Solutions to H1 Mathematics (8864) J2 Preliminary Examination 2013

1(a) 2 2 3 0kx x k

0k and 2

2 4( )(3 ) 0k k

2

2

2

4 12 0

1 3 0

3 1 0

3 1 3 1 0

k

k

k

k k

1

3k or

1

3k

1

3k

1(b) 33

2e 1

e

x

x

Let 3e xy

2

3 3

21

2 0

( 2)( 1) 0

2 or 1

e 2 or e 1 (reject)

3 ln 2

1ln 2

3

x x

yy

y y

y y

y y

x

x

2. 3 22 3 3 8x x x

3 22 3 8 3 0x x x

1

3

1

3

k

x

y

-3 1 1

2

O

2

8864/J2 Prelim 2013

13 or 1

2x x

Hence, solve 3 22(ln ) 3(ln ) 8ln 3 0x x x

3 22(ln ) 3(ln ) 3 8lnx x x

Replace x by ln x ,

1

3 2

1ln 3 or ln 1

2

0 e or e e

x x

x x

3. Area:

2

2

12 10

2

12 10

2

5 1

4

x xy

xy x

y xx

Cost 15 2 2 20

5 130 30 20

4

150 2530

2

150 60 25

2

P x y x

x x xx

xx

xx

For cost to be minimum, d

0d

P

x

2

d 150 60 25

d 2

P

x x

=0

2

2

150 60 25

2

300

60 25

x

x

2 60

12 5

60 60, ( . .)

12 5 12 5

x

x N A

3

8864/J2 Prelim 2013

2

2 3

d 3000

d

P

x x when

60

12 5x

Cost is minimum when 60

12 5x

.

150 60 25 60min

2 12 560

12 5

P

$204 (nearest dollar)

4(a)

3

1f ( )

(1 2 )x

x

1

3 2 2

3

1 (1 2 ) 1f ( )d d (1 2 ) d

1 1 2(1 2 ) ( 2)2

xx x x x x C C

xx

30

0

1d 1.5

(1 2 )

11.5

1 2

11 1.5

1 2

12.5

1 2

1 25

1 2 4

41 2

25

21 (or 0.42)

50

c

c

xx

x

c

c

c

c

c

4

8864/J2 Prelim 2013

4(b)

Points of intersection (1.4721, 2.6656) and (3.6635, 4.5597)

(1.47, 2.67), (3.66, 4.56)

3.6635 2 0.5

1.47215 3 2 1 d 2.0141 2.01xx x

5(i)

14 4 2 1

2 1

ay a x

x

d

d

y

x

2

2

22 2 1

2 1

aa x

x

At P, 3x , 2

d 2 2

d (2 3 1) 5

y a

x

2 2

25 5

5 [AG]

a

a

(ii) At P, 3, 3x y , gradient of normal is

5

2

53 ( 3)

2

5 110

2 2

y x

y x

(iii) When y = 0,

50 4

2 1x

52 1

4x

2

5 3y x

0.52 1xy

1y

5

8864/J2 Prelim 2013

9

8x

9,0

8Q

(iii) 5 110

2 2y x

When1

0, 102

x y

98

98

3

3

5Required Area Area of trapezium 4 d

2 1

1 1 510 3 3 4 ln 2 1

2 2 2

xx

x x

81 5 9 5 512 ln 5 ln

4 2 2 2 4

51 5ln 5 ln 5 ln 4

4 2

51 5ln 4

4 2

(iii) Alternative:

5 110

2 2y x

When1

0, 102

x y

98

98

3 3

0

3 3

2

0

Required Area

5 1 510 d 4 d

2 2 2 1

5 1 510 4 ln 2 1

4 2 2

x x xx

x x x x

25 5 9 5 53 10.5 3 0 12 ln 5 ln

4 2 2 2 4

51 5ln 5 ln 5 ln 4

4 2

51 5ln 4

4 2

6

8864/J2 Prelim 2013

6(i) Select every 10th

(10% means every one out of 10

orpopulation size

10sample size 0.1

Nk

N ) electrical component from a

production line, starting from say the 5th

component, where the

number 5 is randomly determined from the first 10 components

produced. i.e choose the 5th

, 15th

, 25th, until 10% of the

components are selected.

(ii) Choosing the sample using systematic sampling ensures that the

sample chosen is evenly spread out and consists of components

from different batches compared to choosing components from the

first batch of sample.

7 (i)

(ii) P(Samuel wins a prize)

= P(B) + P(RB) + P(RRB)

= 1 24 1 24 23 1

25 25 24 25 24 23

= 3

25 (or 0.12)

(iii)

P took more than 1 ball Samuel won a prize

P took more than 1 ball AND won a prize

P won a prize

P RB or RRB

3

25

24 1 24 23 1

25 24 25 24 23

3

25

2 (or 0.667 to 3 s.f.)

