(x

x < - 2, or 0 < x < 4

For the inequality ,6x1x

11x

replace x above by 1x

Hence, 21x or 41x0

1x (n.a.) or 1 < x < 5 1 < x < 25

3.(i) Let the first three terms of the GP be

2arara ,, .

17283332 arraarara [M1]1217283 ar

112

ar

When the third term is reduced by 2, the resulting AP is 22 arara ,,

So nnnn TTTT 11

araraar 22 0222 aarar

012

2212

2

a

aa

aa

0144262 aa

So 18a or 8a


18/111

3

2r

2

3r

(NA as series is convergent)

Therefore the three terms of the GP are 18, 12, 8.

(ii) For convergent series,1r

54

31

18

321

18

1

r

aS

(iii) The AP has first term 18a

and common difference

6183

2181

aarTTd nn

dnan

Sn

122

16362

nn

nn

6422

2321 nn

5Sn > S

543215 2 )( nn 05410515 2 nn

44165590 .. n

n= 1,2,3,4,5,6

4. (i)

Since every horizontal line cuts the graph of f(x) at most once, therefore f is one-

one. Since f is one-one, hence 1f exists.

(ii)

yx

ye

ye

ey

x

x

x

11ln

11

11

1

1


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1,1

1ln)(1

xx

xf

(iii) Since ),0()0,( fg DR ,fgdoes not exist.

Since ),0(),1( gf DR ,gfexists.

0,1

11

1

1)(

xee

gxgfxx

(iv) k=1

0,1)(

1

1

1

xxxg

yx

xy

5. (i) Using GC, the root = 0.16744919 0.16745 (to 5 decimal places)

(ii) Usingx1 = 0,x2 =16

(x3

1 + 1) =16

( [0]3 + 1) = 0.166 666 666 7

x3 =16

( [0.166 666 666 7]3 + 1) = 0.167 438 271 6

Using GC command: 16 ( [Ans]3 + 1), we generate the rest of thexi values

x4 = 0.167 449 038x5 = 0.167 449 189 (stabilised at 5 decimal places)

So the sequence converges to 0.16745(iii) Since when n ,xn a and alsoxn + 1

Hence, the recurrence relationxn + 1 =16

(x3

n + 1) tends to =16

(3 + 1)

6 = 3 + 1 3 6 + 1 = 0

But this is just the equation where is its root.

Hence the sequence can be used to estimate the root .

6. (a)

For no real roots, 3 < m < 0

(b)(i) y = f (3x1)

y

x0

2

1

(1,3) 2 4

4

y =(x 4)2

x24

y

x0

2

(1, 3)


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7(i) t

dt

dxcos tk

dt

dysin

tkdx

dytan

(ii) For tangents parallel to y-axis, t = 2,2

For tangents parallel to x-axis, t = ,0,

(iii) At kdx

dyt ,

4

Grad of normal =k

1

2

1

4sin kkx

24cos kky

Equation of normal:

2

11

2kx

k

ky

Since normal has a y-intercept of -1,

2

10

1

21 k

k

k

1k

(iv) When k= 1, equation of normal:

1

2

11

2

1

xy

xy

(ii)y2 = f (x)

(iii)y = f(x )

y

x01

y

x01 2


21/111

At point P,

4

3

1tan

sincos

1sin1cos

t

t

tt

tt

When ,4

3t

2

11x

2

1y

Intersection point P =

2

1,

2

11 (Ans.)

8. Solution: (a)(i) dx

x

xxxdxx

2

11

41

22sin2sin

= cxxx 21 412

12sin

(a)(ii)

dx

xdx

xx

xdx

xx

x

2222 2)1(

1

32

22

2

1

32

=

cx

xx

2

1tan

2

132ln

2

1 12

(b)

ddx

x 1tan

sec

14

2

2

2

2

= dsec = c tansecln

= cxx 241ln 2

(c) 4

22

1

224

1

22

a

a

dxaxdxxadxxa

=

4

23

1

32

33a

a

xaxx

xa

=3

655

3

4 23

aa

9. (a)i

bi

ai 12

1)

2sin

2(cos2

2

)1(2)1(20

ibiai


22/111

ia

bba

i )2

()2

(20

2

ab and 2

2

ab

Solving a = -2, b = 1

(b) The other root isi

)1)(1())(1()(222

zzizzzP

123 zzz

(c) 814 iz

ii

i

iz 81

814

1,0,1,2,8181)

22(

22

neeein

ini

1,0,1,2,3)

82(

nezi

n

i

ez)

8

7(

1 3

,

i

ez)

8

3(

2 3

,

i

ez)

8(

3 3

,i

ez)

8

5(

4 3

10.Differentiating 2xvy , with respect to x, x

dx

dv

dx

dy2 .

221

x

yx

dx

dy

2

2

212x

xvxx

dx

dv Correct substitution and simplification.

2211

x

v

x

v

dx

dv (Shown)

2x

dx

v

dv

2xdx

v

dv

cx

cxv

11

ln1

When x = 1, y = 2, v = 1.

11

11ln c

11

ln x

v

11

xev

11

2

xexy

11

2

xexy


23/111

11

AQ = q a= a+ (1 )ba= (1 )( ba)

= (1 )AB

AQ //

AB and having a common point A.

Hence, the vectorqis collinear with points A and B.

Given a=

1

10

and b=

4

26

Given also angle between OA and OQ is 60

|OA | = 2

OQ = a+ (1 )b=

1

10

+ (1 )

4

26

=

43

3266

|OQ | = (43)2+(32)2+(66)2 = 542108 + 56

Weve OA . OQ = | OA || OQ | cos 60

1

10

.

43

3266

= 2 542 108 + 56 (12

)

4 = 1082216 + 112

42 = 1082216 + 112 0 = 1082 216 + 96 (3 4)(3 2) = 0

=4

3or =

2

3

12. (a) 2)2( zizi

)02()20( iziz

22

2

4sin2)2(

z

12.(b) (i)

OPOQ

=|z|

|1/z*|

=|z|

|1/z|

2

2

x2


24/111

= |z|So OP : OQ = |z|2 : 1

(ii) w =z+1

z*

= 2(cos + i sin ) +1

2(cos i sin )

= 2(cos + i sin ) +12

1

(cos i sin )cos + i sin cos + i sin

= 2(cos + i sin ) +12

cos + i sin (cos2 + sin2)

=52

(cos + i sin )

Re(w) =52

cos , Im(w) =52

sin


25/111

2008 CJC H2 Maths Prelim Paper 2

1a. xx 1sin1 Sincex is small.By standard Maclaurins Series expansion,

221

!2

12

1

2

1

2

1111 xxxx

2

81

211 xx (Shown)

1b. Given 3x

e x n a = 0 3x = ex ln a 3x = ax 3 = a

Given that ex

1 +x +x

2

2+

x3

3!

3x = ex ln a

3x 1 +xln3 +(xln3)2

2+

(xln3)3

3!

1c. 1)0()( fxfy

2

1

2

sin1)0('sin12

2

2

y

xfx

dx

dyy

By implicit differentiation,

04

2

141

2

4cos

)0("cos42

2

2

2

2

2

22

y

dx

dyyx

fxdx

dyy

dx

ydy

By implicit differentiation,

4

1

2

124sin

)0("'sin41222

2

23

3

2

2

3

32

y

dx

yd

dx

dyy

dx

dyx

fxdx

dy

dx

yd

dx

dyy

dx

ydy

By Maclaurins Expansion,

3

32

24

1

2

11

.....!3

)0("'

!2

)0(")0(')0(

xx

fx

fxxffy

2.Let Pn be the statement

Zn

nr

rn

r

,32

3ln

32

12ln

1

When n = 1,

LHS of P1 =5

3ln

3)1(2

1)1(2ln

RHS of P1 =5

3ln

3)1(2

3ln

= LHS of P1

P1 is true.

Assume that Pkis true for someZk , i.e.

