m2 mock exam (te) - munsang college

New Progress in Senior Mathematics Module 2 (Extended Part)

M2 Mock Exam 13 (2019) (Teacher’s Edition)

© Hong Kong Educational Publishing Company 1

M2 Mock Exam 13 (2019)

Section A (50 marks)

1. Let 2( ) ( 4) lnf x x x . Find f '(2) from first principles.

(4 marks) Answer:

4 ln 2

Solution:

Public Exam Reference: HKDSE 2018 Math M2 Section A Q1 f '(2)

=0

(2 ) (2)limh

f h f

h

1M

=

2

0

((2 ) 4)ln(2 ) 0limh

h h

h

=

2

0

( 4 )ln(2 )limh

h h h

h

1M

=0

lim( 4)ln(2 )h

h h

1M

= 4ln 2 1A

(4)

2. Expand 5(2 3 )x . Hence, find the constant term in the expansion of

2

5

2

1(2 3 ) 3x

x

.

(5 marks) Answer:

2 3 4 532 240 720 1080 810 243x x x x x ; 5418




Solution:

Public Exam Reference: HKDSE 2016 Math M2 Section A Q1

5(2 3 )x

=5 4 3 2 2 3 4 52 5(2 )(3 ) 10(2 )(3 ) 10(2 )(3 ) 5(2)(3 ) (3 )x x x x x 1M

=2 3 4 532 240 720 1080 810 243x x x x x 1A

2

2

13

x

=2 4

6 19

x x 1M

The constant term

= 32(9) 720(6) 810(1) 1M

= 5418 1A

(5)

3. (a) Prove the identity sin sin

tan2 cos cos

x y x y

x y

.

(b) Using (a), express tan37.5 in surd form. Rationalize the denominator of the answer if

necessary.

(5 marks) Answer:

(a) (The answer is skipped.)

(b) 6 3 2 2

Solution:


(a) R.H.S.

=sin sin

cos cos

x y

x y

2sin cos2 2

2cos cos2 2

x y x y

x y x y

1M

sin2

cos2

x y

x y

tan2

x y 1

L.H.S.




(b) tan 37.5

=75

tan2

=45 30

tan2

sin 45 sin30

cos45 cos30

(by (a)) 1M

2 1

2 2

2 3

2 2

2 1

2 3

=2 1 2 3

2 3 2 3

1M

= 6 3 2 2 1A

(5)

4. (a) Using integration by parts, find 2 (2 )uu du .

(b) Define 2 3( ) (2 )xf x x for all real numbers x. Find the area of the region bounded by the

graph of ( )y f x , the x-axis, the straight lines 1x and 2x .

(6 marks) Answer:

(a)

2 2

3

2 [ (ln 2) 2 ln 2 2]

(ln 2)

u u uC

(b)

2

3

2232(ln2) 720ln2 112

27(ln2)

Solution:


(a) 2(2 )uu du

= 21(2 )

ln 2

uu d

21(2 ) 2 (2 )

ln 2

u uu u du 1M

21 2(2 ) (2 )

ln 2 ln 2

u uu ud

21 2(2 ) (2 ) 2

ln 2 ln 2

u u uu u du

2 2

3

2 [ (ln 2) 2 ln 2 2]

(ln 2)

u u uC

, where C is a constant 1A




(b) The required area

=2

2 3

1(2 )xx dx 1M

=6

2

3

1(2 )

27

uu du (by letting u = 3x) 1M

6

2 2

3 3

12 ( (ln 2) 2 ln 2 2)

27(ln 2)

u u u (by (a)) 1M

2

3

2232(ln 2) 720ln 2 112

27(ln 2)

1A

(6)

5. (a) Using integration by substitution, find 5 3 9x x dx , where 3 9x .

(b) At any point (x, y) on the curve , the slope of the tangent to is 5 310 9x x . The

y-intercept of is 100. Find the equation of .

(7 marks) Answer:

(a) 3 5 3 32

( 9) 2( 9)15

x x C

(b) 3 5 3 34( 9) 20( 9) 116

3y x x

Solution:


(a) Let 3 9u x , then 23du x dx . 1M

5 3 9x x dx

=

1

21

( 9)3

u u du 1M

3 1

2 21

93

u du u du

3 5 3 32

( 9) 2( 9)15

x x C , where C is a constant 1A

(b) y

=5 310 9x x dx 1M

= 3 5 3 3210 ( 9) 2( 9)

15x x C

(by (a)) 1M

= 3 5 3 34( 9) 20( 9)

3x x C , where C is a constant

Since the y-intercept of is 100, we have 5 34(3) 20(3) 100

3C . 1M

Solving, we have 116C .

