YJC H1Maths 2013 Prelim Soln

YISHUN JC 2013 JC 2 MATHEMATICS 8864 PRELIMINARY EXAM

Page 1 of 8

Qn Solution

1 235

( 2) 2 5 0k x kx has no real solution i.e. discriminant 2 3

5(2 ) 4( 2)( 5) 0D k k

2 35

5( 2) 0k k 2 3 10 0 ( 5)( 2) 0k k k k

Therefore 5 2k .

2 3 62

6

4 24 2

64 2

x xy y

i.e. 6 6 2 3 3x y y x

lg( ) 1 lg( 1)y x x

lg(2 3) 1 lg( 1)x x

lg(2 3) lg( 1) 1x x

lg (2 3)( 1) lg10x x (2 3)( 1) 10x x

22 5 7 0x x (2 7)( 1) 0x x

1x or 3.5

Reject 1x since lg( 1)x is undefined.

When 3.5x , from 3 3 10.5 3 7.5y x

Therefore 3.5x and 7.5y

3i

3ii 1.000x and 1.501x (3 d.p.)

k 5 2

(0, 1)

y

x

( 1,0)

(0,ln2)

(1,0)

2 1y x

ln( 2)y x

2x


Page 2 of 8

3iii Area bounded by the curves

1.5010175392

1ln( 2) ( 1) dx x x

2.926338462

2.926 (3 d.p.)

4 212 1 2

x xy y x x

x

2

2

d 11 1

d

yx

x x

When 3

2x ,

1

6y and

d 4 51

d 9 9

y

x

Equation of tangent: 1 5 3

6 9 2y x

5 2

9 3y x or 9 5 6y x

Equation of normal: 1 9 3

6 5 2y x

9 43

5 15y x or 15 27 43y x

Point T : When

20

3x y

Point N : When 43

015

x y

Area of 1 43 2 3

2 15 3 2PNT

53

20 or 2.65

5a 2

2 2 4 43 d 3 2 dx x x x x xe e e x e e e x

5 33 2 dx x xe e e x 5 3

3 25 3

x xxe ee C


Page 3 of 8

5b

1 12 22 2d 13 2 7 3 2 7 6 2

d 2x x x x x

x

1223 2 7 3 1x x x

or

122

3 1

3 2 7

x

x x

3

32

2 00

3 1d 3 2 7

3 2 7

xx x x

x x

28 7

7 3 3

2 20 0

9 3 3 1d 3 d

3 2 7 3 2 7

x xx x

x x x x

3 7

5c At

11,

2

, 2d

3 0 3d

yx ax b a b

x

At 2

,3

k

, 2 4

3 3a b

Solving, 5 5

13 3

a a and 2b

2d 3 2d

yx x

x

23 2 y x x dx 3 21

22y x x x C

At 12

(1, ) , 1 12 2

1 2 2C C

Therefore, equation of curve is 3 21

2 22

y x x x

At 2

,3

k

, 8 2 4 76

227 9 3 27

k

6

A B

0.13 0.28


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6i ( ) ( ) ( ')P A P A B P A B

0.35 ( ) 0.28P A B

Therefore ( ) 0.07P A B

6ii ( ) ( ) ( ' )P A B P A P A B

0.35 0.13 0.48

Therefore ( ' ') 1 ( ) 0.52P A B P A B

6iii ( ) ( ) ( ')P B P B A P B A

0.07 0.13

0.2

6iv ( )( | )

( )

P A BP A B

P B

0.07

0.2

0.35

Since ( | ) ( ) 0.35P A B P A , events A and B are independent.

7i

7ii P(correctly diagnosed)

0.08(0.95) 0.92(0.98)

0.9776 or 611

625

7iii P(has diabetes given incorrectly diagnosed)

( ') 0.08(0.05)

1 0.9776 0.0224

P A B

5

28

A

A

B

B

B

B

0.08

0.92

0.95

0.05

0.02

0.98


Page 5 of 8

8a

Age Group

(Yrs)

Uses

Public

Transport

Uses

Private

Transport

Below 26 87 11

26 to 54 124 32

Above 54 43 3

The required number of people are to be selected randomly from each stratum.

8b Advantage: Sample obtained is a good representation of the population.

Disadvantage: Requires precise and accurate information about the population.

9i Let X be the number of ripe watermelons out of 30 shipped out. ~ (30,0.85)X B

P(all 30 are ripe) ( 30)P X

0.0076307596

0.00763

9ii P(fewer than 25 are ripe) ( 24)P X

0.289424

0.289

Let Y be the number of crates, each with fewer than 25 ripe watermelons, out of 50

crates.

