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)
2017-DSE-MATH-EP(M2)-2
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)
2017-DSE-MATH-EP(M2)-3
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-4
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-5
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-6
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-7
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-8
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-9
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-10
Solution Marks Remarks
(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)
2017-DSE-MATH-EP(M2)-11
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-12
Solution Marks Remarks
(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)
2017-DSE-MATH-EP(M2)-13
Solution Marks Remarks
(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)
2017-DSE-MATH-EP(M2)-14
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-15
Solution Marks Remarks
(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)
2017-DSE-MATH-EP(M2)-16
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-17
Solution Marks Remarks
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)
2017-DSE-MATH-EP(M2)-18
Solution Marks Remarks
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.