32

3ln

32

12ln

1 kr

rk

r


26/111

We need to prove that P k+1 is true, i.e.

3)1(2

3ln

32

12ln

1

1 kr

rk

r

LHS of P k+1

1

1

1

1

3)1(2

3ln

3)1(2

32

32

3ln

3)1(2

32ln

32

3ln

3)1(2

1)1(2ln

32

12ln

32

12ln

k

k

r

k

r

PofRHSk

k

k

k

k

k

k

k

k

r

r

r

r

.1 truePtrueP kk By Mathematical Induction, Pn is true for all .Zn

43

13ln3ln15

3)5(2

3ln

3)20(2

3ln3ln)1620(

32

12ln

32

12ln3ln

32

)12(3ln

5

1

20

1

20

6

20

6

rrr

r

r

r

r

r

r

r

A = 15,B = 13, C= 43

3. (i) Since O lies on the line l, thus the plane contains OA and parallel to (i + 4j3k).

Thus a normal to the plane is

3

4

1

x

0

0

1

=

4

3

0

Eqn of the plane is r.

4

3

0

= 0

(ii) Let be the acute angle between OA and l.

cos =

3

4

1

.

0

0

1

26=

1

26Shortest distance fromA to lisAN.

AN= OA sin = 1126

= 5//26

(iii) Perpendicular distance fromB to the plane is

| nOB

. | = |

1

1

1

.

4

3

0

5

1| =

75

(iv) LetBCintersect the plane atMandBM:MC = 1 : m


27/111

By ratio theorem, OM =OC + mOB

1 + m=

11 + m

m

m

m

3

3

6

Mlies on the plane , so OM .

4

3

0

= 0

1

1 + m( 9 + 3m12 + 4m) = 0

m= 3

4. A = (1, 1) and B = (0, 222 )

Area R = dxxdxx

0

1

2

0

1

)22(

= 0.448

Volume = dyydyy 1

22

22

1

0

)2)2((

= 1.10

5. . (i) Only motorists in that particular residential estate areinterviewed. Hence it is a biased sample.

(ii) Systematic sampling

(iii) A better sampling method is stratified sampling.The group is first divided into non-overlapping age groups. Then motorists are selectedrandomly from each stratum (age group) to be interviewed. The number of peopleinterviewed in each stratum will be proportionate to the size of the stratum. In this way,the sample is more representative of the population

6. (a)

No. of different 4-digit numbers = 3! x 3 = 18

(b) CUCUMBER

Case 1: all letters are different.

No. of 4-letter code words = 46 P = 360

Case 2: one pair of repeated letters.

No. of 4-letter code words = 240!2

!4xCxC 2

5

1

2

Case 3: two pairs of repeated letters.

No. of 4-letter code words = 6!2!2

!4

Total no. of 4-letter code words = 606


28/111

7.(a) P[ the 3 guests are seated separately] =

12

5

!9

x!6 37

P

[B1 each for 6!, 7P3, 9! ]

(b)P[MRT/Late] =

P(51

31

(0.1x0.01)x0.06)5.0((0.4x0.05)

(0.1x0.01)x0.06)5.0()

Late

LateCarorLateBus

[M1, B1 for numerator, A1 for final answer]P[Late] = 0.4x0.05 + 0.5x0.06 + 0.1x0.01 = 0.051 [B1]

P[Late for second time on last day]= P[Late once during the first 4 days and late on the last day] [M1or implied]= 4C1(0.051)(0.949)

3(0.051) = 0.00889 [A1 for any two of4C1,(0.051),(0.949)3; A1

for final answer]

8. (i) Let W be the random variable no. of air bubbles in 1 randomly chosen plastic sheet.

Then W ~ Po(12

)

P(W 3) = 1 P(W 2)= 0.0143877 0.0144 (to 3 s.f.) (shown)

(ii) Let V be the random variable no. of air bubbles in 5 randomly chosen plastic sheets.

Then V ~ Po(12 5) = Po(2.5)

Using GC, P(V = 1) = poissonpdf(2.5,1) = 0.2052P(V = 2) = poissonpdf(2.5,2) = 0.2565 (highest probability)P(V = 3) = poissonpdf(2.5,3) = 0.2138

Hence the most likely number of air bubbles is 2.

(iii) Let X be the random variable no. of plastic sheets with at least 3 air bubbles out of 15 plastic sheets.

Then X ~ B(15, 0.0144)P(X 2) = 1 P(X 1)

= 1 binomcdf(15, 0.0144, 1)= 0.0192244 0.0192

(iv) Let Y be the random variable no. of rejected crates out of 100 crates.Then Y ~ B(100, 0.0192244)Since n = 100 > 50 and np < 5So Y ~ Po(1.92244) approximatelyP(Y < 2) = P(Y 1)

= poissoncdf(1.92244, 1)= 0.42741 0.427

9. Let X be the waiting time on a randomly chosen day. X~ N(8,5)Let Y be the journey time on a randomly chosen day. Y~N(11,4)Let T be the total time taken on a randomly chosen day. T=X+Y~N(19,9)

(i) P(T>20) = 0.369 (to 3 sf)

(ii) Expected no. of days late in a month = 30 x 0.369 = 11.07 11 days.

(iii) Let c be the time taken.

P(T > c) < 0.05P(T < c) > 0.95c > 23.9346


29/111

Least value of c = 24 min

Thus, the latest time he can leave his house = 7:40am24 min = 7:16am

(iv) Let T be the average time taken in a month.

T~N(19,30

9)

P(15 0.0642Smallest level of significance = 6.42%

11. (i)

Equation of regression line ofx on tis:x = 0.26484t+ 76.503r= 0.877

(ii)

[Correct number of points with axis labeledB1][Straight line passing through one point with correct number of points on top and belowB1](iii) When t= 130,x = 0.26484(130) + 76.503 = $110.93

The linear model is not very suitable since the estimated value differs from the actual value of $95quite substantially.

(iv) Forx = aebt, where a,b > 0

Taking ln on both sides:lnx = ln a + bt


30/111

Store tvalues in L1

Storex values in L2Store lnx values in L3

Find reg. Line of lnx on t r= 0.9092 [A1]

Forx = c + dt2 where c, d> 0

Store tvalues in L1Storex values in L2Store t2 values in L4

Find reg. Line ofx on t2

[M1]

r= 0.9607 [A1]

(v) The regression line to use isx = c + dt2 where c, d> 0 since the rvalue is much closer to +1. SoUniversity Zs model is more accurate.

Comparing Scatter Plot (Not Needed):

x = 0.26484t+ 76.503 lnx = ln a + bt x = c + dt2r= 0.877 r= 0.9092 r= 0.9607


31/111

DUNMAN HIGH SCHOOLH2 Mathematics

Year 6 Prelims P1 Solutions

Solution

1(i) 12

5 5

n

n nS

1

1

1 2

1

1

1

2 25 5

5 5

22 5

5

23 (1)

5

n n n

n n

n n

n

n

n

T S S

1

1

2325

523

5

n

n

n

n

T

T

(constant)

series is a geometric series.

Alt:1

23

5

n

nT

is in the form 1nnT ar

.

series is a geometric series.1(ii) 1

2

5n

n

Tr

T

Since 1r , the series converges.

1 3 521 15

TS

r

2 Let Pn be the proposition

3

2

2

12

3

2

2

1

1

1

0

nnrrr

n

r

When n = 0, L.H.S. of P0 =6

5

3

2

2

1

1

1

R.H.S. of P0 =6

5

3

2

2

12

P0 is true.

Assume Pkis true for some k 0, i.e.3

2

2

12

3

2

2

1

1

1

0

kkrrr

k

r

Consider Pk+1,

R.H.S. of Pk+1 =4

2

3

12

kk

L.H.S. of Pk+1 =

1

0 3

2

2

1

1

1k

r rrr


32/111

PofR.H.S.4

2

3

12

4

2

3

1

2

1

3

2

2

12

4

2

3

1

2

1

3

2

2

1

1

1

1+k

0

kk

kkkkk

kkkrrr

k

r

Pkis true Pk+ 1 is true

Since P0 is true and Pkis true Pk+ 1 is true, then by induction,

3

2

2

12

3

2

2

1

1

1

0

nnrrr

n

r

, n 0.

