2012 Maths Practice Paper(M1 & M2) Marking Scheme
Transcript of 2012 Maths Practice Paper(M1 & M2) Marking Scheme
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© 香港考試及評核局 保留版權
Hong Kong Examinations and Assessment Authority
All Rights Reserved 2012
PP-DSE-MATH-EP(M1)–1
只限教師參閱 FOR TEACHERS’ USE ONLY
香 港 考 試 及 評 核 局
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
香 港 中 學 文 憑 考 試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION
練習卷
PRACTICE PAPER
數學 延伸部分
單元一(微積分與統計)
MATHEMATICS Extended Part
Module 1 (Calculus and Statistics)
評卷參考(暫定稿)
PROVISIONAL MARKING SCHEME
本評卷參考乃香港考試及評核局專為本科練習卷而編寫,供教師參考之用。教師應提醒學生,不應將評卷參考視為標準答案,硬背死記,活剝生吞。這種學習態度,既無助學生改善學習,學懂應對及解難,亦有違考試着重理解能力與運用技巧之旨。因此,本局籲請各位教師通力合作,堅守上述原則。
This marking scheme has been prepared by the Hong Kong Examinations and Assessment Authority for
teachers’ reference. Teachers should remind their students NOT to regard this marking scheme as a set
of model answers. Our examinations emphasise the testing of understanding, the practical application
of knowledge and the use of processing skills. Hence the use of model answers, or anything else which
encourages rote memorisation, will not help students to improve their learning nor develop their
abilities in addressing and solving problems. The Authority is counting on the co-operation of teachers
in this regard.
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PP-DSE-MATH-EP(M1)–2
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General Notes for Teachers on Marking
Adherence to marking scheme
1. This marking scheme is the preliminary version before the normal standardisation process and some revisions may be necessary
after actual samples of performance have been collected and scrutinised by the HKEAA. Teachers are strongly advised to conduct
their own internal standardisation procedures before applying the marking schemes. After standardisation, teachers should adhere
to the marking scheme to ensure a uniform standard of marking within the school.
2. It is very important that all teachers should adhere as closely as possible to the marking scheme. In many cases, however, students
may have arrived at a correct answer by an alternative method not specified in the marking scheme. In general, a correct alternative
solution merits all the marks allocated to that part, unless a particular method has been specified in the question. Teachers should
be patient in marking alternative solutions not specified in the marking scheme.
Acceptance of alternative answers
3. For the convenience of teachers, the marking scheme was written as detailed as possible. However, it is likely that students would
not present their solution in the same explicit manner, e.g. some steps would either be omitted or stated implicitly. In such cases,
teachers should exercise their discretion in marking students’ work. In general, marks for a certain step should be awarded if
students’ solution indicate that the relevant concept / technique has been used.
4. In marking students’ work, the benefit of doubt should be given in students’ favour.
5. Unless the form of the answer is specified in the question, alternative simplified forms of answers different from those in the
marking scheme should be accepted if they are correct.
6. Unless otherwise specified in the question, use of notations different from those in the marking scheme should not be penalised.
Defining symbols used in the marking scheme
7. In the marking scheme, marks are classified into the following three categories:
‘M’ marks – awarded for applying correct methods
‘A’ marks – awarded for the accuracy of the answers
Marks without ‘M’ or ‘A’ – awarded for correctly completing a proof or arriving at an answer given in the question.
In a question consisting of several parts each depending on the previous parts, ‘M’ marks should be awarded to steps or methods
correctly deduced from previous answers, even if these answers are erroneous. ( I.e. Teachers should follow through students’ work
in awarding ‘M’ marks.) However, ‘A’ marks for the corresponding answers should NOT be awarded, unless otherwise specified.
8. In the marking scheme, steps which can be skipped are enclosed by dotted rectangles , whereas alternative answers are enclosed
by solid rectangles .
Others
9. Marks may be deducted for poor presentation (pp), including wrong / no unit. Note the following points:
(a) At most deduct 1 mark for pp in each section.
(b) In any case, do not deduct any marks for pp in those steps where students could not score any marks.
10. (a) Unless otherwise specified in the question, numerical answers not given in exact values or 4 decimal places should not be
accepted.
(b) Answers not accurate up to specified degree of accuracy should not be accepted. For answers with an excess degree of
accuracy, deduct 1 mark for pp. In any case, do not deduct any marks for excess degree of accuracy in those steps where
students could not score any marks.