The equation of is 3 5 3 34( 9) 20( 9) 116

3y x x . 1A

(7)




6. (a) Using mathematical induction, prove that 2

1

( 1) (2 1)(2 1) 1( 1) (2 1)

2

nnk

k

n nk

for

all positive integers n.

(b) Using (a), evaluate 200

1 2

99

( 1) (2 1)k

k

k

.

(7 marks) Answer:

(a) (The answer is skipped.)

(b) 60 792

Solution:


(a) Note that

11 2 ( 1) (2(1) 1)(2(1) 1) 1

( 1) (2(1) 1) 12

.

The statement is true for n = 1. 1

Assume that 2

1

( 1) (2 1)(2 1) 1( 1) (2 1)

2

mmk

k

m mk

is true, where m is a

positive integer. 1M

12

1

( 1) (2 1)m

k

k

k

=2 1 2

1

( 1) (2 1) ( 1) (2( 1) 1)m

k m

k

k m

=1 2( 1) (2 1)(2 1) 1

( 1) (2 1)2

mmm m

m (by induction assumption) 1M

=( 1) (2 1)[2 1 2(2 1)] 1

2

m m m m

=

1( 1) (2 1)(2 3) 1

2

m m m

The statement is true for n = m + 1 if it is true for n = m.

By mathematical induction, 2

1

( 1) (2 1)(2 1) 1( 1) (2 1)

2

nnk

k

n nk

for all

positive integers n. 1

(b) Putting n = 98 and n = 200 in (a) respectively, we have

98982

1

( 1) (2(98) 1)(2(98) 1) 1( 1) (2 1) 19 208

2

k

k

k

and 1M

2002002

1

( 1) (2(200) 1)(2(200) 1) 1( 1) (2 1) 80 000

2

k

k

k

. 1M

2001 2

99

( 1) (2 1)k

k

k

=

200 981 2 1 2

1 1

( 1) (2 1) ( 1) (2 1)k k

k k

k k

=

200 982 2

1 1

( 1) (2 1) ( 1) (2 1)k k

k k

k k

= 80 000 19 208

= 60 792 1A

(7)




7. Let n be a positive integer.

(a) Define 1

0 1

cM

, where c is a real number. Evaluate

(i) 2M ,

(ii) nM ,

(iii) 1( )nM .

(b) Evaluate

(i) 1

0

1

3

n

kk

,

(ii)

1 2

10

3

n

.

(8 marks) Answer:

(a) (i) 1 2

0 1

c

(ii) 1

0 1

nc

(iii) 1 0

1nc

(b) (i) 3 1

12 3n

(ii) 1

11 3

3

10

3

n

n

Solution:


(a) (i) 2M

=1 1

0 1 0 1

c c

1 2

0 1

c

1A




(ii) Note that

3M

=1 2 1

0 1 0 1

c c

1M

1 3

0 1

c

So, we have

nM

1

0 1

nc

1A

(iii) 1

det( ) 10 1

nnc

M

1( )nM

=1( )nM

=1 01

1det( )n ncM

1M

1 0

1nc

1A

(b) (i) 1

0

1

3

n

kk

=

11

3

11

3

n

3 1

12 3n

1A

(ii) Note that

2

0 1

2

1 11 21 2 1 2 1 2

3 31 1 1

0 0 0 103 3 3

3

3

0 1 0 1 2

2 3

1 1 1 1 11 2 1 21 2 1 2

3 3 3 3 31 1

0 01 10 03 3

3 3

1M

So, we have

1

0

11 21 2

31

0 103

3

nn

kk

n

By (b)(i), we have 1

11 2 1 3

31

1003

3

n

n

n

1A

(8)




8. Define 2

( )8 12

kf x

x x

, where k is a constant. It is given that the extreme value of f (x) is

2.

(a) Find ( )f x .

(b) Find the asymptote(s) of the graph of ( )y f x .

(c) Someone claims that there is at least one point of inflexion of the graph of ( )y f x . Do

you agree? Explain your answer.

(8 marks) Answer:

(a) 2 2

16( 4)( )

( 8 12)

xf x

x x

(b) Vertical asymptotes: 2x and 6x ; Horizontal asymptote: 0y .

(c) No

Solution:


(a) Note that k 0.

( )f x

=2 2

(2 8)

( 8 12)

k x

x x

1M

( ) 0f x if and only if x = 4. 1M

Since the equation ( ) 0f x has only one solution x = 4 and the extreme value

of f (x) is 2, we have (4) 2f .