As n, 03

2,0

2

1

nn

1 0

5

6

7

6

1 1 2 1 1 2 1 1 2

1 2 3 1 2 3 1 2 3

(2 0 0)

r rr r r r r r

3 Letx,y,zbe the price rate of electricity, gas and water respectively.

22 10 19 100

30 16 35 155

25 12 20 113

x y z

x y z

x y z

augmented matrix =

22 10 19 100

30 16 35 15525 12 20 113

rref matrix =

1 0 0 2.2

0 1 0 2.5

0 0 1 1.4

August's monthly utility bill = 28 2.2 19 2.5 21 1.4 $138.50


33/111

4

5(a)2 22 4 4 0y x x ---- (1)

2 22 2 4y x x

22 2 1 1 4y x

22 2 1 2y x

2

2

21 1

2

yx


34/111

1

asymptotes: 12

1 1 1 1or

2 2 2 2

y x

y x y x

20 : (1) 4 0

2

x y

y

From GC,

(b) 228 2 0y kx y For real values ofy,

discriminant = 4

4 8 2 0kx

4

64kx

4 244 4

4

2 2x

k

2 2x

k (shown)

Alt:

4 28 0kx

2 28 8 0kx kx

2 2

22

2

8 or 8

2 2(rej)

2 2

(shown)

kx kx

xk

x k

6(i)

Since any horizontal liney = kcuts the graph ofy = e2x at most once, f is one-one.

Lety = e

2x

yx

xy

y x

ln2

1

2ln

elnln 2

8 8

y = e2xy = k

xO

y


35/111

,ln2

1:f1- xx 0


36/111

1

, 22

b a

2 1

2 2

2 2 2

2

(1 ) 3(1 ...)(1 )

1 2

3(1 ...)(1 ...)

2

31 ...

2

71 2

2

baxx x x

x

x x x x

x x x x x

x x

2

2 1

2 1

2

11

2

1

2 2

21 ...2 2 2

1 1 1 2 2...

2 4 8

1 1 9 2(shown)

2 4 8

5 45 4cos

2 sin 2

(1 2 )(2 )

(1 2 )(1 )

( ) 1

x

x

x x

x x x

x x

x

x x

x x

x

x

9(a) Since 1 2u , 2nu for all n , i.e.

22 22

2 1

a

6 6a 6 36a 30a

(b)(i) 0a 2

1 12

1 1

u uu

u

For 2u to be defined,2

1 1 0u u and 1 1u

1 1 1 0u u

1 11 or 0u u

Since 1 1u , 1 11 or 0u u

(ii) As n , nu and 1nu ,2

1

21

22 21

22 1 1 0

21 1 0


37/111

10(a)Given w* =z 2i,z= w* + 2i.

Therefore, |w|2 = w* + 2i + 6.Let w = ia b .

2 2 i 2i + 6

6 i 2

a b a b

a b

Comparing real and imaginary parts,

2 b = 0 b = 2

2

2

4 6

2 0

2 1 0

2 or 1(N.A.)

a a

a a

a a

a

Therefore w = 2 + 2i.10(b)

arg(a) =4

a = 1 + i

11(a)2

2d

2 3

xx

x x

2 2

2 2

2 1

2 1

2 2 2d

2 3 2 3

2 2 2d

2 3 ( 1) 2

2 1ln( 2 3) tan2 2

1ln( 2 3) 2 tan

2

xx

x x x x

xx

x x x

xx x C

xx x C

(b)(i) 2sinx d 2cos dx

3

2

3sin3sin2 1 x

6

2

1sin1sin2 1 x

332 2

16

4 d 4 4sin 2cos dx x

3 2

6

2 1 sin 2cos d

21 1 4 1 10 or 1

2

1 5

2

Since 0.618 ,1 5

2

Locus of (i)

Im(z)

Re(z)

Locus of (ii)

A

arg(a)|a|


38/111

3 2

6

4cos d

3

6

2(cos2 1) d

36

sin 2 2

3 2 3 ( ) ( )

2 3 2 3

3

(b)(ii)3 3

2

1 1

3Area 4 d dx x x

x

3

1

3 ln3

x

( 3 ln 3 0)3

3 ln 33

(iii)3 2

1Volume dx y

23

1

3

3 d

4.0 units

yy

12(i)

2

d e e 1

d ee

x x

xx

y x x

x

d0 1

d

yx

x

Maxy =1

e

(ii)1

eexx

1ln ln e ln

e

ln 1

ln 1, 0

xx

x x

x x x

(iii)

2

2 2 2

d e e (1 ) 2

d e e

x x

x x

y x x

x

Concave upwards:

2

2

2

d0

d

20

e

2

x

y

x

x

x


39/111

y

x

(iv)

2 ( 1)y k x a) k> 0

b) k< 0


40/111

DUNMAN HIGH SCHOOLH2 Mathematics

Yr 6 Prelim P2 Solution

Solution:

1

22

2

14 144 2 4

3 32

2 03

x x x

x x

3 2

2

3 2 2

2

2

2

6 15 21 283

2 7 15

6 15 21 28 3 (2 7 15)0

2 7 15

146( 4 )

3 02 7 15

x x xx

x x

x x x x x x

x x

x x

x x

Since 214

43

x x is always positive, we only require 22 7 15 0x x .

2 3 5 0x x 3

or 52

x x

2 4 2 4 2i 32 32 8

arg( 4 2 4 2i ) =4

i(2 )3 3 44 2 4 2i 2 e

n

z

2

i3 122e

n

z

, n = 1, 0, 1.

Cube roots are:3

i42e

,i

122e

,

7i

122e

.

The points 1 2 3, andP P P lie on a circle of radius two centred at origin. The

P2

2

22

Re(z)

Im(z)

P3

P1


41/111

vectors 1 2 3, andOP OP OP are each separated by equal angles of2

3

.

3 Surface area of stem = 22 2 2rx kx x k x

2

2

d

d 2

where (shown)2

x cqx

t k x

p cqx p

x k

p = 2q:

qxx

q

t

x

2

2

d

d

3

2

2

3

3

3

3 3

3 3

d 2

d

d d2

1 ln | 2 |3

ln | 2 | 3 , where 3

2 e

2 e

qt

qt

x xq

t x

xx q t

x

x qt C

x qt D D C

x A

x A

t= 0,x = 0:

0 = 2Ae0 A = 2

3 3

33

e22

e22

qt

qt

x

x

As t , 0since0e 3 qqt 33 202 x .

4i) Annual salary = 10 1

36000 1.04 $51239

4ii)

Amount saved in 1st

year = 1

36000 $180002

9

2904.1

9

20104.1

1000000104.1

)104.1(18000

n

n

n

nS

30

8.2904.1ln9

29ln

n

n

4iii) 9 years = 108 months


42/111

108 = 21(5) + 3 n = 21

total amount = 21

10000 2 25 20 32

= $11155

5i)

1l :

1 3

5 2 ,

12 2

r = ,

2l :

1 8

5 11 ,

12 6

r =

Consider

3 8 34 2

2 11 34 17 2

2 6 17 1

A normal to 1 is

2

2

1

.

