JJ 9647 H2 Maths Prelim 2010 Plus Solutions

JURONG JUNIOR COLLEGE

J2 Preliminary Examination

MATHEMATICS 9740/01 Higher 2 20 August 2010

Paper 1 3 hours

Additional materials: Answer Paper Graph paper List of Formulae (MF15) Cover Page

READ THESE INSTRUCTIONS FIRST

Write your name and civics 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 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 a 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, with the cover page in front.

This document consists of 6 printed pages.

[Turn over

2

1 (i) Express 3

1

1 x− in the form

21 1

A Bx C

x x x

++

− + +, where A, B, and C are constants to be

found. [2]

(ii) Hence find the coefficient of rx in the expansion of

2

2,

1

x

x x

+

+ + in ascending powers

of .x [4]

2 (i) Show that 1

,2 3 3

k k k+ = −

where ,k +∈� 3k ≥ .

[2]

(ii) Hence find

3 2

n

r

r

=

∑ . [4]

3 A curve is defined by the parametric equations

3(sin 2 cos 2 ),x t t= − 3(2 cos 2 ).y t t= −

Find the equations of the tangent and the normal to the curve at the point P, where4

tπ

= .

The tangent and normal to the curve at P meet the y-axis at R and S respectively. Find the area of the triangle PRS. [6]

4 A sequence 1u , 2u , 3 , u K is such that 1 0u = and

( )1 3 4 1 ,n nu u n n+ = + − for all .n+∈�

(i) Prove by induction that 23 2 1n

nu n= − − for all n

+∈� . [4]

(ii) Given 0 0u = , find

2

0

N

n

n

u

=

∑ in terms of .N

( )( )2

1

1Given 1 2 1

6

n

r

r n n n

=

= + +

∑

[4]

3

5 Given that 1 ln(1 )y x= + + , show that

(i)

22

2 2

d d 10.

d d 2(1 )

y yy

x x x

+ + =

+

[2]

(ii) the Maclaurin’s series for y in ascending powers of x, up to and including the

term in 3,x is

2 31 3 171 .

2 8 48x x x+ − +

[3]

(iii) Expand

2

2 31 3 171

2 8 48x x x

+ − +

in powers of x up to and including 3,x simplifying

your answer. Explain briefly how the result can be used as a check on the correctness

of the first four terms in the series for y. [3]

6 (a) A geometric progression has first term 1 and common ratio r. The sum of the first four

terms is less than the sum to infinity of the remaining terms. Without the use of a

graphic calculator, find the range of values of r. [4]

(b) Adam decided to save some money each day to buy his favorite toys. On the first day,

he saved one twenty-cent coin; on the second day, he saved two twenty-cent coins;

and so on. Find the total amount of money saved at the end of one year.

(Assume 365 days in a year). [3]

After spending all his savings in the first year to buy his favorite toys, Adam started

saving in the same manner again in the following year. As an encouragement, his

mother contributed the same amount that he saved every Saturday and Sunday.

Assuming the second year started on a Monday, find the total amount of money saved

at the end of the second year. [3]

7 The line l passes through the points A and B with coordinates ( )1, 2, 3− and

( )5, 14, 11 respectively. The plane p has equation 2 3 6 7.x y z+ − = −

(i) Show that the line l is parallel but not contained on the plane p.

[3]

(ii) Find the distance of the line l from the plane p.

[3]

(iii) Find a cartesian equation of the plane which contains l and is perpendicular to p. [2]

[Turn over

4

8 (a) The complex numbers p and q are such that

2 ip a= + , iq b= − ,

where a and b are real numbers.

Given that 13 13ipq = + , find the possible values of a and b .

[4]

(b) Sketch the locus of z which satisfies

4 3i 2 and Re(z) 4z − − = ≥ .

(i) Find the least and greatest value of z .

(ii) Show that the greatest value of arg( i)z − is 4

π.

[2]

[2]

[1]

9 The function f is defined by 2

f:x

xxλ −

a for , 0,x xλ∈ − < <� where λ is a

positive constant.

(i) Show by differentiation that ( )f x increases as x increases. [2]

(ii) Find ( )1f ,x− stating the domain of 1f .− [4]

(iii) Find the value of λ such that 1 1 1 1f f .

2 2

− − − = −

[3]

10

(i) Find the integral 2

1 d .

1 4

xx

x

−

+∫

[3]

(ii) By sketching the graphs of 3exy = and 3,y x= + or otherwise, solve the inequality

3e 3.xx> + [3]

Hence find 2

2 3e 3 d ,x x x

−− −∫

giving your answer in an exact form.