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–3
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1. (a) 16128)12( 233 +++=+ xxxx 1A
(b) L−+−=−
21
22xaaxe
ax 1A
(c)
−+−+++=
+L
21)16128(
)12( 2223
3 xaaxxxx
e
xax
1M
The coefficient of 2
)1()(6)1(122
2 aax +−+= 1M
41262
2
−=+−∴ aa
032122 =+− aa
4=a or 8 1A
(5)
2. (a) 12 2
1
3 ++=
−
yyt
2
3
23d
d−
−= yyy
t 1A
(b) 12 += xt xe
xxt ln)1( 2 += 1A
xxx
x
x
tln2
1
d
d 2
++
= 1A
(c) y
t
x
t
x
y
d
d
d
d
d
d÷= 1M
−
++=
13
)ln21(
2
7
2
3
22
yx
yxxx 1A OR
2
3
2
2
3
ln21
−
−
++
yy
xxx
x
(5)
3. (a) By similar triangles, we have 15
20=
r
h . 1M
3
4rh =
=∴
3
4
3
1 2 rrV π
3
9
4rπ= 1A
22
3
4
+=
rrrA π
2
3
5rπ= 1A
h cm 20 cm
15 cm
r cm
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–4
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(b) (i) t
r
r
V
t
V
d
d
d
d
d
d⋅= 1M
t
rr
d
d
3
4 2π=
t
r
d
d)3(
3
42
2ππ =−
6
1
d
d −=
t
r 1A
Hence the rate of change of the radius of the water surface is 6
1−cm/s.
(ii) t
r
r
A
t
A
d
d
d
d
d
d⋅=
t
rr
d
d
3
10π=
−=
6
1)3(
3
10π
π3
5−= 1A
Hence the rate of change of the area of the wet surface is π3
5− 2cm /s.
(6)
4. (a) 2
1
)12( −= xxy
)2()12(2
1)12(
d
d2
1
2
1 −
−⋅+−= xxxx
y 1M For product rule
2
1
)12(
13
−
−=
x
x 1A
(b) For tangents parallel to 02 =− yx , we need 2d
d=
x
y .
2
)12(
13
2
1=
−
−
x
x 1M
)12(4169 2 −=+− xxx
05149 2 =+− xx
1=x or 9
5 1A
For 1=x , 1=y and hence the equation of the tangent is
)1(21 −=− xy
012 =−− yx 1A
For 9
5=x ,
27
5=y and hence the equation of the tangent is
−=−
9
52
27
5xy
0252754 =−− yx 1A
(6)
Either one
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–5
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5. (a) eee
e x
x−=−1
0)1()( 2 =++− eeee xx 1A
1=xe or e
0=x or 1 1A
(b) The area of the region bounded by 1C and 2C
xeee
e x
xd)(1
1
0∫
−−−= 1M For lower and upper limits
[ ]1
0exeeexxx +−⋅+= −
1M Accept [ ]1
0xx
eexexe−⋅−−−
111 +−+−+= eee
e−= 3 1A
(5)
6. (a) )(Var4)72(Var XX =+
=
10
84 1M For
n
XX
)(Var)(Var =
2.3= 1A
(b) A 97% confidence interval for µ
×+×−=
10
817.250,
10
817.250 1M+1A
)9409.51,0591.48(= 1A
(5)
7. (a) P(a player is rewarded)5
1
2
1
5
2
2
1⋅+⋅=
3.0= 1A
(b) P(both players are rewarded | one player is rewarded)27.03.03.03.0
3.03.0
××+×
×= 1M OR
7.07.01
3.03.0
×−
×
17
3= 1A OR 0.1765
(c) E(no. of players having drawn a blue ball from A )3.0
60 52
21 ×
×= 1M
40= 1A
(5)
8. (a) P(a box contains more than 1 rotten eggs)
)04.0()96.0()96.0(12930
130
C−−= 1M+1M
338820302.0≈
3388.0≈ 1A
(b) (i) P(the 1st box containing more than 1 rotten egg is the 6
th box inspected)
)338820302.0()338820302.01( 5−= 1M
0428.0≈ 1A
1M for d±50
1A for 2.17
1M for binomial prob
1M for correct cases
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–6
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(ii) E(no. of boxes inspected until a box containing more than 1 rotten egg is found)
338820302.0
1= 1M
9514.2≈ 1A
(7)
9. (a) )(P)(P)(P BABAA ′∩+∩=
k+= 12.0 1A
)(P
)(P)|(P
B
BABA
′
′∩=′
)(P1
6.0B
k
−=
3
51)(P
kB −= 1A
)(P)(P)(P)(P BABABA ∩−+=∪
12.03
51)12.0( −
−++=
kk 1M
3
21
k−= 1A
(b) If A and B are independent, )(P)(P)(P BABA ∩= .
