Mock Exam 2 - hkep. · PDF fileMock Exam 2 Section A 1. ... real number λ, det ......

Mock Exam 2

– 1 – © Hong Kong Educational Publishing Company

Mock Exam 2

Section A

1. Reference: HKDSE Math M2 2013 Q2

(1 + ax)n

= + + + +

= + +−

+− −

+

C ax C ax C ax

nax n n a x n n n a x

1 ( ) ( ) ( )

1 ( 1)2

( 1)( 2)6

n n n1 2

23

3

22

33

1M

\ = −

−+

− −= −

nan n a n n n a

16( 1)

2( 1)( 2)

6336

2 3

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

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

From (1): = −an

16 ................................................................... (3)

Substituting (3) into (2),

−−

− −= −

− − − − = −

− + − =− − − =

n nn

n n nn

n n n n n n

n n nn n n

128 ( 1) 2048 ( 1)( 2)3

336

384 ( 1) 2048 ( 1)( 2) 1008656 5760 4096 0

16 ( 8)(41 32) 0

2 3

2 3

3 2

Since n is a positive integer, n = 8. 1A Substituting n = 8 into (3),

= −

= −

a168

2

1A (4)


−

+

=+ −

+

=+ −

+

×+ + + +

+ + + +

=+ −

+ + + + +

=

+ + + + +

x x h

x h x

x x h

x h x

x x h

x h x x h x

x h x x h xx h x

x x h x h x x h xh

x x h x x h x x h

1 1

( )

( )

( )

( )

( )

( ) ( )

( ) ( )( )

( ) [( ) ( ) ]

( ) ( ) ( )

13

13

13

13

13

13

13

13

13

13

23

13

13

23

23

13

13

23

13

13

23

13

13

23

13

23

23

13

1M

1

(a - b)(a2 + ab + b2)= a3 - b3

Mathematics: Mock Exam Papers Module 2 (Extended Part) Second Edition Solution Guide


=

+

−

=−

+ + + + +

=−

+ + + + +

=−

+ +

= −

→

→

→

ddx

xh

x h xh

h x x h x x h x x h

x x h x x h x x h

x x x

x

6 lim 1 6

( )

6

lim 6

[ ( ) ( ) ( ) ]

lim 6

( ) ( ) ( )6

2

h

h

h

13

0 13

13

0 13

23

23

13

0 13

23

23

13

43

43

43

43

(by (a))

1M

1M

1A

(5)

3. (a) By long division,

)+

+

−

− −

+−

− +x x

x x

x

x x

x

x xx

x x x12 7

7 5

5

5 3

3

3

5 3

\

+=

− + + −+

= − + −+

xx

x x x x xx

x x xx

x

1( )( 1)

1

1

7

2

5 3 2

2

5 32

1M

1A

(b) ∫

∫

∫

∫

∫

∫

∫ ∫

∫ ∫

+

= +

=+

− +

=+

− ×+

=+

−+

=+

− − + −+

=+

− − + ++

=+

− − + +++

=+

++

− + − +

x x dx

x d x

x x x d x

x x x xx

dx

x x xx

dx

x x x x x xx

dx

x x x x x dx xx

dx

x x x x x dx d xx

x x x x x x C

ln( 1)

ln( 1)6

ln( 1)6

16

[ln( 1)]

ln( 1)6

16

21

ln( 1)6

13 1

ln( 1)6

13 1

ln( 1)6

13

( ) 13 1

ln( 1)6

13

( ) 16

( 1)1

ln( 1)6

ln( 1)6 18 12 6

5 2

26

6 26 2

6 26

2

6 2 7

2

6 25 3

2

6 25 3

2

6 25 3

2

2

6 2 2 6 4 2

1M

1M

1M

, where C is a constant 1A

(6)

(by (a))

Dividend= Quotient × Divisor

+ Remainder

Mock Exam 2


4. (a) Let u = π - x. Then du = -dx. 1M

When x = 0, u = π; when x = π, u = 0.

