2009 CJC

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CATHOLIC JUNIOR COLLEGEGeneral Certificate of Education Advanced LevelHigher 2

MATHEMATICS 9740/01Paper 1 3 September 2009

3 hoursAdditional Materials: Answer Paper

Graph PaperList of Formulae (MF 15)

READ THESE INSTRUCTIONS FIRST

Write your name and class on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.You may use a soft pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer allthe questions.Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case ofangles in degrees, unless a different level of accuracy is specified in the question.You are expected to use a graphic calculator.Unsupported answers from a graphic calculator are allowed unless a question specifically statesotherwise.Where unsupported answers from a graphic calculator are not allowed in a question, you arerequired to present the mathematical steps using mathematical notations and not calculator

commands.You are reminded of the need for clear presentation in your answers.

At the end of the examination, arrange your answers in NUMERICAL ORDER. Place thiscover sheet in front and fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.

Question No. Marks Question No. Marks

1 / 4 7 / 10

2 / 6 8 / 11

3 / 6 9 / 11

4 / 7 10 / 14

5 / 7 11 / 14

6 / 10 TOTAL / 100

-----------------------------------------------------------------------------------------------------------------------------------

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CJC Preliminary Examination 2009 / 9740 / I 2 of 5

1 Solve the inequality 0432 xx .

Hence find the exact value of 23

1

2d43 xxx .

[1]

[3]

2 Referred to the origin O, the position vectors of the points A and B are3i+ 9k and3i + 3j3krespectively.

(i) Find the position vector of the point Mon the line segment AB such thatAM :AB= 2 : 3.

(ii) Show that OM is perpendicular toAB.(iii) The point C is the reflection of the origin O in the line AB. Calculate the

area of triangle OAC.

[2][2]

[2]

3 Find

(i) xxx dsin ,

(ii) xxx

d1

1

, where 10 x , using the substitution ux 2cos .

[2]

[4]

4 The rth term of a sequence is given by ur= r(3r+ 1), r= 1, 2, 3, .

(i) Write down the values of

n

r

ru

1

for n= 1, 2, 3, and 4.

(ii) Make a conjecture for a formula for

n

r

ru

1

, giving your answer in the form

nf(n), where f(n) is a function of n.

(iii) Prove by induction a formula for

n

r

ru

1

.

[1]

[1]

[5]

5 Use the substitutionx

uy to show that the differential equation

xx

yxy

x

y

2

122

d

d2

reduces to the differential equation

2

13

d

d

x

u

x

u. Hence find

the general solution foryin terms ofx. [7]


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6 Given that 21

e x

y , prove that

x

yxy

x

y

d

d2

d

d2

2

.

Hence expand 21e x in ascending powers of xup to and including the term in

x4.

Use this series expansion to estimate the value of

x

x

de21

0

4

4 2

, correct to 3 decimal places.

[2]

[6]

[2]

7 A candy maker is interested in using containers in the shape shown below topackage his candies. The container is made up of an open cylinder of height hcm and radius rcm, with a hollow hemispherical lid of radius rcm.

In order to minimise production cost in this difficult time, the candy makerwants to use containers with the least surface area while maintaining thevolume of each container at 500 cm3. If the material used to construct thecontainer cost $0.015 per cm2, find, using differentiation, how much a containerwith minimum surface area costs to the candy maker. Leave your answer to 2decimal places.

[Volume of sphere, 3

3

4rV ; Surface area of sphere, 24 rS ]

[10]

8(a) One of the roots of the equation 034 234 pzzzz is 12i, wherepisa constant.

A student says that One of the other roots must be 1 + 2i.Explain why this statement is not entirely correct.

(i) Determine the value ofp.(ii) Find the exact values of all the other roots of the equation.

[1]

[1][3]

r

h


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8(b) The fixed complex number ahas modulusRand argument , and a* denotes

the conjugate of a. Given that*

3

a

aq , express qin trigonometric form.

If 61

q is purely imaginary, determine the least positive value of .

[3]

[3]

9 A curve is defined by the parametric equations24tx ,

at

ty

2

2, where ais

a positive constant.

