ACJC Math Prelim 10

ANGLO-CHINESE JUNIOR COLLEGE

MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 1 19 August 2010

JC 2 PRELIMINARY EXAMINATION

Time allowed: 3 hours

Additional Materials: List of Formulae (MF15)

READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles 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 states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.

This document consists of 6 printed pages.

[Turn Over

9740 / 01

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Anglo-Chinese Junior College H2 Mathematics 9740: 2010 JC 2 Preliminary Examination Paper 1

Page 2 of 6


MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010

MATHEMATICS 9740 Higher 2 Paper 1

Calculator model: _____________________

Arrange your answers in the same numerical order.

Place this cover sheet on top of them and tie them together with the string provided.

Question no. Marks 1

2

3

4

5

6

7

8

9

10

11

12

13

14

/ 100 Index No: Form Class: ___________ Name: _________________________


Page 3 of 6

1 The nth term of a sequence is given by � � 1 21 n

nu n� � , for 1n t . The sum of the first n terms is denoted by Sn. Use the method of mathematical induction to show that

� � � �1 11

2n

nn n

S � � � for all positive integers n. [4]

2 The 3 flavours of puddings produced by a dessert shop are mango, durian and strawberry. A

mango pudding requires 5g of sugar and 36ml of water. A durian pudding requires 6g of sugar and 38ml of water. A strawberry pudding requires 4g of sugar and 40ml of water. The puddings are sold in pairs of the same type at $1.60, $2.20 and $1.80 for mango, durian and strawberry respectively. On a particular day, 754g of sugar and 5972ml of water were used to make the puddings and all the puddings made were sold except for a pair of strawberry puddings. The collection from the sale of puddings was $142.40. Formulate the equations required to determine the number of each type of pudding made on that day. [4]

3 The diagram below shows the graph of f( )y x . The curve passes through the origin and

has a maximum point at � �4 , 4A and asymptotes 2x � and 2y . Sketch on separate diagrams, the graphs of

(i) 1f( )y x [3]

(ii) f '( )y x [3] showing clearly asymptotes, intercepts and coordinates of turning points where possible.

4 The complex number w has modulus 3 and argument 23S . Find the modulus and argument of

*i

w� , where w* is the complex conjugate of w. Hence express

*i

w� in the form a ib� , where

a and b are real, giving the exact values of a and b in non-trigonometrical form. [4]

Find the possible values of n such that *

niw�§ ·

¨ ¸© ¹

is purely imaginary. [2]

[Turn Over

y

x

f( )y x

2

2�

� �4 , 4 A


Page 4 of 6

5 Solve the equation 4 0z i� , giving the roots in the form ire D , where 0r ! and S D S� � d . [3]

The roots represented by 1z and 2z are such that � � � �1 2arg arg 0z z! ! . Show 1z , 2z and

1 2z z� on an Argand diagram. Deduce the exact value of � �1 2arg z z� . [3]

6 An economist is studying how the annual economic growth of 2 countries varies with time. The annual economic growth of a country is measured in percentage and is denoted by G and the time in years after 1980 is denoted by t. Both G and t are taken to be continuous variables.

(i) Country A is a developing country and the economist found that G and t are can be modeled

by the differential equation 12

dG Gdt

� . Given that, when 0t , 0G , find G in terms of t.

[4] (ii) Comment on the suitability of the above differential equation model to forecast the future

economic growth of Country A. [1] (iii) Country B is a developed country and the economist found that G and t can be modeled by

the differential equation 12

dG Gdt

�§ · �¨ ¸© ¹

.

Given that Country B has been experiencing decreasing economic growth during the period of study, sketch a member of the family of solution curves of the differential equation model for Country B. Hence, comment on the economic growth of Country B in the long term. [2]

7 (i) Given that � � 1cosf xx e

�

, where 1 1x� d d , find � �f 0 , � �f 0c and � �f 0cc . Hence write down

the first three non-zero terms in the Maclaurin series for � �f x . Give the coefficients in terms

of ke S , where k�� . [4] (ii) Given that � �g tan secx x x � , where x is sufficiently small for 3x and higher powers of x

to be neglected. Deduce the first three non-zero terms in the series expansion of � �g x .

Hence, show that � � � � 22 2 2f g 2x e x e x eS S S

� | � . [3]

(iii) Explain clearly why it is inappropriate to state that � � � � 22 2 2f g 2a a

a ax e x dx e x e dx

S S S

� �� | �³ ³ ,

where a�� . [1]

8 (i) Show that � � � � � �21 2 1

! 1 ! 2 ! 2 !Ar Br C

r r r r� ��

� � �, where A, B and C are constants to be

found. [2]

(ii) Hence find � �2

1

3 3 32 !

n

r

r rr

� ��¦ . [3]

(iii) Give a reason why the series � �2

0

3 3 32 !r

r rr

f

� ��¦ converges, and write down its value. [2]

y


Page 5 of 6

9 Relative to the origin O, two points A and B have position vectors given by 3 3a i j k � � and 5 4 3b i j k � � respectively.

