BT2 Revision Package Solutions (2009 Prelims) hj

7/30/2019 BT2 Revision Package Solutions (2009 Prelims) hj

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Anglo-Chinese Junior CollegeH2 Mathematics 9740

2009 JC 2 PRELIM Solution

Paper 1:

1

Solving for point of intersection:

25 2 4 202 2 5 0 1 62

x x x xx

For5

2xx

, 1 6 0 or 1 6x x

Using previous results,For

52x

x , 1 6 0 (rej.) or 1 6x x

1 6 or 1 6x x 2

1

1

1 1

1

1 1

1

2

2 21

1

( 1)

1

11

1

n nr

r r

r r

nr

n

r

n

nn

n

n

U U e e

U U e e

e eU e e e

e

U e

3 4 5 1

2 3 4

11 4

2

... nn

n n

U U U U e e e e

e

3 Let x, y and z be the cost of the flights for each parent, eachgrandparent and each child respectively.

2 2 1959 ----- (1)

2 3 2196 ----- (2)

2 4 2640 ----- (3)

2 ?

x y z

x y z

x y z

x y z

Solving (1), (2) & (3) using GC,x = 453,y = 354 andz= 345Cost for Tan family = $[453 + 354 + 2(345)] = $1497

4 (i) 11

2S , 2

2

3S , 3

3

4S , 4

4

5S

4 (ii) 1

f 12

, 1

f 23

, 1

f 34

, 1

f 45

1f ( )1

nn

4 (iii) LetPn denote the statement1

11

nS

n

, n .


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Forn = 1, LHS =1

1

2S RHS =

1 11

1 1 2

LHS = RHS.P1 is true.

AssumePk true for some k , i.e.

11

1k

Sk

Prove thatPk+1 is true, i.e.1

11

2kS

k

1

1LHS

1 2

1 11

1 1 2

11

1 2

11 RHS

2

k kS S

k k

k k k

k

k k

k

Pk+1 is true ifPk is true. ButP1 is true. By the principle of mathematical induction, Pn is true

n .

5 (i) fy x

5 (ii) 1

fy

x

5 (iii) 2 fy x

y

x

O

O

y

x


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6 (i)

Since f is one-one for 0 1x , 1f exists.

2

2

2 2

1 1 1 1 0 1

y x x

x x y x

1f 1 1, 1 2x x x

6 (ii) Range of f = 1,2 Domain of g = 0,3

Since f gR D , gf exists.

2

2

2 4

gf , 0 12 3

x x

x xx x

6 (iii) Restricted Domain of g = 1,2

Range of gf =4 3

,3 2

7 (i)Differentiating

1 1 1

x y a w.r.t. x ,

2 2

1 10

dy

x y dx

2

2

dy y

dx x

Since 0dy

dx , y is a decreasing function.

7 (ii)There are no stationary points on the curve because 0

dy

dx as

0y 7 (iii)Equation of tangent is

2

2

221

2 2

ay a

x a a

y

x

O

f

O

1

1

2


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Therefore, 4y x a

Substituting 4y x a into eqn of curve

1 1 1

4x a x a

2 24 4 0x ax a

2

2 0x a

2x a

Since only one value ofx is obtained,

the tangent does not meet the curve again.

8 (a)

Domain of 1cos x : 1 1x

Range of 1cos x : 10 cos x Domain of 1tan x : x

Range of 1tan x : 1tan2 2

x

8 (b)

1 1sin sin3 3

sin or sin3 3

x x

x x

3

2x

8 (c) The graph ofxy e can be obtained from the graph of

lny x by a reflection in the line y x .

1

1

O 11


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For lnx

x e

, 0 x 9 (i)1

du dyu x y

dx dx

d.e. becomes 22 0du

udx

after replacingdy

dxand x y

Therefore, 22du

udx

9 (ii)2

1

2

du dx

u

11 tan2 2

ux C

tan 22

ux D where D = 2C

2 tan 2y x D x

When x = 0, y = 2 . Therefore, 2 2 tanD . 1tan 2D

The particular solution is 12 tan 2 tan 2y x x 9 (iii) 12 tan 2 tan 2A x x x

d xydAdz dz

=dy dx

x ydz dz

Whenx = 0, y = 2 therefore,3

2

dx

dz

10 (a) (2 ) 9 16z i w i (1)* 3z w i (2)

Substitute *3w i z into equation (1)*(2 )(3 ) 9 16z i i z i

*( 2 ) ( 3 6 ) 9 16z i z i i *( 2 ) 6 10z i z i

Letz x iy ( ) ( 2 )( ) 6 10x iy i x iy i ( ) ( 3 ) 6 10x y i x y i


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Equating real parts: 6 6x y x y (3)Equating imaginary parts: 3 10x y (4)Solving equations (3) and (4): 2x and 4y

2 4z i 3 (2 4 ) 2 7w i i i

10 (b) 3z i

3

exp{ (2 )}2z i n

where n Z

2exp{ ( )}

3 6

nz i

1,n exp{ ( )}2

z i

0,n exp{ ( )}6

z i

1,n 5

exp{ ( )}6

z i

3

w i

* *3w i

3

*w i

From the results of (a) above,

* exp{ ( )}2

w i

* exp{ ( )}6

w i

* 5

exp{ ( )}6w i

Hence exp{ }2

w i

, exp{ ( )}6

i

,5

exp{ ( )}6

i

11 (i)cot 2

4y x

2 2cos 2 .2 2 14

dyec x y

dx

2

2 4

d y dy

ydx dx 23 2

3 24

d y d y dyy

dx dx dx

. Therefore, 4k

11 (ii)When x = 0, y = 1, 4

dy

dx ,

2

216

d y

dx ,

3

3128

d y

dx

Therefore, 2 364

cot 2 1 4 84 3

x x x x


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11 (iii) Differentiating the expansion above,

2 22cos 2 4 16 644

ec x x x

Therefore, 2 2cos 2 2 8 324

ec x x x

Let13

2 4 50x

. Then 200x

Hence,2

2 13cos 250 25 1250

ec

where

1 12, ,

25 1250a b c

12 (i) Let N be the foot of perpendicular from A to the plane1p .

1

1

1

AN

1 1 1

2 1 2

4 1 4

ON OA AN

SinceNis a point on the plane 1p ,

1

. 1 7

1

ON

1 1

2 . 1 7

4 1

( 1 ) (2 ) (4 ) 7 2

21

31 12 1

2 2 83 34 14

24

3

ON

12 (ii) 4 1

3 . 1 4 3 0 7

0 1

(satisfies equation of plane 1p )

4 1

3 . 1 4 3 0 10 a

(satisfies equation of plane 2p )

Hence the point (4, 3, 0) is on both planes1p and 2p .

(4, 3, 0) is then a point on the line where the 2 planes


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intersect.

1 1 1

1 1 1

1 2

a

a

a

is the direction vector of the line ,

Hence r

4 1

3 10 2

a

a

is the equation of line .

12 (iii)

If the planes3p with equation r

1

. 2

3

b

intersects with1p and

2p at line , then

1 1

2 . 1 0

3 2

a

a

(1 ) 2(1 ) 6 0a a 3a

The line lies in the plane3p and thus

4

3

0

lies in the plane3p

also.

4 1

3 . 2 4 6 0 10

0 3

b b

12 (iv) 0 1 1

2 2 4

5 4 1

AB OB OA

.

Given the length of projection of line segment AB on the line

is1

6.

1

. 12 1

1 6

1

2

a

AB a

a

a

2 2 2

(1 ) 4(1 ) 2 1

6(1 ) (1 ) ( 2)

a a

a a

25( 1) 1

3( 3)aa


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2

2

25( 1) 1

( 3) 3

a

a

2 275( 1) 3a a 274 150 72 0a a

150 1188 75 247

2(74) 74a

0.781a or 1.25a


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Anglo-Chinese Junior CollegeH2 Mathematics 9740

2009 JC 2 PRELIM Marking Scheme

Paper 2:1

Equation of lineAB: r

1 3

2 4

1 1

orr

4 3

2 4

2 1

C is on the lineAB:

1 3

2 4

1

OC

Let AOC BOC

. .cos

OA OC OB OC

OA OC OB OC

Hence, . .OAOC OB OC

OA OB

1 1 3 4 1 3

2 . 2 4 2 . 2 4

1 1 2 1

1 1

2 2 2

1 1

2 2(1 3 ) (2 4 ) (1 )(1 3 ) 2(2 4 ) (1 )6 2 6

(3 8 1) (1 4 1) (6 4 1) (2 2 1) 15 5

1

3

32

132

OC

31

43

1

AC OC OA

32

43

1

CB OB OC

12

ACCB

: 1: 2AC CB


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2 (i)

2

22

2 2 12

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

2 2! 2

( 1)2 1 ...

2 8

bb b

b

b

axax

ax b b axb

ab a b bx x

Comparing constant term:1

2 38

b b

Comparing coeff ofx:1 9

38 2 16

aba

Comparing coeff ofx2:21 ( 1) 27

8 8 16

a b bc

2 (ii)

2

( )( 1)...( 1)coefficient of 2 ( )

! 2

1 ( 3)( 4)...( 2) 3( )8 ! 2

1 (3)(4)...( 2) 3( 1) ( )

8 ! 2

1 ( 1)( 2) 3( )

8 2 2

1 3( 1)( 2)( )

16 2

r b r

r

r r

r

r

b b b r ax

r

r

r

r

r

r r

r r

1 3,

16 2p q

3 (i)

3 (ii) Required area =5

2y dx

= 5

2ln 2t t dt

= 2

52 2

5

22

1ln .