(i) Using differentiation, find the turning points of this curve in terms of a.

(ii) Show that the cartesian equation of the curve is2

2

)4(

16

ax

xy

.

(iii) Sketch 22

)4(

16

ax

x

y , indicating clearly the coordinates of the points

where the graph crosses the axes, turning points and the equations of anyasymptotes.

(iv) Describe a sequence of geometrical transformations that map the graph of

2

2

)4(

16

ax

xy

onto the graph of

2

2

9

)43(16

x

axy

.

[4]

[2]

[3]

[2]

10(a) Susan baked 2500 cookies. She decided to give part of them to her friends and

sell the remainder for charity.

(i) To pack the cookies meant for her friends, Susan placed 5 cookies in thefirst bag. Each subsequent bag she packed contained double the number ofcookies in the previous bag. How many complete bags of cookies didSusan have to offer her friends?

(ii) To pack the remaining cookies meant for sale, Susan placed 4 cookies inthe first bag. Each subsequent bag she packed contained 3 cookies morethan the previous bag. If she charged $0.50 for each cookie, how muchwould the last complete bag of cookies cost?

[3]

[5]

(b) Given that rbrT

22 , where bis a constant,

(i) show that the terms of the series

n

r

rT1

ln form an arithmetic progression,

(ii) express Sn , sum of the first nterms of this arithmetic progression, in termsof nand b,

(iii) hence find an inequality satisfied by the constant bsuch that the differencebetween S13and S14is not more than 0.5.

[2]

[2]

[2]


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11 The planesp1andp2, which meet in the line l, have vector equations

r =

1

1

0

1

0

1

6

4

2

11 ,

r =

1

0

1

0

3

2

6

4

2

22

respectively, where121 ,, and 2 are real constants.

(i) Show that lis parallel to the vector 5i+ 6j+ k.(ii) Calculate the acute angle betweenp1andp2.(iii) Find, in exact form, the perpendicular distance from the point with

coordinates (4, 2, 2) top2.

The planep3has equation ax2y+ 2z= b, where a, b .

(iv) Find bin terms of a such that all three planes meet at the single common

point with position vector

6

4

2

.

(v) If given instead that a = 2, find the values of b, such that the distance

between the planesp1andp3is3

1units.

[3][2]

[2]

[4]

[3]


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JC 2 H2 Mathematics Preliminary Examination 2009Paper 1 -- Marking Scheme

1

041

0432

xx

xx

41 x

23

4

24

1

223

1

2 434343 dxxxdxxxdxxx

[M ark awarded for spli tting correctly in to 2 sections either using

modulus or negative sign]

23

4

234

1

2323

1

2 42

3

34

2

3

343

x

xxx

xxdxxx

[M ark awarded for correct integration]

262

11243

23

1

2 dxxx OR

2

2122543

23

1

2 dxxx

[B1]

[M1]

[M1]

[A1]

2(i)

(ii)

32 OAOBOM

1

2

1

3

9

0

3

3

3

3

2

If OMis perpendicular toAB, ABOM = 0

AB

12

3

6

ABOM

12

3

6

1

2

1

= 0

[M1]

[A1]

[M1]

[M1]


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(iii) Area of triange OAC= OMOA

=

1

2

1

9

0

3

=

6

12

18

= 22.4 units2(or 146 units2)

[M1]

[A1]

3

xdxxx

dxxx

coscos

sin

cxxx cossin

uduudx cossin2

cx

x

x

cuu

udu

duu

uduu

uu

11ln2

tansecln2

sec2

cos

12

cossin2cos1cos

1

22

[M1]

[A1]

[M1]

[M1]

[M1]

[A1]

4 u1= 4, u2= 14, u3= 30, u4= 52

(i) 41

1

r

ru , 182

1

r

ru , 483

1

r

ru , 1004

1

r

ru

(ii) 2

1

1

nnun

r

r

(iii) Let Pnbe the statement 2

1

1

nnun

r

r ,Zn

LHS of P1= 41

1

1

uur

r

RHS of P1 = 1(1 + 1)2= 4

P1is true and forms the basis for induction.