3

23 22

P only one of them wins a prize = 3 0.27878425 25

R

B

R

B

R

B 24

25

1

25

23

24

1

24

22

23

1

23

7

8864/J2 Prelim 2013

8 (i)

P A B2 2P A B

3 P B 3

P A B P A P B P A B

1 P A' B' P A P B P A B

81 2 P A B

15

53P A B 1

15

p p

p p p

p

2

2

2

P A B 2

P B 3

531

215

3

53 21

15 3

53 15 10

10 53 15 0

p

p

pp

p p

p p

p p

210 53 15 0p p

3 or 5 (NA since 0 1)

10p p p

P exactly one of A and B occurs

P A B P A B

8 3 53 31 1

15 10 15 10

39

50

9(i)

Let X be the no. of seeds that germinates (out of 12).

(12,0.65)X B

P( 7) 1 P( 6) 1 0.212735 0.78726 0.787 (to 3 s.f.)X X

(ii) Let X be the no. of seeds that germinates (out of 12).

(12,0.65)X B

8

8864/J2 Prelim 2013

P(4 9)

P( 9) P( 3)

X

X X

= 0.8471 0.005698

= 0.84310

0.843 (to 3 s.f.)

(ii) Alternative:

P(4 9)

P( 4) P( 5) P( 6) P( 7) P( 8) P( 9)

X

X X X X X X

= 0.84310

0.843 (to 3 s.f.)

Let Y be the no. of packets of seeds with fewer than 7 seeds that

germinates (out of 25 packets).

(25,0.212735)Y B

P( 3) 0.18982 0.190 (to 3 s.f.)Y

Let W be the no. of seeds that will germinate (out of 120 seeds).

(120,0.65)W B

Since 120n is large,

120(0.65) 78 5, 120(0.35) 42 5np nq

(120(0.65),120(0.65)(0.35)) approximately

(78,27.3) approximately

W N

W N

. .P( 90) P( 90.5) 0.0083699 0.00837 (to 3 s.f.)c cW W

10 (i) Let X = mass of jam in a jar

2N ,6.5

1P 320

1000

320 1P

6.5 1000

X m

X

mZ

9

8864/J2 Prelim 2013

3203.0902

6.5

320 20.0863

m

m

340.0863 340.09 m Least m = 340.09 (Shown)

(ii) Let Y = mass of an empty jar

2

2

2 2

N 160,4

N 340.09,6.5

N 340.09 160,6.5 4 N 500.09,58.25

P 505 0.26001 0.260 (to 3 s.f.)

Y

X

X Y

X Y

(iii) Let J = X +Y . So .

P 498 0.85556 0.856J (iv) Let C = mass of an empty large carton.

Let D = mass of an empty small carton.

Let L = mass of a large carton of jam,

ie. L =C + J1 + J2 + ...+ J12

Let S = mass of a small carton of jam,

ie. 1 2 6...S D J J J

P -10 < L - 2S

10

8864/J2 Prelim 2013

11 (i) Unbiased estimate of the population mean

= 12 105

12 12 14.150

xx

n

Unbiased estimate of the population variance

2

22 1

1

x cs x c

n n

21 1052250

49 50

=

2029.5

49 = 41.4 (3 s.f.)

(ii) Let X be the percentage of fat content in burgers

0H : 12

1H : 12

Since n is large, by Central Limit Theorem,

2029.5

N 12,2450

X

approximately.

Test statistic 2029.52450

12(0,1)

XZ N

approximately

= 0.02

From GC, z = 2029.52450

14.1 12= 2.307 (5 sf)

p-value = P( 2.307)Z =0.0105 (3 sf)

Since p-value = 0.0105 (3 sf) < 0.02, we reject 0H at 2% level

of significance and conclude there is sufficient evidence to refute

the companys claim.

(iii) 2% level of significance means there is a probability of 0.02 of

wrongly concluding that the mean fat content is more than 12 %

(rejecting Ho) when in fact it is 12%.

(iv) It is not necessary to assume a the percentage fat content have a

normal distribution for the test to be valid, since sample size is

large, the Central Limit Theorem applies.

0H : 12

1H : 12

Since n is large, by Central Limit Theorem,

26.5

N 12,50

X

approximately.

Test statistic 26.5

50

12(0,1)

XZ N

approximately

Claim is untrue reject Ho Reject Ho at 5% level of significance

11

8864/J2 Prelim 2013

[using GC: invNorm(0.95) = 1.64485]

26.550

1.64485

121.64485

13.5

z

F

F

Least possible value of F is 13.5

12(i)

(ii) r = 0.993

The value of r is close to 1. There is strong negative linear correlation between the amount of catalyst and the time taken for

the chemical reaction. This means that the time taken for the

chemical reaction decreases as the amount of catalyst present

increases.

(iii) t = 119.4385 40.5813x

(iv) t on x line is used since x is controlled and t is random.

60 = 119.4385 40.5813 x

x =1.4647 1.46 grams The estimate is reliable because t = 60 is within the data range

hence this is an interpolation and r is close to 1, indicating a strong negative correlation between the amount of catalyst and

time taken for the reaction.

0

5%

1.64485

95%

t

x

,x t = (1, 78.9)