12.03
51)12.0( =
−+
kk 1M
03
58.0
2
=−k
k
48.0=k or 0 (rejected) 1A
Alternative solution 1
If A and B are independent, )|(P)(P BAA ′= .
6.012.0 =+ k 1M
48.0=k 1A
Alternative solution 2
If A and B are independent, )(P)(P)(P BABA ′∩=′ .
kk
k =
+
3
5)12.0( 1M
08.03
5 2
=− kk
48.0=k or 0 (rejected) 1A
Alternative solution 3
If A and B are independent, )|(P)|(P BABA ′= .
)|(P)(P
)(PBA
B
BA′=
∩∴
6.0
3
51
12.0=
−k
1M
48.0=k 1A
(6)
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–7
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10. (a)
2
5
)1(
61
d
d
+
=
t
t
t
x
Let 1+= tu and hence tu dd = .
The amount of alloy produced by A
t
t
td
)1(
6110
02
5∫+
= 1A
u
u
ud
)1(6111
12
5∫−
=
uuu d616111
1
2
5
2
3
∫
−=
−−
1M
11
1
2
3
2
1
3
122122
+−=
−−
uu 1A For primitive function
Alternative Solution
t
t
tx d
)1(
61
2
5∫+
=
u
u
ud
)1(61
2
5∫−
= 1A
uuu d6161 2
5
2
3
∫
−=
−−
1M
Cuu ++−=
−−
2
3
2
1
3
122122
Ctt ++++−=
−−
2
3
2
1
)1(3
122)1(122 1A
The amount of alloy produced by A
++−−
++++−=
−−
CC3
122122)110(
3
122)110(122 2
3
2
1
6636.45≈ 1A OR 1133
3904
3
244−=
(4)
(b) The amount of alloy produced by B
tt
d16
)100ln(1510
0
2
∫+
=
)1002[ln(2)10010ln()1000{ln(16
15
2
2 22 +++++⋅≈
)]}1008ln()1006ln()1004ln( 222 ++++++
6792.45≈ 1A
(2)
1M
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–8
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(c) 16
)100ln(15
d
d
d
d
d
d2 +
=
t
tt
y
t
)100(8
152 +
=t
t 1A
22
2
2
2
)100(
)2()100(
8
15
d
d
d
d
+
−+⋅=
t
ttt
t
y
t
22
2
)100(8
)100(15
+
−=
t
t 1A
0d
d
d
d2
2
>
∴
t
y
t for 100 << t
Thus, 45.6792 is an over-estimate of the amount of alloy produced by B . 1A
Hence it is uncertain whether machine B is more productive than machine A by
the results of (a) and (b). The engineer cannot be agreed with.
(4)
11. (a) t
a
ktet 20)(P =′
kta
t
tln
20
)(Pln +=
′ 1A
(1)
(b)
t 1 2 3 4
)(P t′ 22.83 43.43 61.97 78.60
t
t)(Pln
′ 3.13 3.08 3.03 2.98
1A
1A
1A
3.3
3.2
3.1
3.0
2.9
1 2 3 4 5
t O
t
t)(Pln
′
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–9
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From the graph, 14
13.398.2
20 −
−≈
a 1M
1−≈a 1A
From the graph, 18.3ln ≈k
24≈k 1A
(5)
(c) (i)
=′
−
2024d
d)(P
d
dt
tet
tt
−=
−
20124 20
te
t
1A
0)(Pd
d=′∴ t
t when 20=t
t 20< 20 20>
)(Pd
dt
t′ +ve 0 -ve
Alternative Solution
−+
−
−=′
−
20
1
201
20
124)(P
d
d20
2
2t
ett
t
−=
−
2205
620
te
t
1M
0)(Pd
d2
2
<′∴ tt
when 20=t
Hence the rate of change of the population size is greatest when 20=t . 1A
(ii) 202020
20
1
d
dttt
teetet
−−−
−=
1A
−=
−−−
202020
d
d48048024
ttt
tet
ete
Cteette
ttt
+−−=
−−−
∫ 202020 4809600d24 1M
2020 9600480)(P
tt
eteCt
−−
−−= 1A
Since 30)0(P = , we have
309600)0(480 00 =−− eeC 1M
9630=C
2020 96004809630)(P
tt
etet
−−
−−=∴ 1A
(iii)
−−=
−−
∞→∞→
2020 96004809630lim)(Plim
tt
ttetet
9630= 1A ∴ the population size after a very long time is estimated to be 9630 thousands.