(i) Find the length of the projection of OA)))&

on OB)))&

. [2] (ii) Hence, or otherwise, find the position vector of the point C on OB such that AC is

perpendicular to OB. [2] (iii) Find a vector equation of the reflection of the line AB in the line AC. [3] 10(a) The first 2 terms of a geometric progression are a and b ( b < a ). If the sum of the first n

terms is equal to twice the sum to infinity of the remaining terms, prove that 3n na b . [3]

(b) The terms 1 2 3, , ,...u u u form an arithmetic sequence with first term a and having non-zero common difference d. (i) Given that the sum of the first 10 terms of the sequence is 105 more than 510u , find

the common difference. [3] (ii) If 26u is the first term in the sequence which is greater than 542, find the range of

values of a. [3] 11 The region R in the first quadrant is bounded by the curve � �22 2y a a x � , where 0a ! ,

and the line joining � �2 ,0a and � �0,a . The region S, lying in the first quadrant, is bounded

by the curve � �22 2y a a x � and the lines 2x a and y a . (i) Draw a sketch showing the regions R and S. [1] (ii) Find, in terms of a, the volume of the solid formed when S is rotated completely about the x-axis. [4]

(iii) By using a suitable translation, find, in terms of a, the volume of the solid formed when R is rotated completely about the line 2x a . [4]

12 The curve C has the equation 23 2x axy x a� � � where a is a constant.

(i) Find dydx and the set of values of a if the curve has 2 stationary points. [4]

(ii) Sketch the curve C for a = 1, stating clearly the exact coordinates of any points of intersection with the axes and the equations of any asymptotes. [3]

Hence, find the range of values of k such that the equation 23 2 ( 1)x x k x� � � has exactly 2 real roots. [2]

13 The curve has the parametric equations

25

1x

t

� , 1tany t�

(i) Sketch the curve for 2 2t� d d . [1] (ii) Find the cartesian equations of the tangent and the normal to the curve at the point

where 1t . [5] (iii) Find the area enclosed by the x-axis, the tangent and the normal at the point where t = 1. [3]

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Page 6 of 6

14 The functions f and g are defined as follows:

f : sin�x x , ,2 2

x S Sª º� �« »¬ ¼,

� �� g : 1 38S

� ��x x x , x�� .

(i) Sketch the graph of the function g, labeling clearly the exact values of the coordinates of turning point(s) and intersections with the axes, if any. [1] State the range of the function g in exact values. [1]

(ii) Given that gf exists as a function. By considering the graphs of f and g, explain why

gf ( ) gf ( )D Ez if 2 2S SD E� d � d . [2]

Hence what can be said about the function gf ? [1] Without sketching the graph of gf , find the range of gf in the form > @,a b , giving the exact values of a and b. [1]

(iii) (a) Give a reason why fg does not exist as a function. [1]

(b) Find the greatest exact value of k for which fg is a function if the domain of g is restricted to the interval > @1, k . [2]

- End of Paper -


MATHEMATICS DEPARTMENT MATHEMATICS Higher 2 Paper 2 23 August 2010

JC 2 PRELIMINARY EXAMINATION

Time allowed: 3 hours

Additional Materials: List of Formulae (MF15)

READ THESE INSTRUCTIONS FIRST Write your Index number, Form Class, graphic and/or scientific calculator model/s on the cover page. Write your Index number and full name on all the work you hand in. Write in dark blue or black pen on your answer scripts. You may use a soft pencil for any diagrams or graphs. Do not use paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles 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 states otherwise. Where unsupported answers from a graphic calculator are not allowed in the question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together.

This document consists of 5 printed pages.

[Turn Over

9740 / 02


Page 2 of 5


MATHEMATICS DEPARTMENT JC2 Preliminary Examination 2010

MATHEMATICS 9740 Higher 2 Paper 2

Calculator model: _____________________

Arrange your answers in the same numerical order.

Place this cover sheet on top of them and tie them together with the string provided.

Question no. Marks

1

2

3

4

5

6

7

8

9

10

11

12

13

14

/ 100 Index No: Form Class: ___________ Name: _________________________


Page 3 of 5

Section A: Pure Mathematics [40 marks]

1 Find the exact value of � �

12

2 11

1 .xxe dx

e ��

�³ [4]

2 The variable complex numbers z and w are such that 2 3z i� � and � �arg 5 3w i S� � .

(i) Illustrate both of these relations on a single Argand diagram. [2] (ii) State the least value of z w� . [1]

(iii) Find the greatest and least possible values of � �arg 3z � , giving your answers in radians correct to 3 decimal places. [4]

3 Find in terms of a, the range of values of x that satisfy the inequality ln 2 0axx

§ ·� t¨ ¸© ¹

,

where 1a! . [4] 4 (a) State the derivative of 3cos x . Hence, find 5 3sin .x x dx³ [4]

(b) Find the exact value of � �2 5 3

2 2

10

5x dx�

�³ , in the form 2 3a b� , using the

substitution 5 secx T . [6]

5 The plane p1 has equation 2 3x y z� � . The line l1 has equation 1 12 4

x zy� � .