∫ ∫∫∫

= − π −

= π −

= π −

π

ππ

π

f x dx f u du

f u du

f x dx

( ) ( )

( )

( )

00

0

0

1

(b) Let = ++

f x xx

( ) 1 cos1 sin

3. By (a), we have

∫ ∫

∫ ∫

∫ ∫ ∫

∫ ∫

∫

++

= +

π −+ π −

++

= −

+

++

= +

+

+ −

+

++

=

++

= π

π π

π π

π π π

π π

π

xx

dx xx

dx

xx

dx xx

dx

xx

dx xx

dx xx

dx

xx

dx dx

xx

dx

1 cos1 sin

1 cos ( )1 sin( )

1 cos1 sin

1 cos1 sin

2 1 cos1 sin

1 cos1 sin

1 cos1 sin

2 1 cos1 sin

2

1 cos1 sin

3

0

3

0

3

0

3

0

3

0

3

0

3

0

3

0 0

3

0

1M

1M

1M

1A

(6)

5. Reference: HKALE P. Math 1997 Paper 1 Q7

(a) det A × det(A-1 - xI) = det[A(A-1 - xI)]

= det(I - xA) 1M

= det[-x(A - x-1I)]

= (-x)3det(A - x-1I)

= -x3det(A - x-1I) 1

(b) (i) − =− −

−− − −

= − + + − − −

= − + + −

A xIx

xx

x x x x xx x x

det( ) det4 11 12

0 42 7 2

(4 )( 2) 88 24 28(4 )2 12 24 ......(*)3 2

When x = 2, det(A - xI) = -23 + 2(2)2 + 12(2) - 24 = 0. 1M

\ x - 2 is a factor of (*).

\ 2 is a root of det(A - xI) = 0. 1

det(A - xI) = (x - 2)(-x2 + 12)

\ Other roots = ±

= ±

122 3 1A

(ii) det A = 88 - 112 = -24 ≠ 0

By (a), det(A-1 - xI) = 0 if and only if det(A - x-1I) = 0. 1M

By (b)(i), x-1 = 2 or ± 2 3

\ =x12

or ±3

6 1A

(7)

For n × n matrix P and real number λ , det(λP) = λ ndet P.




(a) cos 3θ = sin 2θ

cos 3θ = cos(90° - 2θ )

3θ = 90° – 2θ

5θ = 90°

θ = 18° 1A

(b) cos 3θ = sin 2θ

cos(2θ + θ ) = sin 2θ

cos 2θ cos θ - sin 2θ sin θ = 2 sin θ cos θ 1M

(2 cos2 θ - 1)cos θ - 2 sin2 θ cos θ = 2 sin θ cos θ

(2 cos2 θ - 1) - 2 sin2 θ = 2 sin θ (cos θ ≠ 0 for 0° < θ < 45°)

2 cos2 θ - 1 - 2(1 - cos2 θ ) = 2 sin θ

4 cos2 θ - 3 = 2 sin θ

16 cos4 θ – 24 cos2 θ + 9 = 4 sin2 θ 1M

16 cos4 θ – 24 cos2 θ + 9 = 4 - 4 cos2 θ

16 cos4 θ - 20 cos2 θ + 5 = 0 1

(c) By (a), θ = 18°. By (b), we have

16 cos4 18° - 20 cos2 18° + 5 = 0

cos2 18° = − − ± − −( 20) ( 20) 4(16)(5)2(16)

2

cos2 18° = +5 58

or -5 58

(rejected) 1M

° = °= °

= ° −

=+

−

=+

sin 54 cos 36cos 2(18 )2 cos 18 1

2 5 58

1

1 54

2

1M

1A

(7)


(a) Note that x = -2 is the vertical asymptote. 1A

=+ ++

=+ + +

+

=+ + + −

+

= + −+

f x x xx

x x xx

x x xx

xx

( ) 5 12

( 2) 3 12

( 2) 3( 2) 52

3 52

2

1M

Note that +

→x

52

0 when → ±∞x .