(9)
Either one
1M
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–10
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12. (a) The estimate of the mean 100
4760 ×++×=
L
21.3= 1A
(1)
(b) (i) The sample proportion of school days with less than 4 visits 100
57= 1A
(ii) An approximate 95% confidence interval for the proportion
×+
×−=
100
43.057.096.157.0,
100
43.057.096.157.0 1M
)6670.0,4730.0(= 1A
(3)
(c) (i) By (a), 21.3=λ .
P(crowded on a day)
+++−= −
!3
21.3
!2
21.321.311
3221.3
e 1M For Poisson probability
399705729.0≈
3997.0≈ 1A
(ii) P(crowded on alternate days | crowded on at least 2 days)
)399705729.0()399705729.01(5)399705729.01(1
)399705729.0()399705729.01()399705729.01()399705729.0(45
2323
−−−−
−+−= 1M+1M+1M
0869.0≈ 1A
(6)
1M for numerator
1M for denominator
1M for binomial probability
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Solution Marks Remarks
PP-DSE-MATH-EP(M1)–11
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13. Let rX minutes and eX minutes be the waiting times for a customer in the regular
and express counter respectively.
(a)
−>=>
2.1
6.66P)6(P ZX r 1M
)5.0(P −>= Z
6915.0≈ 1A
(2)
(b) (i) P(more than 10 from 12 customers with 6>rX )
121112
11 )6915.0()6915.01()6915.0( +−= C 1M+1M
0759.0≈ 1A
(ii) Let Y minutes be the average waiting time of the 12 customers
Y ~ N =
12
2.1,6.6
2
N(6.6 , 0.12) 1A
−>=>
12.0
6.66P)6(P ZY
)73.1(P −>≈ Z OR )732.1(P −>Z
9582.0≈ 1A OR 0.9584
(5)
(c) (i) 2119.0)(P =< kX r
2119.02.1
6.6P =
−<
kZ 1M
8.02.1
6.6−=
−k
64.5=k 1A
0359.0)(P => kX e
0359.08.0
64.5P =
−>
µZ 1M
8.18.0
64.5=
− µ
2.4=µ 1A
(ii)
−>=>
2.1
6.62.4P)(P ZX r µ
9772.0≈ 1A
P(1 customer pays at regular counter | 2 customers wait more than µ min)
2)]5.0)(12.0()9772.0)(88.0[(
)5.0)(12.0)(9772.0)(88.0(2
+≈ 1M+1M
1219.0≈ 1A
(8)
1M for numerator
1M for denominator
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© 香港考試及評核局 保留版權
Hong Kong Examinations and Assessment Authority
All Rights Reserved 2012
PP-DSE-MATH-EP(M2)–1
只限教師參閱 FOR TEACHERS’ USE ONLY
香 港 考 試 及 評 核 局
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
香 港 中 學 文 憑 考 試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION
練習卷
PRACTICE PAPER
數學 延伸部分
單元二(代數與微積分)
MATHEMATICS Extended Part
Module 2 (Algebra and Calculus)
評卷參考(暫定稿)
PROVISIONAL MARKING SCHEME
本評卷參考乃香港考試及評核局專為本科練習卷而編寫,供教師參考之用。教師應提醒學生,不應將評卷參考視為標準答案,硬背死記,活剝生吞。這種學習態度,既無助學生改善學習,學懂應對及解難,亦有違考試着重理解能力與運用技巧之旨。因此,本局籲請各位教師通力合作,堅守上述原則。
This marking scheme has been prepared by the Hong Kong Examinations and Assessment Authority for
teachers’ reference. Teachers should remind their students NOT to regard this marking scheme as a set
of model answers. Our examinations emphasise the testing of understanding, the practical application
of knowledge and the use of processing skills. Hence the use of model answers, or anything else which
encourages rote memorisation, will not help students to improve their learning nor develop their
abilities in addressing and solving problems. The Authority is counting on the co-operation of teachers
in this regard.