[4]

5

11 (a)

The function f is defined by f(x) = (ln x)2 for x > 0. The diagram shows a sketch of the graph of

y = f(x). The minimum point of the graph is A.

(i) State the x-coordinate of A. [1]

(ii) Use the substitution x = eu to show that the area of the region bounded by the

x-axis, the line x = e and the curve is given by

12

0d .u

u e u∫

Hence, find the exact value of this area. [5]

(b)

Find the volume of the solid formed when the region R, bounded by the lines 3

22

y x= − + ,

1y = and the curve cos ,y x= is rotated 2π radians about the y-axis, giving your answer

correct to 3 decimal places. [4]

x

y

O A

( )2lny x=

[Turn over

x

y

O

1y =

32

2y x= − +

cosy x=

R

6

12 The curve C has equation

2

2,

ax ky

x kx k

−=

− +

where a, k are constants and 0a > .

(i) Find the values of k such that C has two asymptotes.

[2]

(ii) The diagram shows the graph of 2

2

5x ky

x kx k

−=

− + for some 0.k >

On separate diagrams, draw sketches of the graphs of

(a) 2

2

25,

25 5

x ky

x kx k

−=

− +

(b) 2

2

2

5,

k xy

x kx k

−=

− +

(c) 2

2,

5

x kx ky

x k

− +=

−

including the coordinates of the points where the graphs cross the axes and the equations of any asymptotes. [8]

y

x 2

5

k

k−

O

7

2010 H2 Mathematics Prelim Exam P1 Solutions

1. (i) Let

3 2

1

1 1 1

A Bx C

x x x x

+≡ +

− − + +

Hence, ( ) ( ) ( )21 1 1A x x Bx C x≡ + + + + −

Using cover-up rule, 1

3A =

Let 0x = , 1 A C= + ,

2

3C =

Comparing coefficients of x , 0 A B C= + − , 1

3B =

( ) ( )3 2

1 1 2

1 3 1 3 1

x

x x x x

+∴ ≡ +

− − + +.

1. (ii)

2

2

1

x

x x

+

+ +

3

3 1

1 1x x= −

− −

( ) ( )1 133 1 1x x

− −= − − −

( ) ( )3 6 23 1 1x x x x= + + + − + + +K K

2 3 4 52 2x x x x x= − − + − − +K

Hence, coefficients of 2 if is a multiple of 3

1 otherwise

rr

x

= −

2. (iii) RHS

( )( ) ( )

1 ! !

3! 1 3 ! 3! 3 !

k k

k k

+= −

+ − −

( )

! 11

3! 3 ! 2

k k

k k

+ = −

− −

( )

! 1 2

3! 3 ! 2

k k k

k k

+ − + =

− −

( )

! 3

3! 3 ! 2

k

k k

=

− −

( )

!

22! 2 !

kk

k

= =

− = LHS

8

2. (iv)

3 2

n

r

r

=

∑1 2 4 3

2 2 2 2 2

n n n− − = + + + + +

K

1

3 3

1

3 3

1 2

3 3

5 4

3 3

4 3

3 3

n n

n n

n n

+ = −

− + −

− − + −

+ −

+ −

M

1

13

n + = −

or ( 1)!

13!( 2)!

n

n

+−

−

3. 3(sin 2 cos 2 )x t t= − , 3(2 cos 2 )y t t= −

3(2cos 2 2sin 2 )dx

t tdt

= +

3(2 2sin 2 )dy

tdt

= +

3(2 2sin 2 ) 1 sin 2

3(2cos 2 2sin 2 ) cos 2 sin 2

dy t t

dx t t t t

+ += =

+ +

When 4

tπ

= , at point P , 3x = , 3

2y

π= and 2

dy

dx=

Equation of tangent at P is 3

2( 3)2

y xπ

− = −

When x = 0, 3

62

yπ

= − . R is (0 , 3

62

π− )

Equation of normal at P is 3 1

( 3)2 2

y xπ −

− = −

When x = 0, 3 3

2 2y

π= + . S is (0 ,

3 3

2 2

π+ )

As PR ⊥ PS,

Area of the triangle PRS =1

.(3)2

RS = 1 3 1

( 6)(3) 11 2 2 4

+ =

9

4. (i) Let nP be the statement 23 2 1n

nu n= − − for all 1n ≥ .

When n = 1, LHS = 1 0u = (By definition)

RHS = ( )223 2 2 1 0− − =

1P∴ is true.