2 2

t tt dt

t

=5 3

ln5 ln 22 2

Therefore,5 3

, 1,2 2

2 5


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3

(iii) Required volume = 2 2

25

2

1 1ln5 5 ln 2 2 ln

2 2x dx

= 5.75 square units4 (a)

2 21 1 1 2z i , 2

23 3 1 2w i

arg( ) arg(1 )4

z i

, arg( ) arg( 3 )6

w i

2

2

zz

w w

arg arg( ) arg( )z

z ww

arg4 6 12

zw

1 1 3 1

3 1 3 143 3 3

z i i ii

w i i i

2cos sin

2 12 12

zi

w

13 1 3 1

4i

Equating real parts:2 1

cos ( 3 1)2 12 4

Equating imaginary parts:2 1

sin ( 3 1)2 12 4

3 1 3 1tan 2 3

12 3 1 3 1

3 1 3 1

3 1 3 1

2 3

4 (b)

-1

y

1

0 x2

A(2, 1)

P

Q


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AQ = 2

2cos 2 2

4 2AN AQ

2sin 2 2

4 2NQ AQ

x-coordinate of Q: 2 2

y-coordinate of Q: 1 2

x-coordinate of P: 2 2

y-coordinate of P: 1 2

Coordinates of Q: ( 2 2 , 1 2 )

Coordinates of P: ( 2 2 ,1 2 )

Hence (2 2) (1 2)z i

or (2 2) (1 2)z i

5 The number of seating arrangements = 4! = 24

Number of ways = 3! 2! = 12

Number of ways = 4! 2! 6 = 2886 (i) LetXbe the random variable for the number of vehicles arriving in 2

minutesX~ Po(20)

P(X= 18) = 0.08446 (ii) Vehicles: 180 min = 3hr1800

Variance = Var(0.7X) = 0.7 x 0.7 x 1800 = 882

6 (iii) Let C be the random variable for the number of carsC ~ B(50, 0.7)Since np = 50 x 0.7 = 35 > 5 and nq = 50 x 0.3 = 15 > 5Using Normal approximation to BinomialC ~ N(35, 35(0.3) i.e N( 35, 10.5)P(C 30) = P( C > 29.5) (with cc)= 0.955

7 (i) Stratified sampling is more representative of the entire population.Random samplingcant ensure that both genders are representedappropriately .(A random sample might produce all boys and their eating

preferences may be different from girls)

7 (ii) Biased or not random. The population of leaves beyond the studentsreach has been excluded .

8 Let Wbe the random variable for the number of matches of tenniswon by Wayne

X~ B(120,2

1)

P(W 70) = 0.0412Probability that Wayne won 70 or more matches is only 0.0412.

Q

A N


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It does not seem likely that Henry and Wayne are equally skilledplayers.

9 (i) n = 50, large using Central Limit theorem_

X ~ N(20 000, )50

22002

P( 00020_

X < 800) = P(19200_

X 20 800) = 0.990

9 (ii) Let Ybe the random variable for number of samples with samplemean within $800 of the population meanY~ B(5, 0.989868)P(Y= 3) = 9.96 x 10-4

9 (c) Assume :P(success ) = 0.989868 is a constant for each trial

10 (i) LetXbe the random variable for the length of a pickleX~ N(9 , 0.82)(a) P(X< 8 ) = 0.106

(b) P(X> 10. 5) = 0.030410 (ii) Since 10.6% of the pickles are discarded because they are too short ,it is impossible to reduce the total rejects to 5%

10 (iii) LetL be the random variable for the number of pickles which are toolong

L ~ B(100, 0.0304)

P(L 8) = 1P(L 8 )Since n = 100 large,p = 0.0304 < 0.1 and np = 3.04 < 5Using Poisson approximation to Binomial

L ~ Po(3.04)

P(L> 8) = 0.00413

10 (iv) P( both pickles to meet his guaranteed standard |X1 +X2 < 16 cm)

= 1 2

1 2

( ' & 16)

( 16)

P both pickles fail his g teed standard X X

P X X

=)16(

))8((

21

2

XXP

XP.(***)

=03854989.0

)1056498.0( 2= 0.290

11 (i)

11(ii)

To test H0: = 3.6against H1: < 3.6 at 1 % level of significance

Under H0 ,_

X ~ N(3.6,n

22.1)

Test statistics: Z=

n

X

0

_

=

n

2.1

6.35.3 =

n

2.1

1.0

p-value = P(_

X < 3.5) = P( Z (2.32635 x 12) = 779.314Hence minimum n = 780

11 (iii) Carry out same test as above at 5% level of significance

n = 15 , x = 34 Hence 13.315

47_

n

xx

1)1( xx = 471(15) = 32

s2 =

]))1((

)1([1

12

2

n

xx

n=

14

1[90 - ]

15

2)32(= 1.55238

using the t-distribution

Test statistics t =

n

s

X 0

_

=

15

55238.1

6.31333.3 = - 1.45

p-value = 0.0845 > 0.05Do not reject at 5% level of significance and conclude that there is

insufficient evidence that the modification has decreased the meandiameter of the ball bearings at 5% level of significance.

11 (iii) If = 3.6 , in fact100p% of tests using the same method will lead to

wrongly rejecting = 3.6 or

p value is the probability of wrongly rejecting the hypothesis that =

3.6

12 (i) r= -0.952

For the depths sampled the moisture content decreases

approximately linearly with an increase in the depth of the sample.

Since ris close to 1 , this implies that the regression line ofm onx andx on m are almost identical or close to each other

12 (ii) m = - 2.2048x + 83.583.

When m = 50 ,x =2048.2

50583.83 = 15.232 = 15.2

Reliable since the value ofm is within the range of the given data. Notextrapolating.

12 (iii) 2yxr = (-0.904)

2 = bd= -2.33 d

Hence d= -0.351Hence regression line ofx ony isx = -0.351y + C

Since 5.17_

x and 225.950)5.17(33.2_

y


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(17.5, 9.225) also falls on the line of regression ofx onyHence C= 20.7Regression line ofx ony is x = -0.351y + 20.7

13 Let U andR be the events that United and Rover won the matchrespectively.P(match is still undecided after 1 round )

= P(R U ) + P(R U) = 0.8 0.9 + 0.2 0.1 = 0.74 =

50

37

13 (i) P(United won in less than 3 rounds| United scores a goal in the firstround)

=roundfirsttheingoalascoresUnitedP

roundfirsttheingoalascoresUnitedandroundsthanlessinwonUnitedP

(

)3(

=8.0

1.08.09.08.01.08.0 xx..(*)

= 0.172

13 (ii) LetXbe the random variable for the number of rounds playedP(match is decided in at most n rounds) > 0.98

P(X )n > 0.98P(X= 1) + P(X= 2) + P(X= 3) + +P(X= n) > 0.98

Hence50

13+

50

37(

50

13)+ 2)

50

37(

50

13++( 1)

50

37 n

50

13> 0.98 (**)

50

13

12

50

37........

50

37

50

371

n

> 0.98

n

50

37< 0.02

Hence n >5037ln

02.0ln= 12.992

Least n = 13


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DUNMAN HIGH SCHOOLH2 Mathematics

2009 Yr 6 Prelim Paper 1 Suggested Solns

1 Letx,y,zbe the number ofB1,B2,B3 bears made in the month.

180x y z 5 4 3 650x y z

20 12 9 2100x y z

Augmented matrix M =

1 1 1 180

5 4 3 650

20 12 9 2100

RREF(M) =

1 0 0 30

0 1 0 50

0 0 1 100

Thusx = 30,y = 50,z= 100

2 From GC, a = 01tan

tan

tan

x y

x y

x y

Since 0x and 02

y

, therefore tanx y

1f : tanx x , 02

x

3(a) 2 29 4 36 0x y x

2 29 2 4 36x y

2 22

14 9

x y i.e Hyperbola

Asymptotes:

2 22 3 3

2 or 24 9 2 2

x yy x y x

(b)(i)Asymptotes: and 1

2

xy x

y

x42

33

2y x

32

2y x


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(ii)

2

d 10

d 2 1

y A

x x

2

1 2x A

Therefore, forCnot to have stationary points,A < 0.

4(a) g( ) 2 g( ) g( )

f ( ) tan e f '( ) sec e g '( )ex x x

x x x 2f '(5) sec (e)( 3)e 9.81

(b) 2 2

2

d (2 1)(2 ) (2)

2 1 d (2 1)

x y x x xy

x x x

2

2

2 2

(2 1)

x x

x

Function is increasing2

2

d 2 20 0

d (2 1)

y x x

x x

2

2 ( 1)0 0 or 1(2 1)

x xx xx

5(a)

By ratio theorem,

2 5

3 2 14

4 21

330

OP

(b) 11 3 8

4 4

14 6 8

AB EC OE

(c) 2

1

1

AD

Angle required = 1

3 2

4 1

6 1cos 102

61 6

(d) 11 5 6 3

1 1 4

14 2 12 6

BC k AB BC k

2 & 8 1 9k


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6(i) d, where and arepositiveconstants.

d

d 2 3Set ( ) 0, .

d 3 2

d 3

d 2

3(1 )

2

(2 3 ) (2 3 )2

mR Bm R B

t

mR B B R

t

mR Rm

t

R m

Rm k m

(ii)

3 3

3

d(2 3 )

d

1 d

(2 3 ) d

1ln | 2 3 |

3

ln | 2 3 | 3 3

d 3(2 3 ) e ( (2 3 ) 0)

d 2 2

2e

3

kt c

kt

mk m

t

mk

m t

m kt c

m kt c

m Rm R Rm m

t

m A

(iii) Any positive value ofk

But A must be a positive value STRICTLY LESS than2

3

e.g. k=1,A=1

3, then 3

1[2 ]

3

tm e

m

2/3

2/3-A

0 t

7

4 2 3 1

1 2 1 2

r

r r r r r r


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(a)

1

1 1

2 3 1

1 2

2 3(1 ) 2 12

r r r

rr r

Coefficient ofr3 =47

16

(b)

1 1

4 2 3 11 2 ( 1) ( 2)

2 3 1

(1) (2) (3)

2 3 1

(2) (3) (4)

2 3 1

(3) (4) (5)

2

(

n n

r r

rr r r r r r

n

3 1

2) ( 1) ( )

2 3 1

( 1) ( ) ( 1)

2 3 1

( 1) ( 2)

n n

n n n

n n n

3 1 3 1

2 ( 1) ( 1) ( 2)

3 2 12 ( 1) ( 2)

3 2 4 ( 1) 3 3

2 ( 1)( 2) 2 1 2

n n n

n n

n n n

n n n n

(shown)

1

4 3

1 2 2

n

r

r

r r r

31 1

31 1

4 4

( 2) 2 2 2

4 4 3

1 2 22

n n

r r

n n

r r

r r

r r r r

r r

r r rr


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8(a)

2 2 2

1 23 32 2

0 1

2 1 2

0 0 14 3 d 4 3 d 4 3 d

2 3 2 33 3

1 8 12 3 8 6 2 3

3 3 3

x x x x x x x x x

x xx x x x

2

(b)1 1

2

1

2

1 2

1tan d tan d

1

1 2tan d

2 1

1tan ln(1 )

2

x x x x x xx

xx x x

x

x x x c

(c)2

22 2

2

1

2

1 1d sec d

tan sec1

cosd

sin

(sin )

1

x t tt tx x

tt

t

t c

xc

x

9(a)

Common ratio =

q

p

2

n n

n

S S S

S S

1

2

1 1

2 1 1

2 1

2 (shown)

n

n

n

n n

qp

p p

q q

p p

q

p

q

p

p q

(b)(i) First term of each row: 2, 4, 8, 14.Difference between the first terms: 2, 4, 6. is APwith a = 2, d= 2.Sum of the first (n1) differences

= 21 2 2 2 22n n n n

First term of nth row = Sum of the first (n1) differences + 1st


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term = 2 2n n (ii) Total no. of terms from 1

st row to (n1)th row = 1 + 2 + 3 +(n1)

1 11 1

2 2

n n nn

Sum of all terms from 1st

row to (n1)th row

21

2 2 2 22

n n

n n

21 24

n n n n

10(i) At pointA, 7 1 2 8 8 1 RHS

(ii)

Normal vector of2 =

1 0 144 2 4

1 4 2

Let be the angle required.