只限教師參閱 FOR TEACHERS’ USE ONLY
PP-DSE-MATH-EP(M2)–2
只限教師參閱 FOR TEACHERS’ USE ONLY
General Notes for Teachers on Marking
Adherence to marking scheme
1. This marking scheme is the preliminary version before the normal standardisation process and some revisions may be necessary
after actual samples of performance have been collected and scrutinised by the HKEAA. Teachers are strongly advised to conduct
their own internal standardisation procedures before applying the marking schemes. After standardisation, teachers should adhere
to the marking scheme to ensure a uniform standard of marking within the school.
2. It is very important that all teachers should adhere as closely as possible to the marking scheme. In many cases, however, students
may have arrived at a correct answer by an alternative method not specified in the marking scheme. In general, a correct alternative
solution merits all the marks allocated to that part, unless a particular method has been specified in the question. Teachers should
be patient in marking alternative solutions not specified in the marking scheme.
Acceptance of alternative answers
3. For the convenience of teachers, the marking scheme was written as detailed as possible. However, it is likely that students would
not present their solution in the same explicit manner, e.g. some steps would either be omitted or stated implicitly. In such cases,
teachers should exercise their discretion in marking students’ work. In general, marks for a certain step should be awarded if
students’ solution indicate that the relevant concept / technique has been used.
4. In marking students’ work, the benefit of doubt should be given in students’ favour.
5. Unless the form of the answer is specified in the question, alternative simplified forms of answers different from those in the
marking scheme should be accepted if they are correct.
6. Unless otherwise specified in the question, use of notations different from those in the marking scheme should not be penalised.
Defining symbols used in the marking scheme
7. In the marking scheme, marks are classified into the following three categories:
‘M’ marks – awarded for applying correct methods
‘A’ marks – awarded for the accuracy of the answers
Marks without ‘M’ or ‘A’ – awarded for correctly completing a proof or arriving at an answer given in the question.
In a question consisting of several parts each depending on the previous parts, ‘M’ marks should be awarded to steps or methods
correctly deduced from previous answers, even if these answers are erroneous. ( I.e. Teachers should follow through students’ work
in awarding ‘M’ marks.) However, ‘A’ marks for the corresponding answers should NOT be awarded, unless otherwise specified.
8. In the marking scheme, steps which can be skipped are enclosed by dotted rectangles , whereas alternative answers are enclosed
by solid rectangles .
Others
9. Marks may be deducted for poor presentation (pp), including wrong / no unit. Note the following points:
(a) At most deduct 1 mark for pp in each section.
(b) In any case, do not deduct any marks for pp in those steps where students could not score any marks.
10. (a) Unless otherwise specified in the question, numerical answers not given in exact values should not be accepted.
(b) In case a certain degree of accuracy had been specified in the question, answers not accurate up to that degree should not be
accepted. For answers with an excess degree of accuracy, deduct 1 mark for pp. In any case, do not deduct any marks for
excess degree of accuracy in those steps where candidates could not score any marks.
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–3
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1. The general term of 9)2( x− is rrr xC )(299 −− 1M
rrrr xC )1(299 −= − 1A
Alternative Solution
L+−+−+−=− 5495
4594
3693
2792
891
99 222222)2( xCxCxCxCxCx 1M+1A
Hence the coefficient of 5x is 495 2C− 1M
2016−= 1A
(4)
2. If the system of homogeneous equations has non-trivial solutions, then
0
12
31
771
=−
−
k
k 1M+1A
037144272 =−++−+− kkk 1M
038212 =+− kk
19=k or 2 1A
(4)
3. For 1=n ,
181)1(1541 =−+ which is divisible by 9 .
∴ the statement is true for 1=n . 1
Assume 1154 −+ kk is divisible by 9 , where k is a positive integer. 1
i.e. let Nkk 91154 =−+ , where N is an integer.
11594 +−=∴ kNk
1)1(154 1 −+++ kk
11515)1159(4 −+++−= kkN (by induction assumption) 1
184536 +−= kN
)254(9 +−= kN which is divisible by 9 1
Hence the statement is true for 1+= kn .
By the principle of mathematical induction, the statement is true for all positive integers n . 1 Follow through
(5)
4. (a) θ
θ22 tan1
tan2
1
2
+=
+ x
x
θ
θ2sec
tan2= 1M
θθ
θ 2cos
cos
sin2 ⋅=
θ2sin= 1
(b) 2
2
2
2
1
21
1
)1(
x
xx
x
x
+
++=
+
+
21
21
x
x
++= 1M
Since x is real, we can let θtan=x for some θ .