24 8(4) 12

k

= 2

k = 8

2 2

16( 4)( )

( 8 12)

xf x

x x

1A

(b) Note that 2 8 12 ( 2)( 6) 0x x x x when 2x or 6x . 1M

Also note that 2

8lim 0

8 12x x x

.

2x and 6x are two vertical asymptotes of ( )y f x , and 0y

is a horizontal asymptote of ( )y f x . 1A

(c) ( )f x

=2 2 2

2 4

( 8 12) (16) 16( 4)(2)( 8 12)(2 8)

( 8 12)

x x x x x x

x x

1M

=2

2 3

16(3 24 52)

( 8 12)

x x

x x

Since 2( 24) 4(3)(52) 48 0 , 23 24 52 0x x for all real numbers x. 1M

There is no solution for ( ) 0f x , hence there is no point of inflexion of the

graph of ( )y f x .

The claim is disagreed. 1A

(8)




Section B (50 marks)

9.

Figure 1

Peter wants to move a heavy box upstairs by using a wooden board as an inclined plane.

Initially, the wooden board touches the floor and the edge of the highest step of the stairs at the

same time as in Figure 1. Let PQ be the side-view of the wooden board, and E is the foot of

perpendicular from P to AB, so that EB = 128 cm and BC = 54 cm. Let PQC = .

(a) Find the length of PQ in terms of .

(1 mark)

(b) Find the shortest length of the wooden board.

(5 marks)

(c)

Figure 2

Suppose the length of the wooden board is 260 cm. To adjust the inclination of the

wooden board, Peter moves the board slowly so that the end of the board, Q, moves

towards D (see Figure 2). The board touches the floor and the edge of the highest step of

the stairs at the same time. Let x cm be the perpendicular distance from P to BC.

(i) When BQ = 162 cm, the rate of change of is 0.03 rad s1

. Find the rate of change

of x at this moment.

(ii) Peter claims that Q is moving towards D at a speed higher than the horizontal speed

of P moving towards BC. Do you agree? Explain your answer.

(6 marks)

A B

C D

E

P

Q

A B

C D

x cm P

Q




Answer:

(a) 128 54

cmcos sin

(b) 250 cm

(c) (i) 11.98 cm s1

(ii) Yes

Solution:

Public Exam Reference: HKDSE 2014 Math M2 Section B Q10

(a) PQ

= PB BQ

=128 54

cmcos sin

1A

(1)

(b) Let L cm be the length of the wooden board.

dL

d

=2 2

128sin 54cos

cos sin

1A

Note that 0dL

d when 1M

2

128sin

cos

=

2

54cos

sin

3tan =27

64

tan =3

4

= 1 3tan

4

1 3

0, tan4

1 3tan

4

1 3 πtan ,

4 2

dL

d 0 +

1M

When 1 3tan

4 , L is minimum. 1A

By (a), the shortest length of the wooden board

=5 5

128 54 cm4 3

= 250 cm 1A

(5)




(c) (i) x cm CQ = cosPQ

x =54

260costan

1M

dx

dt =

2

54260sin

sin

d d

dt dt

1M

When BQ = 162 cm,

dx

dt =

254 162

260 ( 0.03) 54 ( 0.03)162 54

= 11.98 1A

The rate of change of x is 11.98 cm s1.

(ii) Let CQ z m. Note that the speed of Q moving towards D is dz

dtcm s1

while the horizontal speed of P moving towards BC is dx

dt cm s1. 1M

z x = 260cos

dz dx

dt dt

= 260sind

dt

Since sin 0 and 0d

dt

, we have 0

dz dx

dt dt

.

dz dx

dt dt 1A

The claim is agreed. 1 (6)




10. (a) (i) Prove that 4 3 21

tan tan tan3

xdx x xdx .

(ii) Evaluate

5π

443π

4

tan xdx .

(5 marks)

(b) (i) Let f (x) be a continuous function for a x b , where a and b are constants, such

that ( ) ( )f x f a b x for all a x b . Prove that ( ) ( )2

b b

a a

a bxf x dx f x dx

.

(ii) Evaluate

5π

443π

4

tanx xdx .

(5 marks)

(c) Consider the curve :2tany x x , where

7π 9π

4 4x . Let R be the region bounded by

, the x-axis, the two lines 7π

4x and

9π

4x . Find the volume of the solid of

revolution generated by revolving R about the x-axis.

(3 marks) Answer:

(a) (i) (The answer is skipped.)

(ii) π 4

2 3

(b) (i) (The answer is skipped.)