Assume that kP is true for some k+∈� , 1.k ≥

That is, 23 2 1k

ku k= − − -----------------------(1)

We want to prove 1+kP , ie ( )

21

1 3 2 1 1.k

ku k

+

+ = − + −

LHS ( )( )3 4 1ku k k= + −

( ) ( )23 3 2 1 4 1k k k k= − − + −

1 2 23 6 3 4 4kk k k

+= − − + −

( )1 23 2 2 1 1k k k+= − + + −

( )213 2 1 1k

k+= − + −

kP∴ is true 1kP +⇒ is true.

Since 1P is true, and k

P is true 1kP +⇒ is true. By Mathematical

Induction, 23 2 1n

nu n= − − is true for all 1n ≥ .

4.

(ii) ( )2 2

2

0 0

3 2 1

N Nn

n

n n

u n

= =

= − −∑ ∑

( )( ) ( ) ( )

( )

( ) ( )

2 1

2 1

2 12

1 3 1 22 2 1 2 2 1 2 1

2 6

3 1 2 12 4 1 3

2 3

2 13 18 2 3

2 3

N

N

N

NN N N

NN N

NN N

+

+

+

−= − + + − +

− += − + +

+−= − + +

5 (i) Method 1

2 1 ln(1 )

12

1

1

2 (1 )

y x

dyy

dx x

dy

dx y x

= + +

=+

=+

10

Method 2

1

2d 1 1

(1 ln(1 ))d 2 1

1

2 (1 )

d 1

d 2(1 )

yx

x x

y x

yy

x x

− = + +

+

=+

=+

222

2 2

22

2 2

d d 1 1 ( 1)(1 )

d d 2 2(1 )

d d 10

d d 2(1 )

y yy x

x x x

y yy

x x x

− − + = − + =

+

+ + =

+

5. (ii)

3 2 2

3 2 2 3

2 3

2 3

2 3

2 3

d d d d d 1. 2 0

d d d d d (1 )

when 0,

d 1 d 3 d 17 1, , ,

d 2 d 4 d 8

1 3 171

2 4 2! 8 3!

1 3 17 1

2 8 48

y y y y yy

x x x x x x

x

y y yy

x x x

x xy x

x x x

+ + − =+

=

= = = − =

∴ ≈ + + − +

≈ + − +

(iii)

2

2 31 3 171

2 8 48x x x

+ − +

2 3 2 3

2 3

1 3 17 1 3 171 1

2 8 48 2 8 48

1 11 ...

2 3

x x x x x x

x x x

= + − + + − +

= + − + +

Given 1 ln(1 )y x= + + 2 1 ln(1 )y x⇒ = + + 2

2 3

2 3

LHS

1 11 ...

2 3

RHS 1 ln(1 )

1 11 ...

2 3

y

x x x

x

x x x

=

= + − + +

= + +

= + − + +

Since LHS = RHS, the first four terms in the series for y is correct.

11

6. (a) 4 4S S S∞< −

42S S∞<

( )42 1 1

1 1

r

r r

−<

− −

( )42 1 1r− < (Since 1 0r− > )

4 1

2r >

1

411

2r

− < < −

or

1

411

2r

< <

.

6. (b) Total sum ( ) ( ) ( )20 2 20 3 20 365 20= + + +K

( )( ) ( )365 1 365

202

+ =

= 1335900 cents Total sum

( ) ( ) ( ) ( ) ( ) ( )1335900 6 20 7 20 13 20 14 20 ... 363 20 364 20= + + + + + + +

( ) ( )( ) ( )( )( )1335900 20 13 13 14 13 2 14 13 51 14= + + + + + + +K

( )52

1335900 20 13 7272

= + +

1720700= cents

7

5 1

14 2

11 3

6 3

12 2 6

8 4

AB

−

= −

= =

uuur

3 2

6 3 6 18 24

4 6

0

• = + − −

=

⇒ l is parallel to p.

1 2

2 3 2 6 18

3 6

14 7

−

• = − + − −

= − ≠ −

∴ l is parallel but not contained on the plane p.

12

7. (i) Let m be the line perpendicular to p and passing through A.

Vector equation of line

1 2

: 2 3 ,

3 6

m µ µ

−

= + ∈ −

�r

Let F be the foot of perpendicular of A to p.

( )

1 2 2

2 3 3 7

3 6 6

2 1 2 3(2 3 ) 6(3 6 ) 7

1

7

µ

µ

µ

µ µ µ

µ

− +

+ • = − − −

− + + + − − = −

=

2 2 2

1 21

2 37

3 6

1 2 11

2 3 27

3 6 3

21

37

6

12 3 ( 6)

7

1

OF

AF

AF

−

= + −

− −

= + − −

= −

= + + −

=

uuur

uuur

uuur

∴Distance of the line l from the plane p=1 unit

7. (ii) Let the plane required be 1p .

