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Anglo-Chinese Junior Col legeH2 Mathematics 9740

2010 JC 2 PRELIM Marking Scheme

Paper 2:

1( )

( ) ( )

( ) ( )

( )

1

2

2 1

1

112

2 2

2 1 2 111

2

1 12

2 1 2 12 2

112

4 2 2

1

1 1

1 1 1 1

2 2 2 2

14 1

2

x

x

x x

x x

x xx x

e dxe

e dx e dxe e

e e e e

e e e e

= +

= + + +

= + + +

2 (i)

2 (ii)

2 (iii)

Least value of z w = 1

Greatest ( )arg 3z + = 1 11 3

tan sin 0.826 (3 dp)5 26

+ =

Least ( )arg 3z + = 1 11 3

tan sin 0.432 (3 dp)5 26

=

3ln 2 0

ax

x

where 1a > .

2

2 1

20

1 1 8 ) 1 1 8 )2

4 40

1 1 8 1 1 8

04 4

ax

x

x x a

x

a ax x

x

a a

x or x

+ + +

+ + +

<

4 (a)( ) ( )3 2 3cos 3 sin

dx x x

dx=

x

y

(2,1)

O3

(5,3)

o

2

3

26

3

z

w

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2

5 3

3 2 3

3 3 3 2

3

3 2 3

33 3

sin

( sin )

1 1cos cos 3

3 3

cos cos3

1cos sin

3 3

x x dx

x x x dx

x x x x dx

xx x x dx

xx x c

=

=

= +

= + +

4 (b)5 sec 10,

4

5 sec tan 2 5,3

x when x

dxwhen x

d

= = =

= = =

( )

( )

[ ]

2 5 32 2

10

332 2

4

3

2

4

3 3

2

4 4

3

3

4

4

5

5sec 5 5 sec tan

1 sec

5 tan

1 cos 1 cot cos5 sin 5

1 1 1cos

5 sin 5

1 2 1 22 3 . .

5 15 5 15

x dx

d

d

d or ec d

ec

i e a b

=

=

=

= =

= = =

5

(i)

(ii)

1

1

: 2 3

1

p r

=

1

1 2

: 0 1 ,

1 4

l r

= +

2 1

1 . 2 2 2 4 0

4 1

= + =

1 1

0 . 2 1 0 1 3

1 1

= +

l1is parallel top1. l1is not contained inp1.

Alternative method:

1 2 1

2 2 3

1 4 1

+ = +

Since no solution for , l1is parallel and not contained top1

1 2 32 1 3 2

1 4 1

=

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No , (i) The weight of a husband and wife may not be independent

Or (ii) Randomness is not there ( a randomly chosen women but spouse is not

randomly chosen)

Or (iii) Distribution of weight of married woman is different from distribution of

adult woman.

Etc

9 Telephone costs are assumed to be normally distributed.

To test H0: = 72

against H1: > 72 at 5% level of significance

Under H0,

ns

x 0

_

t(5-1)

Test statistics : T=ns

x 0

_

=

82.6 72

10.6677 / 5

= 2.2219

pvalue = P(T > 2.2219) = 0.0452 < 0.05

Reject H0at the 5% level of significance. We conclude that there is sufficient

evidenceat the 5% level of significancethat there is evidence of an increase in

mean monthly costs.

To test H0: = 0

against H1: > 0 at 5% level of significance

Under H0,n

x

89.9

0

_

N(0,1)

Test statistics : Z =n

x

0

_

= 082.6

9.89 / 5

= (82.6 - 0 )89.9

5

Do not reject H0if P(Z > (82.6 - 0 )89.9

5) > 0.05

(82.6 - 0 )89.9

5< 1.64485...........(1)

0> 75.310

(i) 161x= (from calculator or computation)

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(ii)

(iii)

when 161x= , 103.6 0.726x y= +

(161 103.6) / 0.726y=

79.06336088=

using y y n= 1

79.06336088 (65.1 73.2 85 80.9 89.9)6

k= + + + + +

80.3k= Use G.C. to find regression line ofyonx:

97.593 1.097y x= +

Useyonxline to predict weight.

When 165x= , 97.593 1.097(165)y= +

83.4y= (1 d.p.) using 3 d.p. of aand bto compute.

or 83.5y= - using full accuracy of aand bto compute.

Using G.C., 0.893r=

C is unusually overweight.

11

(i)

Breakages occur randomly or

Breakages occur independently or

Meannumber of breakages is a constant

Let A be the r.v for the number of broken cups per day

A Po(2.1)

P(A 3) = 1 P(A 2) = 0.350369 = 0.350 (3 sig figs)

Let X be the r.v for the number of days with a least 3 broken cups out of n cups.

X

y

65.1

73.2

80.380.9

85.0

89.9

150 157 160 162 167 170

x

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(ii)

P(X 2) > 0.9991 P(X 1) > 0.999............(1)P(X 1) < 0.001n 22least n is 22

T Po(2.1x7 + 1.6 x 7)

T Po(25.9)Method 1:

Since n is large,_

T (25.9,100

9.25) approx by central limit theorem

P(_

T 26) = 0.578 (to 3 sig fig)Method 2:

100 weeks, Y Po(2590)

= 2590 > 10 . Normal approx to Poisson

Y N(2590,2590) approx

P(Y 2600) = P( Y < 2600.5) (With cc)= 0.578 (to 3 sig fig)

12

(i)

(ii)

Let X be the r.v for the number of fish which measures less than 8 cm long.

9 5 4

5

1 4 1( ) ( ) ( )5 5 5

C = 0.00330 ( to 3 sf)

X B(n,0.2)

Since n large and p= 0.2 , X N(0.2n,(0.2)(0.8)n) approx

P(X 10) 0.0227P(X < 10.5) 0.0227

P(Z