(ii) 2π 4π

2 3

(c) 2

3 8ππ

3

Solution:


(a) (i) 4tan xdx

= 2 2tan (sec 1)x x dx

= 2 2 2tan sec tanx xdx xdx

= 2 2tan (tan ) tanxd x xdx 1M

= 3 21tan tan

3x xdx 1




(ii)

5π

443π

4

tan xdx

=

5π5π

43 24

3π3π

44

1tan tan

3x xdx

(by (a)(i)) 1M

=

5π

243π

4

2(sec 1)

3x dx

=5π

43π

4

2tan

3x x 1M

=π 4

2 3 1A

(5)

(b) (i) Let x a b u , then dx du .

( )b

axf x dx

= ( ) ( )a

ba b u f a b u du

= ( ) ( )b

aa b u f u du 1M

= ( ) ( ) ( )b b

a aa b f x dx xf x dx

2 ( ) ( ) ( )b b

a axf x dx a b f x dx

Hence, we have ( ) ( )2

b b

a a

a bxf x dx f x dx

. 1

(ii) Note that 4 4 43π 5πtan tan (2π ) tan

4 4x x x

for

3π 5π

4 4x . 1M

5π

443π

4

tanx xdx

=

5π

443π

4

π tan xdx (by (b)(i)) 1M

=π 4

π2 3

(by (a)(ii))

=2π 4π

2 3 1A

(5)




(c) The required volume

=

9π

2 247π

4

π( tan )x x dx 1M

=

9π

447π

4

π tanx xdx

=

5π

443π

4

π (π ) tan (π )y y dy (by letting πx y ) 1M

=

5π

4 443π

4

π (π tan tan )y y y dy

=

5π 5π

2 4 44 43π 3π

4 4

π tan π tanxdx x xdx

=2

2 π 4 π 4ππ π

2 3 2 3

(by (a)(ii) and (b)(ii))

=2

3 8ππ

3 1A

(3)

11. (a) Consider the system of linear equations in real variables x, y, z

2 (2 1) 2

( ) : 3 5 2

3 (3 ) 1

hx y h z

E x y z k

x y h z k

, where h and k are real numbers.

(i) Assume that (E) has a unique solution.

(1) Prove that 2h and 8h .

(2) Solve (E).

(ii) Assume that 2h and (E) is consistent.

(1) Find k.

(2) Solve (E).

(9 marks)

(b) Is there a real solution of the system of linear equations

3 6

2 2 3 2

3 5 10

x y z

x y z

x y z

satisfying 22 1x y z ? Explain your answer.

(3 marks)




Answer:

(a) (i) (1) (The answer is skipped.)

(2) 18 29 35

,( 2)( 8)

hk kx

h h

22 5 5 5,

( 2)( 8)

h k hk h ky

h h

7 10 22

( 2)( 8)

hk h kz

h h

(ii) (1) 5

(2) {(7 20, , 4 14) : }t t t t R

(b) No

Solution:


(a) (i) (1) (E) has a unique solution if and only if

2 2 1

3 1 5 0

1 3 3

h h

h

. 1M

2 2 1

3 1 5

1 3 3

h h

h

( 1)(3 ) ( 2)(5) (2 1)(3)( 3) (5)( 3) ( 2)(3)(3 ) (2 1)( 1)h h h h h h 1A

2 10 16h h

( 2)( 8)h h

2h and 8h 1

(2) Since (E) has a unique solution, by Cramer’s rule, we have

x

=

2 2 2 1

2 1 5

1 3 3

( 2)( 8)

h

k

k h

h h

1M

18 29 35

( 2)( 8)

hk k

h h

1A

y

=

2 2 1

3 2 5

1 1 3

( 2)( 8)

h h

k

k h

h h

22 5 5 5

( 2)( 8)

h k hk h k

h h

z

=

2 2

3 1 2

1 3 1

( 2)( 8)

h

k

k

h h

7 10 22

( 2)( 8)

hk h k

h h

1A




(ii) (1) When 2h , the augmented matrix of (E) is

2 2 3 2

3 1 5 2

1 3 1 1

k

k

~

1 3 1 1

3 1 5 2

2 2 3 2

k

k

(R1 R3)

~

1 3 1 1

0 8 2 5 3

0 4 1 2 4

k

k

k

(R2 3R1 R2; R3 2R1 R3)

~

1 3 1 1

0 8 2 5 3

0 8 2 4 8

k

k

k

(2R3 R3)

~

1 3 1 1

0 8 2 5 3

0 0 0 5

k

k

k

(R3 R2 R3) 1M

Since (E) is consistent, we have 5k . 1A

(2) When 2h and 5k , the augmented matrix of (E) is

1 3 1 6

0 8 2 28

0 0 0 0

.