θ2sin11
)1(2
2
+=+
+∴
x
x by (a) 1M
Withdraw the last mark if
“N is an integer” was omitted
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–4
只限教師參閱 FOR TEACHERS’ USE ONLY
Since the maximum value of θ2sin is 1 , the maximum value of 2
2
1
)1(
x
x
+
+ is 2 . 1A
(5)
5. (a) 2
11cos
2
11cos2)1cos()1cos(
+−+−++=−++
xxxxxx 1M
xcos1cos2=
Alternative Solution
0cos)10cos()10cos( k=−++ 1M
i.e. 1cos2=k 1A
(b)
9cos8cos9cos7cos
6cos5cos6cos4cos
3cos2cos3cos1cos
9cos8cos7cos
6cos5cos4cos
3cos2cos1cos
+
+
+
= 1M For column (or row) operations
9cos8cos8cos1cos2
6cos5cos5cos1cos2
3cos2cos2cos1cos2
= by (a) 1M
9cos8cos8cos
6cos5cos5cos
3cos2cos2cos
1cos2= 1M
0= 1A
(6)
6. h
xhx
xx h
11
lim1
d
d
0
−+=
→
1M+1A
xhxh
hxx
h )(lim
0 +
−−=
→
xhxh )(
1lim
0 +
−=
→ 1A
2
1
x
−= 1A
(4)
7. (a) )cos(sin)(f xxex x +=
)sin(cos)cos(sin)(f xxexxex xx −++=′
xe x cos2= 1A
xexex xx sin2cos2)(f −=′′
)sin(cos2 xxe x −= 1A
(b) 0)(f)(f)(f =+′−′′ xxx
0)cos(sincos2)sin(cos2 =++−− xxexexxe xxx 1M
0)sin(cos =− xxe x
xx cossin = or 0=xe (rejected) 1A
1tan =x
4
π=x for π≤≤ x0 1A
(5)
For using (a) or sum-to-
product formula of cosine
OR 1sinsin1coscos xx −
1sinsin1coscos xx ++
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–5
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8. (a) Let θsin2=x . 1M OR θcos2=x
θθ dcos2d =x
θθ
θd
sin44
2cos
4
d
22 ∫∫−
=− x
x
θd1∫= 1A
C+= θ
Cx
+= −
2sin
1 1A
(b) xxxxxx lndlndln ∫∫ −= 1M
xx
xxx d1
ln ⋅−= ∫
Cxxx +−= ln 1A
(5)
9. 012 22 =−−− yxyx --------------------------------------------- (*)
0d
d4
d
d2 =−−−
x
yyy
x
yxx 1A
For the tangents parallel to 12 += xy , 2d
d=
x
y .
0)2(4)2(2 =−−−∴ yyxx 1M
0=y 1A
By (*), 012 =−x 1M
1±=x
Hence the tangents are )]1([20 ±−=− xy 1M
i.e. 22 += xy and 22 −= xy 1A For both
(6)
10. (a) ∫∫ −− = 2d
2
1d
22
xexxexx
1M OR ∫ −−−
)(d2
1 22
xex
Cex +
−= − 2
2
1 1A
(b) The volume of the solid
∫
−= −
2
1
2
d2
22
xex
xxπ 1M+1A 1M for ∫= xxyV d2π
∫
−= −
2
1
3
d2
22
xxex xπ
2
1
42
2
1
82
+= −xe
xπ 1M For using (a)
π
−+= −− 14
4
15ee 1A
(6)
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–6
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11. (a)
−+
−+=
0101
2 αββααββαA
−+
+−−+=
αββα
βααβαββα )()(2
1A
−
−++=−+
10
01
01)()( αβ
αββαβααββα IA
−+
+−−+=
αββα
βααβαββα )()(2
i.e. IAA αββα −+= )(2 1
(2)
(b) IAAIA 222 2)( ααα +−=−
IAIA 22)( αααββα +−−+= by (a) 1M
IA )()( 2 αβααβ −+−=
))(( IA ααβ −−= 1
Alternative Solution
2
2
10
01
01)(
−
−+=− α
αββααIA
−
−
−
−=
α
αββ
α
αββ
11
−−
−−=
αβααβ
αββααββ2
222
1A
−
−+−=−−
10
01
01)())(( α
αββααβααβ IA
−
−−=
α
αββαβ
1)(
−−
−−=
αβααβ
αββααββ2
222
i.e. ))(()( 2 IAIA ααβα −−=− 1