(i) Show that the line 1l is parallel to, but not contained in the plane p1. [2] (ii) Find the cartesian equation of the plane p2 which contains the line l1 and is perpendicular to the plane p1. [3]

(iii) Find, in scalar product form, the vector equation of the plane p3 which contains the point � �4,1, 1� and is perpendicular to both p1 and p2. [2]

Another line l2 which is parallel to the vector 203

§ ·¨ ¸¨ ¸¨ ¸�© ¹

intersects the line l1 at the

point � �1,0,1�A . (iv) Given that the line l2 meets the plane p1 at the point B, find the coordinates of B. [4] (v) Find the sine of the acute angle between the line l2 and the plane p1, and hence, find the

length of the projection of the line segment AB on the plane p1, giving your answer in surd form. [4]

Section B: Statistics [60 marks]

6 Mr Raju, who owns a supermarket wishes to find out what customers think about the goods

that he sells. He has been advised that he should take a random sample of his customers for this purpose. State, with reasons, which of the following sampling procedures is preferable. A. Select every 10th customer on each day in a typical week. B. Select the first 20 customers on each day in a typical week [2]

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Page 4 of 5

7 One day, Pinocchio went shopping and bought a pair of size 30 Wood Shoes. The right

shoes have lengths which are normally distributed with mean 20 cm and standard deviation 0.14 cm. The left shoes have lengths which are normally distributed with mean 20.1 cm and standard deviation 0.11 cm. The length of the right shoe is independent of the length of the left shoe. When wearing the pair of shoes, Pinocchio takes six steps, heel to toe as shown in the diagram. Calculate the probability that the distance AB, from the back of the first step to the front of the sixth step, exceeds 120 cm. [3]

8 On Ulu Island the weights of adult men and women may both be taken to be independent

normal random variables with means 75kg and 65 kg and standard deviations 4 kg and 3 kg respectively. Find the probability that the weight of a randomly chosen man and the weight of a randomly chosen woman differ by more than 1 kg. [3] Explain if this is equal to the probability that the difference in weight between a randomly chosen married woman and her husband is more than 1kg. [1]

9 Research has shown that before using an Internet service, the mean monthly family

telephone costs is $72. A random sample of families which had started to use an Internet service was taken and their monthly telephone costs were :

$70, $84, $89, $96, $74 Stating a necessary assumption about the population, carry out a test at the 5% significance level, whether there is an increase in the mean monthly telephone costs. [5] If the assumption stated above still holds, and if the standard deviation of the monthly telephone costs is $9.89, find the range of values of the mean monthly family telephone costs 0P that would lead to a reverse in the decision to the above test. [3]

10 Six overweight men registered at a slimming centre for a slimming programme. The

following table records x, the height (to the nearest cm) and y, the weight (to the nearest 0.1 kg) of these six men.

(i) Given that the least square regression line of x on y line is 103.6 0.726x y � , show that the value of k to the nearest 0.1 kg is 80.3. Hence or otherwise, find the least square regression line of y on x in the form y ax b � , giving the values of a and b to the nearest 3 decimal places. [5] (ii) Based on the data given, use an appropriate regression line to predict the weight of an overweight man who is 165 cm tall. [2] (iii) Find the value of the product moment correlation coefficient between x and y and sketch the scatter diagram of y against x. A particular man among the 6 men who registered for the slimming programme is unusually overweight. Indicate who this man is. [3]

Man A B C D E F x (height in cm) 150 157 160 162 167 170 y (weight in kg) 65.1 73.2 85 k 80.9 89.9

A B


Page 5 of 5

11 In a hotel, large number of cups and saucers are washed each day. The number of cups that

are broken each day while washing averages 2.1. State in context, a condition under which a Poisson distribution would be a suitable probability model. [1] Assume that the number of broken cups and saucers follow a Poisson distribution.

(i) Show that on any randomly chosen day, the probability that at least 3 cups are broken is 0.350 correct to 3 significant figures. [1] The probability that there will be at least two days in n days with at least 3 broken cups is more than 0.999. Find the least value of n. [3]

(ii) The number of saucers broken each day averages 1.6, independently of the number of cups broken. The total number of cups broken and saucers broken during a week of 7 days is denoted by T. State a possible model for the distribution of T. [2]

A random sample of 100 weeks is chosen. Using a suitable approximation, find the probability that the average weekly total number of broken cups and saucers does not exceed 26. [3]

12 Fish are bred in large batches and allowed to grow until they are caught at random for sale.

When caught, only 20% of the fish measure less than 8 cm long. (i) What is the probability that the 10th fish caught is the sixth fish that is less than 8 cm long? [2] (ii) A large number, n, of fish are caught and the probability of there being 10 or fewer fish in

the catch which measures less than 8 cm long is at most 0.0227 . Using a suitable approximation, derive the approximate inequality 10.5 0.2 0.8n n� d � . [4] Hence find the least possible number of fish to be caught. [2]

13 An automated blood pressure machine is being tested. Members of the public, %p of whom

have high blood pressure (hypertension), try it out and are then seen by a doctor. She finds that 80% of those with hypertension and 10% of those with normal blood pressure have been diagnosed as hypertensive by the machine. The probability that a randomly chosen patient who was diagnosed as hypertensive by the machine actually has hypertension is

32 .

