2017 HKDSE Math M2 CTL(20170422)(20170427-V1) Mathematics Module … · 2018-10-26 · 2017 HKDSE...

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)

2017-DSE-MATH-EP(M2)-1

Mathematics Module 2 (Algebra and Calculus) Solution Marks Remarks

1.

6sec

d

d

h

h

h

6sec6seclim

0

6cos6cos

6cos6coslim

0 hh

h

h

6cos6cos2

666sin

2

666sin2

lim0 hh

hh

h

6cos6cos

3sin36sin2lim

0 hh

hh

h

h

h

h

h

h 3

3sin

6cos6cos

36sin6lim

0

)1(

6cos6cos

6sin6

6sec6tan6

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)


Solution Marks Remarks

2.

8

0

88)1(k

kk

k xaCax

9

0

999)(k

kk

k xbCxb

k

kk aC8 for k = 0, 1, 2, 3, … ,8

k

kk bC 99 for k = 0, 1, 2, 3, … ,9

4

7

7

2

4

7799

7

28

2 bC

aC

22 94 ba

0681

06899

8

18

1 bCaC

698 ba

9

86 ab

2

2

9

8694

aa

0369628 2 aa

3a or

7

3a

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



3. (a) OP

23

23

OAOB

ba

5

3

5

2

(b) (i) ba

ba

AOB cosba

4

12045

225

(ii)

2

OP

OPOP

baba

5

3

5

2

5

3

5

2

abbabbaa

5

3

5

2

5

3

5

2

5

3

5

222

baba

5

3

5

22

5

3

5

2 22

22

225

5

3

5

2220

5

345

5

2 2

2

2

2

576

24OP

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



4. (a) xex xd2

xex d2

22 dxeex xx

xxeex xx d22

xx exex

d22

xexeex xxx d222

constant 222 xxx exeex

(b) Required Area

6

0

2 dxex x

6

0

2 dxex x

602 22 xxx exeex

6

502

e

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



5. (a) (i)

h32

1183

121

Δ

h32

1183

121

210

820

121

h

)8)(1(22 h

42 h

(E) has unique solution

042 h

2h

Range of values of h:

2h or 2h

(ii)

Δ

32

4983

1121

kz

Δ

2210

1620

1121

k

42

16442

h

k

2

16

h

k

(b) (E) has infinitely solution 2-h

k

-

-

232

491183

11121

14000

8410

11121

~

k

(E) has infinitely solution 2-h and 14k

),48,57(),,( tttzyx for Rt

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



6. (a) Let r cm be the radius of the water inside the container

2015

hr

hr

4

3

2

2

4

3hhhA

2

16

15hA

(b)

t

hh

t

A

d

d2

16

15

d

d

When the volume of the container is 96 ,

hh

2

4

3

3

196

8h

8d

d

ht

A

38

16

15

45

The rate of change of the set curved surface = 45 cm2 s-1

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



7. (a) x3sin

xxxx sin2coscos2sin

xxxxx sinsin21coscossin2 2

xxxx sinsin21)sin1(sin2 22

xx 3sin4sin3

(b) (i)

x

x

sin

3sin

sincoscossin

3cos3

sin3

cos3sin

xx

xx

sincoscossin

3cos3

sin3

cos3sin

xx

xx

xx

xx

cos2

2sin

2

2

3cos2

23sin

2

2

xx

xx

sincos

3cos3sin

(ii) 2

sincos

3cos3sin

xx

xx

2

sin

sin4sin3 3

x

xx

2sin43 2

x

2

1sin

x (2

2

4sin0

x )

x (

440

x )

12

5x

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



8. (a) The equation of the tangent to Γ at P:

)(f

7 3

3e

ex

y

23

33ln

17e

eex

y

1

63

xe

y

(b) xx

xy dln

1 2

xx

xy dln

2

xxy lndln2

2ln xy + C , where C is a constant

Since Γ passes through point (e3, 7) , we have Ce 23ln7

C = 2

Thus, the equation of Γ is 2ln2 xy

(c) x

xx ln

2)(f

)(f x

2

ln1

2x

xx

x

2

ln12

x

x

0)(f x

0

ln12

2

x

x

ex

ex

ex

ex

x

,

,

0,

0

0

0

)(f

The point of inflexion of Γ = )1,( e

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



9. (a)

4

369)f(

xxx

The equation of vertical asymptote : 4x

The equation of oblique asymptote : 9 xy

(b) )(f' x

2

2

4

15524

x

xxxx

22

4

208

x

xx

24

102

x

xx

)(f' x

2)4(

361

x

22

4

208

x

xx

24

102

x

xx

(c) 0)(f' x 2x or 10x

x )10,( 10- )2,10( 2 ),2(

)(f' x + 0 0 +

The maximum point )25,10(

The minimum point )1,2(

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



(d) The volume of the solid

5

0

22

d4

5x

x

xx

5

0

22

d4

5x

x

xx

9

4

22

4d454

yy

yy (Let 4 xy )