\ y = x + 3 is the oblique asymptote. 1A

Mock Exam 2


(b) =f (0) 12

′ = ++

f xx

( ) 1 5( 2)2 1M

′ = ++

=f (0) 1 5(0 2)

942 1M

Slope of the normal = -49

1M

The equation of the normal is = − +y x49

12

. 1A

(7)


(a) When n = 1,

L.H.S. = sin x cos 2(1)x = sin x cos 2x

R.H.S. = sin x cos (1 + 1)x = sin x cos 2x

\ The proposition is true for n = 1. 1

Next, assume the proposition is true for n = m, where m is a positive integer, that is,

∑ = +=

x kx mx m xsin cos 2 sin cos( 1)k

m

1, 1M

when n = m + 1,

∑=

+x kxsin cos 2k

m

1

1

∑= + +

= + + +

=+ + −

++ + − −

=+ −

=+

+

= + + +

=x kx x m x

mx m x x m xm x x m x m x

m x x

m x m x

m x m x

sin cos 2 sin cos 2( 1)

sin cos( 1) sin cos 2( 1)sin(2 1) sin( )

2sin(2 3) sin( 2 1)

2sin(2 3) sin

2

cos 2 42

sin 2 22

sin( 1) cos[( 1) 1]

k

m

1

\ The proposition is true for n = m + 1.

By the principle of mathematical induction, the proposition is true for all positive integers n. 1

(by the assumption) 1M

1M

1M



(b) Let =π

x18

and n = 2025.

By (a), ∑π π=

π π

=

ksin18

cos9

sin 202518

cos 202618k 1

2025

∑ π=

π + π

π + π

π

=

π π + π

π

=− π

π

= −

=

kcos9

sin 1122

cos 112 59

sin18

sin2

cos2 18

sin18

sin18

sin18

1

k 1

2025

1M

1A

(8)

Section B


(a) Since (0, -1) is a point on C,

= −

+ + = −

+ = −= −

f

e ae

b

aa

(0) 1

(0) 1

1 12

00

1M

1A

′ = − + = + +f x e ae

b ee

b( ) 2xx

xx 1M

Since (0, -1) is a stationary point of C,

′ =

+ + =

= −

f

ee

b

b

(0) 02 0

3

00

(3)

(b) By (a), ′ = + −f x ee

( ) 2 3xx

′′ = −

′′ = − = − = − <

f x ee

f ee

( ) 2

(0) 2 1 2 1 0

xx

00

1M

\ P is a maximum point of C.

\ The claim is agreed. 1A (2)

Mock Exam 2


(c) When f ′(x) = 0,

ex + e2x - 3 = 0

(ex)2 - 3ex + 2 = 0

(ex - 1)(ex - 2) = 0

ex = 1 or 2 1M

x = 0 or ln 2

′′ = − = >f (ln 2) 2 22

1 0 1M

\ C has a minimum point at x = ln 2.

= − − = −f (ln 2) 2 22

3ln 2 1 3ln 2

\ The coordinates of Q are (ln 2, 1 - 3 ln 2). 1A (3)

(d) When f″(x) = 0,

− =

=

=

=

ee

ee

e

x

2 0

2

2ln 22

xx

xx

x2

x <xln 22

=xln 22

>xln 22

f″(x) - 0 +

1M

= − − = −f

ln 22

2 22

3ln 22

3ln 22

\ −

ln 22

, 3ln 22

is the point of inflexion. 1A

(2)

(e) Note that the equation of L is y = 1 - 3 ln 2.