(i) Find the value of p [3] (ii) Hence, find the probability that a randomly chosen patient does not have hypertension,

given that the machine diagnosed him as having normal blood pressure. [2] Comment on the usefulness of the machine. [1]

14 Ten balls are identical in size and shape of which 2 are red, 3 are blue and 5 are green. The two red balls are labeled ‘1’ and ‘2’, the three blue balls are labeled ‘1’, ‘2’ and ‘3’, and the five green balls are labeled ‘1’, ‘2’, ‘3’, ‘4’ and ‘5’.

(i) Find the number of ways of choosing 2 balls of identical colour. [2] (ii) Find the number of ways of choosing 6 balls if it includes at least one ball of each colour. [4] (iii) A person arranged 3 balls in a row with the numbered sides facing him forming a 3-digit

number. Among these 3 balls, none of them are green. Find the number of possible 3-digit numbers facing that person.

[The number formed is independent of the colours of the balls used. i.e. the number 112 is counted as one number whether the colour of the ball labeled ‘2’ is red or blue.] [3]

- End of Paper -

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xx

()

()

22

22

22

2

22

2

fg

12

2

2 (

Show

n)

ex

xe

xe

ex

xe

x

ee

x

ππ

ππ

π

ππ

+≈

−+

++

+

=+

The

stat

emen

t()

()

22

22

fg

2a

a

aa

xe

xdx

ex

edx

ππ

π

−−

+≈

+∫

∫ is

inap

prop

riate

as

()

()

22

22

fg

2x

ex

ex

eπ

ππ

+≈

+ o

nly

whe

n a

is su

ffic

ient

ly sm

all.

8 (i) (ii)

()

()

()(

)(

)(

)

()

2

12

1!

1!

2!

12

22

12

!

12

!

rr

r

rr

rr

rr

r−+

++

++

−+

+=

+

+−

=+

1,1,

1.A

BC

==

=−

()

() (

)(

)

2

1

2

1 133

32

!

13

2!

12

13

!1

!2

!

n rn r n r

rr

r rr

r rr

r

=

= =

+−

+ +−

=+

=−

+

+

+

∑

∑ ∑

5

(iii)

()

()

()

()

()

()

12

11!

2!3!

12

12!

3!4!

12

13!

4!5!

12

14!

5!6!

3

12

12

!1

!!

12

11

!!

1!

12

1!

1!

2!

nn

n

nn

n

nn

n

−+

+−

+

+

−+

+−

+

=

+−

+

−

−

+

−+

−+

+−

+

+

+

⋮

(

)(

)1

11

32

1!

2!

nn

=−

+

+

+

()

()

()

2

0

2

1

33

32

!

33

33

2!2

!

n rn r

rr

r

rr

r

=

=

+−

+

+−

=−

++

∑

∑

()

()

()

31

11

32

21

!2

!n

n

=−

+−

+

+

+

As

()

()

11

, 0

and

0.1

!n+

2!

nn

→∞

→→

+

the

serie

s con

verg

es to

0.

∴

9(i) (ii)

Leng

th o

f the

pro

ject

ion

of O

A

on

OB

= 3

51

201

42

250

503

3

⋅−

==

or

4 2

Met

hod

1:

From

(i),

22

OC=

55

12

22

44

550

33

OC

=−

=−

Met

hod

2:

Line

OB

: 0

50

40

3λ

=+

−

r

53

53

41

41

33

33

ACO

CO

Aλ

λλ λ

−

=−

=−

−=

−−

−

Sinc

e AC

OB

⊥

,

A

C

B O

6

(iii)

53

54

1.

40

33

32 5λ λ λ

λ

−

−

−−

=

−

−

=

5

24

53

OC

=−

Sinc

e 2 5

OC

OB

=

, :

2:3

OC

CB=

5

11

'4

55

3O

BO

B

=−

=−

−

Or u

se m

idpo

int t

heor

em,

()

1'

2O

CO

BO

B=

+

55

52

1'

22

44

45

53

33

OB

OC

OB

=−

=−

−−

=−

−

5

320

11

''

41

15

53

318

ABO

BO

A

=−

=−

−−

=−

Vec

tor e

quat

ion

of li

ne'

AB is

3

201

1,

318

r

αα

=+

∈

ℝɶ

10

(a)

(b) (

i)

GP

: a =

a, r

= b a

The

sum

to in

finity

of t

he re

mai

ning

term

s, 1

nar

Sr

∞=

−

()

()

2

12

11

12

31 1 3

1 3

3nn

n

nn

n

n

n

nn

SS

ar

arr

rr

r

r r b a ba∞

= −=

−−

−=

= =

= =

7

(ii

)

[]

[]

105

105

1010

29

105

104

2 1045

105

1040

21

Su

ad

ad

ad

ad

d

=+

+=

++

+=

++

=

2554

224

(21)

542

38

u a a

≤

+≤

≤

a

nd

2654

225

(21)

542

17

u a a

>

+>

>

17

38a

∴<

≤

11

(i) (ii)

(iii)

Whe

n S

is ro

tate

d co

mpl

etel

y ab

out t

he x

-axi

s,

()

