HKDSE Practice Paper 2012 Mathematics_Module 2

8/3/2019 HKDSE Practice Paper 2012 Mathematics_Module 2

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HKD SE Practice Paper 2012 Mathematics (Module 2)

1

Section A

1. Find the coefficient of 5x in the expansion of9(2 ) .

(4 marks)

2. Consider the following system of linear equation inx,y,z

7 7 0

3 0

2 0

x y z

x ky z

x y kz

, where kis a real number.

If the system has non-trivial solutions, find the two possible values ofk.

(4 marks)

3. Prove by mathematical induction that 4 15 1n

n is divisible by 9 for all positive integers n.

(5 marks)

4. (a) Let tanx , show that 22

sin21

x

x .

(b) Using (a), find the greatest value of

2

2

(1 )

1

, wherex is real.

(5 marks)

5. (a) It is given that cos( 1) cos( 1) cos x k x for any realx. Find the value ofk.

(b) Without using a calculator, find the value of

cos1 cos2 cos3

cos4 cos5 cos6

cos7 cos8 cos9

.

(6 marks)

6. Find1d

dx x

from the first principle.

(4 marks)


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2

7. Let ( ) (sin cos )x f x e x x

(a) Find '( )f x and ''( )f x .

(b) Find the value ofx such that ''( ) '( ) ( ) 0f x f x f x .

(4 marks)

8. (a) Using integration by substitution, find24

dx

.

(b) Using integration by parts, find ln dx .

(5 marks)

9. Find the equations of the two tangents to the curve 2 22 1 0 x xy y which are parallel to

the straight line 2 1y x .

(5 marks)

10. (a) Find2

x xe dx .

(b)

In Figure 1, the shaded region is bounded by the curves2

2

xy and

2xe , where

1 2x . Find the volume of the solid generated by revolving the shaded region about

they-axis.

(5 marks)


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3

Section B

11. Let1 0

A

where and are distinct numbers. LetIbe the 2 2

identity matrix.

(a) Show that 2 ( ) A A I .

(2 marks)

(b) Using (a), or otherwise, show that

2( ) ( )( ) A I A I and 2( ) ( )( ) A I A I .

(3 marks)

(c) Let ( )X s A I and ( )Y t A I wheres and tare real numbers.

Suppose A X Y .

(i) Finds and tin terms of and .

(ii) For any positive integern, prove that

( )n

n X A I

and ( )n

nY A I

.

(iii) For any positive integer n, expressn

A in the form of pA qI , wherep and q are

real numbers.

[Note: It is known that for any 2 2 matricesHandK, if0 0

0 0 HK KH

,

then ( )n n n H K H K for any positive integern.]

(9 marks)


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4

12.

Let OA i

, OB

and OC i j k

(see Figure 2).

Let MandNbe the points on the straight linesAB and OCrespectively such that

: : (1 ) AM MB a a and : : (1 )ON NC b a , where 0 1a and 0 1b .

Suppose that MNis perpendicular to bothAB and OC.

(a) (i) Show that ( 1) ( )MN a b b a b i j k

.

(ii) Find the values ofa and b.

(iii) Find the shortest distance between the straight linesAB and OC.

(8 marks)

(b) (i) Find AB AC

.

(ii) Let G be the projection ofO on the planeABC, find the coordinates of the

intersecting point of the two straight lines OG and MN.

(5 marks)


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5

13. (a) Let ( )f x be an odd function for p x p , wherep is a positive constant.

Prove that2

0( ) 0

p

f x p dx .

Hence evaluate 20

( )p f x p q dx , where q is a constant.

(4 marks)

(b) Prove that

3 tan1 3 tan6

23 tan

6

xx

x

.

(2 marks)

(c) Using (a) and (b), or otherwise, evaluate 30

ln 1 3 tan x dx

.

(4 marks)

14. (a)

In Figure 3, the shaded region enclosed by the circle 2 2 25x y , thex-axis and the

straight line y = h (where 0 5h ) is revolved about the y-axis. Show that the volume

of the solid of revolution is3

253

hh

.

(2 marks)


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6

14. (cont)

(b) In Figure 4, an empty coffee cup consists of two portions. The lower portion is in the

shape of the solid described in (a) with height 4 cm. The upper portion is a frustum of

a circular cone. The height of the frustum is 8 cm. The radius of the top if the cup is 6

cm. Hot coffee is poured into the cup to a depth h cm at a rate of 8 cm3s-1, where

0 12h . Let Vcm3 be the volume of coffee in the cup.

(i) Find the rate of increase of the depth of coffee when the depth is 3 cm.

(ii) Show that 4164 3

( 4)3 64

V h

for 4 12h .