9

4

22

d3613

yy

yy

9

4 2

234

d129693624126

yy

yyyy

9

4 2

234

d129693624126

yy

yyyy

9

4 2

2 d1296936

24126 yyy

yy

9

4

23 1296

ln936241133

yyyy

y

2

3ln1872

3

2285

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



10. (a) AC ji 66

AB kji 2

AE

1

r

ABrAC

kji rrr

r

662

1

1

AD

AC

2

1

ji 33

AE

11

10 AFAD

AF

ADAE 1011

jikji 3030662

1

11

rrr

r

AFAB

1

1

11

1

)6(1

11

2

301

)211(6r

rr

rr

r

rr

rr

4136418

11418

5

6r satisfies both equations

Thus,

5

6r

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



(b) (i) kji

11

6

11

24

11

42AE

DE

AEDA

kjiji

11

6

11

24

11

4233

kji

11

6

11

9

11

9

DEAD

kjiji

11

6

11

9

11

933

11

60

11

93

11

93

0

(ii) BCAB

kjikji 742

117142

0

90FBC

By (b)(i) 0DEAD , we have 90FDC

90FBCFDC

Yes, B, D, C, F are concyclic. (converse of s in the same

segment)

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



(c) Since 90FBC , Q is the mid-point of CF

BF

ABF A

ABADAE 1011

kji 5510

BQ

BFBC 2

1

kji 27

AQ

BQAB

ki 39

AP

kji 27

Volume of the tetrahedron

APAQAB 6

1

AP

309

1126

1kji

kjikji 27)933(

6

1

7

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



11. (a) Let tan21x

So, we have dsec2d 2x

1

0 2d

32

1x

xx

1

0 2d

2)1(

1x

x

2tan

2

2tan

2

2

-1

1-dsec2

2)1tan21(

1

2tan

2

2tan

2

2

-1

1-dsec2

tan12

1

2tan

2

2tan

2

2

-1

1-dsec2

sec2

1

2tan

2

2tan

-1

1-d

2

2

2tan

2

2tan

1-

1-2

2

2

2tan2tan

2

2 1-1-

4

2tan

2

2 1

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



(b) (i) Let tant

21

1cos

t

21sin

t

t

2sin

cossin2

22 1

1

12

tt

t

21

2

t

t

2tan1

tan2

2cos

1cos2 2

11

12

2

2

t

1

1

22

t

2

2

1

1

t

t

2

2

tan1

tan1

(b) (ii) Let tant

So, we have dsecd 2t

t

td

1

1d

2

0

d2cos2sin

1

1

0 2

2

2

2

d1

1

1

1

1

2

1t

t

t

t

t

t

1

0 2d

32

1t

tt

4

2tan

2

2 1

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



11. (c) Let

So, we have dd

0

d2cos2sin

1sin

0

d1

cossin

1sin

0d

22

cos22

sin

122

sin

0

d2cos2sin

12cos

0

d2cos2sin

12cos

(d)

00

d2cos2sin

12sind

2cos2sin

12cos

0

d2cos2sin

2cos2sin

0

d

By (c),

00

d2cos2sin

12cosd

2cos2sin

12sin

So, we have

0d

2cos2sin

12sin

0

d2cos2sin

92sin8

00

d2cos2sin

1d

2cos2sin

12sin8

4

2tan

2

2 1

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



12. (a) Let P(n):

00

1033 1nIA nnn

For 1n ,

00

10133 01I

00

10

30

03

30

13

A

P(1) is true

Assume P(k) is true for some positive integers k,

i.e.

00

1033 1kIA kkk

For 1 kn ,

1kA

AAk

30

13

00

1033 1kI kk

30

13

103

13

kk

30

133

kk

00

103

30

033

kkk

00

10133 11 kI kk

P( 1k ) is true

By M.I.,

00

1033 1nIA nnn is true for all positive integers

n.

2017 HKDSE Math M2 CTL(20170422)(20170427-V1)



12. (b) (i) 1-P

=

12

01

BPP-1

12

01

14

15

12

01

12

01

36

15

30

13

(ii) ABPP- 1

nn- APBP 1

00

1033 11 nIPBP nnn-

nB

111

00

1033 -n-n PnPPP

12

01

00

10

12

0133 1nI nn

12

01

20

1033 1nI nn

24

1233 1nI nn for any positive integer n

(iii) mm BA

24

1233

00

1033 11 mImI mmmm

20

023 1mm

20

023

21

mm

22234 mm

Since 03 22 m and 02 m for all positive integers m,

So, 0 mm BA for all positive integers m.

However, 04 2 m for all positive integers m.

No, there does not exist.

2017 HKDSE Math M2 CTL(20170422)(20170427-V1) Mathematics Module … · 2018-10-26 · 2017 HKDSE...

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