1

2 1 2 2

2 5 2 2 4

1 12 1 1

: r r

5ii) Method 1Line passing through Q and perpendicular to 1 ,

4 2

: 0 2 ,

8 1

l

r

SinceR lies on l,

4 2

2

8

OR

for some

At point of intersection ofland 1 ,R

4 2 2

2 2 4

8 1

43

8 4 4 8 4

43

83

203

OR

(shown)

Method 2

Q

R


43/111

RQ = Projection ofPQ

onto the normal

= 1 1

1 1

PQ

n n

n n

3 2 2

5 2 2

4 1 1

83

83

43

4 4 1 4 4 1

24

23

1

8 43 3

8 83 3

2043 3

4

0

8

OR OQ RQ

(shown)

5iii) 2 possible scenarios

1 : 2 2 4 x y z

2 : 2 5 x ay bz

3 : 3 7 x y z

Direction vector of line of intersection of 1 & 3

=

1 2 5

3 2 31 1 4

5 2

3 0 10 3 4 0

4

3 4 10

a a b

b

a b

6 No. It is not possible to obtain the sampling frame, as list of all readers whopurchased magazines is not available.

Method is unsatisfactory as sample is not random but depends on the readers

responses, e.g. readers who want the token (or have the time) are more likely torespond.


44/111

The publisher decides on the strata, e.g. based on 4 different age groups, thendetermines the number of readers to be surveyed from each age group.The publisher then sends interviewers to newsstands to interview the requirednumber of readers from each age group.

7i) YY E E S T R D A

No. of arrangements = 20160!2

!8

7ii) YY S T R D A

No. of arrangements = 151202

7!6

or

No. of arrangements = 15120!7!2

!8

8 LetXbe no. of accidents involving cyclistsX~ B (50,p)

50p (1p ) = 2.82p = 0.06 or 0.94 (rej asp < 0.5)

x P(X=x)2 0.22623 0.2314 0.1732 most probable number is 3

p = 0.2

P (10 220

One-tail test at 5% sig levelReject H0 ifp-value < 0.05

Since n = 50 is large, under H0

)50

9955.899,220(~ NX

Carry outztest: pvalue = 0.00308

Since p-value < 0.05, we reject H0 and conclude that there is sufficient evidenceat the 5% sig level that the manufacturers claim that the mean weight will

increase is correct.

The test remains valid. As sample size of 50 is large, by central limit theorem,the distribution of the sample mean is still normal.Thep-value is the probability of obtaining a sample mean as extreme as 231.62


45/111

if the population mean is 220 g.

10i) X~ N(800. 250) Y~ N(600. 200)

P(X> 800) P(Y< 600) =4

1

2

1

2

1

10ii) )1550,0(N~)()( 4321321 YYYYXXX

102.0

)50)()((P 4321321

YYYYXXX

10iii) The distribution of the weight of durian cakes is independent of the distributionof the weight of mango cakes.

11i) P(drawing 2 cards each scoring more than 5 points)

=

32 20

2 1

52

3

=9920 496

0.44922100 1105

or

=1105

496

50

31

51

32

52

20

1

3

11ii) P(all three cards are of different suits)

=4

3

313

1

52

3

=169

425= 0.398 or

=

4

3

13 13 13

3!52 51 50 = 0.398 or

=425

169

50

26

51

39

52

52

11iii) P(total score of 3 cards is more than 28 points given they are of different suits)

=suits)differentofareP(cards

suits)diffofarecardsandpoints28scoreP(ttl

=

scores are 10, 10, 9 scores are 10, 10, 10P + P

and of diff suits and of diff suits

169425

=

4 3 44 4 1 4 4 43! + 3!

3 1 52 51 50 3 52 51 50

169

425

=

48 64 112+

1125525 5525 5525 0.0510169 169 2197

425 425

or


46/111

=2197

112

425

169

!2

!3

50

4

51

12

52

16

50

8

51

12

52

16

Method 2P(total score of 3 cards is more than 28 points given they are of different suits)

=2197112

1

13

3

4

1

1

1

4!3

3

4

1

4

3

4

3

23

12i) LetXbe no. of larvae in a sample of 30 ml of waterX~ Po(6)

P(X 4) = 1P(X 3) = 0.849

12ii) Let Ybe no. of vials with 30 ml of water and at least 4 larvae eachY~ B (50, 0.84880)

Since n =50 is large, np = 42.4>5, n(1p)=7.56>5Y~ N(42.44, 6.4169) approxP(Y> 40) = P(Y> 40.5) (continuity correction)

= 0.778

12iii) Since n = 50 is large, by Central Limit Theorem

approx)50

6,6(N~X

00195.0)5P( X

Method 2Let Sbe no. of larvae in 50 vialsS~ Po(300)

Since = 300 > 10, S~ N (300, 300) approx

00177.0

)5.249P(

)250P()5P(

S

SX

13i)

The scatter diagram shows a strong positive linear correlation betweenx andy.

13ii) r= 0.955

0.2 0.4 0.6 0.8 1.0 x

y

1.0

0.8

0.6

0.4

0.2


47/111

axa

b

y

ax

xb

y

xb

axy

111

1

relationship between y

1

and x

1

is linear

13iii)

34972.0268755.0

30124.176850.01

76850272.01

268755.01

ba

b

aa

xy

13iv)

0704.0

76850272.002.0

1268755.0

1

y

y

The calculated growth rate is not reliable as it lies outside the data range and the

relationship betweeny

1and

x

1may not be linear outside the data range.


48/111

HCI 2008 Prelim Paper 1 Solution

Qn Solution

1 Letx, y andzbe the no of 10 cent, 20 cents and 50 cents coins respectively.The 3 equations are 10 20 50 1500x y z ,x y z

50 15002

x y z

Aug matrixA =

10 20 50 1500

1 1 1 0

1 1 1 60

rrefA =

1 0 0 30

0 1 0 10

0 0 1 20

30, 10, 20x y z

2 24 3 4x x 2(2 1) 2x 2 2

2

( 1)(4 3 4 )0

( 5)

x x x

x

2

2

10

5

x

x

2(4 3 4 ) 0x x for all x

5 1x or1 5x .3 Volume of waterV

2 21 1

3 9 93 3 r h

2

1 127 9 9

3 3h h

3

27 927

h

2

9d

d 9

hV

h

d d d

d d d

h h V

t V t

2

9 d.

d9

V

th

Given 3 1d

2 cm s when 6 cmd

Vh

t

12

d 9 22 cm s

d 3

h

t

4i 2

2 2

2

1(2 1)

4

1(2) (2 1)(2 )

4

x

x x

x

de x

dx

e x e

xe

4ii

2 2

2

2 2

2 2

1(2 1)

4

ln(2 -1)

1 1(2 1) ln(2 1)

4 2

1 1(2 1) ln(2 1)

4 4

x x

x

x x

x x

xe dx e x

xe x dx

e x x e dx

e x x e C

5 i 33 i i34 i 4 1 e e ez

3 34 i 4 1z z 3 3i4 4 iz z

3 4 iz i1 4i

i 23 2z e , 1,0,1

k

k

i21: e ik z

i6 3 10 : e + i

2 2k z

5i

6 3 11: e + i

2 2

k z

6

11 1

1

nn nx

x xx

=

2

2

11 ( ) ...

2

11 ( ) ...

2

n nn x x

n nn x x

= 2 21 2 2nx n x

Expansion is valid if : 1x x .

Let x = 120 , n =14

& sub into above expansion

1 2 2419 1 1 1 1

1 2 221 4 20 4 20

=

3121

3200

750

dxkx

dt

150

dx dt kx

1ln 50 kx t C

k

ln 50 kx kt B

When 0, 0,t x 50A

When1

200, 20x k , 20200 20 50 50

t

e

4.46t h

20

dx x

dt


49/111

50 kt Bkx e

50 ktkx Ae

1

50 ktx Aek

1 1

20dx dt

x ln 20

tx C

When 0, 80 ln80t x C

When 20,x ln 20 ln8020

t

27.7t h

8

Any line y k ( 1k ) cuts the graph at 2points.f is not one-to-one.

Therefore, f does not have an inverse. 3b

Let (3 )xy e 3 ln( )x y 1f : 3 ln( ), 1x x x

For 1f g to exist, 1g fR D , 1 gR ,a Therefore, 1a

Largest 1a and 1f gR 3,

9a Let Pn be 2

2 1 4nu n for all n

P1: 1u = 5 & 21 2 1 1 4 5u P1 is true

Assume that Pkis true for some n

We want to show that Pk+1 is true

2

1

2

2

8 8 2 1 4 8 8

4 12 9 4

2 1 1 4

k ku u k k k

k k

k

Pk+1 is trueSince P1 is true & Pkis true Pk+1 is true ,Pn is true for all n

.