(iii) After the cup is fully filled, suddenly it cracks at the bottom. The coffee leaks at

a rate of 2 cm3s

-1. Find the rate of decrease of the depth of coffee after 15

seconds of leaking, giving your answer correct to 3 significant figures.

(11 marks)

End of Paper


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7

Solution

1. The term with 5 is 9 4 55(2) ( )C x

Coefficient of

5

x is

9 4

5 (2) 2016C

2.

1 7 7 1 7 77 4

1 3 0 7 4 05 14

2 1 0 15 14

kk k

kk k

( 7)( 14) 60 0k k

2

21 38 0k k

2k or 19k

3. 14 15(1) 1 18 2 9 , which is divisible by 9.

Assume for some positive integers k, 4 15( ) 1 9k k M where Mis an integer.

When 1n k ,

14 15( 1) 1 4(9 1 15 ) 15 15 1 9(4 5 2)k k M k k M k

, which is divisible by 9.

Proved by induction, 4 15 1n n is divisible by 9 for all positive integers n.

4. (a)22 2 2

2

sin2

2 tan 2sin coscosL.H.S. sin 2 R.H.S.sin1 tan sin cos

1cos

(b)2 2

2 2 2

(1 ) 1 2 211 1 1

x x x

x x

Using (a), consider2

2

(1 )1 sin 2

1

x

x

where tanx

Sincex is real, is also real, so 1 sin 2 1 for all real

2

2

(1 )1 (1) 2

1

x

. Hence the greatest value of

2

2

(1 )

1

x

x

is 2.


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8

5. (a)( 1) ( 1) ( 1) ( 1)

cos( 1) cos( 1) 2cos cos 2 cos1cos2 2

x x x x x x x

Since 2 cos1cos cos x k x , 2cos1k

(b)

cos1 cos 2 cos3 cos1 cos3 cos2 cos3

cos 4 cos5 cos6 cos 4 cos6 cos5 cos6

cos7 cos8 cos9 cos 7 cos9 cos8 cos9

1 3 1C C C

2 cos1cos 2 cos 2 cos 3

2 cos1cos 5 cos 5 cos 6

2 cos1cos 8 cos8 cos 9

(Using (a))

= 0

6.20 0 0

1 1

1 1 ( ) 1 1 1lim lim lim

( ) ( ) ( 0)h h h

d x x hx h x

dx x h h x x h x x h x x x

7. (a) '( ) (cos sin ) (sin cos ) 2 cos x x x f x e x x x x e e x

''( ) 2 sin 2 cosx x f x e x e x

(b) 2 sin 2 cosx xe x e x 2 cosxe x sin cos 0x xe x e x

xe cos xx e sin x [ 0xe for all realx]

tan 1x

4x

8. (a) Let 2sinx , 2cosdx d

1

2 2

2cos 2 cossin

2cos 24 4 4sin

dx d d xd C C

x

(b)1

ln ln ln ln ln lndx x x xd x x x x d x x x dx x x x C


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9

9. 2 4 0dy dy

x x y ydx dx

2

4

dy x y

dx x y

Let (a, b) be the contact points,

( , )

22

4a b

dy a b

dx a b

0b

To find a, 2 2(0) 2(0) 1 0a a 1a

The contact points are (1,0) and ( 1, 0) .

The equation of tangent at (1,0) is0

21

y

x

2 2 0x y

The equation of tangent at ( 1, 0) is0

2

1

y

x

2 2 0x y

10. (a) 2 2 221 1

2 2

x x x xe dx e d x e C

(b)2 2

2 32 2 2

1 1 1Volume 2 2 2

2 2

x xx xe dx dx xe dx

c

2

22

4

11

4 1

1

4

1 1(16) (1)

4 4

x

x e

e e

4 115

4e e

sq. units

11. (a)2

1 0 1 0A

2

20

00

1 0

1 0 0 1

( )A I


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10

11. (b) 2( ) ( )( )A I A I A I

2

2 2

2

2

2

( ) 2

( ) ( )

( )( )

AA AI IA II

A A I

A I A I

A A I I

A I

A I

Similarly,

2 2 2

( ) 2 A I A A I 2( ) 2

( ) ( )

( )( )

A I A I

A I

A I

(c) (i) ( ) ( )A X Y s A I t A I

1 0 1 1

s t

1 0

s t s t

s t s t

...........(1)

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

........................(3)

s t

s t

s t

From (1) and (3),t

t

2 2

( )t

From (2), 1 1s t


11/16


11

11. (c) (ii) ( ) ( )X s A I A I

and ( ) ( )Y t A I A I

Assume for some positive integers n,

( )k

k X A I

and ( )k

kY A I

.