By interchanging α and β , we have ))(()( 2 IAIA ββαβ −−=− . 1
Alternative Solution 1
IAAIA 222 2)( βββ +−=−
IAIA 22)( ββαββα +−−+= by (a)
IA )()( 2 αβββα −+−=
))(( IA ββα −−= 1
Either one
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–7
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Alternative Solution 2
2
2
10
01
01)(
−
−+=− β
αββαβIA
−
−
−
−=
β
αβα
β
αβα
11
−−
−−=
αβββα
βααβαβα2
222
−
−+−=−−
10
01
01)())(( β
αββαβαββα IA
−
−−=
β
αβαβα
1)(
−−
−−=
αβββα
βααβαβα2
222
i.e. ))(()( 2 IAIA ββαβ −−=− 1
(3)
(c) (i) YXA +=
−
−+
−
−=
−+
β
αβα
α
αββαββα
1101ts
−−+
+−+=
βα
αβαβ
tsts
tsts )( 1M
Comparing the entries, we have
=+
=+
+=+
0
1
βα
βααβ
ts
ts
ts
Solving, αβ
β
−=s and
βα
α
−=t 1A For both
(ii) Consider the statement “ )( IAXn
n ααβ
β−
−= and )( IAY
nn β
βα
α−
−= ”.
When 1=n , )( IAX ααβ
β−
−= and )( IAY β
βα
α−
−= are true by (c)(i). 1
Assume )( IAXk
k ααβ
β−
−= and )( IAY
kk β
βα
α−
−= , where k is a
positive integer.
)()(1 IAIAXk
k ααβ
βα
αβ
β−
−−
−=+ by the assumption
))(()(
2
1
IAk
ααβαβ
β−−
−=
+
by (b) 1
)(1
IAk
ααβ
β−
−=
+
)()(1 IAIAYk
k ββα
αβ
βα
α−
−−
−=+ by the assumption
))(()( 2
1
IAk
ββαβα
α−−
−=
+
by (b)
)(1
IAk
ββα
α−
−=
+
1
Either one
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–8
只限教師參閱 FOR TEACHERS’ USE ONLY
Hence the statement is true for 1+= kn .
By the principle of mathematical induction, the statement is true for all
positive integers n . 1 Follow through
(iii) )()( IAtIAsXY βα −−=
])([ 2 IAAst αββα ++−= by (a)
=
00
00
)()( IAsIAtYX αβ −−=
])([ 2 IAAst αββα ++−= by (a)
=
00
00 1
nn YXA )( +=
nnYX += by the note given 1M
)()( IAIAnn
ββα
αα
αβ
β−
−+−
−= by (ii)
IAnnnn
βα
βααβ
βα
βα
−
−+
−
−= 1A
(9)
12. (a) (i) ji aaOM +−= )1( 1A
)( kji ++= bON
])1[()( jikji aabMN +−−++=∴
kji babba +−+−+= )()1( 1
(ii) ij −=AB
0=⋅ ABMN
0)(])()1[( =−⋅+−+−+ ijkji babba 1M
01 =−++−− abba
2
1=a 1A
0=⋅OCMN
0)(])()1[( =++⋅+−+−+ kjikji babba 1M
01 =+−+−+ babba
3
1=b 1A
Alternative Solution
)()( kjiji ++×+−=×OCAB 1M
kji 2−+=
( )OCABMN ×//
211
1
−=
−=
−+∴
babba 1M
Solving, we get 2
1=a and
3
1=b . 1A+1A
For both
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–9
只限教師參閱 FOR TEACHERS’ USE ONLY
(iii) kji3
1
6
1
6
1+−
−=MN
The shortest distance between the lines AB and OC
MN=
222
3
1
6
1
6
1
+
−+
−= 1M
6
6= 1A
(8)
(b) (i) )()( kjji +×+−=× ACAB
kji −+= 1A
(ii) Let the intersecting point of the two lines OG and MN be P .
Since P lies on MN , let MNMP λ= . 1M
MPOMOP +=
+−
−++= kjiji
3
1
6
1
6
1
2
1
2
1λ
kji36
3
6
3 λλλ+
−+
−= 1A
Since P lies on OG , OP // ( )ACAB × .
36
3 λλ−=
−∴ 1M
3−=λ
Alternative Solution
Since P lies on OG , OP // ( )ACAB × .