Assume Pkis true for someZk , i.e. 2

1

1

kkuk

r

r

Required to show Pk+1is true, i.e. 2

1

121

kku

k

r

r

[B1]

[B1]

[B1/2]

[M1]

[M1]

ORArea of triangle

=

8

2

4

1

2

1

= 846

=22.4 units2


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1

1

k

r

ru

LHS

11

k

k

r r

uu

4311 2 kkkk

431 2 kkkk 441 2 kkk 221 kk

=RHS

Since P1is true and Pk is truePk+1 is true.

Therefore, by mathematical induction, Pnis true Zn .

[M1][M1]

[B1/2]

5

2x

udx

dux

dx

dy

Substituting,)2(

122

2

xx

x

uu

x

udx

dux

)2(

22

x

xuuxu

dx

dux

)2(

3

x

xux

dx

dux

)2(

13

x

u

dx

du

dxxduu 21

13

1

cxu ln2ln13ln3

1

3)2(ln13ln xcxy

3)2(13 xkxy

x

xky

3

1)2( 3

[B1]

[M1]

[M1]

[M1]

[M1]

[B1]

[A1]


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6 21 xey

xyxedx

dy x 2)2(21

222

2

ydxdyx

dxyd

dx

dyxy2 (shown)

dx

dy

dx

ydx

dx

dy

dx

yd2

2

3

3

2

2

2

22dx

ydx

dx

dy

2

2

3

3

2

2

2

2

4

4

24

24

dx

yd

dx

ydx

dx

yd

dx

ydx

dx

dy

dx

d

dx

yd

3

3

2

2

26dx

ydx

dx

yd

ey )0( , 0)0(' y , ey 2)0('' , 0)0(''' y and ey 12)0()4(

[deduct B for every mistake]

4321

!4

120

!2

20

2

xe

xxe

xee x

42

2x

eexe

dxedxe

xx

21

0

21

2

1

0

4

4 22

dxxex

ee

2

1

0

42

222

=1.388 (3dp) [use GC]

[M1]

[M1]

[M1]

[M1]

[B2]

[M1]

[A1]

[M1]

[A1]


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7500

3

4

2

1 32

rhr

rr

h

rhr

32500

5003

2

2

2

22 42

12 rrhrS

r

rrr

3

250023

2

2

rr

1000

3

5 2

2

3

2

3

300010

1000

3

10

r

r

rr

dr

dS

03000100 3 rdr

dS

5708.4300

3 orr

(4 dp)

02000

3

1032

2

rdr

Sd when 3

300

r

Minimum surface area 33

2

3001000

300

3

5

= 328.1715 (4 dp)

Cost of box with minimum surface area 92.4$9226.4015.01715.328 (2 dp)

[M1]

[M1]

[M1]

[M1]

[M1]

[M1]

[A1]

[M1]

[M1]

[A1]

8a The statement is only true ifpis real.

(i) Using GC,p= 5.

(ii) We have )))(21())(21((534 2234 bazzizizzzzz ,

where aand b are real.

= ))(52( 22 bazzzz

Comparing coefficients of similar terms, we have a = b= 1

[B1]

[B1]

[M1]

[B1]


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For 012 zz , we have iz2

3

2

1 [A1]

(b) 23

3

Raa

aa

arg(q) = arg (a )3 - arg (a )

= 3 arg(a) + arg(a)

= 4

Thus, )4sin()4cos(2 iRq

6

1

q =

3

2sin

3

2cos3

1

iR

Given that 03

2cos

,

4

3

23

2

or 2.36 radians

[B1]

[M1]

[A1]

[M1] - arg

[M1]

[A1]

9(i) ,

)(

2222

2

at

at

dt

dy

tdtdx 8

22

2

)(4.

att

ta

dx

dt

dt

dy

dx

dy

=0

t = a

,4axa

a

a

ay 2

)2(

Turning points: ),4(a

aa , ),4(

a

aa .