The solution set of (E) is {(7 20, , 4 14) : }t t t t R . 1A

(9)

(b) Putting 2h and 5k into (E), we have

3 6

2 2 3 2

3 5 10

x y z

x y z

x y z

By (a)(ii)(2), the solution set is {(7 20, , 4 14) : }t t t t R . 1M

22x y z

=22(7 20) ( 4 14)t t t

= 2 10 26t t

=2( 10 25) 25 26t t

=2( 5) 1t 1M

1

There is no real solution of the system of linear equations satisfying 22 1x y z . 1A

(3)




12. OAB is a triangle. It is given that 4 3OA i j k and 2 3 7OB i j k . P is a point on OA

such that OA PB.

(a) (i) Find OA OB .

(ii) Find :OP OA .

(4 marks)

(b) Q is a point on BP such that PQ : QB = 1 : t, where t is positive.

(i) Express OQ in terms of t.

(ii) Suppose that Q is the orthocentre of OAB.

(1) Find the value of t.

(2) Find the area of OPQ : the area of AQB.

(9 marks) Answer:

(a) (i) 6

(ii) 3 :13

(b) (i) (12 26) (9 39) (3 91)

13(1 )

t t t

t

i j k

(ii) (1) 182

15

(2) 9 : 364

Solution:

(a) (i) OA OB

= (4)(2) (3)( 3) (1)(7)

= 6 1A

(ii) OP

OA

=cosOB AOB

OA

=2

OA OB

OA

1M

=2 2 2

6

4 3 1 (by (a)(i)) 1M

=3

13 1A

: 3 :13OP OA

(4)




(b) (i) OQ

=1

1 1

tOP OB

t t

1M

=3 1

1 13 1

tOA OB

t t

(by (a)(ii)) 1M

=3 1

(4 3 ) (2 3 7 )1 13 1

t

t t

i j k i j k

=(12 26) (9 39) (3 91)

13(1 )

t t t

t

i j k 1A

(ii) (1) Since Q is the orthocentre of OAB, we have 0OQ AB . 1M

AB

OB OA

2 6 6 i j k 1A

OQ AB

(12 26)( 2) (9 39)( 6) (3 91)(6)

13(1 )

t t t

t

1M

60 728

13(1 )

t

t

60 728

13(1 )

t

t

= 0

t =182

15 1A

(2) Area of

Area of

OPQ

AQB

= Area of Area of

Area of Area of

OPQ APQ

APQ AQB

OP PQ

AP QB

3 15

13 3 182

(by (a)(ii) and (b)(ii)(1)) 1M

9

364 1A

The area of OPQ : the area of AQB = 9 : 364

(9)

END OF PAPER




Answers

1. 4 ln 2

2. 2 3 4 532 240 720 1080 810 243x x x x x ; 5418

3. (b) 6 3 2 2

4. (a) 2 2

3

2 [ (ln 2) 2 ln 2 2]

(ln 2)

u u uC

(b) 2

3

2232(ln2) 720ln2 112

27(ln2)

5. (a) 3 5 3 32

( 9) 2( 9)15

x x C

(b) 3 5 3 34

( 9) 20( 9) 1163

y x x

6. (b) 60 792

7. (a) (i) 1 2

0 1

c

(ii) 1

0 1

nc

(iii) 1 0

1nc

(b) (i) 3 1

12 3n

(ii) 1

11 3

3

10

3

n

n

8. (a) 2 2

16( 4)( )

( 8 12)

xf x

x x

(b) Vertical asymptotes: 2x and 6x

Horizontal asymptote: 0y

(c) No

9. (a) 128 54

cmcos sin

(b) 250 cm

(c) (i) 11.98 cm s1

(ii) Yes

10. (a) (ii) π 4

2 3

(b) (ii) 2π 4π

2 3

(c) 2

3 8ππ

3

11. (a) (i) (2) 18 29 35

,( 2)( 8)

hk kx

h h

22 5 5 5,

( 2)( 8)

h k hk h ky

h h

7 10 22

( 2)( 8)

hk h kz

h h

(ii) (1) 5

(2) {(7 20, , 4 14) : }t t t t R

(b) No

12. (a) (i) 6

(ii) 3 :13

(b) (i) (12 26) (9 39) (3 91)

13(1 )

t t t

t

i j k

(ii) (1) 182

15

(2) 9 : 364

m2 mock exam (te) - munsang college

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