~ (50,0.28942)Y B

Since 50n is large, 50(0.28942) 5np and (1 ) 50(1 0.28942) 5n p

~ (14.471,10.28280318)Y N approx.

Required probability

(15 45) (15.5 44.5)ccP Y P Y

0.374146

0.374


Page 6 of 8

10i Let X be the mass (kg) of an adult trout in the lake.

The unbiased estimate for the population mean is ( 3.2)

3.2100

xx

31.23.2

100

3.512 kg (4 sfs)

The unbiased estimate for the population variance is

2

2 2( 3.2)1

( 3.2)100 1 100

xs x

21 31.215.551

99 100

0.05875354

0.05875 (4 sfs)

10ii 0 :H 3.47 kg

1 :H 3.47 kg

Suppose 0H is true, test statistic ~ (0,1)X

Z Ns

n

approx. by CLT where

3.512x kg, 0.05875s kg and 100n .

Level of significance: 5% From GC, p-value 0.04157

Since p-value100

0.05 , reject 0H .

There is significant evidence at 5% level that the marine biologist has understated the

mean mass of the adult trouts in the lake.

0 :H 3.47 kg

1 :H 3.47 kg

Suppose 0H is true, test statistic ~ (0,1)X

Z Ns

n

approx. by CLT where

0.0625s kg and 100n .

Level of significance: 1% From GC, critical or rejection region is where 2.32635z

Since 0H is not rejected, the test-value is such that 3.47

2.326350.25

10

x

0 3.52815875x

Maximum 3.528x kg (3 d.p.)


Page 7 of 8

11i Let X and Y be the mass (g) of a cheese biscuit and a ginger biscuit respectively.

By Central Limit Theorem, since sample size is large, 2

1 2 44... ~ (44 10,44 0.3 )Y Y Y N approx.

i.e. 1 2 44... ~ (440,3.96)Y Y Y N approx. 2

1 2 74... ~ (74 6,74 0.2 )X X X N approx.

i.e. 1 2 74... ~ (444,2.96)X X X N approx.

1 2 44 1 2 74... ...Y Y Y X X X ~ ( 4,6.92)N approx.

1 2 44 1 2 74( ... ... )P Y Y Y X X X

1 2 44 1 2 74( ... ... 0)P Y Y Y X X X

0.9358163198

0.936

11ii 1 2 7... 4X X X Y ~ (7 6 4 10,7 0.04 16 0.09)N

i.e. 1 2 7... 4X X X Y ~ (2,1.72)N

1 2 7( ... 4 )P X X X Y

1 2 7( ... 4 0)P X X X Y

0.9363686479

0.936

12i Let X be the mass (g) of a bag of cashew nuts and 2~ (128.5,1.5 )X N .

( 130) 0.1586552596 0.159P X

12ii Let W be the mass (g) of a carton of 50 bags of cashew nuts. 2~ (50 128.5 550,50 1.5 )W N

i.e. ~ (6975,112.5)W N

( 7000)P W

0.0092110461 0.00921

12iii ~ (6975,112.5)W N

112.5~ 6975,

30W N

i.e. ~ 6975,3.75W N

(6970 6980) 0.9901766644P W

0.990


Page 8 of 8

12iv Let Y be the mass (g) of a bag of cashew nuts and 2~ ( , )Y N .

Given ( 130) 0.18P Y

130 0.18P Z or 130 0.82P Z

Using GC, 130

0.915365082

0.915365082 130 ------------ (1)

Given ( 126) 0.03P Y

126 0.03P Z

Using GC, 126

1.88079361

1.88079361 126 ---------- (2)

Solving, 128.691 128.7 g (4 sfs)

and 1.43053 1.431 g (4 sfs)

13i

The scatter diagram reveals a negative linear correlation between the temperature and the

pH of skimmed milk where the pH of skimmed milk decreases with an increase in

temperature.

13ii Linear correlation coefficient, 0.9647177563 0.965r

Least square regression line of y on x is

6.84992649 0.0071895151y x

i.e. 6.85 0.00719y x

13iii When 20 Cx , 6.71y

This estimate is reliable as 20 Cx is within the range of data given and the correlation

coefficient is close to 1 .

13iv The equation obtained in (ii) can be used to estimate the temperature when given the pH

of skimmed milk because:

a) temperature, x , is the only independent variable,

b) the linear correlation coefficient, r , is close to 1 .

Temp. ( C)x

(4,6.85)

(63,6.36)

pH ( )y

Temperature (x)

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YJC H1Maths 2013 Prelim Soln