14 7

4 2

2 1 108sin 1 90 or

2216 54 11664

(iii)

Normal vector of1=72

1

is parallel to direction vector ofl

Since lis both perpendicular to 1 and 2, 1 and 2 must beparallel.

Alternative:

Normal vector of1=

7

2

1

=

141

42

2

Since the normal vectors of1 and 2 are parallel, 1 and 2 arealso parallel.

(iv) Observe that 2 contains origin,

Shortest distance required =

22 2

1 1

547 2 1

Alternative:

Let 0,1,2B


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0 1 1

1 8 7

2 8 6

AB

Shortest distance required =

22 2

1 7

7 2

6 1 1

547 2 1

(v) LetMbe the foot of perpendicular ofPon 1.

3 7 7

7 2 27

8 1 1 1 2

254 541

PM PA

n n

' '

7 4 31

2 2 1 32

1 0 1

OP PP OP

Alternative:Let l2 be the line passing through pointPand parallel to normal

vector of1.

2

4 7

: 1 2

0 1

l s

r

4 7 7

1 2 2 1

0 1 1

s

128 49 2 4 1

2s s s s

LetMbe the foot of perpendicular ofPon 1 andP be thepoint of reflection required.

12

2 & '

12

a

OM OP b

c


24/136

Using midpoint theorem:4 1 1 0 1

, 2,2 2 2 2 2

a b c

3

Thus ' 3

1

OP

11(a)

2

d2 2

d

d 2

d

d d d 1

d d d

CC r

r

AA r r

r

C C A

A r r r

When cm,

d d d 2

d d d

d 23

d 3

r k

C C A

t A t k

Ck

t

(b) 2

2 2

2

2

1 1 12 2 ( )(2 ) 800

2 2 2

12 3 800

8

56 800

4

5 24 3200

xy xy x x x

xy x x

xy x

x xy

2

2 3

22 3

3

2

1 (2 )

8

12

4

3200 5 12

24 4

800 1

3 6

1 1600

6

V xy x x

x y x

xx x

x

x x

x x

For max. V,

2d 1 (1600 3 ) 0d 6

23.094 (reject negative ans.)

Vx

x

x

y = 3.0230Thus the values ofx andy are 23.1 and 3.02cm resply.

2

2

d 1( 6 ) 0d 6

Vx x

x Thus volume is maximum.


25/136

12(a) i i ii 2 2 21 e e (e e )

=i22cos e , .

2

4 4

i i

44ii

22

4 4i i

4 4 4 42 2

4 4 i2 i2

4

4

1 e 1 e

2cos e 2cos e2 2

2 cos e 2 cos e2 2

2 cos (e e )2

16cos (2isin 2 )2

32i cos ( ) sin(2 ).2

(b) 1arg i

2 4z

1arg i[ ]

2i 4z

iarg(i) arg

2 4z

i 3

:arg2 4

L z

L

cos1 4

| ( 1) |2

1min | i |

2 2

h

z h

From the graph above, observe that in order for the two loci

1arg i

2 4z

and iz c to have 2 intersection points,


26/136

1 1 1 1we have | ( 1) |

2 22 2 2 2c c


27/136

DUNMAN HIGH SCHOOLH2 Mathematics

2009 Yr 6 Prelim Paper 2 Suggested Solns

1(i)

(ii)

(iii)

2 3 23 25 0z az bz Since 3 4i is a root and the coefficients of the equation are real,

f 2y x

y

x3

14,

2

y

x1

2,1

fy ' x

2

x

1

fy

x

y

2

1

1

2, 2

O

O

O


28/136

3 4i is also a root.

2 23 4i 3 4i ( 3) 16 6 25z z z z z

Thus3 2 23 25 ( )( 6 25)z az bz pz q z z

Comparing coefficient of 3z : p = 3

Comparing constant term: 25q = 25 1q Thus 6 17a q p

25 6 69b p q

The other roots of the equation are 3 4i and1

3 .

Let 2 3arg( )z z

Then

3 1arg( ) z z

2 3 3 1arg( ) arg( ) radz z z z

3(a)(i) 1 1tan tan2

d ee

d 1

xxy y c

x x

Whenx = 0,y = 1 01 e 0c c

Im(z)

Re(z)3

4

z1

z2

z3


29/136

Thus1tane xy

(ii)

1tan

2 2

2

22

2

22

2

d e

d 1 1

d1

d

d d d

1 2 d dd

d d1 (2 1) 0

dd

xy y

x x x

yx y

x

y y y

x x x xx

y yx x

xx

(iii)

22

2

3 2 22

3 2 2

3 22

3 2

d d1 (2 1) 0

dd

d d d d1 2 (2 1) 2 0

dd d d

d d d1 (4 1) 2 0

dd d

y yx x

xx

y y y yx x x

xx x x

y y yx x

xx x

Whenx = 0,y = 1 (given)2 3

2 3

d d d1, 1, 1

d d d

y y y

x x x

Thus Maclaurin series is

2 31 112 6

y x x x

(iv) 1tan2

2

e d 11

d 21

x yx x

xx

(b)(i) Using GC, volume required = 1 22 tan

0 e d 30.607 30.6x x

(ii)Using GC, volume required =

22 2 3

0

1 1 1 d

2 6x x x x

= 38.317 = 38.3

Percentage error =38.317 30.607

100% 25.2%30.607

4(i) 1 1 1

1 13 2 5 5 2 ,

3 3n n n nu u u u u

n nu na b 1 1

3 231 2 13

9 2

231


30/136

3 1927

32

31

Conjecture:

231 ,

2Thus

3

n

nu n

a

(ii) Let Pn be the statement

231n

nu for all n .

When n = 1, LHS 11

(given)3

u

RHS 231

13

= LHS

P1 is true.

Assume Pk is true for some k .

i.e.

2

3

1k

k

u

To prove Pk+1 is also true, i.e. 1

21 31

k

ku

.

LHS = 11

5 23

k ku u

1 25 2 1

3 3

k

1

1 23 2

3 3

21 RHS

3

k

k

Thus 1P is true P is truek k .

Since 1P is true, and 1P is true P is truek k , by Mathematical

Induction, Pn is true for all n .

(iii) 23

1 1 1

1

1

1 as

diverges.

n n nr

r

r r r

n

r

n

u

n

S

(iv)

23

23

23

2

3

1

As , 0, 1

20.005

3

2 ln(0.005)ln ln(0.005) 13.1

3 ln

Minimum integer 14

n

n

n

n

kk

k

u

n u L

u L

k k

k


31/136

5(i) Stratified Sampling

CategoryYear Number of students No. of students in sample

1 340 3401200 60 17

2 300 15

3 280 14

4 280 14

Choose 17, 15, 14 and 14 students from Year 1, 2, 3 and 4respectively.

RandomnessWithin each year group, assign a number to every student andrandomly select the required number of students.


32/136

(ii) The cafeteria can sell the preferred food of the respectivegroup of students during their designated breaks.

The cafeteria can prepare the quantity of preferred food ofthe respective group of students in the same proportion ofthe student cohort of each group.

6(i) No of ways required = 4! 5! 2880

(ii) No of ways required = 4 1 4! 4! 2 2 9216C

4

1 : choose one of the four possible slots for M and AC 4!: once M and A are fixed, all the girls are fixed

4!: arrange the remaining boys

2: M and A can permute

2: there are two sides of the table

G M A B G

B G B G B

X X X X X


33/136

7(i) Required probability=1 P( all seeds selected do not germinate)

16 15 14 13 121

20 19 18 17 16

232

323

(ii) LetXbe the number of seeds that germinate, out of 5.X ~ B(5, 0.10)

Required probability P( 1) 1 P( 1)X X = 10.91854= 0.0815

Binomial distribution is used in part (ii) based on theassumption that probability of success is a constant when thesample is small as compared to a large population.


34/136

8(i) P |G H =

1

4

(ii) P |H G =

1

2

(iii) P R E =

5

8

Since P 0R E ,

R &Eare mutually exclusive.

P |G H =1

4and

2 1P

8 4G

Since P | PG H G , G andHare independent.

Alternative

P |H G =1

2and

4 1P

8 2H

Since P | PH G H , G andHare independent.

Alternative

1

P P(card is blue and has number 5)8

G H


35/136

2 1P

8 4

4 1P

8 2

P = P P

G

H

G H G H

Thus G andHare independent.

9(i)

(ii) Outlier :

17,1500.999r

3.5953 58.639 3.60 58.6y x x (iii) When 17x , 3.5953 17 58.639 119.76 120y

From GC, 0.999r or 2 1r Since the product moment correlation coefficient is almost +1, theregression lines is almost an exact fit to the data. Thus the regressionlines YonXandXon Ywill almost coincide.

y

x

100

120

140

160

10 15 20 25


36/136

10 LetXrepresent the weight loss with population mean .

From GC,x 0.7Unbiased estimate of population variance:

s2 = (2.3118)2 = 5.3444.