2

2 1 4 2 3 2 1nu n n n b

1

1 1

2 3 2 5 2 1 2 3

44

2 1 2 3 2 5r

f r f r

r r r r

ur r r

1

14

ru f r f r

1 1........

3 5 7 5 7 9nS

to n terms

= 1

11

4

n

r

f r f r

=

1 0

2 1

3 21

4 ............

1 2

1

f f

f f

f f

f n f n

f n f n

= 1

04

f n f

=

1 1 1

4 15 2 3 2 5n n

As n ,

1

2 3 2 5n n

0 1lim60nn

S

10i 21 sin sin cos2

dyy

d

21

4 420 0

24

0

1sin

2 sin cos1 2 cos

1sin

2

ydy d

y

d

4

0

4

0

1 1 cos 2

22

1 sin 2

22 2

1 1

4 22 2

d

Ii When 4 41 1 1, ( )42 2

x y x

When2

2

11 12,

2 1 1 1 42

xx y

x

.

Thus two curves intersect at the point 1 1,42

.(Verified)


50/111

111 5

144 4

10

14

4

4 1 1 4 9 2 2Area

1 2 5 4 2 5 40 162 2

yy dy dy y

y

Alternatively

1 21

4 21 202

11

5 21 20

21 1

2 20 0 2 2

12 1

1 1 11

5 2 2(2 1)

4 1 1 1

15 2 4( )

2

xx dx dx

x

x dxx

x dx

x

1

1 2

0

4 1 1 2tan 2

5 420 2 2 2

4 9 2 2

5 40 16

x

10iii

Volume1 1

14 2

10 41 2

fnInt(( / (1 2 ), , 0, 0.25)

fnInt( ^ (0.5), ,0.25,1)

[0.631620]

1.984

ydy y dy

y

x x x

x x

11i 21

2

x t

dxt

dt

ln(2 )

1

2

y t

dy

dt t

1

2 (2 )

dy

dx t t

ii Equation of normal at ( 1, ln2) :21 1

0

x t

t

,

dy

dxis undefined

Therefore, equation of normal: ln 2y

Equation of tangent at ( 5, 2 ln2) :25 1

2

x t

t

,

1 1

2 (2 ) 16

dy

dx t t

2ln 2 1

5 16

5

2ln216 16

y

x

x

y

iii 5ln(2) 2ln 2

16 16

x

16ln 2 5x (16 ln 2 5, ln 2)P

ivLet the angle be the acute angle between normal and tangent. 1tan 0.0624

16

12i ( 1)a+ b + c = 0 = 1 (b a) + (c a) = 0

0

1where

AB AC

AB k AC k

A, B, C are collinearii p = 4a - 3b

4a = p + 3ba = (p + 3b) / 4

Alternatively

4 3 3 3a p a a b a bPA


51/111

A divides PB in the ratio 3 : 1

Plies onBA produced with ratio 34

PA

PB

4 3 4 4b p b a b a bPB

Plies onBA produced with ratio 34

PA

PB

bi 1

1 11 13 42

7 221 21 21

10 4

a r nd

n

Alternatively

Take a point in 1 , say C( 13,0,0) which satisfies12 13

4

r

12

7

10

AC OC OA

Proj AC on

1

2

4

=12 1

17 2

2110 4

n

n

AC

2 21 Alternatively

Using a long method

Vector equation of a line throughA and parallel to

1

2

4

:

1 1

7 2

10 4

r

R

Let Cbe the foot of the perpendicular fromA to 1

Then

1

7 210 4

OC

Clies on 1 :1 1

7 2 2 13

10 4 4

= 2

1

7 2

10 4

OC

=

1

3

2

;

1 1

7 2 7

10 4 10

AC

d = 84 2 21AC

bii Since d is positive, the angle between 1a r & n is acute

1 1 3

12 7 2 2 21 2 1

2110 4 6

nd

nOB OA

Alternatively

Using a long method

Vector equation of a line throughA and parallel to

1

2

4

:

1 1

7 2

10 4

r

R

Let Cbe the foot of the perpendicular fromA to 1

Then

1

7 2

10 4

OC


52/111

Clies on 1 :1 1

7 2 2 13

10 4 4

= 2

1

7 2

10 4

OC

=1

3

2

By ratio theorem :

12

2OC OA OB OB OC OA

3

16

OB


53/111

iii1 : x + 2y 4z= 13 . (1)

2 : x + 3y + 3z= 8 . (2)By G.C. solve equations (1) & (2)The vector equation of the line of intersection is

l :

55 18

21 7

0 1

r

where . or1 18

0 7

3 1

r

etc

iv Since B and l lie on the image plane of 2 so the equation of image plane is3 18 55 3

1 7 21 1

6 1 0 6

r

3 18 29

1 7 10

6 1 3

r

where and

Alternatively

Using

55 3 29

21 1 2 100 6 3

as a dir vector

n =

29 18 31

10 7 83

3 1 23

31 3 31

83 1 83 38

23 6 23

r


54/111

HCI 2008 Prelim Paper 2 Solution

Qn Solution

11 22 tan

3

xy

x

2

2 2

2 2( 3) 22 1/ 1

( 3) ( 3)

dy x x x

dx x x

2

2

2 3

dy

dx x x

2 2 3 2dyx xdx

2

2

22 3 2( 1) 0

d y dyx x x

dxdx

3 2

2

3 22 3 4( 1) 2 0

d y d y dyx x x

dxdx dx

When 0,x 2 3

2 3

2 4 40, , ,

3 9 27

dy d y d yy

dx dx dx

2 32 2 2

3 9 81y x x x (up term in 3x )

2

2

1 1 1 2 4 2

2 2 3 9 272 3

dy

x xdxx x

= 21 2 1

3 9 27x x

2i

2ii

2

iii

3i Amount owed before interest is added at end of nth year 1 12(1000)nu (as $12000 is paid for 1 yr)Thus amt owed with interested added

11.03( 12000)n nu u

a1

1

1 aa

(a1, 4a)

(1 a, 0)


55/111

3ii 1

2

2 2

2 1

(200000 12000)1.03

((200000 12000)1.03 12000)1.03

200000(1.03 ) 12000(1.03) 12000(1.03 )

200000(1.03 ) 12000(1.03 1.03 ... 1.03 )

1.03(1.03 1)1.03 (200000) 12000

0.03

1.03 (200000

n n

n

nn

n

u

u

u

) 412000(1.03 ) 412000

412000 212000(1.03 )

n

n

3

iii

0 when 23nu n (Use GC, Sequence mode) or

412000 212000(1.03 ) 0

212000(1.03 ) 412000

4121.03

212

412lg

212 22.48lg1.03

least 23

n

n

n

n

n

Needs 23 years

3iv

Total interest paid22

16

0.03

0.03(277247.491)

8317.42

r

r

u

15

15 412000 212000(1.03 ) 81710.90768u

Penalty = 150.05 4085.55u As total interest paid is greater than penalty,

its to Johns benefit to terminate his loan early.

4i 1z

arg 2 34 3

z

32

2

2

4ii

cos isin2 2

z

cos isin2 2

n n nz

cos 0 2 1,2

or 2 1,

nn m m

n m m

4

iii

3 iarg arg 3 i

2 2w

3 iarg

2 2 6w

4iv

3 i 2 3 i 2u u 3 i

12 2


56/111

Greatest 3 i 1 4 52 2

u

5i Quota Sampling

5ii A list, in alphabetical order of the male and female employees can be generated separately. The list ofmale employees will be numbered from 1 to 580 in that order while the list of female employees will benumbered from 1 to 420 in that order. A number between 1 to 20, inclusive is then selected at random. Thecorresponding employee and every 20th employee thereafter will be included in the sample for interview.This method of selection is done for both the list of male and female employees.