(ii)

at

ty

2

2

2

2

2 2

at

ty

[M1/2]

[M1/2]

[M1/2]dy/dx[M1/2]=0

[A1]

[A1]

[M1]


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2

2

4

ax

xy

22

4

16

ax

xy

(shown)

(iii) 2

2

4

16

ax

xy

(iv) Method 1: A translation of 4aunit in the direction of positive x-axisfollowed by a scaling of 1/3 units along the x-axis.

Method 2: Scaling of 1/3 units along the x-axis followed by atranslation of 4a/3 unit in the direction of positive x-axis.

(Accept use of the word shift)

[M1]

ShapeB1

[-1/2 if

sharp at

origin]

Turning ptsB1/2 each(0, 0)B1/2y= 0 B1/2

Each transfB1

[No mark]

10a(i)

GP : a= 5, r = 2

2500125 nnS 97.8n (3 sf)

No. of complete bags to offer her friends = 8

[M1] - Sn[M1] - ineq

[A1]

(ii) No. of cookies left for sale = 25005(2 -1) = 1225

AP : a= 4, d = 3

12251382

nn

Sn

8.274.29 n (3 sf)

Last complete bag = 27thbag

82)3(26427 T

[M1]

[M1]

[B1]

[M1]

10 20 30 40 50 60 70 80

2

1

1

2

x

y

(0, 0) y=0

),4(a

aa

),4(a

aa


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Cost of last complete bag = 82(0.5) = $41[A1]

b(i) 2ln122ln2lnln 1 brrbTT rr = bln 2, a constant (shown)

[M1]

[B1]bln 2

(ii)

n

r

rn bnbn

TS1

2ln12ln222

ln

412

2ln bn

n

[M1]

[A1]

(iii) 5.0414

2

2ln134152ln75.01314 bbSS

0913.0194.0 b (3 sf)[No marks if modulus sign i s left out]

[M1]

[A1]

11i

(ii)

(iii)

Let n1and n2be the normals ofp1andp2respectively.

n1=

1

1

0

1

0

1

1

1

1

n2 =

3

2

3

1

0

1

0

3

2

1

6

5

3

2

3

1

1

1

Therefore lis parallel to 5i+ 6j+ k.

Acute angle betweenp1andp2=223

3

2

3

1

1

1

cos 1

= 75.7o

Perpendicular distance =

22

3

2

3

6

4

2

2

2

4

[M1]

[M1]

[M1]

[M1]

[A1]

[M1]


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(iv)

=22

3

2

3

4

2

2

=22

22

= 22

The point (2, 4, 6) must lie onp3So 2a8 + 12 = b

b= 2a+ 4

Since the three planes meet at a single point, we must exclude the case

where they meet along a line.

Ifp3meets in the line l, the normal ofp3is perpendicular to l.

So

1

6

5

2

2

a

= 0

a= 2

Therefore b= 2a+ 4, where a 2

[A1]

[M1][A1]

[M1]

[A1]

(v)

p1: 4

1

1-

1

.

r and p3: b

2

2-

2

.r

Distance betweenp1andp2=3

1

3

1

323

4

b

3

1

323

4

b or

3

1

323

4

b

b= 6 or 10

[M1]

[M1]

[A1]


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CATHOLIC JUNIOR COLLEGEGeneral Certificate of Education Advanced LevelHigher 2

MATHEMATICS 9740/02Paper 2 16 September 2009

3 hoursAdditional Materials: Answer Paper

Graph PaperList of Formulae (MF 15)

READ THESE INSTRUCTIONS FIRST

Write your name and class on all the work you hand in.

Write in dark blue or black pen on both sides of the paper.You may use a soft pencil for any diagrams or graphs.Do not use staples, paper clips, highlighters, glue or correction fluid.

Answer allthe questions.Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case ofangles in degrees, unless a different level of accuracy is specified in the question.You are expected to use a graphic calculator.Unsupported answers from a graphic calculator are allowed unless a question specifically statesotherwise.Where unsupported answers from a graphic calculator are not allowed in a question, you are requiredto present the mathematical steps using mathematical notations and not calculator commands.

