2017 hkdse math m1 ctl(20170423) mathematics module 1 ... · 2 3 0.8 4 0.8 k4 1 k 3 k 16 note that...

2017 HKDSE Math M1 CTL(20170423)

2017-DSE-MATH-EP(M1)-1

Mathematics Module 1 (Calculus and Statistics) Solution Marks Remarks

1. (a) 112.03.018.016.02 kk

0.2k or -1.2k (rejected)

0.2k

(b) E(X)

0.1290.280.350.1840.1620.20 2

5.22

(c) )E( 2X

0.1290.280.350.1840.1620.202222222

33.54

)Var( X

22 )(E)(E XX

222.533.54

6.2916

)3Var(2 X

)9Var( X

6244.56



Solution Marks Remarks

2. (a) )|(P AB

)(P

)(P

A

BA

)(P

)(P)|(P

A

BBA

)(P

)(P)|(P

A

BBA

)(P

)(P1)|(P

A

BBA

2.0

3.016.0

9.0

(b) Since 0)|(P AB , A and B are NOT mutually inclusive.

(c) P(A)P(B)

0.7)0.2(1

06.0

)(P BA

So, A and B are NOT independent.




3. (a) Let X kg be the weight of chickens

X~N( ,2 )

0.89043.43)P(1.83

3.43)P(1.83)P(

X

XX

8904.013.43)P(1.83)P( XX

0.10961.83)2P( X

0.05481.83)P( X

0.0548)

1.83P(Z

4452.0)

1.83ZP(0

1.6

1.83

0.05483.43)P( X

4452.0)

3.43ZP(0

1.6

3.43

43.31.6

Solving, 63.2 and 5.0

(b) Let 9X kg be the mean weight of 9 chickens

9X ~N(2.63,

9

5.0 2

)

Requited Probability

)1.35.2(P 9 X

)

9

5.0

63.21.3

9

5.0

63.25.2(P

29

2

X

)82.278.0(P 9 X

4976.02823.0

7799.0




4. (a) Required Probability

6.04.0 3

0384.0

(b) 95.06.04.0...6.04.06.0 9 k

95.06.04.0

9

0

k

x

x

95.06.0

4.01

4.01 10

k

05.04.0 10 k

4.0ln

05.0ln10 k

730587608.60 k

The greatest value of k is 6

(c) The expected amount of money

6.0

115

25$

5. (a) ...

2

331

2

3 x

xe x

...

2

931

23

xxe x

...

2

9321

23

xxe x

231 xe

22

...2

932

xx

...27124 2 xx

(b)

4

0

444 )1(5)5(k

kkk

k xCx

The coefficient of x2

0044

0

1144

1

2244

2 )1(527)1(512)1(54 CCC

11475




6. (a) 33)6f(

33615666423

nm

2526 nm

nmxxx 212)(f 2

0)6(f

0626122

nm

43212 nm

Solving, 30m and 72n

(b) 726012)(f 2 xxx

0)(f x 6x or 1x

x )1,( 1 )6,1( 6 ),6(

)(f x + 0 - 0 +

The minimum value of )f( x

= )6f(

= 33

The maximum value of )f( x

= )1f(

= 653




7. (a)

x

y

d

d

2

)2(2

12 2

1

x

xxx

2

3

2

2

12

x

xx

2

3

22

4

x

x

(b) Let (a, b) be the point of contact (a > 2)

),(9

02

badx

dy

a

ba

ab

2

3

)2(2

4

9

02

a

a

a

ba

ab

2

3

)2(2

94

2

a

aa

a

a

)9)(4()2(2 aaaa

03692 aa

3a or 12a (rejected)

The slope of tangent

3

xdx

dy

2

1




8. (a) Let

x

ey ln

x

ex

e

x

y 2

d

d

x

xy d

1d

xx d)g(

x

x

edln

x

1

yyd

2

2y + constant

2

ln2

1

x

e+ constant

(b) (i) x-intercept of Γ = 0

(ii) The required area

2

d)(gd)(g1

e

e

e

xxxx

2

2

1

2

ln2

1ln

2

1e

e

e

x

e

x

e

1




9. (a) (i) Estimate value of

3

2

0.35.25

2

5.20.28

2

0.25.113

2

5.10.111

2

0.15.0

40

1

45.1

90% confidence interval for

40

4.0645.11.45,

40

4.0645.11.45

554038935.1,345961065.1

5540.1,3460.1

(ii) Suppose there are n samples

3.0

4.017.22

n

48551111.33n

The least sample size required is 34.