()

()

()

22

0

22

3

0

32

3

Req

uire

d vo

lum

e2

22

22

22

22

2 c

u. u

nits

a

a

aa

aa

xdx

ax

aa

aa

a

a

ππ

ππ

ππ π

=−

−

−=

−

−

=−

=

∫

Afte

r a tr

ansl

atio

n of

2a

units

in th

e ne

gativ

e x-

dire

ctio

n,

New

equ

atio

n is

(

)2

22

22

(2

)y

ya

ax

ax

a=

−+

⇒=−

Whe

n R

is ro

tate

d co

mpl

etel

y ab

out t

he li

ne

2x

a=

,

()()

22

2

0

53

20

33

3

12

Req

uire

d vo

lum

e2

3 44

35

44

8 c

u. u

nits

35

15

a

a

ya

ady

a

ya

a

aa

a

ππ

ππ

ππ

π

=−

−

=−

=−

=

∫

12

x

y a

2a

O

R S

()

22

2y

aa

x=

−

8

(i) (ii)

()(

)(

)(

)

()

()

2

2

2

22

2

32 6

32

36

2

xax

yx

ax

ax

ax

axdy dx

xa

xax

a

xa

++

=+ +

+−

++

=+

++

−=

+

For s

tatio

nary

poi

nts,

()

22

0

36

20

dy dx xax

a

= ++

−=

For 2

stat

iona

ry p

oint

s, (

)() (

)2

2

2

264

32

0

2424

1

aa

a

a

−−

>

>−

>−

2 is a

lway

s pos

itive

.a

.

a ∴∈ℝ

2

32

43

21

1x

xy

xx

x+

+=

=−

++

+

Po

ints

of i

nter

sect

ion

with

the

axes

: (0

,2)

Asy

mpt

otes

: 3

2

and

1y

xx

=−

=−

. Th

e ra

nge

of v

alue

s of

k s

uch

that

the

equ

atio

n 2

32

(1)

xx

kx

++

=+

has

exa

ctly

2

real

root

s :

1.93

or

11.9

kk

><−

13

(i)

9

(ii)

(iii)

()

()

()

()2

22

22

2

2

-52t

1

and

=

11

. 1+t

1.

101

1=

10

dydx

dtdt

tt

dydy

dtdx

dtdx

tt

t t

=+

+

=

=−

+

+−

Whe

n 1,

1 5tdy dx

=

=−

Equa

tion

of ta

ngen

t :

()

14 5

52

11

52

4

πy x

πy

x

−=−

− =−

++

Gra

dien

t of n

orm

al is

5

Equa

tion

of n

orm

al:

()

45

5 225

52

4

πy x

πy

x

−=

− =−

−

()

()

()

2

11

125

Are

a =

52

24

52

44

131.

60 (3

s.f )

80

ππ

π

π

×+

−−

×

==

14

Func

tions

G

iven

:

sin

,f

xx

֏

,2

2x

ππ

∈−

5

10

(i) (ii)

and

:(

1)(3

),8

gx

xx

π+

−֏

x∈ℝ

.

(,

] 2gR

π=−∞

Fr

om th

e gr

aph

of

()

yf

x=

,

()

()

ff

αβ

≠ if

2

2π

πα

β−

≤<

≤.

In fa

ct,

1(

)(

)1

ff

αβ

−≤

<≤

.

From

the

grap

h of

(

)y

gx

=,

()

()

gfgf

αβ

<

or

()

()

gfgf

αβ

≠ 0

x

y

-1

1 3

(1,) 2

π

2π

38

π

(

)y

gx

=

0 x

y

2π

2π−

-1

1

α

β

()

fα

()

fβ

()

yf

x=

0 x

y

1 -1

2π

38

π

()

yg

x=

11

(iii)

(a)

(iii)

(b)

if 1

()

()

1f

fα

β−≤

<≤

. gf

is o

ne-o

ne (o

r inc

reas

ing)

.

[0,

] 2gfR

π=

.

(,

][

.]

22

2g

fR

Dπ

ππ

=−∞

⊄−

=.

Alte

rnat

ive:

5

(4)

[(4

)][

]8

fgf

gf

π=

=−

is u

ndef

ined

bec

ause

5

[,

]8

22

fD

ππ

π−

∉−

=.

Hen

ce, f

g do

es n

ot e

xist

as a

func

tion.

()

2g

xπ

=

()(

)1

38

2x

xπ

π+

−=−

()(

)1

34

xx

+−

=−

2

23

4x

x−

++

=−

2

27

0x

x−

−=

2

322

x±

=

18

x=

±

Sinc

e 1,

x≥

1

8x=

+.

11

8k

≤≤

+

Gre

ates

t val

ue o

f k is

18

+.