You are reminded of the need for clear presentation in your answers.

At the end of the examination, arrange your answers in NUMERICAL ORDER. Place this coversheet in front and fasten all your work securely together.The number of marks is given in brackets [ ] at the end of each question or part question.

Name : Class :

Question No. Marks Question No. Marks

1 / 9 7 / 8

2 / 9 8 / 8

3 / 11 9 / 9

4 / 11 10 / 11

5 / 4 11 / 13

6 / 7 TOTAL / 100

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CJC Preliminary Examination 2009 / 9740 / II 2 of 5

Section A: Pure Mathematics [40 marks]

1 (i) Express)2)(1(

4

rrr

rin partial fractions.

(ii) Hence find n

r rrr

r

3 )2)(1(

4.

(iii) Use your answer in (ii) to find

n

r rrr

r

1 )2)(1(

2.

[2]

[4]

[3]

2 (a) Calculate the area of the region bounded by the curves 1e 4 xy

and 1ln xy .

(b) The region S is bounded by the curve 12 xy and the lines 1y and bx ,

where 0b .

xV is the volume of the solid of revolution formed when S is rotated completely

about 1y . yV is the volume of the solid of revolution when S is rotated

completely about they-axis. Find the value of bsuch that yx VV .

[3]

[6]

y

x

12 xy

b0

1y S


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3 (a) Find the eighth roots of the complex number 838 i, giving your answers in

the form ier , where ris a positive real constant and .

(b) The fixed complex number wis such that 0 < arg (w) 0

g :x sinx, x[0 , 2]

(i) With the aid of a diagram, show that f 1exists.(ii) Define f 1in a similar form.(iii) Sketch the graphs of f and f 1on the same axes. Your sketch should show clearly

the axial intercepts, and the geometrical relation between the two graphs and the

liney=x.(iv) Show that the composite function fg does not exist.(v) Determine the largest possible domain of g for which the composite function fg

exists. Hence, define fg in a similar form.

[1][4]

[2][1]

[3]

Section B: Statistics [60 marks]

5 The Student Council of Catholic Junior College is organizing Rockella, a musicconcert, to raise funds for the needy students in school. The Council intends to survey asample of 300 students to find out the music genre preferences of students in theschool.

(i) Given that the school has 1500 students, describe how the sample could be chosenusing systematic sampling.

(ii) Suggest why it would be possible to use stratified sampling instead.

[2]

[2]


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6 In July 2009, Company J introduced the new jPhone 3GS, the fastest, most powerfuljPhone, packed with improved performance up to twice as fast as the previous modelwith longer battery life. The battery life in a randomly chosen jPhone 3GS has a normaldistribution and the battery life of the phone is supposed to be 120 hours.

A random sample of 90 jPhones 3GS is taken, and the battery life of each phone, xhours, is recorded. The data are summarised by

29)120( x , 419)120( 2

x .

Test, at the 5% significance level, whether the mean battery life of a jPhone 3GS is lessthan 120 hours.

Explain, in the context of the question, the meaning of at the 5% significance level.

[6]

[1]

7 (a) EventsAandBare such that31)(,

41)( BPAP and

41)'|( ABP , where 'A is

the complement of A. Investigate whether A and B are mutually exclusive,justifying your answer.

(b) A game is played using a fair die as follows:

If a six is thrown, the game ends and the score is 6.

If a three is thrown, the player throws the die one more time and the score is thetotal of the two numbers thrown.

If any other number is thrown, the player throws the die two more times unlesshis second throw is 6, in which case the game ends. The score is the total of thenumbers thrown. For example, if the first throw is 4 and the second throw is 6, thegame ends and the score is 10. If the first throw is 5, second throw is 2 and thirdthrow is 1, the score is 8.

(i) Find the probability that the score is 5.(ii) Given that the score is at most 5, find the probability that the score is 4.

[3]

[2][3]

8 A group of 10 people consists of 3 married couples and 4 single men.(a) A committee of 4 is to be formed from the 10 people.