To test H0: 0

against H1: 0

One-tail test at 10% level ( = 0.10)

Since population variance is unknown and the sample size of 10is small, assuming X is normally distributed and s2 is a good

estimate of2,

under H0, T=0

2.3118/ 10

X~ t(9).

From GC,p-value = P( 0.7 when 0)X = 0.18166 = 0.182.

Sincep-value =0.182 > 0.10, we do not reject H0 and concludethat there is insufficient evidence at 10% level of significancethat the slimming companys claim is justified.

The probability of concluding that the pill helps people to lose

weight when it actually does not is 0.10.p-value = 0.18166 2 = 0.36332 = 0.363

For H0 to be rejected,p-value <

Thus > 0.363


37/136

11(i) X~N(40, 4.5 ), Y~N(20, 6.9 )X+ Y1 + Y2 + Y3 ~N(100, 163.08)

1 2 3P( 120) 0.058658 0.0587X Y Y Y

(ii) Let T= 20X8Y

T~N(640, 11147.04)P( ) 0.3 584.63T k k

(iii) Let Wbe the no. of observations ofTthat is more than k.W~ B(120, 0.7)Since n = 120 is large, np = 84 > 5 and nq = 36 > 5,W~ N(84, 25.2) approximately.P(W 100) = P(W> 99.5) after continuity correction

= 0.00101


38/136

12(i) LetXbe the no. of earthquakes in 10 years.

X~ Po(0.0000236510) i.e.X~ Po(0.073)

P(X 2) = 1P(X 1) = 10.99746 = 0.00254

(ii) One earthquake in the year 2000 to 2005 and none in the year2004 2009 One earthquake in the year 2000 to 2003 andnone in the year 20042009.

Let Y and Wbe the no. of earthquakes in 4 years and 6 yearsrespectively.Y~ Po(0.0292), W~ Po(0.0438)

Required probability = P(Y = 1)P(W= 0)

= 0.028360 0.95715= 0.027145 = 0.0271

(iii) Let V be the no. of periods where there are at least twoearthquakes, out of 200.V~ B(200, 0.00254)V~ Po(0.508) approximately.

P(V q) > 0.999

Using the GC, q 4

q P(V q)

3 0.99815 < 0.999

4 0.99981 > 0.999

Since n =200 is large, np = 0.508 < 5.

Let Tbe the mid-day temperature of the city.E(T) = 18, Var(T) = 16Since n = 100 is large, by Central Limit Theorem,

16~ N 18, approximately100

T

P(17 19) 0.988T


39/136

2009 HCI Preliminary Examinations H2 Mathematics Paper 1Solutions

1.

Let( 1)

be the proposition that for .

( 1)!n n

a n nP u n

n

Forn = 1, LHS of 1P = 2a

RHS of 1P=(1 1)1 2

= 2(1 1)! 1

a aa

LHS of1P = RHS of 1P

Thus,1P is true.

Assume thatk

P is true for some k .

i.e.( 1)

( 1)!

k

a k ku

k

To prove that1kP 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.

Since1P is true, and kP true 1kP is true. By Mathematical

Induction, is true for all .nP n

Limit of sequence is 0.

2.Given curve C is defined by:

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

x t t y t t t

(i) The graph of Cis


40/136

(ii)

Area of the shaded region

2

2

2

2

d 4 1

2cos cos 2 2cos 2cos 2 d 4

1.42478

1.4 correct to1d.p.

y x

t t t t t

x

y

O

22

1

1

2

y

O

2

x


41/136

3(i)Concave downwards f ''( ) 0x so the gradient of f '( )x


42/136


43/136

5

1

g( )3 2 3 1 3 4 3 2 3 1 3 1 3 4

1 (3 4) (3 2)

1Solving, we get

6

a br

r r r r r r r

a r b r

a b

(i)1

g( )n

r

r

=

f (1) f (2)

+f (2) f (3)1 1f (1) f ( 1)

...................6 6

f ( ) f ( 1)

n

n n

1

g( )n

r

r

=

1 1 1

6 4 3 1 3 4n n

(ii)

1 1 ...3 5 3 2 3 1 3 2 3 1 3 4n n n n n n

2

1 1 1

g( ) g( ) g( )

1 1 1 1 1 1

6 4 6 4 3 5 3 2 6 3 5 3 2

n

r n r r

r r r

n n n n

1

g( )n

r

r

represents the sum of area of rectangles drawn with breadth

of unit 1 and height g(r) forr= 1, 2, 3, ., n as shown in diagramwhile

1

1g d

n

x x

represents the area under the curve bounded byr= 1 to r= n + 1 which is smaller than the sum of area of rectangles.

1 2 3 4 n n + 1

y = g(x)

xO

y


44/136

6(i) 2( ) (4 )ay x x --- (1)

2 ( ) 4 2dy

a ay xdx

2 2

4 2 2

2

dy x x

dx a y a y

For tangents // to they-axis, is undefineddy

dx.

0y

Sub. 0y in (1), 0 (4 )x x

x = 0,x = 4

(ii) A = Area ofPQRS= 2y (PQ)

2 (4 )y x x , 4 2PQ x

(symmetrical abtx = 2)

A =

2

8(2 ) 4x x x

2

2

1 4 28 4 (2 )

2 4

dA xx x x

dx x x

2 22

84 (2 )

4x x x

x x

For 0dA

dx , 2 24 (2 ) 0x x x 2 2

4 4 4 0x x x x 22 8 4 0x x 2 4 2 0x x

4 82 2

2x

Sincex < 2,x = 2 2 .

Check maximum using first derivative test.

x ( 2 2 ) 2 2 ( 2 2 )+

f '( )x +ve 0 ve

A is maximum whenx = 2 2 .

7.

(i)

2

2

3 3 4f

2 1 3

x xx

x x


45/136

2

22

2 2

2

2

22

12

1

3 3 4Let

2 1 32 1 3

3 3 4 3 2 1

1Put 1

2

Equate coefficients of : 1Equate constant terms : 1

3 3 4 1 1Ans

2 1 32 1 3

1 2 1 3 13

1 1

x x A Bx C

x xx x

x x A x Bx C x

x A

x BC

x x x

x xx x

xx x

2

2

22

2

1 22 2

2!

11 1 1

3 3

1 11 2 4

3 3 9

4 5 35Ans

3 3 9

x x

xx

xx x x

xx

The expansion is valid when

2

2 1 and 1 [1]

31

i.e. and 32

1 1i.e. [2] Ans

2 2

xx

x x

x

2 2

1

2

1 1Coefficient of 2

3 3

12 Ans

3

n

n n

n

n

x


46/136

9(b)

nthbirthday

Amount of Mrs Lees contribution ($) on her daughtersnth birthday

0 1000

1 1000(1.05)+1000

2 (1000(1.05)+1000)(1.05) + 1000 =

21000 1.05 1000 1.05 1000

8(i)

5 4

f 11 1

xx

x x

Since any horizontal line,y = kcuts the graph of f at at most 1point,

f is 1-1. f1 exists.

Let4 4

1 11 1

y xx y

.

14 4

f 1 1 , \{1}1 1

x xx x

f1= f .

(ii)

51 50

2 1

f 4 = f f 4

f 4 f f f

5 4 1

1 4 3

x x x

(iii)

2

2

g 2 4

= 2 1 2 2

x x x

x

For fg to be defined, Rg Df = \{1}

2 + > 1 > 3.

g f 7

2, 2 , , 13

9(a) 3n n nS S S 3 2n nS S

3

2 3 1 2 12

na n n a n

7 1

2

na


47/136

3 21000 1.05 1000 1.05 1000 1.051000

= 3 2

1000 1.05 1.05 1.05 1

17

17 16 15

2

1000 1.05 1.05 1.05

1.05 1.05 1

18

17 16 15

2

1.05 1000 1.05 1.05 1.05

1.05 1.05 1

Mrs. Lee: 1000(1.05), 1.05a r

18

1050 1.05 1Total amount $

1.05 1

= $29500 (3 s.f.)

nth birthday Amount of Mr Lees contribution ($)on his daughters nth birthday

1 1.05 500 1.05 500

2 21.05 2 1.05 500 1.05 500

= 23 1.05 500 3 2 31.05 3 1.05 500 1.05 500

= 34 1.05 500

17 17

18 1.05 500

18 17

$1.05 18 1.05 500

18

Total amount $ 18 1.05 500

= $21700 (3 s.f.)

10(i) 22 2

d 11 2

d2 1 1

xx x

xx x

1 1

2

1

2

1 2

111 1 2

-1 -1

1cos d cos d

1

1 2cos d

2 1

cos 1 Ans

cos d cos 1 Ans

x x x x x xx

xx x x

x

x x x C

x x x x x

(a)

1

cos 2 , 1 3y x x


48/136

Translate the graph of 1cosy x two units in the positivedirection ofx-axis to obtain the graph of

1cos 2y x .

Area of the region R = 3

1

1cos 2 d x x

11

-1cos d where 2 ;

1 1 ; 3 1

2 Ans

t t t x dt dx

x t x t

(b) Volume generated

22

0 0

2

0

0

2

0

d cos 2 d

cos 4cos 4 d

1cos 2 1 4cos 4 d

2

1 9 9sin 2 4sin Ans

4 2 2

x y y y

y y y

y y y

y y y

11(i) 1d 2

1 1

A x x Ay x

x x

1, 1a b

(ii) When 0,x y c

When2

0, 01

x x cyx

1 1 4

2

1 1 4

2

cx

c

Since 0y when 0, 0 0x c c

(iii)

y

xO

3

R


49/136

Sketch graph of 1

fy

x

1 1

f( ) 1x x

2

2 2

11

2 1

1

x x cx

x

x x c x x

x c

1

1 1 4 12

c x or 1

1 1 4 12

c x c


50/136

12

(i)

16 16 12

0 , 8 , 8

4 6 7

OE OF OG

(ii)

Area ofOEG =1

2OE OG =

16 121

0 8 16 212

4 7

Area ofOEG =1

( )( ) 16 21

2

OE d , where dis the

perpendicular distance from G to OEd = 8.89

(iii) Equation of the plane containing pointsE,Fand G:

0 4

8 ; 8

2 3

0 4 4 1

4 8 4 4 11 3 16 4

1 16 1

equation of the plane is 1 0 . 1 32

4 4 4

EF EG

n

r

(iv) From (iii), equation of plane containing the pointsE,F

and G is11 32

4

r .

i

jk

O

E

F

G

16

8

7

A

BC

D


51/136

Since

0

0

0

does not satisfy the equation of planeEFG,

the points O,E,Fand G do not lie on the same plane.