0 x

y

-1

1 3

(1,

) 2π

()

yg

x=

2y

π=−

12

1

Ang

lo-C

hine

se J

unio

r Col

lege

H

2 M

athe

mat

ics 9

740

2010

JC

2 P

RE

LIM

Mar

king

Sch

eme

Pape

r 2:

1

() (

)(

)

()

()

()

12

21

1

11

22

22

12

11

12

11

22

12

12

2

11

2

42

2

1

11

11

11

22

22

14

12

xx

xx

xx

xx

xx

edx

e

edx

edx

ee

ee

ee

ee

ee

−−

−−

−

−−

−−

−

−

−

=−

−+

−

=−

++

+

=+

−+

+

∫

∫∫

2

(i)

2

(ii)

2 (ii

i)

Leas

t val

ue o

f z

w−

= 1

G

reat

est

()

arg

3z+

=

11

13

tan

sin

0.82

6 (3

dp)

526

−−

+=

Leas

t (

)ar

g3

z+ =

1

11

3ta

nsi

n0.

432

(3 d

p)5

26−

−

−=−

3

ln2

0a

xx

−≥

whe

re

1a>

.

2

21

20

11

8)

11

8)

24

40

11

81

18

04

4

ax

xx

xa

xa

ax

x

xa

ax

orx

−≥

−−

≥

+

+−

+−

−

≥

−+

++

≤<

≥

4 (a

) (

)(

)3

23

cos

3si

nd

xx

xdx

=−

x

y

i(2,

1)

O

−3

(5,−

3)

o

2

3

263

3

z

w

2

5

3

32

3

33

32

33

23

33

3

sin (

sin

)

11

cos

cos

33

3

cos

cos

31

cos

sin

33

xx

dx

xx

xdx

xx

xx

dx

xx

xx

dx

xx

xc

=

=−

−−

=−

+

=−

++

∫ ∫

∫

∫

4

(b)

5se

c10

,4

5se

cta

n2

5,3

xw

hen

x

dxw

hen

xd

πθ

θ

πθ

θθ

θ==

=

==

=

()

()

[]

25

32

2

10

33

22

4

3

2

4 33

2

44

33 4

4

5

5sec

55

sec

tan

1se

c5

tan

1co

s1

cot

cos

5si

n5

11

1co

s5

sin

5

12

12

23

..5

155

15

xdx

d

d dor

ecd

ec

iea

b

π π

π π ππ

ππ

ππ π

π

θθ

θθ

θθ

θ θθ

θθ

θθ

θθ

−

−

−

=−

= =

=−

=−

=−

==−

∫ ∫

∫ ∫∫

5 (i) (ii)

1

1:

23

1p

r

⋅

=

−

ɶ

1

12

:0

1,

14

l

rλ

λ−

=+

∈

ℝɶ

21

1.

22

24

04

1

=+

−=

−

11

0.

21

01

31

1

−

=−+

−≠

−

⇒ l 1

is p

aral

lel t

o p 1

.

⇒

l 1 is

not

con

tain

ed in

p1.

Alte

rnat

ive

met

hod:

1

21 2

23

14

1

λλλ

−+

=−

≠

+

−

i

Sinc

e no

solu

tion

for λ

, ∴ l 1

is p

aral

lel a

nd n

ot c

onta

ined

to p

1 1

23

21

32

14

1−

×=−

−

3

(ii

i) (iv

) (v)

2

31

3:

20

23

01

41

11

pr

−−

−

⋅

==

++

=

iɶ

2

:3

24

px

yz

−+

+=

3

24

2:

11

18

14

54

14

pr

⋅=

⋅=

+−

=

−

ɶ

3

2:

15

4p

r

⋅=

ɶ

2

12

:0

0,

13

l

rµ

µ−

=+

∈

−

ℝɶ

1

1:

23

1p

r

⋅

=

−

ɶ

()

12

10

.2

3

25

3

1 po

int

is

1,0,

21

31

µµ

µµ

−+

=⇒

−+

=⇒

=∴

−

−

−

B

2

11

15

sin

02

136

783

1θ

=⋅

=

−

−

Leng

th o

f the

pro

ject

ion

of A

B on

p1 =

co

sAB

θ

= 2

2553

531

01

1331

878

786

63

−=

==

−

6

Proc

edur

e A

is p

refe

rabl

e as

it is

unb

iase

d.

Early

cus

tom

ers m

ay n

ot b

e ty

pica

l cus

tom

ers i

n ge

nera

l.

7 Le

t R b

e th

e r.v

for t

he le

ngth

of a

righ

t sho

e an

d L

for t

he le

ft sh

oe

R

N(2

0, 0

.142 ) a

nd L

N

(20.

1 , 0

.112 )

Met

hod

1

X =

R +

L

N(4

0.1,

0.0

317)

3X

N

(40.

1 x

3, 0

.031

7 x

32 ) P(

X >

120

) = 0

.713

M

etho

d 2

R +

L

N(4

0.1,

0.0

317)

P(

R +

L >

3120

) =

0.71

3 8

Let M

and

W b

e th

e rv

for t

he w

eigh

t of a

n ad

ult m

an a

nd w

oman

resp

ectiv

ely.