(i) How many different committees can be formed?(ii) How many different committees can be formed if the committee can consist

of at most 1 married couple?(b) The group sits at a round table with 10 seats each of a different colour.

(i) If each man sits next to his wife, how many ways can they be seated?(ii) If one man is absent and the rest are allowed to sit without any restrictions,

how many ways can they be seated?

[1]

[2]

[3]

[2]


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9 After the Monetary Authority of Singapores report on the structured financial productssold by financial institutions in Singapore, a survey found that 95% of the respondentswould favour greater consumer protection.(i) Find the probability that out of 30 respondents, 25 of them would favour greater

consumer protection.

(ii) Find the least number of respondents surveyed such that the probability that atleast 40 of them would favour greater consumer protection exceeds 0.980.

(iii) Using a suitable approximation, find the probability that out of 60 respondents, atmost 58 of them would favour greater consumer protection.

[2]

[3]

[4]

10 The ages,xyears, and heights,ycm, of 11 boys are as follows:

x 6.6 6.8 6.9 7.5 7.8 8.2 9.0 10.1 11.4 12.8 13.5

y 112 116 119 123 125 130 135 139 140 141 141

(i) Sketch the scatter diagram for the data and comment on the suitability of a linearmodel betweenxandy.

(ii) State, with a reason, which of the following models is more appropriate to fit thedata points:

(a)y2= a + bx2,where a> 0, b> 0

(b)y = axb where a > 0,0< b < 1

(c)xb

axy

where a> 0, b< 0

(iii) For the appropriate model chosen, state the product moment correlationcoefficient and estimate the values of aand b, for the transformed data.

(iv) Estimate the age of a boy when his height is found to be 110cm. Comment on thereliability of this estimate.

[3]

[2]

[4]

[2]

11 The weight of a bar of BrandAchocolate is normally distributed with mean 180g andstandard deviation 10g. The weight of a bar of Brand B chocolate is normallydistributed with mean 240g and standard deviation 20g.(a) Find the probability that the weight of 3 randomly chosen bars of Brand A

chocolate is more than twice the weight of a randomly chosen bar of Brand B

chocolate.(b) A sample of 60 bars of BrandAchocolate is sent for inspection.(i) Find the probability that the sample mean exceeds 179g.(ii) Explain whether you need to use Central Limit Theorem in your working.

(c) BrandAchocolates are sold at $2 per 100g.(i) Find the probability that a bar of BrandAchocolate costs more than $3.80.(ii) 100 bars of Brand A chocolate are packed in a box. Using a suitable

approximation, find the probability that, in one box, more than 5 bars costmore than $3.80 each.

[3]

[2][1]

[2]

[5]


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JC 2 H2 Mathematics Preliminary Examination 2009Paper 2 -- Marking Scheme

Section A: Pure Mathematics [40 marks]

1(i)

21)2)(1(

4

r

C

r

B

r

A

rrr

r

By Cover Up rule or any other methods,A= 2,B= 3 and C= 1

[Subtract 1 mark for every mistake]

(ii)

n

r

n

r rrrrrr

r

3 3 2

1

1

32

)2)(1(

4

2

1

1

32

3

1

2

3

1

2

4

1

3

3

2

2

26

1

16

3

6

2

25

1

15

3

5

2

24

1

14

3

4

2

23

1

13

3

3

2

nnn

nnn

nnn

=nnn

2

1

3

1

2

2

11

2

3

=

1

12

nn

= 1

2

nn

n

(iii)

2

3

2

1 )22)(12)(2(

)2(2

)2)(1(

2 n

k

rkn

r kkk

k

rrr

r

2

3 ))(1)(2(

4n

k kkk

k

2

3 ))(1)(2(

4n

r rrr

r

[B2]

[M1]

[M1]

[M1]

[A1]

[M1]


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1)2()2(2)2(

nn

n

)1)(2(

nn

n [Accept answer i n partial fr actions]

[M1]

[A1]

2 (i) By G.CIntersection point(1.05395, -0.947453), (4.3919, 0.47976)