1

3 2 48

Normal of = 6 3 26

4 6 3

1 48

2 26 48 52 9 91

3 3

p

−

× = − −

− −

• = + − = −

1

48

Equation of : 26 91 48 26 3 91

3

p r x y z

−

∴ • = ⇒ − + − = −

13

8 (a) 13 13pq i= +

( ) ( )2 13 13

2 2 13 13

( 2 ) ( 2) 13 13

ia b i i

b i abi a i

a b i ab i

+ − = +

− + + = +

+ + − = +

Comparing real and imaginary parts,

2 13 - (1)

2 13 - (2)

a b

ab

+ =

− =

(2): 15

ba

=

Subst. 15

ba

= into (1):

( ) ( )

2

152 13

13 30 0

3 10 0

3 or 10

35 or

2

aa

a a

a a

a

b

+ =

− + =

− − =

=

=

(i)

2 2

Least

4 1

17

z OP=

= +

=

2 2

Greatest

4 3 2

7

z OQ=

= + +

=

(4, 3)

x

y

O 4

3

Locus of z

P (4, 1)

Q

A(0, 1)

R (4,5)

14

(ii)

1

Max arg( )

4tan

4

4

z i PAR

π

−

− = ∠

=

=

9 (i) ( )

2f

xx

xλ=

−

( )( ) ( )

( )

12 2 2

'

3 22 2

12

2f 0

x x x x

xx x

λ λλ

λ λ

− − − − −

= = >− −

(ii) domain of 1f − = range of f = ( ),0−∞ .

( )

( )

2

2 2 2

2 2 2

2

2

1

1

xy

x

y x x

x y y

yx

y

λ

λ

λ

λ

=−

− =

+ =

= ±+

Since 0,xλ− < < that is, 0,y <21

yx

y

λ=

+.

That is, 1

2f ( )

1

xx

x

λ− =+

, for 0.x <

(iii)

( )1 1

2f f ( )

1 1

xx

x

λ

λ

− − =+ +

( )

( )( )

( )( )

2

2

0.50.5

1 1 0.5

4 5 0

4 5 1 0

λ

λ

λ λ

λ λ

−= −

+ + −

− − =

− + =

Since 0,λ >5

4λ = .

Alternatively:

( )

1 1

1

f f ( 0.5) 0.5

f ( 0.5) f 0.5

− −

−

− = −

− = −

(Note: students could also consider to solve ( )f 0.5 0.5− = − )

15

( )

( )

( )

( )

( )( )

2 2

2

0.5 0.5

1 0.5 0.5

4 5 0

4 5 1 0

λ

λ

λ λ

λ λ

− −=

+ − − −

− − =

− + =

Since 0,λ >5

4λ = .

10 (i)

( )

2

2 2

2 1

1

1 4

1 8 1

8 1 4 1 4

1 1ln 1 4 tan (2 )

8 2

xdx

x

xdx dx

x x

x x c−

−

+

= −+ +

= + − +

∫

∫ ∫

(ii) From the graphs, 2.82x < − or 0x > .

For 2.82x < − or 0x > , 3 3xe x> +

i.e. 3 3 0xe x− − > if 2.82x < − or 0x >

For 2.82 0x− < < , 3 3xe x< +

i.e. 3 3 0xe x− − < if 2.82 0x− < <

2

2 3 3 xe x dx

−− −∫

= 0 2

2 0(3 3) (3 3)x x

e x dx e x dx−

− − − + − −∫ ∫

= 2 2

0 2

2 0[3 3 ] [3 3 ]2 2

x xx xe x e x−− − − + − −

= 2 2[3 (3 2 6)] [(3 2 6) 3]e e−− − − + + − − −

= 2 23 3 10e e− + −

x

y

O

3y x= +

3 xy e=

-2.82

3

16

11 (a) (i) x-coordinate of A = 1

(ii) Area of region bounded = ∫

edxx

1

2)(ln

Let x = eu ⇒ ln x = u, u

edu

dx=

When x = 1, u = 0 & When x = e, u = 1

∴ Area of region bounded = ∫1

0

2 dueu u

Area of region bounded = ∫

1

0

2 dueu u

= 1 12

002[ ]

uu ue duu e − ∫

= ]2)2[(1

0

10 ∫−− dueuee uu

= e – 2e +10)2( u

e

= 222 −=−+− eee

11 (b) Let P be the point of intersection of

32

2y x= − + and cosy x= .

Using GC, the coordinates of point of intersection = (0.94031, 0.58954)

Volume formed about the y –axis

= ∫−−

−

1

58954.0

212

)(cos)2(3

2dyyyπ

= π (0.0907) = 0.285 (3 d.p)

12 (i) Horizontal asymptote is 0y = .