M

N

(75,

42 ) an

d W

N

(65,

32 ) W

M

~ N

(-10

, 52 )

P(1

>−

MW

) = P

(W –

M >

1) +

P(W

– M

< -1

) = 0

.978

O

r P(

1>

−M

W) =

1 -

P(1

<−

MW

) = 1

– P

(-1<

W –

M<1

) = 0

.978

4

No

, (i)

The

wei

ght o

f a h

usba

nd a

nd w

ife m

ay n

ot b

e in

depe

nden

t O

r (ii)

Ran

dom

ness

is n

ot th

ere

( a ra

ndom

ly c

hose

n w

omen

but

spou

se is

not

ra

ndom

ly c

hose

n)

Or (

iii) D

istri

butio

n of

wei

ght o

f mar

ried

wom

an is

diff

eren

t fro

m d

istri

butio

n of

ad

ult w

oman

. Et

c 9

Tele

phon

e co

sts a

re a

ssum

ed to

be

norm

ally

dis

tribu

ted.

To

test

H0 :

µ=

72

agai

nst H

1: µ

> 72

at 5

% le

vel o

f sig

nific

ance

Und

er H

0 , T

= n

sx0

_µ

−t(5

-1)

Test

stat

istic

s : T

= n

sx0

_µ

−=

82.6

7210

.667

7/5

−=

2.22

19

p va

lue

= P(

T >

2.22

19) =

0.0

452

< 0.

05

Rej

ect H

0 at t

he 5

% le

vel o

f sig

nific

ance

. We

conc

lude

that

ther

e is

suff

icie

nt

evid

ence

at t

he 5

% le

vel o

f sig

nific

ance

that

ther

e is

evi

denc

e of

an

incr

ease

in

mea

n m

onth

ly c

osts

. To

test

H0 :

µ=

0µ

agai

nst H

1: µ

> 0µ a

t 5%

leve

l of s

igni

fican

ce

Und

er H

0 , Z

=

nx 89.9

0

_µ

− N

(0,1

)

Test

stat

istic

s : Z

=

nx σ

µ 0_−

= 0

82.6

9.89

/5µ

−=

(82.

6 -

0µ)

89.95

Do

not r

ejec

t H0 i

f P(Z

> (8

2.6

- 0µ)

89.95

) > 0

.05

(82.

6 -

0µ)

89.95

< 1.

6448

5....

......

.(1)

µ 0 >

75.

3 10

(i)

16

1x=

(fro

m c

alcu

lato

r or c

ompu

tatio

n)

5

(ii)

(iii)

whe

n 16

1x=

, 10

3.6

0.72

6x

y=

+

(1

6110

3.6)

/0.7

26y=

−

79.0

6336

088

=

usin

g y

yn

=∑

1

79.0

6336

088

(65.

173

.285

80.9

89.9

)6

k=

++

++

+

80.3

k=

U

se G

.C. t

o fin

d re

gres

sion

line

of y

on

x:

97

.593

1.09

7y

x=−

+

Use

y o

n x

line

to p

redi

ct w

eigh

t. W

hen

165

x=

, 97

.593

1.09

7(16

5)y=−

+

83.4

y=

(1 d

.p.)

– us

ing

3 d.

p. o

f a a

nd b

to c

ompu

te.

or

83.5

y=

- us

ing

full

accu

racy

of a

and

b to

com

pute

. U

sing

G.C

., 0.

893

r=

C

is u

nusu

ally

ove

rwei

ght.

11

(i)

Bre

akag

es o

ccur

rand

omly

or

Bre

akag

es o

ccur

inde

pend

ently

or

Mea

n nu

mbe

r of b

reak

ages

is a

con

stan

t Le

t A b

e th

e r.v

for t

he n

umbe

r of b

roke

n cu

ps p

er d

ay

A

Po(2

.1)

P(A≥

3) =

1 –

P(A

≤2)

= 0

.350

369

= 0.

350

(3 si

g fig

s)

Let X

be

the

r.v fo

r the

num

ber o

f day

s with

a le

ast 3

bro

ken

cups

out

of n

cup

s. X

y

65.1

73.2

80.3

80

.9

85.0

89.9

150

157

160 1

62

167

170

x

6

(ii)

P(X≥

2) >

0.9

99

1 –

P(X≤

1) >

0.9

99...

......

...(1

) P(

X≤

1) <

0.0

01

Usi

ng G

C,

n P(

X≤

1)

21

0.00

145

22

0.00

0984

le

ast n

is 2

2 T

Po(2

.1x7

+ 1

.6 x

7)

TPo

(25.

9)

Met

hod

1:

Sinc

e n

is la

rge,

_ T

(2

5.9,

1009.

25) a

ppro

x by

cen

tral l

imit

theo

rem

P(_ T≤

26)

= 0

.578

(to

3 si

g fig

) M

etho

d 2:

10

0 w

eeks

, Y

Po(2

590)

λ

= 2

590

> 10

. N

orm

al a

ppro

x to

Poi

sson

Y

N

(259

0,25

90) a

ppro

x P(

Y≤

2600

) = P

( Y <

260

0.5)

(With

cc)

=

0.57

8 (to

3 si

g fig

)

12 (i) (ii)

Let X

be

the

r.v fo

r the

num

ber o

f fis

h w

hich

mea

sure

s les

s tha

n 8

cm lo

ng.