5iii In (ii), the sample selected would be random whereas in (i), the sample is non-random as the interviewers

were free to choose who they wanted to interview thus creating biasness.6 0H : 55 vs 1H : 55

101467.6

15

xx

n

2 2

2 21 1 (1014)83568 1072.97141 15 1 15

x

xs x

n n

Under 0,T t (14)H . Using GC, select t- test, p - value = 0.0792

Sincep - value = 0.0792 < 0.10, we reject H0 andconclude that there is sufficient evidence at the 10% level of significance that the promotional campaignwas successful in encouraging customers to spend more money at the store.

7General comments for Q7

There is a general lack of explanation on the approach to the qn. When solving a qn by consideringcases, candidates should state the cases clearly.

Many students just wrote a 'string of numbers' without explanation. Solution only has all numerical computations without any words; not even the phrase 'number of ways". Use of diagrams is lacking. Diagrams are helpful for marker's understanding (if you are not good in

words)

7 No. of ways the committee can be formed= 4 10 4 103 5 4 4C C C C = 1218

7i No. of ways the committee can be seated without restriction = (8 1)! = 5040

7ii No. of ways the committee can be seated such that no two girls are seated next to each other = 4! (5)(4)(3)= 1440

No. of ways the committee can be seated when there are two absentees = (8 1)!2! = 2520

or 6 26 1 ! 7 1 !C = 2520

8i Solving both equations: 18.5 0.1y x and 16.6 0.4x y

25, 21x y

5 25 125, 5 21 105x y 8ii

2 2

125 1052634

5 or 0.4 0.1 0.2125 105

3215 2227.55 5

r

8iii Letx = 26.y =18.5+0.1 (26) = 21.1 21The above equation is used becausex is given andy has to be found.

The estimate is not reliable because r is small, i.e. it is close to 0, hence the linear correlation betweenx

andy is very weak.

9i P (player wins the grand prize) = P (WWW)

=2 4 6

8 10 12

=1

20

9ii P (player wins a consolation prize)= P (BWW or WBW or WWB)

=6 2 4 2 6 4 2 4 6

8 10 12 8 10 12 8 10 12

=3

20

9iii P (player wins a consolation prize/first draw was white) =P (player wins a consolation prize and 1st draw was white)

P (1st draw was white)


57/111

=P (WWB or WBW)

P (W)=

110

14

=2

5

P (player wins at least 2 grand prizes in his 4 attempts)

= 1 P (player did not win any grand prize) P (player wins only 1 grand prize)

=

4 319 1 19

1 420 20 20

= 0.0140

Alternative:

Let random variableXbe the number of grand prizes a player wins in his 4 attempts at the game. Then1

~ 4,20

X B

.

P (player wins at least 2 grand prizes in his 4 attempts)

= P ( 2)X

= 1 P ( 1)X

= 1 0.98598 = 0.014010i X B(108,0.04)

Assume independence of customers not turning up.P(X= 0) = 0.96108 = 0.0122

10

ii

Since n > 50, by CLT, the sample mean4.1472

(4.32, )60

X N approx.

P 4 0.112X 10

iii

Since n is large, np = 4.32 0.999P(Y 5) < 0.001Using G.C, m > 3.8Or when m = 3, P = 0.011 > 0.001

when m = 4, P = 0.00055 < 0.001Least number of days = 4

11

a

P(X>500) 0.01

Using GC, graphically 453 (3 s.f.) Alternatively,

P(X>500) 0.01500

0.0120

P Z

5002.326 453 (3 s.f.)

20

11

bi

Let Xdenote the amount of soft drink that is dispensed into a cup of 500 ml. XN(470,202)P(X>500) = 0.0668 (3s.f.)

bii Required Probability = (10.066807)2 = 0.871

b

iii

X1+X2N(940,800) 1 2 1000 0.983 (3 s.f.)P X X

biv Let Ydenote the number of cups that overflowed out of 300 cups.Y B(300,0.066807), np = 20.04217 > 5, nq = 279.95783 > 5,Y N(20.04217, 18.70321) (approximately)

P(Y> 20).c c

P(Y > 20.5) = 0.458 (3 s.f.)


58/111

2008MJCH2MATHS (9740)JC2PRELIM EXAM P1SOLUTIONS

Qn Solution

1 Functions

(i)

(ii) g f, , 0,R D

For fg to exist, g fR D , largest domain of g is g 1,D .

g g fg

fg

1, 0, 1,1

1,1

fD R D R

R

Qn Solution

2 Complex Number 1

Qn Solution

3 Integration Techniques

Ox

y

1


59/111

2

3 32

32 2

3 3 32

3 3 32

ln d

1ln 2ln d

3 3

2ln ln d

3 3

2 1ln ln d

3 3 3 32 2

ln ln3 9 27

x x x

x xx x x

x

xx x x x

x x xx x x

xx x x

x x C

2

2 2

2 1

1d

1 2

1 4 1d d

4 1 2 1 2

1 1ln 1 2 tan 2

4 2

xx

x

xx x

x x

x x C

Qn Solution

4 Mathematical Induction

LetnPbe the statement

2

1

1(2 1) (2 1)(2 1)

3

n

r

r n n n

for n For 1,n

LHS = 21 1

RHS = 1 (2 1)(2 1) 13

i.e LHS = RHS 1 is true.P

Assume thatkP is true for some k

, i.e 21

1(2 1) (2 1)(2 1)

3

k

r

r k k k

.

Need to show 1kP is true, i.e1

2

1

1(2 1) ( 1)(2 1)(2 3)

3

k

r

r k k k

.For 1,n k

LHS = 1

22 2

1 1

(2 1) (2 1) 2( 1) 1k k

r r

r r k

21

(2 1)(2 1) 2 13

k k k k

21 (2 1) 2 3(2 1)3

k k k k

21 (2 1) 2 5 33

k k k

1(2 1)(2 3)( 1)

3k k k =RHS

Thus 1is true is also true.k kP P

Therefore since 1 1is true and is true is also true,k kP P P then by mathematical

induction,nP is true .n


60/111

Qn Solution

5 Curve Sketching & Transformation

(i)

(ii)

y

0

y

1

x


61/111

Qn Solution

6 Vectors 1 & 2

(a)1 1

2 2

3 3

,

a b

a b

a b

a b

1 1 2 2 3 3

1 1 2 2 3 3| || | cos

a b a b a b

a b a b a b a b

a b

2 2 2 2 2 2

1 2 3 1 2 3 1 1 2 2 3 3

2 2 2 2 2 2 2 2

1 2 3 1 2 3 1 1 2 2 3 3

cos

( )( )cos ( )

a a a b b b a b a b a b

a a a b b b a b a b a b

22 1 1 2 2 3 3

2 2 2 2 2 2

1 2 3 1 2 3

2 2 2 2 2 2 21 1 2 2 3 3 1 2 3 1 2 3

( )cos 1

( )( )

( ) ( )( )

a b a b a b

a a a b b b

a b a b a b a a a b b b

Alternative Method

| || | cos a b a b

| || | cos 1 a b a b 2 2 2| | | | a b a b

If

1 1

2 2

3 3

,

a b

a b

a b

a b ,

2 2 2 2 2 2 21 1 2 2 3 3 1 2 3 1 2 3( ) ( )( )a b a b a b a a a b b b

x0


62/111

(b) p 3q = 2p rp 3q 2p r = 0p (3q2r) = 0p // 3q2r

p = k(3q2r), k

Qn Solution

7 Complex Numbers 2

Qn Solution

8(a) Recurrence relations

(i)

(ii)

(iii)

10.9 90, 1, 2,3,...n nx x n

10.9 90n nx x

2

2

2

2

3

3 2

3

1 2

0

0

0

0

0.9 0.9 90 90

0.9 0.9(90) 90

0.9 0.9 90 0.9(90) 90

0.9 0.9 (90) 0.9(90) 90

...