(v) The required plane will be parallel to OFand the lineof intersection of planes OEG andEFG.

i.e. the plane is parallel to OFandEG.Thus, the equation is

8 4

4 8 ,

3 3

where , .

r


52/136

2009 HCI Preliminary Examinations H2 Mathematics Paper 2

No. Solution

1

2

d d3

d d

d3 3 3

d

d d3

d 3 d 3

y ux u

x x

ux x u x x u

x

u x u xx u u

x x x x

Method 1: (Partial Fractions)

2 2

2

d 3 1

d 33 3

3 1d

33

3

ln 33

3 ln 3 3

u x

x xx x

u xxx

x cx

y x x c

Method 2: (Splitting of numerator)

2 2

d 1 2 6 6

d 2 9 63

u x x

x x xx

22

d 1 2 6 6

d 2 9 6 3

u x

x x x x

21 6 3ln 3 or ln 32 3 3u x c x c

x x

21 33 ln 3

2 3

3 ln 3 3

y x x cx

y x x c

Whenx = 0,y = 0,

3 ln3 3 0 1 ln3c c

3 ln 3 1 ln3 3y x x

As 3, 3 0 0 3 3x x y .

Graph of 3 ln 3 1 ln3 3, 3y x x x .

2 1 3

Given 2 , 2

3 1

OA OB

y

x


53/136

Equation ofl:

3

2 ,

1

m m

r .

Method 1:

LetFbe the foot of perpendicular from Cto the line l,

3

2 0

1

CF

.

Thus,

3

( ) 2 0

1

OF OC

3 3

2 2 . 2 0

1

m k

m k

m k

39 3 4 4 0

7m k m k m k m k

SinceFis the mid-point ofCand 'C , ' 2OC OF OC

=

111

2713

k

(Shown)

Method 2:

Let OFbe the projection vector of OCon the line l.

Then

1 3 3

2 . 2 2

1 1 1

14 14

k

OB OBOF OC

OB OB

=

33

27

1

k

SinceFis the mid-point of C and 'C , using Ratio Theorem

SinceFis the mid-point of C and 'C ,

' 2OC OF OC

=

3 1 116 1

2 2 27 7

1 1 13

k k k

(shown)

WhenBCis perpendicular to OA,

0

3 1

2 2 . 2 0

1 3

3 1

2 2 . 2 0

1 3

3 4 4 3 3 0

1

3

BC OA

k

k

k

k

k

k

k k k

k


54/136

When k=1

3,

13 11

12 and ' 23 21

1313

OC OC

.

Hence,

131 11

2LHS = 6 15 6 2 15 23

3 1313

11

RHS = 21 ' 2 LHS

13

OA OC

OC

(Verified)

Method 1:

6 15 21 '

6 6 ' 15 15 '

6( ') 15( ')

6 ' 15 '

OA OC OC

OA OC OC OC

OA OC OC OC

C A CC

0

0

0

Thus, 'C A is parallel to 'CC . Since 'C is the common point, the pointsA, C

and 'C are collinear.

Method 2:

6 15 21 '

6 15'

21

OA OC OC

OA OC OC

Hence by Ratio Theorem, we can deduce thatA, Cand 'C are collinear.

3(i) Complex No

(ii)

(iii)

(iv)

(v)

4f(x) =

3

...sin 3!

xx

xe e


55/136

2 3 43 3 3

3 3! 3! 3!1 ...

3! 2! 3! 4!

x x xx x x

xx

3 4 3 421 21 ...

6 2 6 6 24

x x x xx x

=

2 4

1 ...2 8

x xx

sin( ) sin cos cos sin sin sin( )x x x x xe e e e

=2 4

1 ...2 8

x xx [Replace x by x]

(i)

2

2 2

2 2

23 2 2

3 2 2

d4 1

d

Differentiating with respect to ,

d d d d

4 2( 1) 2 ( 1) (1)d d d d

d d d d d d2 ( 1) ( 1) (2)

d d d d d d

yy

x

x

y y y y

y yx x x x

y y y y y yy y

x x x x x x

Subx = 0, 1y ,4 2 3 2

4 2 3 2

d d d d d d2 ( 1) 2

d d d d d d

y y y y y yy

x x x x x x

2

2 2

2 2

d d4 1 1 1

d d

d d2 (1 1)1 2 1d

y y

x x

y ydx x

3 32

3 3

4 4

4 4

d d 32 (1 1)1 1

d d 2

d 3 d2 1 (1 1) 2 3

d 2 d

y y

x x

y y

x x

Using the Maclaurins formula,

g(x) =2 3 43 3

1 ...2! 2 3! 4!

x x xx

=

2 3 4

1 ...2 4 8

x x xx

2 3 4 2 4

3 4

g( ) f ( ) 1 ... 1 ...2 4 8 2 8

4

x x x x xx x x x

x x

(ii)Asx 0 ,

3 4

g( ) f ( ) 04

x xx x

.

Whenx is close 0, the graph of g( )y x is close to the graph of f ( )y x OR

the graph ofy= g( )x is a good approximation to the graph ofy= f ( )x .


56/136

(iii)

Estimated error =3 4 4 5 51

4 4 4 5 10

k

k

k

k

x x x x kdx

5210

10

k 5 110k 0.631k 0 0.631k .

5(i) Let X denote the number of emergency telephone boxes in 1 km . Then

Po(0.6)X .

(ii) Poisson distribution model is not suitable since the average rate of occurrenceof the emergency telephone boxes in the rural area (including the city) is notconstant throughout this area as city and rural area have different density fordistribution of emergency telephone boxes

6a) No.Because not every resident will visit the shopping mall, or Consider total

population size, n, and take sample interval to be50

ninstead of choosing every

10th shopper (or any other equivalent / logical reason).

(b) The surveyor is given quotas to fill. An example is the table shown below:

GenderMale Female

25 25

The surveyor is given a free choice in picking the shoppers to fill the quotas.

(c) Increase sample size.Interview people from different shopping malls.Increase no. of categories or strata.

7 Case 1: Only 1 female does not know hip-hop

Prob =10 8 12

2 1 1

20 202 2

C C C

C C

or

10 9 8 122

20 19 20 19

= 0.119668 or216

1805

Case 2: Only 1 male does not know hip-hop

Prob =8 10 10

2 1 1

20 20

2 2

C C C

C C

or

8 7 10 102

20 19 20 19

= 0.0775623 or28

361

Case 3: All 4 people know hip-hop

Prob =10 8

2 2

20 20

2 2

C CC C

or 10 9 8 720 19 20 19

= 0.034903 or63

1805

P(A)=216

1805+

28

361+

63

1805=

419

1805or 0.232


57/136

P(B |A) = P(all 4 married | at least 3 know hip-hop)

=P(all 4 married at least 3 know hip hop)

P(at least 3 know hip hop)

=P(all 4 married all 4 know hip hop)


=P(all 4 married )


=

2 4

2 220 20

2 2

419

1805

C C

C C

or

2 1 4 3

20 19 20 19

419

1805

= 0.000716

P(B) =2 4

2 2

20 20

2 2

C C

C Cor

2 1 4 3

20 19 20 19 = 0.000166

Since P(B |A) P(B),A andB are not independent events.8

(i)

Assumption: The total no. of tickets in the box has to be large so that theprobability of getting a winning ticket is approximately constant after eachticket is drawn.LetXbe the number of winning tickets in a draw of 5 tickets.

1B(5, )

6X

P( 1)X = 1 P( 0)X = 0.5981224

Method 1:

Required probability = 23 P( 1)[P( 0)]X X

= 23 (0.5981224)(0.4018776)

= 0.2897996 0.290

Method 2:

Let Ybe the number of customers out of 3 with at least one winning ticket.B(3, 0.5981224 )Y

Required probability = P( 1)Y = 0.2897996 0.29

ii) Since n = 60 is large, by the Central Limit Theorem,

5 5~ N( , )

6 432

X approximately

Required probability = P( 1)X = 0.0607The event that all 60 customers each obtains at least one winning ticket is asubset of the event that the average no. of winning tickets obtained by the 60customers is at least one. Hence the probability is smaller.

9i) Case 1: I_ _ _ _ _ _ _ _ _ I:

No. of ways =9!

2! 2!= 90 720

Case 2: I_ _ _ _ _ _ _ _ _ O or O_ _ _ _ _ _ _ _ _ I:

No. of ways = 2 9!2! 2! 2! = 90 720

total no. of ways = 181 440

(ii)No. of arrangements =

11!

3! 2! 2!

8!

2! 2!


58/136

= 1 663 20010 080= 1 653 120

(iii) The first letter T can be placed at the 1st, 2nd, 3rd, ..., 7th position.The second T is then placed 4 positions after the first T.

No. of ways = 7 9!

3! 2!1 = 211 679

(case where it is correct word is excluded)

10(i)

r= 0.972Even though r is close to 1, it does not mean that there is a cause and effectrelationship betweenx andy.

(ii)

The points on the scatter diagram lie close to a straight line with positivegradient. This agrees with the value ofrobtained in part (i).

(iii)

For the modely = abx, i.e. lny = ln a +x ln b, r= 0.98759

For the modely = axb, i.e. lny = ln a + b lnx, r= 0.97789Since | r| = 0.98759 is closest to 1 for the modely = abx, this is the most

suitable model.(iv) lny = 0.128367x + 0.9994

ln a = 0.9994 a = 2.72

ln b = 0.128367 b = 1.14

11a) H0 : = m

H1 : > m

X ~ N(m, )100

2sapproximately.