(b) (i) Let T hours be the daily time spent

T~N(1.48, 0.42)

The required probability

2)P( T

)

4.0

48.12P(Z

)3.1P(Z

4032.05.0

0968.0

(ii) The required probability

0968.00968.010968.011

0968.00968.00968.011415

1

15

89

1

C

C

086102962.0

0861.0 (corr to 4 d.p.)




10. (a) The required probability

4

0

2

!

2

k

k

k

e

947346982.0

9473.0 (corr to 4 d.p.)

(b) P( 5000 x )

1.02.045.01

25.0

The required probability

2.045.02.025.01.025.0

!3

2 23

1

23

1

23

1

32

CCCe

030721109.0

0307.0 (corr to 4 d.p.)

(c) The required probability

44

0

2224

2

34

1

45.0

2.025.045.0!1!2!1

!42.025.01.025.0

C

CC

18375625.0

1838.0 (corr to 4 d.p.)

(d) The required probability

4

0

2

4222

1

2212

!

2

18375625.0!4

2030721109.02.01.025.0

!2

21.0

!1

2

k

k

k

e

eC

ee

104214881.0

1042.0 (corr to 4 d.p.)




11. (a) For Ada,

I

=

1

5.0d)g( xx

)1f()9.0f(2)8.0f(2)7.0f(2)6.0(f2)5.0(f

5

5.01

2

1

10.74755967

0.7476 (corr to 4 d.p.)

For Billy,

I

=

1

5.0d)g( xx

1

5.0

2

d005.00.11

xx

xx

1

5.0d0.0050.1

1xx

x

1

5.0

2

2

005.01.0ln

xxx

0.74502218

0.7450 (corr to 4 d.p.)




(b) For Ada,

)(f x

2

1.01.01.0

x

eex xx

2

1.0 11.0

xxe x

)(f x

2

1.0

32

1.0 11.01.0

21.0

xxe

xxe xx

3

21.0 22.001.0

x

xxe x

3

2

1.0 11001.0

x

xe x

For 15.0 x , 01.0 xe , 03 x and 011001.02

x

0)(f x , for 15.0 x

)f( x is concave upward for 15.0 x

It is an over-estimate.

For Billy,

0

1.0

!

1.0

k

k

x

k

xe

3

21.0

!

1.0005.01.01

k

kk

x xk

xxe

2

21.0

!

1.0005.01.01

k

kkx

xkx

xx

x

e

1

5.02

1

5.0

21.0

d!

1.0d

005.01.01xx

kx

x

xx

x

e

k

kkx

For 15.0 x and k = 2, 3, …, 01.0 k, 0kx , 0!k

0d

!

1.01

5.02

xxkk

kk

1

5.0

21

5.0

1.0

d005.01.01

d xx

xxx

x

e x

It is an under-estimate.

(c) 747559671.074502218.0 I

001559672.0746.000097782.0 I

0.002001559672.0746.00 I

Yes, I agree with the claim




12. (a) (i) )1)(4( xx

1

2

344

2

34

k

k

k

ktt

k

kk

k

kt

t

t

2

3323

2

3

22

29

k

kt

t

(ii) For 0t , 0k , 02 t , 022 kt

0)1)(4( xx , for 0t

1x or 4x , for 0t

Yes, I agree the claim

(b) (i)

t

x

d

d

22

2ln3

kk

t

22

9

3

2ln

k

kt

14

3

2ln xx

So, we have

3

2ln

24

2ln




(ii) (1) When t = 0, x = 0.8

k

k

0

2

340.8

k3k14k10.8

16k

Note that 0

d

d

t

x, x is a strict decreasing function of t.

As x > 0 when t = 0, the crocodiles will be extinct

only if x = 0 when t > 0

Put x = 0, we have

162

4840

t

42

t

16t

So, the crocodiles will become extinct after 16 years

elapsed since the start of the research.

(ii) (2) When t = 0, x = 7

k

k

0

2

347

k3k14k17

2

1k

Put x = 0, we have

5.02

1.540

t

8

128

1

t

024t

So, the crocodiles will NOT become extinct.

The number of crocodiles after a long time

5.02

5.14lim

8

1tt

t

t

t8

1

8

1

22

23

4lim

02

034

4 thousands

2017 hkdse math m1 ctl(20170423) mathematics module 1 ... · 2 3 0.8 4 0.8 k4 1 k 3 k 16 note that...

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