95

45

14

1(

)(

)(

)5

55

C =

0.0

0330

( to

3 sf

)

X

B(n

,0.2

) Si

nce

n la

rge

and

p= 0

.2 ,

X

N(0

.2n,

(0.2

)(0.

8)n)

app

rox

P(X≤

10) ≤

0.02

27

P(X

< 1

0.5)

≤0.

0227

P(Z

< n

n4.0

2.05.

10−

) ≤

0.02

27

nn

4.02.0

5.10

−≤

-2.0

0092

9....

......

(1)

10.5

– 0

.2n ≤

-0.8

0037

2n

H

ence

10.

5 –

0.2n

≤-0

.8n

app

rox

Met

hod

1

Usi

ng G

C Y

1 = 1

0.5

-0.2

x +

0.8

x

→

Tabl

e

A

ns :

91

Met

hod

2 : U

se G

C a

nd g

raph

7

Met

hod

3:

Hen

ce (1

0.5

– 0.

2n)2 ≥

(-0.

8n

)2 H

ence

4n2 -

484n

+ 1

1025

≥0

appr

ox…

…(2

) Fr

om G

C :

n≥

90.6

or n

≤ 3

0.4

n ≥91

or n

≤30

(N

A b

ecau

se d

oes n

ot sa

tisfy

(1) )

Le

ast n

= 9

1

13 (i) (ii)

Let H

be

the

even

t tha

t the

mem

ber o

f pub

lic h

as h

yper

tens

ion

Let D

be

the

even

t tha

t the

mac

hine

dia

gnos

ed h

yper

tens

ion

(i)

P(H

/D) =

32=

)(

)(

DP

DH

P∩

32=

)1.0)(

1(8.0

8.0k

kk −

+

(

1)

k =

0.2

p

% =

20

%

p =

20

P(H

’/D’)

=

)(

)(

DP

DH

P′′

∩′=

)9.0)(

1(2.0

)9.0)(

1(k

kk −

+−=

0.94

7

Exam

ples

of p

ossi

ble

com

men

ts:

(1)

If it

doe

s not

find

you

hyp

erte

nsiv

e th

en y

ou c

an b

e re

ason

ably

con

fiden

t th

at y

our b

lood

pre

ssur

e is

nor

mal

. (2

) If

it d

iagn

oses

hyp

erte

nsio

n, th

en y

ou sh

ould

con

sult

your

doc

tor f

or fu

rther

te

sts.

(3

) A

ny o

ther

logi

cal c

omm

ents

with

refe

renc

e to

the

cont

ext o

f the

que

stio

n

14

(i) (ii)

Cas

e 1:

2 re

d ba

lls -

21

2 =

Cas

e 2:

2 b

lue

balls

- 3

32 =

Cas

e 3:

2 g

reen

bal

ls -

5

102 =

No.

of w

ays =

13

1014

++

=

Cas

e 1:

No

red

ball.

C

hoos

e 6

balls

from

a to

tal o

f 8 (b

lue

and

gree

n)

bal

ls:

8 6

.

Cas

e 2:

No

blue

bal

l.

C

hoos

e 6

balls

rom

a to

tal o

f 7 (r

ed a

nd g

reen

)

H

H’

D

D

D ’ D’

0.8 0.2

0.1

0.9

k 1-k

8

(ii

i)

bal

ls:

7 6

[N

ote:

We

can’

t exc

lude

gre

en b

alls

bec

ause

tota

l

n

umbe

r of r

ed a

nd b

lue

balls

is o

nly

5.]

No.

of w

ays =

10

87

210

(28

7)17

56

66

−+

=−

+=

Alte

rnat

ive

Met

hod:

C

ase

Gre

en

Blu

e R

ed

No.

of w

ays

1 4

1 1

53

230

41

1

××

=

2 3

2 1

53

260

32

1

××

=

3 3

1 2

53

230

31

2

××

=

4 2

3 1

53

260

32

1

××

=

5 2

2 2

53

220

22

2

××

=

6 1

3 2

53

25

13

2

××

=

Tota

l 17

5

Ex

clud

ing

the

gree

n ba

lls, w

e on

ly h

ave

1,

1, 2

, 2, 3

. Sin

ce w

e ar

e ig

norin

g th

e co

lour

s of

the

balls

, we

are

form

ing

3-di

git n

umbe

rs fr

om th

e

5 di

gits

1, 1

, 2, 2

, 3.

Cas

e 1:

All

3 di

gits

are

dis

tinct

.

T

he 3

dig

its a

re 1

, 2, 3

and

the

num

ber o

f way

s of

arr

angi

ng th

em a

re 3

!

C

ase

2: 2

dig

its a

re id

entic

al.

Ste

p 1:

Cho

ose

2 di

gits

that

are

iden

tical

(1, 1

or 2

, 2):

2

S

tep

2: C

hoos

e a

digi

t fro

m th

e re

mai

ning

dig

its

(1

, 3 o

r 2, 3

): 2

Ste

p 3:

Arr

ange

the

3 ch

osen

dig

its in

a ro

w:

3!

32!=

N

o. o

f way

s =

()(

)3!

3!2

26

1218

2!

+=

+=

Documents

ACJC Math Prelim 10