Using z-test with n = 100,

x =48005

100

+ 120 = 600.05; 600.05 100 60005x

s2 =99

1[36 990 985

260005

100] = 9949.340909

105 x

10

5

y


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Hence, Z =9949.340909 /100

X m~ N(0,1)

To reject H0 at 5% level of significance,

600.05

9949.340909 /100

m 1.644853626

m 584 (3s.f.)b(i) If the claim that the mean SAT score has not improved is in fact true, 100p% of

tests using same method will lead to wrongly accepting the claim that the meanSAT score has improved.

(ii) t-test should be carried out because sample size = 10 is small and populationvariance is not given.Assumptions:The SAT score of students in the school follows a normal distribution.The SAT scores of students are independent (or randomly selected) from oneanother.


60/136

(iii) Not necessarily true that the same conclusion as t-test is reached (i.e., the nullhypothesis is not rejected) ifz-test is carried out.

Reason:

For a fixed level of significance level , z < t (refer to graph above).

When H0 is not rejected for a t-test, there are two possible conclusions for az-test.

1) z < test statistic < t : H0 will be rejected ifz-test is used.

2) test statistic < z

< t : H0 will not be rejected ifz-test is used.

OR

Let andt zp p be the p-values obtained using a t-test and z-test respectively.

Note that for the same test statistic, >t zp p

Since H0 is not rejected for a t-test, pt >100

. There are two possible

conclusions for az-test.

1)tp >

100

zp : H0 will be rejected ifz-test is used.

2)t

p > zp >100

: H0 will not be rejected ifz-test is used.

12(i) Let Tbe the travel time taken by Andy to reach the pick-up point.2N(18, 4 )T

P(Andy takes bus B) = P(15 25)T

= 0.733313 0.733

(ii) P(Andy misses the buses) = P( 25)T = 0.040059LetXbe the number of times in 8 work weeks that Andy misses the bus.

B(40, 0.040059)X

test statistic

N(0,1)

t( )

N(0,1)

t( )


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Since n is large and 1.60236 5np ,Po(1.60236)X

Required probability = P( 4)X = 0.0791

(iii) Let2T be the average time spent on travelling to the pick-up point on the first

two days of the week and3T be the average time spent on travelling to the

pick-up point on the last three days of the week.2

2

4N(18, )

2T and

2

3

4N(18, )

3T

2 3

40N(0, )

3T T

2 3P(0 5) 0.414548T T

0.415

(iv) Let the number of minutes that Andy has to leave the home earlier be aminutes.

P( 15 ) 0.90T a or P( ) 0.90T k

From GC, P( 23.126) 0.90T Hence 15 23.126a 8.126a Thus he should leave his home at least 9 mins earlier.i.e his latest time to leave home is 6.51 a.m.


62/136

2009MJCH2MATHS (9740)JC2PRELIM EXAM P1SOLUTIONS

Qn Solution

1 MOD

2

2 1 1

6 8 2 4r r r r

21 1

1 153

1 14 6

1 11 3

1 1

2 41 1 1 13 4 3 4

7 1 112 3 4

2 1 1

6 8 2 4

n n

r r

n n

n n

n n

n n

r r r r

2

2 21 1

21

7 1 1

12 3 4

70 1 13 3 4

6 88 801

6 8 6 8

240

6 8

40

40

n n

r r

n

r

n n

n n

r r

r r r r

nr r

n

n

2 Mathematical InductionRecurrence Relation

LetnPbe the statement

2

11

1

nn

xS for all n .

To prove 1P is true:

2

1

1

1

1

1:LHS

1

1

1

1

x

xS

i i

LHS2

1

2

11

243

11

2

11:RHS

1

2

1

2

xx

x

(sub in 31 x )

true.is1P


63/136

Assume i.e.,somefortrueis kPk2

11

1

kk

xS .

To prove i.e.,truealsois1kP2

11

2

1

k

kx

S .

RHS.2

11

243

11

23

11

12

211

1

1

2

11

11

211

1

1

1

1

1

1:LHS

2

1

2

1

1

2

1

11

11

11

11

1 1

1

1

1

k

kk

kk

kk

kk

kk

kk

k

i ki

k

i i

k

x

xx

xx

xx

xx

xx

xx

xx

xS

1

1

1

is true is also true.

Since is true

and is true is also true,by Mathematical Induction, is true for all .

k k

k k

n

P P

P

P P

P n

Q3 Integration by substitution

xxx

xd

22lnln

2lnln tx e2

tt

x

tt

t

dd

d

22lne2lne2

2lne2ln t

t

xe2

d

d

ttt t

tde2

22ln2lne22ln2ln

ttt

d2


64/136

2 2d

2

tt

t

1

22

2 d2

t tt

(shown)

4

2

e2

e2

d

22lnln

2lnlnx

xx

x

4

2

2

1

d2

22 t

tt

4

2

2

1

2

3

2423

2

tt

2

1

2

3

244243

2

21

2

3

2423

2

423

22 21

3

216

Therefore 16a , 3b Q4 DE

LetB denote the number of births andD denote the number of deaths.

d d

Given d d

B B

x axt t

and 2 2d d

d d

D Dx bx

t t where a, b are positive constants

2

2

2

d d d

d d d

d(shown)

d

dGiven that 0 when 200,d

200 200 0

200

d(200 )

d

x B D

t t t

xax bx

t

xxt

a b

a b

xax bx bx x

t

1d d

(200 )x b t

x x


65/136

1 1 1d

200 (200 )

1ln ln(200 )

200

1ln

200 200

x bt dx x

x x bt d

xbt d

x

ln200

e200

e , e200

200e

200e 1

kt r

kt r

kt

kt

xkt r

x

x

x

xA A

x

xB

x

xB

As200

, 200e 1kt

t xB

Hence there is a limit to the size of the population. (shown)

200 is the maximum number of ToTo bears the environment can support.

Q5 VectorsExclude Planes and Lines(i)

(ii)

(iii)

Since AB b a andA, B and Care collinear,2k

2 2

3 2

OC OB BC

b+ b a

= b a

2

2

3

43

44

16

3

4 4 3

33

.

aa b

a

a

a

a

Alternatively,


66/136

(iv)

^ ^2

2

2

2

3

4

43

4

34

4

16

3

4 4 3

33

b a aa b a a

a

aa

a

a

a

a

2 2 2

0

b 2 2 2 2 3

a.bcos

a b4

4 32 3

3

1

2

or 603

A

B

CO

N

D

E


67/136

2

3

2 2 3

3

2

2

2

DC DADE

a a b

b

=

Alternatively,

2 2

3 2 2 23 3

2 3 2 2

2

2

2

AE AE

DE DA AE CO AE

b a a b a

a b b a

b

Q6 Complex NumbersLoci

(i)

(ii)

(iii)(iv)

Q7 Complex NumbersExclude Loci

(a)

(b)

Q8 Binomial Expansion and Maclaurins Series

(a) Method 1

Let f ( ) 1 en

xx .

1

f ( ) 1 e en

x xx n

2 2 1f ( ) 1 1 e e 1 e en nx x x xx n n n

Therefore, f (0) 2n ; 1

f (0) 2n

n

and 2 1 2

f (0) 1 2 2 1 2n n n

n n n n n

2

1 2

1 3 2

1 2f ( ) 2 2 ...

2!

2 2 1 2

n

n n

n n n

n nx n x x

n x n n x

Method 2

Using standard series2

e 12

x xx (taking the first three terms only)


68/136

2 2

1 1 22 2

n n

x xx x

2

2 12 4

n

n x x

2

2 212 1

2 4 2 2 4

nn nx x x x

n

22 12 12 4 2 2

nn nn n x

x x

212 12 4 8

nn nn n

x x

23

12 1

2 2

nn nn

x x

1 3 22 2 1 2n n nn x n n x

(bi) Method 1

3 2

xy

x

1

1

2

2

2 3

(3 2 )

2(1 )

3 3

1 22 21 1 ...

3 3 2! 3

2 41 ...

3 3 9

1 2 4...

3 9 27

y x x

xx

xx x

xx x

x x x

Method 2

3 2

1 3 1

2 2 3 2

xy

x

x

1

1 3 1

2 2 3 2

1 33 2

2 2

x

x


69/136

1

2 3

2 3

2 3

1 3 1 21

2 2 3 3

1 2 1 2 31 1 2 2 21 1 ...

2 2 3 2 3 3 3

1 1 1 2 4...

2 2 3 9 27

1 2 4...

3 9 27

x

x x x

x x x

x x x

(bii) 21

3x

Therefore the set of values ofx =3 3

: ,2 2

x x x


70/136

(biii

) 3 2

xy

x

2 31 2 4 ...3 9 27x x x

Clearly the coefficient of nx follows a GP with first term1

3and common ratio

2

3

Therefore coefficient of nx = 1nnT ar =

11 2

3 3

n

Let f ( )y x .

n2f 0 f 0f f 0 f 02! !

nx x x xn

By considering the coefficient of nx for the Maclaurins expansion, at 0x ,

1nf 0 1 2

! 3 3

n

n

1

n 1 2f 0 !

3 3

n

n

.

Q9 VectorsPlanes and Lines

(i)

(ii)

(iii)

Sub , ,0 into 1 and 2.

1

2

: 1( ) 3( ) (0) 8

: 3( ) 1( ) (0) 0

a

b

Using GC, 1, 3

1 3 3

3 1 3

8

b a

a b

a b

1

1 3

: r 3 3 ,

0 8

b a

l a b

1 3

3 3

0 8

b a

a b

5

2

2

+

4

1

0

1

4

Sub a b and1

4 , we have


71/136

(iv)

11 4 5 4

4

13 4 2

4

a

a

22

a

b

Since the 3 planes have no point in common, l1 cannot intersect 3.

1

1 1

: r 3 1 ,

0 1

l

Condition 1:

1,3,0 is not on 3.Hence,( 1) (3) (0) 11

4

p q

p

Condition 2:

1

p

q

and

1

1

1

are perpendicular.

1 1

1 0

1

1 0

1

.p

q

p q

p q

Q10 AP, GP

(a)Given

3

31 26 26

1 0, 11 27 1 27

a r ar a r

r r

3 1

27

1

3

r

r

Since we have 2 1ar ar d ar r d

1 23 3

2

9

a d

a d


72/136

Therefore the sum of the first fifty-five terms of the A.P.