Discriminant of ( )2 20 4 4x kx k k k k k− + = = − = − .

Therefore when 0k = or 4, C has two asymptotes.

(ii) (a)

y

x 2 25k

k−

O

17

(ii) (b)

(c)

y

O 2 5k

k−

k

2 5k

21

5 25 5

k ky x= + −

1 k− O

18

JURONG JUNIOR COLLEGE

J2 Preliminary Examination

MATHEMATICS 9740/02 Higher 2 26 August 2010

Paper 2 3 hours

Additional materials: Answer Paper Graph paper List of Formulae (MF15) Cover Page

READ THESE INSTRUCTIONS FIRST

Write your name and civics 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 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 a 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, with the cover page in front.

This document consists of 6 printed pages.

[Turn over

19

Section A: Pure Mathematics [40 marks]

1 (i)

The diagram shows a rectangle inscribed in a semicircle of centre, O, and fixed radius a.

The length OP is denoted by .x Show that, as x varies, the perimeter of the rectangle is a maximum when its sides are in the ratio 4 : 1. [6]

(ii) Variables x and y are related by the equation

2 2 23y xy x

x+ = − + , where 0y > .

Given that x is increasing at the rate of 1

5 units 1

s− , find the rate of increase of y when x is

1.

[4]

2 A particular solution of a differential equation is given by 2 32( ) 2 .

3x y xy y+ = − Show that

2( )dy

y y xdx

+ = −

[2] A second, related, family of curves is given by the differential equation

2d

d

yx y y

x= +

By means of the substitution y ux= , show that the general solution for y, in terms of x, is

xy

x c

−=

+,

O x P

20

where c is an arbitrary constant. [3] Sketch, on a single diagram, three distinct members of the second family of solution curves, stating clearly the coordinates of the points where the curves cross the axes and the equations of any asymptotes. [5]

21

3 Relative to the origin O, the position vectors of A, B and C are

α i + 3j + 4k, 2i – 2k and 4i + β j – k respectively.

(i) Point M lies on the line segment AB such that : 1: 2AM MB = . Given that the position

vector of M is 2j + 2k, findα . [2]

(ii) Given that the length of projection of BCuuur

onto the line OM is 3

2units, find β ,

where β is a positive constant.

[3]

(iii) The line l has vector equation

2

0 0

2 1

a

µ

= + −

r , , aµ ∈ ∈� � . Using the values of

α and β found above, determine the value of a if l makes an angle of 6

π radians with the

plane ABC. [4]

4 (i) Given that 2 2z = and arg( i )4

zπ

− = , find w in the form ia b+ , where ,a b ∈� , if

2 2wz = and 2 5

arg6

z

wπ

= −

.

[4]

(ii) Solve the equation 4 1 3 i 0z + − = ,

giving the roots in the form re iθ where r > 0 and − π < θ ≤ π. [3]

Show the roots on an Argand diagram. [2] Describe the geometrical shape formed by the points representing the roots and justify

your answer. [2]

22

Section B: Statistics [60 marks]

5 From the letters of the word DISTRIBUTION, find

(i) the number of 4-letter code-words that can be formed if the code-word contains exactly three ‘I’s. [2]

(ii) the number of code-words that can be formed using all the letters such that all the three

‘I’s are separated. [2]

6 In a badminton team of 8 players, 5 are boys and 3 are girls. Boy A and Girl B are the only 2

left-handed players in the team. In a particular practice, 4 players are chosen to play doubles.

Find the probability that

(i) exactly 1 left-handed player is chosen,

[2]

(ii) 2 girls are chosen given that exactly 1 left-handed player is chosen,

[3]

(iii) either Boy A or Girl B is chosen (or both).

[2]

7 The number of guitars sold by a music shop per day follows a Poisson distribution with mean

λ . It is known that on 2 in 7 days, there are no guitars sold. Show that 1.253λ = , correct to 3

decimal places. [2]

(i) Calculate the probability that less than 4 guitars are sold in a day.

[2]

(ii) Using a suitable approximation, find the probability that, in a random sample of 100 days,

there will be more than 95 days in which less than 4 guitars are sold per day.

[4]

23

(iii) Calculate the probability that in a period of 90 days, the mean number of guitars sold per

day is more than 1.5.

[3]

24

8 The random variable X has a normal distribution with mean 15 and variance 5. The random

variable T is the sum of 2 independent observations of X.

(i) Find ( 2 3 )P T X> + .

[3]

(ii) Three independent observations of X are obtained. Find the probability that exactly two

of the observations have value less than 20.

[3]

The random variable Y has a normal distribution with mean µ and variance 2σ .