When 1n k ,

11 2

2( ) ( ) ( )

( )

k kk k X X X A I A I A I

1

2( )( )

( )

k

A I

[Using (b)]

1

( )k

A I

11 2

2( ) ( ) ( )

( )

k kk kY Y Y A I A I A I

1

2( )( )

( )

k

A I

[Using (b)]

1

( )k

A I

Proved by induction, ( )n

n X A I

and ( )n

nY A I

for all

positive integern.

(iii) Check 2( )( ) ( ) XY st A I A I st A I

( ) ( )

0 0

0 0

st A I I

Similarly, 0 00 0

YX

( ) ( ) ( )n n

n n n nA X Y X Y A I A I

n n n n

A I

n n n n

A I


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12

12. (a) (i) MN ON OM s

( ) ( )( 1) ( )

bOC OA a AB

b a

a b b a b

i j k i j i

i j k

(ii) Since MN AB

, 0MN AB

( 1) ( ) ( ) 0a b b a b i j k j i

( 1) ( ) 0a b b a

1

2

a

Since MN OC

, 0MN OC

( 1) ( ) ( ) 0a b b a b i j k i j k

( 1) ( ) 0a b b a b

1

3b

(iii) Since MNis perpendicular to bothAB and OC,

the required shortest distance is MN

.

2 2 21 1 1 1 1 6

12 3 3 2 3 6

MN

units

(b) (i) AB j i

, AC OC OA i j k i j k

1 1 0

0 1 1

AB AC

i j k

i j k


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13

12. (b) (ii) ( ) i j k is a vector perpendicular to the planeABC,

Note that OG

is also perpendicular to the planeABC, therefore //( )OG i j k

LetR be the point of intersection ofOG and MN, and we let ( )OR i j k

,

where is a non-zero real constant.

RN ON OR

1( ) ( )

3

1 1 1

3 3 3

i j k i j k

i j k

1 1 1 1 11

2 3 3 2 3MN

i j k

1 1 1

6 6 3 i j k

Consider // RN MN

(sinceR is a point on MN)

1 1

3 3

1 1

6 3

1

HenceR = (1,1, 1)

13. (a) Let u x p , du dx . When 2x p , u p ;when 0x , u p

2 0

0 0( ) ( ) ( ) ( )

p p p

p p f x p dx f u du f u du f u du

For0

( )p

f u du , let v u , dv du , When 0u , 0v ;when u p , v p

0 0 0

( ) ( )( ) ( ) p p p

f u du f v dv f v dv

(since ( )f x is odd for p x p )

Hence2 0

0 0( ) ( ) ( )

p p

p f x p dx f v dv f u du

0 0( ) ( ) 0

p p

f x dx f x dx

2 2

0 0( ) 0 2

p p

f x p q dx q dx pq


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14

13. (b)

tan tan63

3 tan 1 tan tan6 6

tan tan3 tan66 3

1 tan tan6

x

x x

xx

x

1 13 1 tan tan

3 3

1 13 1 tan tan

3 3

x x

x x

13 2 tan

31

33

x

3 1 2 3 tan

3 1

1 3 tan

2

x

(c) 3 tan

1 3 tan 6ln 1 3 tan ln ln 2 ln ln 2

23 tan

6

xx

x

x

Let3 tan

( ) ln3 tan

xf x

, where

6 6x

Since

3 tan 3 tan( ) ln ln

3 tan 3 tan

x xf x

x

1

3 tan 3 tanln ln ( )

3 tan 3 tan

x xf x

x x

,

( )f x is odd function for6 6

x

Hence 3 30 0

3 tan6

ln 1 3 tan ln ln 2

3 tan6

x x dx dx

x

3

0ln 2

6 f x dx

2 ln 2 ln 26 3


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15

14. (a)3 3

2 2

0 00

Volume (25 ) 25 253 3

hh h y h x dy y dy y h

(b) (i)3

25

3

hV h

for 0 4h

225dV dh

hdt dt

23

8 25 3h

dh

dt

3

1

2h

dh

dt

The required rate of increase of the depth of coffee is1

2cm s-1.

(ii)

From the figure, the required volume Vis given by

32 24 1 125(4) ( ) ( 4) (3) (8)

3 3 3V r h

where

( 4)

3 8

r h

2164 1 3( 4)

( 4)3 3 8

hh

3164 3 ( 4)3 64

h

where 0 12h


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15. (b) (iii) The total volume of the cup is 3164 3 740

(12 4)3 64 3

cm3

After 15 seconds of leaking, the depth of coffee is given by

3740 164 330 ( 4)3 3 64

h

33192 30 ( 4)64

h

11.73021845h cm

29 ( 4)64

dV dhh

dt dt

2

15

92 (11.73021845 4)64 t

dh

dt

15

0.018295659t

dh

dt cm s-1

The required rate of decrease of the coffee is 0.0183 cm s-1

End of Solution

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HKDSE Practice Paper 2012 Mathematics_Module 2