3919.4

05395.1

4lnArea dxex x [M 1 - correct limits ; M1 - correct form]

68.1Area

(ii)

dxxVb

x

0

22

55

5 5

0

bxV

b

x

dyybbVb

y 2

0

22 [M 1 - Vol of cylinder ; M1 - Vol of revolution abt y-axis]

22

2 4

0

4

2

bybV

b

y

02

1

5

25

4

45

bb

bb

)(0 rejectedb

2

5b

Alternative Solution

dxxVb

x

0

22

5

55

05

bVx

b

x

dyybbVb

y 11

1

222

[M 1 - vol of cylinder, M1 - vol of revolu tion abt y-

axis]

1

2

11

2

1

2

22

224

1

1

4

2

bb

byy

bV

b

y

[M1+M1]

[A1]

[M1]

[M1]

[M1+M1]

[M1]

[A1]

[M1]

[M1]

[M1+M1]


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222

2 444

1

1

4

2

bbby

ybV

b

y

)(0 rejectedb or2

5b

[M1]

[A1]

3(a) 838 i =

6

5

42

i

e

kii

eez 265

48 2 =)

6

52(

42ki

e

)6

52(

82

ki

ez

, k= 0, 4,3,2,1

[M - mod,

M - arg]

[M1]

[M1 taking8throot, A1]

(b) M1/2, M1/2 - Correct marking of Vand W(i) B1 - correct locus

B if the perpendicular bisector does not pass through origin(ii) B1 correct locus

B1 - showing awareness that the halfline is parallel to iw

Let Qbe the point of intersection.

Using vector addition, )( iwwWQOWOQ It is a square.

[M1,A1]

[B1]

w

W

arg (w)

-iw

V(i)

arg(-iw)

(ii)

Q

O

y

x


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1 x

y4 (i)

Since any horizontal liney= k, kR will cut the graph off at most once, hence f is one-one. Thus, f -1exists.

(ii) Letxxy 1

42

1

2

2 yyx

Butx> 0, 42

1

2

2 yyx

Hence, f -1:x 421

2

2 xx ,x(-, )

(iii) [B1] correct graph of f-1withy=x[B1] - intercepts

(iv) Df : (0,) and Rg: [-1 , 1].

Since RgDf , therefore fg does not exist.

(v) Largest Dg: (0 ,)

fg(x) = f(sinx)

=x

xsin

1sin

Hence, fg :xx

xsin

1sin ,x(0 ,).

[M1]

[M1]

[M1]

[M1]

[A - rule,A - domain]

[B1]

[B1]

[M1]

[A1]

1 x

y

1


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Section B: Statistics [60 marks]

5(i)

(ii)

Mention of sampling frame/list. Random starting point

(i) from first 5 on the list (either explicitly or by giving an example),

(ii)

otherwise student must mention that once they reach the end of the list,they have to go back to the beginning of the list to pick.

Select every 5thstudent

E.g. From the schools registration list[B], randomlyselect a student from the first

5 on the list as a starting poin t [B1]. Select every 5thstudent[B]down the list from

that starting point, until 300 names are obtained.

A sampling frame is easily accessible.

The population can be divided into non-overlapping subgroups(e.g. boys andgirls)

[demonstrate knowledge of subgroups being non-overlappingby explicitlymentioning or giving examples]

[B ][B1]

[B]

[B1][B1]

6 LetXbe the battery life of a jPhone 3GS in hours.

Unbiased est. of popn mean, 6777778.11912090

29120

)120(

n

xx

Unbiased est. of popn var,

602871411.490

29419

89

1

2

90

602871411.4,120~ NX

120:0 H

120:1 H

Test statistic,

90

602871411.4

1206777778.119 z

Use GC, p-value = 0.0771035664 = 0.0771 [accept 0.07710.0785]

Since p-value = 0.0771 > 0.05, we do not reject0

H .

Hence, there is insufficient evidence at 5% level to indicate that the mean battery lifeof a jPhone is less than 120 hours.