(iii) If 22.5σ = , find the greatest probability of (15.1 29.9)P Y< < , stating the value of µ .

[2]

(iv) If 10µ = and ( 27) 0.25P X Y+ > = , calculate the value of σ and state an assumption

needed to carry out the calculation.

[4]

9 A sample of 60 customers is to be chosen to take part in a survey conducted by a restaurant

owner.

(i) Explain briefly how the restaurant owner could use quota sampling.

[1]

(ii) The purpose of the survey is to investigate customers’ opinions about the different lunch

and dinner menus.

Give a reason why a stratified sample might be preferable in this context. [1]

Explain clearly how the restaurant owner could use stratified sampling from his list of

regular customers if the ratio of regular customers for lunch and dinner is 2 : 3.

[2]

25

[Turn over

26

10 Water in a reservoir undergoes a purification process before it can be consumed. The

effectiveness, y %, of the process for various flow rates, x m3 s-1, is shown below.

x 1 2 4 6 8 10 20 30 40 y 80 60 45 40 30 25 18 15 10

The variables x and y are thought to be related by the equation ey bax= , where a and b

are constants.

(i) Give a sketch of the scatter diagram of y against ln x . Comment on whether a linear

model would be appropriate referring to the scatter diagram. [2]

(ii) Find the value of the product moment correlation coefficient between y and ln x and

explain whether it supports your comment in part (i). [2]

(iii) Find the least squares regression line of y on ln x and estimate the values of a and b .

[3]

(iv) Predict the effectiveness of the process when water flows at 0.5m3s-1. Comment on the

reliability of your prediction.

[2]

(v) Explain why in this context, the above model would not be appropriate for large values of

x . [1]

11 The past records of a supermarket show that the mean amount spent per customer was $59 with standard deviation $8. The supermarket’s management suspects that the mean amount spent per customer has decreased. A random sample of 8 customers was taken and the amount spent per customer , $x, was recorded. The following result was obtained.

432x =∑

Stating a necessary assumption about the population, test the supermarket’s management’s suspicion at the 5% significance level. [6]

To encourage customers to spend more at the supermarket, the management initiated a

promotional campaign whereby each customer will receive a voucher which can be used to redeem products at the supermarket. A week after the start of the campaign, the manager of the supermarket took a sample of 9 customers and the amount spent per customer, $y, is summarised by

2( 70) 72, ( 70) 1234y y− = − − =∑ ∑ .

The actual mean amount spent per customer is $ µ . In a test at the 5% level of significance,

the hypotheses are:

27

Null hypothesis : 0µ µ=

Alternative hypothesis : 0µ µ≠ .

Given that the null hypothesis is rejected in favour of the alternative hypothesis, find the set

of possible values of 0µ .

[6]

28

2010 JJC Prelim P2 Solutions

1(i) Let the length and the width of the rectangle be 2x and y The perimeter S = 2y + 4x ……(1)

2 2 2y a x= − …………………(2)

Subst (2) into (1)

2 22 4S a x x= − +

2 2

24 0

dS x

dx a x= − + =

−

2 2

24 0

x

a x− + =

−

2 2

2x

a x=

−

2 21 y

5a⇒ =

⇒ 4 1

and 5 5

x a y a= =

32

2 2

2 2 2

20

( )

d s a

dx a x= − <

−

Length : width = 2x : y = 4:1

a

x

y

29

1(ii) 2 2 23y xy x

x+ = − + ……(1)

Differentiate w.r.t x

2

22 2

dy dyy x y x

dx dx x+ + = +

2

2(2 ) 2

dyy x x y

dx x+ = + −

2

22

2

x ydy x

dx y x

+ −=

+ … (2)

Subst. x = 1 into

(1): 2 2 y y+ = ⇒ y= 1 and y ≠ − 2

(2): 1dy

dx=

1

.

1(1)( )

5

1units

5

dy dy dx

dt dx dt

s−

=

=

=

2

2 3

2 3

2 2 3

2

2

2( ) 2

3

2( ) 2

3

2

3

2 2 2

( ) (shown)

x y xy y

x y xy y

y x y

dy dyy x y

dx dx

dyy y x

dx

+ = −

+ − = −

+ = −

+ = −

+ = −

Alternative:

2

2

2( ) 1 2 2 2

( ) (shown)

dy dy dyx y x y y

dx dx dx

dyy y x

dx

+ + = + −

+ = −

30

2

( )

