M2 Mock Exam (TE) - Munsang College
Transcript of 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
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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:
Public Exam Reference: HKDSE 2013 Math M2 Section A Q7
(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.
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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(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:
Public Exam Reference: HKDSE 2018 Math M2 Section A Q4
(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
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
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(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:
Public Exam Reference: HKDSE 2018 Math M2 Section A Q5
(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)
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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:
Public Exam Reference: HKDSE 2016 Math M2 Section A Q5
(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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:
Public Exam Reference: HKDSE 2016 Math M2 Section A Q8
(a) (i) 2M
=1 1
0 1 0 1
c c
1 2
0 1
c
1A
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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:
Public Exam Reference: HKDSE 2018 Math M2 Section A Q8
(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 9
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
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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)
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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(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)
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M2 Mock Exam 13 (2019) (Teacher’s Edition)
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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:
Public Exam Reference: HKDSE 2018 Math M2 Section B Q10
(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
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 13
(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
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(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 15
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:
Public Exam Reference: HKDSE 2016 Math M2 Section B Q11
(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
New Progress in Senior Mathematics Module 2 (Extended Part)
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© Hong Kong Educational Publishing Company 16
(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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 17
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)
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 18
(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
New Progress in Senior Mathematics Module 2 (Extended Part)
M2 Mock Exam 13 (2019) (Teacher’s Edition)
© Hong Kong Educational Publishing Company 19
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