0.9 0.9 (90) ... 0.9 (90) 0.9(90) 90

1 0.90.9 90

1 0.9

0.9 900(1 0.9 )

0.9 ( 900) 90

n

n

n

n

n n

nn

n n

n

x

x

x

x

x

x

x

x

0(shown)

As ,0.9 0. 900n nn x


63/111

(b) Let the sequences tend to .Thus 0.3 7 10

Also100

a

10 (10)

90

a

a

Qn Solution

9 Complex Numbers 3

Qn Solution10 Integration Applications


64/111

(a) 1

0

1

20

12

0

1area of d (1)(1)

2

2 1d

21

1ln 1

2

1ln 2

2

R y x

xx

x

x

(b)

1 2 2

0

21

20

21

20 2

22

4 20 2

240

40

40

4

0

2

1volume of solid d (1) (1)

3

2d

31

4d

31

4tan

sec d 3sec

4 sin d3

1-cos24 d

2 3

2 (1-cos2 ) d3

sin22

2 3

4

2 3

y x

xx

x

xx

x

Qn Solution

11 Maclaurins Expansion

(a) 2 3e 1 ...

2! 3!

x x xx

Letx = 1,


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2 31

1

1

1 1e 1 1 ...

2! 3!

1e 1

!

1e 1

!

r

r

r

r

(b)(i) 1tan 2y x

tan 2y x Diff wrtx,

2 2d d

sec 2 1 4 2d d

y yy x

x x

Diff wrtx,

2

2

2

d d1 4 8 0

d d

y yx x

x x

Diff wrtx,

3 22

3 2

d d d

1 4 16 8 0d d d

y y y

x xx x x

Whenx = 0,

0,y d

2d

y

x ,

2

2

d0

d

y

x ,

3

3

d16

d

y

x

3823

y x x

(b)(ii) 1 11 35 5

0 0

15

2 4

0

8tan 2 d 2 d

3

2

3

73

1875

x x x x x

x x

Alternative Method

Using GC, and converting answer to fraction

Qn Solution

12 Vectors 3

(i) Method 1

Vector equation (1) in scalar product form is

0

3 2

1

r

(3)

Vector equation (2) in scalar product form is

1

1 9

2

r

(4)


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1 2

i j k 5 5

0 3 1 1 1

1 1 2 3 3

n n

a direction vector for line lis

5

1

3

m

.

Hence equation of line lis3 52 1 ,

4 3

r

.

Method 2

3 2y z (1)

1

1 9

2

r in cartesian form is 2 9x y z (2)

Using GC Simult Eqn Solver:

Equation of line lis

29

3 52

1 ,3

30

r

.

(ii) A line throughB perpendicular to plane 2 is

2 1

3 1 ,

1 2

r

. (5)

Subst (5) into (4):

2 1

3 1 9

1 2 2

2 3 2 4 9 4

3

Hence position vector OD

of foot of perpendicular fromB to 2 is

42

3 104 1

3 53 311

81

3

OD


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Now

10 3 11 1

5 2 13 3

11 4 1

AD

Hence length of projection ofAB

onto2 is

2 2 21 3

1 ( 1) 13 3AD

.

(iii) i j k 6 4 2

3 2 4 (9 20) 29

5 1 3 3 10 13

a m

vector equation of plane containing the line land the origin is2

29 0

13

r

or

2

29 0

13

r

Qn Solution

13 Differentiation and Applications

(a)

12 2

1 2

2

2

2

2

2

2

2

11 4 8

d 2cos 1 4d 1 1 4

4

1 4

4

4

2 1 4

2

1 4

2, 0

1 4

2, 0

1 4

x x

xx x

x

x

x

x

x x

x

x x

xx

xx

(b)

r


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Length of circle = 2 r Perimeter of hexagon = 2d r

total areaA 2

2 1 26 sin2 6 3

d rr

22 3

224r d r

d 3

2 2 2d 12

Ar d r

r

d 3

2 2d 6

Ar d r

r

d 3

0 2 2 0d 6

Ar d r

r

3 32

3 6

dr

0.075692r d 0.076r d

2 2

2

d 32 0

d 3

A

r

Thus, the sum of the areas enclosed by the two shapes is a minimum when the radiusof the circle is approximately 0.076d units.


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2008 MJC H2 MATHS (9740) JC2 PRELIMINARY EXAM P2- SOLUTION

Qn Solution

1 Inequalities

2

2

2

2

2

2

2

2

2

61

4 5

6 4 50

4 5

6 4 50

4 5

4 10

4 5

4 10

4 5 1

Since 4 1 0 for ,

4 5 1 0

51

4

x

x x

x x x

x x

x x x

x x

x

x x

x

x x

x x

x x

x

Qn Solution

2 AP & GP

For series to converge,2 3 2

11

a a

a

( 1)( 2)1

1

a a

a

1 2 1a 1 3a

Also, ( 1)( 2) 0a a ( all terms in GP positive)1 or 2a a

set of values of : 2 3a a a

Alternative Solution:

For sum to infinity to exist and all terms positive,

0 1

0 2 1

2 3

r

a

a

Given11

4a .

11 32

4 4r

7 314 4

6.9993

14

n


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3 6.999

14 7

n

3 6.9991

4 7

n

4ln(1.4286 10 )

3ln

4

n

30.776n

least 31n

Qn Solution

3 Summation

1 1 1 ( 1)( 2) 2 ( 2) ( 1)

2 1 2 2 2 1 2

x x x x x x

x x x x x x

=

2 23 2 3

2 1 2

x x x x

x x x

=

2

2 1 2x x x

1

1 2x x x

(verified)

3 3

1 1 1 2 1

1 2 2 1 2

1 2 1

3 2 1

1 2 1

4 3 2

1 2 11

5 4 32

1 2 11 2 3

1 2 1

1 2

N N

n nn n n n n n

N N N

N N N

1 1 1 1 2

2 2 1 1

1 1 1 1

2 2 1

1 1

4 2 1

N N N

N N

N N

Note that 3

1 1 1 1 1

1 2 2 1 2 2n n n n n n n


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3 3 31 3

31

1 1 1 1

1 2

1 1 1< 1 .

8 4 2 1

1 1 1 111 .

8 4 8

N N

n n

n

n n

N N

n

Qn Solution

4 Binomial Expansions

(a) 8 2 3

2 3

(8)(7) (8)(7)(6)(1 ) 1 8 ...

2! 3!

1 8 28 56 ...

x x x x

x x x

let 2x y ky

2 8 2 2 2 2 3(1 ) 1 8( ) 28( ) 56( ) ...y ky y ky y ky y ky

From 2 228( )y ky , term in 3 3 3: 28(2 ) 56y ky ky

From 2 356( )y ky , term in 3 3: 56y y

Given the coefficient of 3y is zero, 56 56 0 1k k

(b)

1 1

11

22

22

2

2 3 1 1

( 1)( 2) 1 2

( 1) ( 2)

1(1 ) 1

2 2

( 1)( 2) 1 ( 1)( 2)1 ( 1) ... 1 ( 1) ...

2! 2 2 2! 2

11 ...

2 4 8

3 5 9...

2 4 8

x

x x x x

x x

xx

x xx x

x xx x

x x


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Alternative Solution:

1 1

11

22

22

2 3(2 3)( 1) ( 2)

( 1)( 2)

1(2 3)(1 ) 1

2 2

1 ( 1)( 2) ( 1)( 2)(2 3) 1 ( 1) ... 1 ( 1) ...

2 2! 2 2! 2

1(2 3)(1 ...)(1 ...)

2 2 4

1(2 3)

2

xx x x

x x

xx x

x xx x x

x xx x x

x

2

2

3 7(1 ...)

2 4

3 5 9...