2

2

2

2

2

d

d

use

d d

d d

d

d

d

d

1 d1

d

1 d 1 d

1 , where c is an arbitrary constant

1

yx y y

x

y ux

y ux u

x x

ux x u ux ux

x

ux u u x u

x

u

u x

u xu

x cu

xx c

y

x cy

x c x c

= +

=

= +

∴ + = +

+ = +

=

=

− = +

− = +

= − = − ++ +

∫ ∫

2 Family of solution curves:

11, 1

1

0, 1

11, 1

1

c yx

c y

c yx

= − = − −−

= = −

= = − ++

y

1y = −

x

0c =

1x = − 1x =

1c =

1c = −

0

31

3(i) Using ratio theorem,

2

3

3

2

0 21

3 2 02

2 2

1

3

4

1

OA OBOM

OM OBOA

α

+=

−=

= − −

−

=

= −

uuur uuuruuuur

uuuur uuuruuur

3(ii) 4 2

0

1 2

2

1

BC β

β

= − − −

=

uuur

Length of projection of BCuuur

onto the line OM = 3

2

2 2

2

3

2

2 0

1

1 1 3

21 1

1 3

2 8 0

2 or 4 (rejected)

2

BC OM

OM

β

β

β β

β

β

•=

•

=+

+ =

+ − =

= −

∴ =

uuur uuuur

uuuur

32

3(iii) 2 1 1

0 3 3 1

2 4 2

4 1 5

2 3 1

1 4 5

1 5

Normal of plane 1 1

2 5

3

5

4

AB

AC

ABC

−

= − = − − −

−

= − = − − −

= − × − − −

= −

uuur

uuur

( ) ( )

2

2 2

2

3 3

0 5 0 5 sin6

1 4 1 4

1 3 4 1 50

2

6 8 50 1

7 48 7 0

1 7 or - (reject)

7

7

a a

a a

a a

a a

a

a

π

• − = −

+ = +

+ = +

− − =

=

∴ =

33

4(i) 2 2 2z z= ⇒ =

( ) ( )arg arg arg( ) 4 4

arg( ) 4 2

3

4

iz i z

z

π π

π π

π

− = ⇒ − + =

= − −

=

2 2

2 2

2

wz

w z

w

=

=

∴ =

2 5

arg6

52arg( ) arg( )

6

3 5arg( ) 2

4 6

7

3

( )3

2 cos sin3 3

1 3

z

w

z w

w

pv

w i

i

π

π

π π

π

π

π π

= −

− = −

= +

=

=

= +

= +

4(ii)

( )

4

4

22

3

1

6 24

11 3

64

2 51 1 1 1

6 3 3 64 4 4 4

1 3 0

1 3

2

2

2 , 0, 1, 2

2 , 2 , 2 , 2

k i

ki

k i

i i i i

z i

z i

e

z e

e k

e e e e

ππ

π π

π

π π π π

+

+

+

− −

+ − =

= − +

=

=

= = ± −

=

y

z2

z3

π

34

The points form a square since the diagonals are perpendicular and of equal length.

35

5(i) The different letters left are D S T R B U O N

No of different code words 8

1

4!32

3!C= =

5(ii) To permute the remaining 9 letters 1st,

9!

2 ways.

To slot in the 3Is, 10

3C ways.

Thus number of code words that can be formed is

=9!

2

10

3C

= 21772800

6(i) P(1 left-handed player) =

2 6

1 3

8

4

4

7

C C

C=

6(ii) R L R R R R L RP(G G B B )+P(G G B B )P(2 G given exactly 1 left-handed)=

P(exactly 1 left-handed)

2 4 2 4

1 2 2 18

4

C C + C C

C=

4

7

8235= =

4 5

7

6(iii) P(Boy A or Girl B is chosen) = 1 – P (Both Boy A and Girl B are not chosen)

= 6

81

C

C− =

Alternative:

P(Boy A or Girl B is chosen) = P(Boy A is chosen) + P(Girl B is chosen)

−P(Boy A and Girl B are chosen)

= 7 7 6

3 3 2

8 8 8

4 4 4

C C C

C C C+ −

36

7 Let X denote the number of guitars sold by a music shop in a day.

( )0X P λ�

Given that 2

( 0)7

P X = =

2

7e

λ−⇒ = [M1]

2

ln 1.2537

λ⇒ = − =

7(i) ( 4) ( 3)P X P X< = ≤ [M1]

0.96145 0.961= ≈

7(ii) Let Y be the number of days out of 100 in which at least 4 guitars were sold per day.

(100,1 0.96145) (100,0.03855)Y B B− =�

Since n is large and 5np < , we use a Poisson approximation.