Probability of concluding the mean battery life of a jPhone is less than 120 hourswhen it is actually 120 hours is 0.05.

[M1]

[M1]

[B1]

[M1]

[A1]

[B]

[A]

[B1]


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7(a)

16

3)(

4

11

)(

4

1 ''

ABPABP

0

48

7

16

3

3

1)()()( ' ABPBPABP

HenceAandBare not mutually exclusive.

(bi) P(score is 5) = P[(3,2), (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1)]

=216

11

216

5

36

1

(bii) 216

1),1,1,1()3( PscoreP

P(score 4) =P[(3,1), (1,1,2),(1,2,1), (2,1,1) ] = 216

9

5

5454

score

scorescorePscorescoreP

5

4

score

scoreP

=7

3

1191

9

[M1] 1ststmt

[M1] 0

[A1]

[M1]

[A1]

B1 for findingP(score 3, 4)

[M1]

[A1]

8 (a)(i)10C4=210(ii) 10C4 - 3C2 = 207

(b)(i) 222(7-1)! 10 =57600

[M1 - considering permutation of wife and husband, M1 - (7-1)! 10]

(ii) (10-1)! 10 = 3628800

[B1][M1, A1]

[M1,M1, A1]

[M1 ,A1]

6

3 1,2,3,4,5,6

1,2,4,56

1,2,3,4,5 1,2,3,4,5,6


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9 (i) Let r.v.Xrbe the number of respondents who favour greater consumer protectionout of rrespondents.X30~ B(30, 0.95)P(X30= 25) = 0.0124

(ii)Xn~ B(n, 0.95)P(Xn 40) > 0.981 P(Xn 39) > 0.98Using GC, least n= 46

(iii)Let r.v. Ybe the number of respondents who do not favour greater consumerprotection out of 60 respondents.Y~ B(60, 0.05)Y~ Po(3) approximately.P(X60 58) = P(Y 2)

= 1 P(Y 1)

= 0.801

[M1] - distn[A1]

[M1]

[M1][A1]

[M1][B1][M1]

[A1]

10 (i)ShapeLabel axes and 2 end points (6.6,119) &(13.5,141)

Based on the scatter diagram, the linearmodel is not suitable even thoughr(= 0.906) is quite close to 1.

(ii) Choose Model (b):y = axbReason: The graph ofy = axbfits the scatter diagram better.

: From the scatter diagram, we see that as x increases, y increases at a

decreasing rate.

(iii) r= 0.932

y = axblny= ln(axb)lny= ln a+ blnx

lny= 4.1912 + 0.3056 lnxln a= 4.1912 a= 66.1b= 0.306

(iv) wheny= 110,x= 5.29. Extrapolation hence not reliable.

[B1][B1]

[B1]

[B1][B1 any 1reason]

[B1]

[M1]

[A1][A1]

[B1 + B1]


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11 (a)LetAbe the random variable that denotes the weight, in grams, of a bar of Brand Achocolate

100,180~ NA LetBbe the random variable that denotes the weight, in grams, of a bar of Brand B

chocolate 400,240~ NB

1900,60~2321 NBAAA 022 321321 BAAAPBAAAP )s.f.3to(916.002321 BAAAP

(b)(i)

60

100,180~ NA

)s.f.3to(781.0179 AP

(b)(ii) No need to use Central Limit Theorem because the population follows aNormal Distribution.

(c)(i)

10002.0,18002.0~02.0 2NA 04.0,6.3~02.0 NA

)s.f.3to(159.01586552596.080.302.0 AP

(c)(ii)LetXbe the random variable that denotes the number of bars of BrandAchocolate, ina box, that cost more than $3.80.

1586552596.0,100~BX 30,100 nn

5,86552596.15 npnp

5,13447404.84 nqnq

34837682.13,86552596.15~NX approx 5.55 XPccXP

)s.f.3to(998.09977237843.05.5 XP

[M1]

[M1]

[A1]

[M1]

[A1]

[B1]

[M1][A1]

[M1]

[M1]

[M1]

[M1]

[A1]


Documents

2009 CJC