55 22 54 275 (Ans)

2 9a a a

(b) Arithmetic Progression with first term, 1 30T , common difference, 5d , in minutes.

(i) 7 1 6 30 6 5 60T T d minutes = 1 hour.

(ii) 77

30 60 3152

S minutes = 5.25 hours

(iii) To determine the nth lesson,where Ann would have completed 60 hours (3600minutes) of lesson. Note n must be a positive integer.

2 30 ( 1) 5 36002

nn

55 5 7200n n 2 11 1440 0n n

32.8 or 43.8n n (Note that students can use GC to get n)

Hence during the 33rd lesson, Ann will have completed a total of 60 hours.

Alternatively,

2 30 ( 1) 5 36002

n

nS n

Using GC,

when n = 32, 3440 3600nS when n = 33, 3630 3600nS

Hence during the 33rd lesson, Ann will have completed a total of 60 hours.

Q11 Differentiation (Applications and Graphs of f)

a

b d dsin ; cos .d d

dcot

d

x ya t a t t t

yt

x

x0B (-3, 0)

A (-1,0)

x = -2

y

y = 1


73/136

When

2

ax ,

3t

and

3

2

ay

Equation of tangent:

3 3

2 3 2

3 3 3

3 3 2 2

3 3 4

3 2 3

3 3 2

ay a x

ay x a

x a

y x a

Atx = 0:2 3

3y a

Aty = 0: 2x a

Therefore, area of triangle OQR = 21 2 3 2 3

22 3 3

a a a

C Let the internal height be h, external length be Yand the external height be Z.

Volume = 1000 3cm

2

2

10001000x h h

x

Also, 2Y x and 1Z h Material needed, 2 2S Y Z x h

2 2

2 2

2

2

1000 10002 1

10004 1 1

x xx x

x xx

3 2

4 3

d 2000 10002 4 1 4 1 0

d

2 4 4000 8000 0

2 (rejected) or 12.6 cm

Sx x

r x x

x x x

x x

x 12.599 12.599 12.599

d

d

V

x

Answers for non-contextual questions have to be exact when using the test.


74/136

2009 MJC JC2 Preliminary Exam

H2 Mathematics (9740) Paper 2 Suggested Solution

Qn Solution

1 Integration by parts

(i).

(ii)

Qn Solution

2 Graphs

(a)

(b)(i)

f 2 1y x

cos cosd e e sind

x x xx

cos

e sin 2 dx

x x cose 2sin cos dx x x x

cose sin 2cos dx x x x cos cos2e cos 2 e sin dx xx x x cos cos2e cos 2ex xx C

x

y

4y

1,

2a

f 2 1y x

1, 0

2

b

1, 0

2

b

y

xO

3x

0y

8, 0.2

11

9

11

2

2

2 11

3

xy

x

x

y

4y 1, a

3x 5x

f 1y x

1, 0b 1, 0b


75/136

(ii)

Qn Solution3 Equations (System of equations) and Inequalities

(a) 2

2

d2

d 2

by a x cx d

x

y a bcx

x xx

Augmented matrix

1 1 1 1 9

0.5 1 2 0 0

1 232 16 1

4 4

1 2173 81 1

9 3

OR

9

12 0

21 23

2 164 4

1 2173 81

9 3

a b c d

a b c

a b c d

a b c d

Using GC, 2, 3 1 15a b c d

The equation of the curve is 23

2 15y x xx

(b)

(i)

224 4 3 2 1 2 0x x x x

Alternative Method

Discriminant = 24 4 4 3 32

Since coefficient of 2x is positive and discriminant < 0,

2 fy x

0, a

0, a , 0b , 0b

4x 4x

2y

2y

y

x


76/136

24 4 3x x is positive(ii)

Therefore or

(iii)

3 2

2

4 ln 4 ln 3 ln0

ln ln 2

x x x

x x

Replace by

The solution is

or

Qn Solution

4 Functions

(i)

Either:

Any line cuts the graph twice; hence f is not 1-1 and does not exist.

OR:

3y cuts the graph twice; hence f is not 1-1 and does not exist.

The largest value ofkis 1.

3 2

2

2

2

4 4 30

2

4 4 30

2 1

0 since 2 1 2 02 1

x x x

x x

x x x

x x

x

x xx x

2x 0 1x

x ln x

2 ln 0

ln 0 ln 0

1

x

x x x

x

1

ln 1

ln 1 or ln 1

0 e or e

x

x x

x x

, 2y k k 1f

1f

0 1

++

-2

Ox

y


77/136

1

2

2

2

2

21

ff

21

1Let ,

1 0

4

2

4

Reject 0 1.2

4

2

4f

2

D R 2,

4f : , 2.

2

y xx

x xy

y yx

y y

x x

y yx

x xx

x xx x

For hg to exists, .

, .

Hence, the minimum value of is 9.

To find the ,

Method 1

Method 2

From the graph,

.

Qn Solution

5 IntegrationArea and Volume

(a) 22 2x y a a

22 2x a y a

g hR D

gR 9, hD 0,

hgR

g g h hgD 2,2 R 0,9 D R ln 2,ln11

hgR ln 2,ln11


78/136

Volume formed 2 22

0d

a

a y a y

2

2 2 2

02 d

a

a y ay a y OR 2

32

0

1

3

a

a y y a

2

2 2 2

02 d

a

a y ay a y 3 3 31 12 0

3 3a a a

2 2

02 da y ay y

343

a

23

2

03

a

yay

3

38 43

aa

34

3a

Volume of sphere with radius a is34

3 a

Therefore volume of the semi-circle obtained when rotated 2 radian about they-axis isequal to the volume of a sphere with radius a

When

2

3lnx 2 t

5

24lnx 5 t

ttt

tx ln1ln

2

11ln

22

1

2

tt

t

t

x 1

1

2

2

1

d

d2

11

d

d2

ttt

x

Area of required region 5

24ln

2

3ln

dxy

5

2

dd

dt

t

xy

5

22

2 d1

185 t

tttt

5

22

2

d1

85t

t

t

5

22

d1

35 t

t

5

2

3 15 ln2 1

ttt

3 4 3 125 ln 10 ln

2 6 2 3


79/136

3 1

15 ln2 2

OR3

15 ln 22

Qn Solution

6 Sampling Methods

A sample is a small representative of the population.

In many practical statistical investigations, it may not be possible to collect information forthe entire population due to time or cost constraints (or any other sensible reasons). Henceit is desirable to use the results obtained from a sample to make statistical inferences aboutthe population.Any sensible real-life example which gives a complete sampling frame and ensures thatevery member of the population has an equal chance of being selected.

Qn Solution

7 Hypothesis Testing

Unbiased estimate of population variance = 280

11.2379

= 127.71

H0: 1000 H1: 1000

Since n = 80 is large, by Central Limit Theorm,2

~ ,s

X Nn

approximately

Test statistic :X

Z

Sn

Level of significance: 4%

Critical Region: Reject H0 ifp-value < 0.04

Assuming H0 is true, from the graphic calculator,p-value = 0.0687

Conclusion: Sincep-value = 0.0687 > 0.04,we do not reject H0 and conclude that there is no significant evidence,

at 4% level, the mean volume is not 1000 ml.


80/136

Qn Solution

8 Permutation and Combination

(a)(i) 1 B, 2 E, 1 A, 3 N, 3 T

8!No. of ways 1680

3!2!2!

(ii) Case 1: _B_E_E_A_

Slot in TTT first in54!

32!

_T_B_E_T_E_T_A_Follow by slotting in NNN.

So no. of ways for case 1 = 5 84!3 32!

= 6720

Case 2: _B_E_E_A_


2!22!

2! Because TT and T can swop.

_T_B_E_T_T_E_A_Follow by slotting in NNN (1 N must slot in between TT)

So no. of ways for case 2 =5 74!

2!2 22!

= 5040

Case 3: _B_E_E_A_


12!

_B_E_T_T_T_E_A_Follow by slotting in NNN (2 N must slot in between TTT to separate them)

So no. of ways for case 3 =5 64!

1 12!

=360.

Total number of ways = 6720 + 5040 +360 = 12 120

(b)

(i) No. of ways =5 10

No. of ways 8! 423360004 4

(ii)3 4 10 9 8!

No. of ways 653184001 2 1 2 2!2!

Qn Solution

9 Probability


81/136

(a)

(i) P(all three cards even) 10 9 8 220 19 18 19

or

10

3

20

3

219

(ii) P(exactly one even) 10 10 9 1520 19 18 38

3

1

or

10 101 2

20

3

1538

(b)

(i)

(ii)

(iii)

16 16 154 4520 19 20 19P B

15 161 420 19 20 19

79380

P( ) P(1st card 5 & 2nd card 5)

P(1st card 5 & 2nd card 5) P(1st card 5 & 2nd card 5)

A B

79P( ) 380 79

951P4

PA B

AB A

Using tree diagram:

16 15 15 154 1 4

520 19 20 19 20 19P B

16 15 794 120 19 20 19 380

P( )A B

79P( ) 380 79

951P4

PA B

AB A

B( 5)

B( 5)

B( 5)

B (< 5)

B (< 5)

B (< 5)

14

5

620


82/136

Qn Solution

10 Binomial and Poisson Distributions (include approximation)

(i) Let CandPbe the number of cheese and peanut waffles sold in a randomly chosen 15-minute period respectively.

C~ Po(4) , P~ Po(5)C+P~ Po(4 + 5) Po(9)

P( 10) 1 P( 9)C P C P = 0.41259

0.413 (ii) Required probability = P all waffles sold are peanut flavour | at least 10 waffles are sold

P all waffles sold are peanut flavour and at least 10 waffles are sold

P at least 10 waffles are sold

P( 10) P( =0)

P( 10)

P C

C P

= 1 P( 9) P( =0)

P( 10)

P C

C P

= 0.031828 0.018316

0.41259

= 0.00141

(iii) P(at least 10 waffles are sold in each of the 4 consecutive 15-minute period in a randomly

selected hour) = 4

0.41259 or 4

0.413

= 0.0290 or 0.0291 (to 3 s.f.)

AlternativelyLet Tbe the number of 15-minute periods in a randomly chosen one-hour period where atleast 10 waffles are sold.