0 (3.855)Y P≈

P(more than 95 days in which less than 4 guitars were sold per day) = P( at most 4 days in which at least 4 guitars were sold per day ) = ( 4)P Y ≤ [A1]

= 0.657 (to 3 sf) [A1]

7(iii) ( )0 1.253X P�

For a large sample of size 90, by Central Limit theorem,

1.253

1.253,90

X N

≈

. [M1]

( )1.5

0.0182

P X >

=.

Alternative : ( ) ( )1 90 0 01.253 90 112.77X X P P+ + × =K � [M1]

( ) ( )1 901.5 135P X P X X> = + + >K [M1]

( )1 901 135P X X= − + + ≤K

0.0183= [A1]

8(i) (15,5)X N�

Let T = 1 2X X+

37

1 2

1 2

( 3 ) ( ) 3 ( ) 15

( 3 ) ( ) 9 ( ) 11(5) 55

E T X E X X E X

Var T X Var X X Var X

− = + − = −

− = + + = =

3 ( 15,55)T X N− −� [A1]

( 2 3 ) ( 3 2)P T X P T X> + = − >

= 0.0109 [A1]

8(ii) ( 20) 0.9873 0.987P X < = ≈ [B1]

Probability = [ ]2

3 ( 20) ( 20)P X P X< >

23(0.9873) (1 0.9873)= −

= 0.0371 [A1]

8(iii) 2( , 22.5 )Y N µ�

For greatest probability, 15.1 29.9 45

22.52 2

µ+

= = =

Greatest (15.1 29.9) 0.258P Y< < = [A1]

8(iv) (iv) (15,5)X N� 2(10, )Y N σ�

2 2(15 10, 5 ) (25, 5 )X Y N Nσ σ+ + + = +�

( 27) 0.25P X Y+ > =

2

27 250.25

5P Z

σ

−> =

+ [M1]

2

2

20.25

5

20.75

5

P Z

P Z

σ

σ

> =

+

< =

+

2

20.6745

5 σ=

+ [A1]

1.95σ = Assumption : The random variables X and Y are independent of each other.

38

9(i) To obtain a quota sampling of 60, divide the diners into two subgroups : male and

female. Select the first 30 males and 30 females who leaves the restaurant.

Or any other relevant answers.

9(ii) By drawing random samples according to the proportion in each stratum, lunch and

dinner, the sample will be a better representation of the population.

Draw random samples from each stratum with sample size proportional to the size of the

strata as follows :

Lunch Dinner

Number of customers to be sampled

260 24

5× =

360 36

5× =

10(i) ey bax=

ln lny a b x= +

The scatter diagram is plotted with y against ln x . y

From the scatter diagram, the points lie close to a straight line, so the linear model is

appropriate.

10(ii) From GC, since 0.982r ≈ − which is very close to -1, it supports the claim in part (i).

10(iii) 73.3 18.4 lny x= −

316.89 10 (3sf )a = ×

18.4b = − (3sf)

10(iv) 86.0% (3sf) Since 0.5x = is out of the given data range of 1 40x≤ ≤ , the prediction is unreliable.

10(v) For large values of x , the model gives 0y < . So the model is not valid for large values

of x .

ln x

39

11 Assume that the amount spent per customer follows a normal distribution.

Let µ be the actual mean amount spent per customer per visit.

0H : 59µ =

1H : 59µ <

Under 0H , 2

~ N( , )Xn

σµ and test statistic

XZ

n

µ

σ

−= ~ N(0, 1)

where 432

54, 59, 8, 8.8

x nµ σ= = = = =

At 5% level of significance, we use a left-tailed z- test and reject H0 if p-value < 0.05.

From the GC, p value 0.0385− = .

Since p value 0.0385 0.05− = < , we reject 0H and conclude that at 5% level of

significance there is sufficient evidence to suggest that the mean amount spent per customer per visit has decreased in recent months.

11 0 0H : µ µ=

1 0H : µ µ≠

( 70)70 62

9

yy

−∑= + =

( )2

2 7211234 82.25

9 1 9s

−= − =

−

Under 0H , 2

0~ N( , )s

Yn

µ and test statistic 0YT

s

n

µ−= ~ t(n− 1)

where 062, , 82.25, 9y s nµ µ= = = =

At 5% level of significance, we use a 2-tailed t- test and reject H0 if 2.306T ≥

Since 0H is rejected at 5% level of significance,

0 2.306Y

s

n

µ−≤ − or 0 2.306

Y

s

n

µ−≥

0

82.2562 2.306

9µ ≥ + or 0

82.2562 2.306

9µ ≤ −

0 55.03µ ≤ or 0 68.97µ ≥

{ }0 0 0: 55.0 or 69.0 µ µ µ∴ ∈ ≤ ≥� or { }0 0 0: 55.0 or >69.0 µ µ µ∈ <