2 4 8

x x

x x

Qn Solution

5 Differential Equations

(a) d

d

yz x

x

2

2

d d d

d d d

z y yx

x x x #

Given2

2

2

d d3

d d

y yx x

x x 2d 3

d

zx

x

Hence 2d 3 dz x x 3 ,z x c c

d

d

yx

x= 3x c

2d

d

y cx

x x

3

ln , ,3

xy c x d c d


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(b)d

(1000 )d

kt

1d d

1000k t

ln(1000 ) = kt+ C

1000

1000 where

kt C

kt C

e

Ae A e

when t= 0, = 40

40 = 1000A A = 960

when t= 1, = 160

160 1000 960

840 7

960 8

71000 9608

k

k

t

e

e

Qn Solution

6 Permutation and Combination

(a)

(b)

No of ways =

11 8 5 2

3 3 3 215400

3!

No of ways = 8 1 ! 4 3 2 =120960

Qn Solution

7 Sampling Method

(i)

Simple random sampling is free from bias relatively small sample size, hence easy to conduct analysis of data relatively easy

Disadvantage: May not be easy to get access to members who have been chosen for thesample


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(ii)

Systematic sampling Simple to use for large number of walk-in customers Sample obtained more evenly spread over the spectrum of walk-in customers Relatively easy to conduct

The hobby shop could decide on an interval to select a walk-in customer for the survey, forexample, every 20th customer. It then chooses a random start between the numbers 1 and 20.This will determine the first customer to be surveyed and every 20th customer thereafter will

be surveyed.

Qn Solution

8 Binomial Distribution

(i) LetXbe the number of students out ofn who use AIKON brand mobile phones.When n = 30,

1~ 30,

5X B

Expected number of students who use AIKON brand phone

= 1

30 65

P 5 0.172X

(ii)

P 1 0.99

1 P 0 0.99

P 0 0.01

4

0.015

lg 0.0120.6

4lg

5

n

X

X

X

n

Hence, least n = 21.

Qn Solution

9 Estimation and Hypothesis Testing

(i)

Unbiased estimate for population mean = 1.9620

x

x

Unbiased estimate for population variance =

2 2

2 2 39.21 1 77.0221 19 20

xs x

n n

= 0.01

(ii)0

1

H : 2

H : 2

Assumption: 2~ ,X N


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Test statistic:X

Ts

n

Level of significance: %

Critical region: Reject0H ifp-value < %

100

Assuming0H is true, from the GC,p-value = 0.044797

= 0.0488 (3 s.f)

Set of values of : 4.48 100

Qn Solution

10 Probability

(i) A B represents the event the card taken is red and numbered 1.

P(A B) =1

40.

Note: If students use the formula that

P = P P ,A B A B they need to prove that

eventA and eventB are independent.

(ii) A B represents the event the card taken is either red or numbered 1.

P(A B) = P(A) + P(B) P(A B)

=10 4 1

40 40 40

=13

40

(iii) ' 'P( ' | ')

( ')

P A BA B

P B

=

131 40

36

40

=3

4

Required probability =4 1 1 1

3! 103 40 39 38

= 0.00405

Qn Solution

11 Poission

(i)

LetXbe the no of requests for the hire of a truck on a day.Po(2)X


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(ii)

Prob at most 1 truck not hired outP( 2)

1 P( 1)

0.59399

0.594(3sf)

X

X

Prob that requests can be met on a given day( 3)

0.85712

0.857(3sf)

P X

Prob that a particular truck is not used2 2 1

P( 0) P( 1) P( 2)3 3 2

0.40601

0.406 (3sf)

X X X

Let the no of trucks be nP( ) 0.01X n P( ) 0.99X n From GC,P( 5) 0.98344 0.99

P( 6) 0.99547 0.99

X

X

least no of trucks is 6

Qn Solution

12 Normal Distribution

LetF= the length, in cm, of a floor tile. 2~ N 30,0.2F

LetL = the length, in cm, of a wall tile. 2~ N 10,0.15L

(i) 1 2 3 4 5 ~ N 150,0.2F F F F F

1 2 3 4 5P 150.5 151.5F F F F F 0.13138 0.131 (3 s.f.)

(ii) 1 2 3 4 5 6 ~ N 60,0.135S L L L L L L

22 ~ N 60,0.4W F

P 1S W

P 1S W ~ N 0,0.295S W 0.96720 0.967

(iii) 21 2 25

...~ N 30,0.04

25

F F FF

P 30 0.02F a

P 0.020.04

aZ


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P P 0.020.04 0.04

a aZ Z

P 0.010.04

aZ

2.326350.04

a . 0.0931a (3 s.f.)

Qn Solution

13 Regression and Correlation

(i) Suitable to use onh s since s is clearly, the controlled [independent] variable.

(ii) Let 20x s and 100y h

Regression line of ony x is:

y y b x x where 1 143 1 3919.5333, =26.06715 15

x x y yn n

2 2

2

143 391484

15and 3.0899143

241315

x yxy

nb

x

xn

26.067 3.0899 9.5333 55.523 3.0899y x y x

Thus the line of onh s is:

100 55.523 3.0899 20 217.32 3.0899 217 3.09 [3 sf]h s h s h s

(iii)

(iv)

(v)

For every extra revolution/minute the life of the drill reduces by about 3 hours.

Find the product moment correlation coefficient, rand if 1r then the calculated regressionline may be a good fit to the data.ORIf the raw data is available then a scatter diagram could be drawn with the regression line on it.If the points lie close to the regression line, then the line is said to fit the data well.

At217.32 125

125, 3.0899 217.32 125 =29.9 30 revs/minute3.0899

h s s


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RJC 2008 Preliminary Exam Paper 1 Solutions

Q1[6] Let

nP be the statement1

2

(3 2 )3 39

( 1)

n r n

r

r

r r n

, , 2n n .

When 2n , LHS2(3 4)3 9

(2)(1) 2

RHS

3

3 992 2

= LHS.

Since LHS = RHS, 2P is true.

Assume thatkP is true for some , 2k k

, i.e.1

2

(3 2 )3 39

( 1)

k r k

r

r

r r k

.

To show1k

P is true, i.e.

1 2

2

(3 2 )3 39

( 1) 1

k r k

r

r

r r k

.

LHS = 1 1

2 2

3 2( 1) 3(3 2 )3 (3 2 )3

( 1) ( 1) ( 1)( 1 1)

k k kr r

r r

kr r

r r r r k k

1 13 (1 2 )39

( 1)( )

k kk

k k k

13 [ 1 (1 2 )]9

( 1)

k k k

k k

13 (3 )9

( 1)

k k

k k

2

391

k

k

= RHS

1kP is true whenever kP is true.

Since2P is true, by mathematical induction, nP is true for all , 2n n

.

Q2[3] Let 2

2

( 1) 1 1

A B C

r r r r r

2 ( 1)( 1) ( 1) ( 1)A r r Br r Cr r 0, 2

1, 1

1, 1

r A

r B

r C


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[4]

2

2 2 1 1

( 1) 1 1r r r r r

2

1 1 1 1...

(4)(15) (5)(24) (6)(35) ( )( 1)N

SN N

2

4

1

( 1)

N

rr r

4

1 1 2 1

2 1 1

N

rr r r

1 1 2 1

[2 3 4 5

1 2 1

4 5 6

1 2 1

5 6 7

1 2 1

6 7 8

+

1 2 12 1N N N

1 2 1]

1 1N N N

1 1 1 1[ ]

2 12 1N N

1 1

24 2 ( 1)N N

As N ,1

02 ( 1)N N

, Limit of1

24N

S

Q3[1]

[4]

[2]

(i) range of f1 = domain of f = (, 3), domain of g = .Since range of f1 domain of g, the function gf1 exists.

(ii) Lety = 3 x. Thenx = 3 y.f1 :x 3 x,x > 0.

gf1(x) = g(3 x) = 2(3 x)2Domain of gf1 = domain of f1 = (0, )Range of gf1 = [0, )

(iii)1

2gf1(x) =

1

2g(3 +x) = (x + 3)2

Translate the graph ofy = g(x)for 0


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[2]

[1]

Alt :From the GC, when

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BT2 Revision Package Solutions (2008 Prelims)

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