Area

∫= − −

− −

= + − − −

= −

ee

x dx

ee

x x

2 3 (1 3ln 2)

2 32

(1 3ln 2)

3(ln 2)2

ln 2

xx

xx

0ln2

2

0

ln2

2

1M

1M

1A

(3)



10. Reference: HKCEE A. Math 1992 Paper 2 Q11

(a) + =

+ + − + − + − =

+ + + + − + + =

+ + + = − − + +

+ + + = − − + +

+ − + +

− + + = −

− + + = −

− + + = − +

− + + = − +

+ =

+ =

PA PB

x y x y

x x y x x y

x x y x x y

x x y x x y

x x y

x x y x

x x y x

x x y x xx x y x x

x yx y

10

( 4) ( 0) ( 4) ( 0) 10

8 16 8 16 10

8 16 10 8 16

8 16 100 20 8 16

8 16

20 8 16 100 16

5 8 16 25 4

25( 8 16 ) 625 200 1625 200 400 25 625 200 16

9 25 225

25 91

2 2 2 2

2 2 2 2

2 2 2 2

2 2 2 2

2 2

2 2

2 2

2 2 2

2 2 2

2 2

2 2

1M

1M

1

(3) (b) (i) Let V cubic units be the volume of water.

∫∫

= π −

= π −

= π −

= π − + −− +

+

= π −

−

− +

−

− +

−

− +

V y dy

y dy

y y

h h

h h

25 19

259

(9 )

259

93

259

9( 3 ) ( 3 )3

18

2527

(9 )

h

h

h

2

3

3

2

3

3

3

3

3

3

2 3

1M

1A

1A

(ii) = π −

= π −

− = π −

=π −

dVdt

h h dhdt

h h dhdt

h h dhdt

dhdt h h

2527

(18 3 )

259

(6 )

30 5 259

(6 )

9(6 )

2

2

2

2

1M

1A

When dhdt

is minimum, 6h - h2 is maximum.

Consider − = − +

= − −

= − − +

+

= − − +

h h h h

h h

h h

h

6 6( 6 )

6 62

62

( 3) 9

2 2

2

22 2

2

1M \ When h = 3, 6h - h2 is the maximum.

\ When the rate of change of the depth of water is minimum, the depth of water is 3 units. 1A

Note that when x = 0, y = ±3. Therefore the lower limit is -3. Furthermore, since h is the depth, the upper limit is -3 + h.

The rate of change of volume of water = The rate of change of volume of water poured into the container - The rate of change of water evaporated.

Mock Exam 2


(iii) = −π

+dVdt

t800

( 100)

V = ∫−π

+t dt800

( 100) 1M

= − π−π

+t t C

1600 8

2, where C is a constant

By (b)(ii), when t = 0, h = 3.

= π − = πV2527

[9(3) 3 ] 502 3

\ C = 50π 1A

= −π

−π

+ πV t t1600 8

502

When V = 0,

−π

−π

+ π =

+ − =− + =

=

t t

t tt t

t

1600 850 0

200 80 000 0( 200)( 400) 0

200

2

2

or -400 (rejected)

1M

\ The time required is 200 minutes. 1A(11)

11. (a) The augmented matrix is

−−

−−

− →− →

− −− −

− →

abc

ab ac a

R R RR R R

ab a

c b aR R R

1 1 01 0 13 2 1

1 1 00 1 10 1 1 3

( ;3 )

1 1 00 1 10 0 0 2

( )

2 1 2

3 1 3

3 2 3

1M

\ If c = 2a + b, the augmented matrix is − −

ab a

1 1 00 1 10 0 0 0

. 1M

\ (E) is consistent. 1 Let z = t, where t is any real number. Then x = b - t, y = a - b + t. 1A (4)

(b) Since (F) is consistent, the system of linear equations formed by the first 3 equations of (F) is also consistent, which is equivalent to (E) for a = 2, b = 2 and c = α .

By (a), α = 2(2) + 2 = 6, 1A

and x = 2 - t, y = t, z = t, where t is any real number.

Substituting x = 2 - t, y = t and z = t into 2x + 3y - z = β ,

2(2 - t) + 3t - t = β 1M

β = 4 1A (3)



(c) Since (G) is consistent, by (a), we have

2p + q = 7

q = 7 - 2p ...... (*)

Let f = x3 + y2 + z.

f = (q - t)3 + (p - q + t)2 + t

= (7 - 2p - t)3 + (3p + t - 7)2 + t (from (*)) 1M

= − − − + + − +dfdt

p t p t3(7 2 ) 2(3 7) 12 1M

= − − +

= − − +

d fdt

p t

p t

6(7 2 ) 2

12 6 44

2

2

Since f attains its local minimum at z = 2 (i.e., t = 2), ==

dfdt

0t 2

.