Let )( kji −+= tOP 1M
+−−+=∴ jikji
2
1
2
1)(tMP
kji ttt
−−
+−
=2
12
2
12 1A
Since P lies on MN , MP // MN .
3
1
6
12
12
t
t
−=
−
−
∴ 1M
1=t
Hence the coordinates of P are )1,1,1( − . 1A
(5)
x
y
z
A B
C
M
N
P
O
G
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–10
只限教師參閱 FOR TEACHERS’ USE ONLY
13. (a) Let pxu −= . 1M
xu dd =∴
When 0=x , pu −= ; when px 2= , pu = .
uuxpxp
p
p
d)(fd)(f2
0 ∫∫ −=−∴ 1A
0= since f is an odd function 1
p
p
qxxqpx20
2
0][0d])(f[ +=+−∴ ∫
pq2= 1A
(4)
(b)
x
x
x
x
x
x
tan1
tan3
tan1
tan3
6tan3
6tan3
3
1
3
1
3
1
3
1
+
−−
+
−+
=
−−
−+
π
π
1M
1tan3tan33
1tan3tan33
+−+
−++=
xx
xx
2
tan31 x+= 1
(2)
(c) ∫∫
⋅
−−
−+
=+ 3
0
3
0d2
6tan3
6tan3
lnd)tan31ln(
ππ
π
π
x
x
x
xx by (b)
∫
+
−−
−+
= 3
0d2ln
6tan3
6tan3
ln
π
π
π
x
x
x
1M
Consider x
xx
tan3
tan3ln)(f
−
+= .
)tan(3
)tan(3ln)(f
x
xx
−−
−+=−
x
x
tan3
tan3ln
+
−=
1
tan3
tan3ln
−
−
+=
x
x
x
x
tan3
tan3ln
−
+−=
)(f x−=
)(f x∴ is an odd function 1A
∫∫×
+
−=+∴ 6
2
0
3
0d2ln
6fd)tan31ln(
πππ
xxxx
3
2lnπ= by (a) 1A
(4)
1M
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–11
只限教師參閱 FOR TEACHERS’ USE ONLY
14. (a) The volume of the solid of revolution
yyh
d)25(0
2∫ −= π 1M
h
yy
0
3
325
−= π
π
−=
325
3hh 1
(2)
(b) (i) By (a), π
−=
325
3hhV for 40 ≤≤ h
π
−=
t
hh
t
h
t
V
d
d
d
d25
d
d 2 1A
When 3=h , t
h
d
d)325(8
2 π−=
π2
1
d
d=
t
h
i.e. the rate of increase of the depth of coffee is 1
scm2
1 −
π . 1A
(ii) Let x , l , r and h be the lengths as shown in the figure.
25422 =+x
3=x 1A
By similar triangles, ll
x
+=
8
6 1M
ll 6324 =+
8=l
By similar triangles, llh
r
+=
+− 8
6
4
8
)4(3 +=
hr 1A
)8()3(3
)4(8
)4(3
33
)4()4(25
223 ππ
π −+
++
−=∴ h
hV 1M
Alternative Solution
Locating the origin at the centre of the base and the x-axis along the base of
the frustum, the equation of a slang edge of the frustum is
36
08
3
0
−
−=
−
−
x
y 1M
)8(8
3+= yx 1A
∫−
++
−=∴
4
0
23
d)8(64
9
3
)4()4(25
h
yyV ππ 1M
4
0
3
3
)8(
64
9
3
236−
++=
h
yππ
i.e. 3
)4(64
3
3
164++= hV
ππ 1
4
8
6
5
x
r
h
l
x
y (6, 8)
3 O
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Solution Marks Remarks
PP-DSE-MATH-EP(M2)–12
只限教師參閱 FOR TEACHERS’ USE ONLY
(iii) After 15 seconds, 33
)4(64
3
3
164152)412(
64
3
3
164++=×−++ h
ππππ 1M
30192)4(64
3 3 −=+ ππ
h
3
1
106444
−=+
π
πh 1A
473.11 >≈h
3
)4(64
3
3
164++= hV
ππ
t
hh
t
V
d
d)4(
64
9
d
d 2+=π
1A
After 15 seconds, t
h
d
d10644
64
92
2
3
1
−=−
π
ππ
3
2
3
1
)1064(9
8
d
d
−
−=
ππt
h
0183.0−≈
i.e. the rate of decrease of the depth of coffee is 1scm0183.0 − . 1A
(11)