T~ B(4, 0.41259)

Required probability P( 4)T = 0.0290 or 0.0291 (to 3 s.f.)

(iv) LetX be the number of waffles sold in an hour.

X~ Po(9 4) oP 36X

Since 36 10 ,

N 36,36X approximately

P 40 P 39.5X X after continuity correction= 0.280 (to 3 s.f.)


83/136

LetA be the event that at least 10 waffles are sold in each of the 15 minute period in anhour.LetB be the event that at least 40 waffles are sold in an hour.Obviously, eventA is a subset ofB and hence the probability ofA is smaller than that ofB

Qn Solution

11 Correlation and Regression

(a)(i) 0.89241 0.892r (3 s.f.)

(a)(ii) 0.95956 0.960r

(3 s.f.)Since 0.960r is closer to 1 than 0.892r , the model in (i) is less suitable than themodel in (ii).

(a)

(iii)

Regression line ofFonx: 0.35903 0.029245F x 0.359 0.0292F x

Regression line ofx onF: 204.51 31.484x F 205 31.5x F

(a)

(iv)

Using 0.35903 0.029245F x ,2100 0.35903 0.029245t

58.37047 58.4t s (3 s.f.)(b)

(i) Regression line ofx ony : 3 2 51x y 51 2

3 3x y

Regression line ofy onx : 8 11 36x y 36 8

11 11y x

Since8 2

11 3bd

is negative and 2r is always positive, therefore 2bd r .

Hence 8 11 36x y is not valid.OR

Since b and d must be both positive or both negative, but8

11b

and

2

3d

,

hence 8 11 36x y is not valid.

(b)

(ii)

From51 2

3 3x y , we have

2

3d .

Gradient of regression line ofx ony in graph ofy againstx is1 1 3

2 2

3

d

Since 0.943r ,Using 2bd r ,

y


84/136

22 ( 0.943)3

31.33

2

b

b

Therefore regression line ony x has more gentle slope than regression line onx y .

Qn Solution

12 Normal Distribution (include Sample Mean CLT)(a)

By symmetry,155 185

1702

P 155 0.025

155 170P 0.025

151.95996

7.6532

X

Z

2

58.6 (3 s.f.)(i) 21 2 32 ~ 0,6X X X N

1 2 3E 2 E E 2E 0X X X X X X

21 2 3Var 2 2Var 4Var 6X X X X X

1 2 3P 2 5X X X

1 2 3P 2 5

0.39484

X X X

= 0.394 (3 s.f.)

(ii) 2~ ,

50X N

(SinceXis normally distributed, there isnt a need to use CLT.)

P 1720.96769

X

= 0.968 (3 s.f.)

(b) Since n is large, by CLT25

~ ,X Nn

approx.


85/136

P 1 0.99

1P 0.99

5

1 1

P 0.995 5

11 2P 0.99

5

1P 0.005

5

0.2 2.57580.2 2.5758

165.87

X

Z

n

Zn n

Z

n

Z

n

n

n

n

Least n = 166.


86/136

2009 RJC H2 Maths Preliminary Examination Paper 1 (Solutions)

1

Fromthegraphiccalculator,thex-coordinateof the3pointsof intersection are:4.4140, 1.3005 and 6.4352. (5 s.f.)

Now, 3 21 11 2 512 6

x x x x

3 21 1

3 1 6 154 2

x x x x .

Hence, from the graph, 4.41 1.30 or 6.44 x x . (3 s.f.)

2

(i)

(ii)

2 e 3 5 (1)xy y

Differentiating (1) with respect tox,

d d2 e 3 0

d dd

(2 3) e 0 (2)d

x

x

y yy

x xy

yx

Differentiating (2) with respect tox,

2

2

2 2

2

d d d2 (2 3) e 0

d d d

d d2 2 3 e 0 (Shown) (3)

d d

x

x

y y yy

x x x

y yy

x x

2 e 3 5 (1)xy y

Whenx = 0, 2 3 4 0

( 1)( 4) 0

y y

y y

(from (1))

y = 1 or4 (NA sincey > 0)

d 1When 0,

d 5

yx

x (from (2))

2

2

d 27

d 125

y

x

(from (3))

So 21 27

1 ...5 250

y x x

3

15

y

x0

(4.41,10.2)

(1.30,6.90)

(6.44,22.3)


87/136

3 Floor area = 1

4 2 2 sin2

4 2sin

Unit cost of carpeting =8000

4 2sin

[MF15(under Maclaurins series) : sin ...

Since is small, sin (as the third and higher powers of are negligible).]

1

1

2

8000

4 2

4000 2

14000 1

2 2

2000 1 (Binomial series)2

$ 2000 1000 /m

So a = 2000, b = 1000

4(i)

(ii)

3 2

3

sec tan d (sec tan )(sec ) d

1sec

3

x x x x x x x

x c

Let u x . Thend

2d

xu

u

2

2

1 d2 ( 9)

1(2 ) d (I)

2 ( 9)

1d

9

1 3ln

6 3

1 3ln +c

6 3

xx x

u uu u

uu

uc

u

x

x

5(i) Since 2(2) 0 2 2 , the point with position vector 2 2i klies on 2.

A direction vector for the line of intersection is

1 2 5

3 1 5

2 1 5

.

An equation of line of intersection, l, is

2 1

0 1 ,

2 1

r


88/136

5(ii) From GC

3 2 6

2 2

3 12

6, 4, 6

x y z

x y z

x y z

x y z

The 3 planes intersect at a common point with position vector64

6

.

5(iii) For the 3 planes to intersect at a common line, line l must lie on plane 3.

2 1 1

0 1 4

2 1 1

(2 ) (2 ) 4

4

4 or 0

b b

b b

b b

b

--- (*)

Equation (*) must be true for all real values of.So b = 0.

Alternative Method 1

2 1

0 4 0

2 1

b b b

and

3 1

1 4 4 0

3 1

b b b b b

Since 2 points on the line l lies on the plane 3, the line lmust then lie on the plane

3. Thus b = 0.

Alternative Method 2

2 1

0 4 0

2 1

b b b

and

1 1

1 0 0

1 1

b b

Since pointPon line l lies on plane 3 and line lis parallel to 3, therefore line l

must then lie on the plane 3. Thus b = 0.

Alternative Method 3If the 3 planes intersect at a common line, the common line of intersection betweenany of the 2 planes must be the same.

For the line of intersection between 1 and 2 to be the same as the line of

intersection between 1 and 3 ,Pmust lie on 3 and the 2 lines must be parallel.


89/136

2 1

0 4 0

2 1

b b b

and

1 1 1

3 1 for some

2 1 1

3 2 1

3 1

3 1

3 and 3 2 3 and 3 30

b

b

b

b bb

Thus b = 0.

6(i)

Since 32

52

3

MN

and

4

3

5

BH

the acute angle between the lines 1l and 2l

32

5 9 2521 1 2 2

32

5

2

3 4

3

512

cos cos 80.350 17.53 4

3

5

6(ii) A vector normal to the plane EFGH is k

The acute angle between the plane EFGH and a line parallel to AP

1 1

12 01

21 010

35 1 3.5sin sin 55.4

12 0 18.1

1 21 0100

35 1

7(i) For the population to grow, population at the end of (n+1)th week must be greaterthan the population at the end ofnth week.

1

1 0

4 0

n n

n n

n n

u u

u u

u qu

Re-arranging, 14

4(1 )1n

u qq

.

Hence there must be initially 14(1 )q fruitflies in the jar.

7(ii) Given that 1 10u and 2 15u , we have


90/136

15 2 10 4q

which gives 0.1q .

The recurrence relation is 1 1.9 4n nu u .

Using the GC to evaluate the terms, we have 20 1100000 1000000u (1 s.f.)

GC key strokes:Press MODE, select seq at the 4thline. Press y=. Enter the following:

Press 2nd, GRAPH.

This is highly impractical as the formulation of the recurrence relation did not takeinto account the physical constraints of the jar.

7(iii) The suggestion should not be taken up as the differential equation gives a continuousmodel as an approximation which only works when the population is large enough.

8 Let Pn be the statement

22

( 1)! 1 ! 2n

r

r r n

where , 2.n n

When 2n ,LHS 22 1! 4 , RHS 3! 2 6 2 4 Hence, P2 is true.

Assume Pk is true for some , 2.k k

22

i.e. ( 1)! 1 ! 2k

r

r r k

.

Want to prove that Pk+1 is true,

1

2

2

i.e. to prove ( 1)! 2 ! 2k

r

r r k

.

LHS = 2 22

( 1)! ( 1) !k

r

r r k k

1 ! 2 1 1 !k k k

1 !(1 1) 2

2 ! 2

k k

k

RHS

Hence Pk+1 is true whenever Pk is true. Since P2 is true, by using mathematical

induction, Pn is true for all , 2.n n

Replacing ( 1) withr r


91/136

12 2

2 3

2 2

2

( 1) ! ( 1)!

( 1)! 2 (2 1)!

( 1)! 2 4

( 1)! 6

n n

r r

n

r

r r r r

r r

n

n

Alternatively

Replacing with ( 1)r r

12 2

2 1

12 2

2

12 2

2 2

( 1)! ( 1) !

( 1) ! 2 (1)!

( 1) ! ( 1)! 2

( 1)! 4 2

( 1)! 6

n n

r r

n

r

n n

r r

r r r r

r r

r r r r

n

n

9

(i)

(ii)

Equating the two equations, we have

4 2 3 04

x x

Solving,1

2x or

1

2 x

1

2y or

1

2y

The coordinates of pointA andB are1 1

( , )22

and1 1

( , )22

respectively.

Area ofR = 2

1 3

22 2 2

10

2

3d d

4x x x x

= 2 ( 0.11785 + 0.05403)= 0.34 (2 d.p.)

Volume =

1

2 2

0

3d

4y y y

=

1

3 2 2

0

34 3 2

y yy

=5

24


92/136

10(i) b = 2

10(ii) 22 3x ax ay

x b

2 2

2 2

d (4 )( 2) (2 3 ) (2 8 5 )

d ( 2) ( 2)

y x a x x ax a x x a

x x x

ForCto have no stationary points, d 0 has no real roots.dyx

2i.e. 2

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