-3(7 - 2p - 2)2 + 2(3p + 2 - 7) + 1 = 0

-3(5 - 2p)2 + 2(3p - 5) + 1 = 0

-12p2 + 66p - 84 = 0

-6(p - 2)(2p - 7) = 0

p = 2 or 72

1M

When p = 2, = − − + = >=

d fdt

12(2) 6(2) 44 8 0t

2

22

, which gives a local minimum.

1M

When =p72

, = −

− + = − <

=

d fdt

12 72

6(2) 44 10 0t

2

22

, which gives a local maximum.

\ p = 2 1A

q = 7 - 2(2)

= 3(5)


(a) Since OC is the angle bisector of ∠AOB, O, A, B and C lie on the same plane.

− =

− − =− − =

m nm nm n

3 4 0325

245

6

15

0

24 18 600 04 3 100 0 ...... (1)

1M

When O, A, B and C lie on the same plane, the volume of the parallelepiped

formed by � ��OA ,

� ��OB

and � ��OC is 0, i.e.,

( ) 0� �� OA OB OC⋅ × = .

Mock Exam 2


Also, ∠AOC = ∠BOC.

∠ = ∠

⋅=

⋅

+ ⋅ + +

+ +=

− +

⋅ + +

+ +

+=

− +

+ = − += −

AOC BOCOA OCOA OC

OB OCOB OC

m n

m n

m n

m n

m nm n

m n m nm n

i j i j ki j k i j k

cos cos

(3 4 ) ( 15 )

5 15

325

245

6 ( 15 )

10 15

3 45

325

245

90

1030 40 32 24 450

32 225 ...... (2)

2 2 2 2 2 2

� ��

� ��

� ��

� ��

1M

Substituting (2) into (1),

4(32n - 225) - 3n - 100 = 0

125n = 1000

n = 8 1A

Substituting n = 8 into (2),

m = 32(8) – 225

= 31 1A (4)

(b) (i) × =−

= − −

OA OB

i j k

i j k

3 4 0325

245

6

24 18 40

� ��

Note that � ��

×DE OA OB // .

\ � ��

= − −DE t t ti j k24 18 40 , where t is a non-zero real number.

� �� = +

= + −

+ −

OE OD DE

t t ti j k24 158

18 173

40

1M

Since � ��

⊥DE OE ,

� �� ⋅ =

− −

− −

=

− =

=

DE OE

t t t t t

t t

t

0

(24 ) 18 158

18 40 173

40 0

2500 312512

0

548

2

2

or 0 (rejected)

1M

\

� ��=

−

−

= − −

DE i j k

i j k

24 548

18 548

40 548

52

158

256

1A



(ii) � ��

=

+ −

+ −

= +

OE i j k

i k

24 548

158

18 548

173

40 548

52

32

From (a), � ��

= + +OC i j k31 8 15 .Note that when E is the incentre of DOAB, E lies on the angle bisector of ∠AOB, i.e., E lies on OC.

Since the j component of � ��OE is 0 while that of

� ��OC is not, it is impossible to

find a real number λ such that � ��

λ=OE OC . 1M

\ E does not lie on OC.

\ The claim is disagreed. 1A

(iii) � ��

=

+

=

OE 52

32

342

2 2

� ��

= −

= + −

EA OA OE

i j k12

4 32

� ��

=

+ + −

=

EA 12

4 32

742

22

2

Since � ��

≠EA OE , E is not the circumcentre of DOAB. 1M

\ The claim is disagreed. 1A (7)

When E is the incentre, E is the point of intersection of the three angle bisectors of the triangle.

When E is the circumcentre of DOAB, a circle with E as the centre and passing through O, A and B can be drawn, i.e., EO = EA = EB.

Mock Exam 2 - hkep. · PDF fileMock Exam 2 Section A 1. ... real number λ, det ......

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