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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
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數學 延伸部分 單元一（微積分與統計） MATHEMATICS MATHEMATICS Extended Extend ed Part Module 1 (Calculus and Statistics)
PROVISIONAL MARKING SCHEME
This marking scheme has been prepared by the Hong Kong E xaminations and Assessment Authority for teachers’ reference. Teachers Teachers should remind their students NOT to regard this marking scheme as a set of model answers. Our examinations emphasise emphasise the testing of understanding, the 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. problems. The Authority is counting on the co-operation of teachers in this regard.
© Hong Kong Examinations and Assessment Authority All Rights Reserved 2012
PP-DSE-MATH-EP(M1)–1
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General Notes for Teachers on Marking M arking Adherence to marking scheme
1.
This marking marking scheme is the preliminary version before the normal standardisation standardisation process process and some revisions revisions may be necessary necessary after actual samples of performance have been collected and scrutinised by the HKEAA. Teachers Teachers are strongly advised to conduct their own internal standardisation procedures before applying the marking schemes. schemes. After standardisation, teachers should adhere to the marking scheme to ensure a uniform standard of marking within t he school.
2.
It is very important that all teachers should adhere adhere as closely as possible to the marking scheme. scheme. In many cases, however, students may have arrived at a correct answer by an alternative method not specified 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 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 written as detailed as possible. However, it is likely that students would would not present their solution in the same explicit manner, e.g. some some steps would either be omitted or stated implicitly. implicitly. In such cases, teachers should exercise their discretion in marking students’ work. In general, marks for a certain step step should be awarded if students’ solution solution indicate that the relevant concept / technique has been used.
4.
In marking students’ work, the the benefit of doubt should be given in students’ favour.
5.
Unless the the form of the answer is specified specified in the question, alternative alternative simplified simplified forms of answers different different from those in the marking scheme should be accepted if they are correct.
6.
Unless otherwise otherwise specified in the question, use of notations different from those in the marking scheme should 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 – ‘A’ ‘A’ marks – Marks without without ‘M’ or ‘A’ ‘A’ –
awarded for applying correct methods awarded for the accuracy of the answers awarded for for correctly completing a proof or arriving at an answer 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 work in awarding ‘M’ marks.) However, ‘A’ ‘A’ marks for the corresponding answers should NOT be awarded, unless otherwise specified.
8.
In the the marking marking scheme, scheme, steps which can be skipped skipped are are enclosed enclosed by dotted rectangles , whereas whereas alternative answers are enclosed 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 marks for pp in those steps where students could not score any marks.
10. (a) (b)
Unless otherwise specified specified in the the question, numerical answers answers not given in exact values or 4 decimal decimal places should not be accepted. Answers not not accurate up up to specified 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 marks for excess degree of accuracy in those steps where students could not score any marks.
PP-DSE-MATH-EP(M1)–2
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Solution 1.
(a)
(2 x + 1) 3 = 8 x 3 + 12 x 2 + 6 x + 1
(b)
e
(c)
−ax
= 1 − ax +
(2 x + 1) 3 e
a 2 x 2
−L
2 3
1A
2
2
The coefficient of x = 12(1) + 6( −a ) + (1) ∴
a
Remarks
1A
= (8 x + 12 x + 6 x + 1)1 − ax +
ax
Marks
a 2 x 2
2 a
− L
1M
2
2
1M
2
2
− 6a + 12 = −4
a 2 − 12a + 32 = 0 a = 4 or 8
1A (5)
−1
2.
(a)
3
t = y + 2 y
dt d y (b)
2 +1 −3
2
= 3 y − y 2
e t = x x
2
1A
+1
t = ( x 2 + 1) ln x
dt x 2 + 1 =
d x (c)
d y
x =
d x
dt
1A
+ 2 x ln x
1A
dt
÷
d x
1M
d y
2
x + 1
3
=
( x 2 + 1 + 2 x 2 ln x ) y 2
1A
7 x 3 y 2 − 1
OR
x
+ 2 x ln x −3
2
3 y − y
2
(5)
3.
(a)
By similar triangles, we have h=
h r
=
20 15
.
1M 15 cm
4r 3
∴ V = =
1 3 4 9
4r 3
r cm
π r 2
π r 3
1A
4r A = π r r + 3
20 cm
2
2
=
5 3
π r 2
1A
PP-DSE-MATH-EP(M1)–3
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h cm
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Solution (b)
dV
(i)
=
Marks
dV dr ⋅
1M
dt
dr dt 4 2 dr = π r 3 dt 4 2 dr − 2π = π (3) 3 dt dr −1 =
dt
1A
6
Hence the rate of change of the radius of the water surface is d A
(ii)
=
dt
Remarks
−1
6
Either one
cm/s.
d A dr ⋅
dr dt 10 dr = π r 3 dt 10 − 1 = π (3) 3 6 =
−5
3
π
1A
Hence the rate of change of the area of the wet surface is
−5
3
π cm 2 /s. (6)
1
4.
(a)
y = x (2 x − 1) 2
d y d x
1
= ( 2 x − 1) 2 + x ⋅
1 2
−1
(2 x − 1)
2
(2)
1M
3 x − 1
=
For product rule
1A
1
(2 x − 1) 2
(b)
For tangents parallel to 2 x − y = 0 , we need 3 x − 1 1
d y d x
=2 .
=2
1M
(2 x − 1) 2 9 x 2 − 6 x + 1 = 4(2 x − 1) 9 x 2 − 14 x + 5 = 0 5 x = 1 or 9 For x = 1 , y = 1 and hence the equation of the tangent is
1A
y − 1 = 2( x − 1)
2 x − y − 1 = 0 For x = y −
5
5 9
1A
, y =
= 2 x −
5 27
and hence the equation of the tangent is
5
27 9 54 x − 27 y − 25 = 0
1A (6)
PP-DSE-MATH-EP(M1)–4
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Solution 5.
(a)
1−
e
Marks
x
x
e
= e −e
x 2
(b)
Remarks
(e ) − (e + 1)e x + e = 0
1A
e x = 1 or e x = 0 or 1
1A
The area of the region bounded by C 1 and C 2 1
∫ 1
=
0
[
e
−
x
e
= x + e ⋅ e
1M
]
1M
− (e − e) d x
x
− x
x
1
− e + ex 0
For lower and upper limits
[
x
Accept e − ex − x − e ⋅ e
− x
= 1+1− e + e − e +1 = 3− e
1A (5)
6.
(a)
Var (2 X + 7) = 4Var ( X ) = =
(b)
8 10
4
1M
3.2
1A
For Var ( X ) =
Var ( X ) n
A 97% confidence interval for µ
8
= 50 − 2.17 × =
10
, 50 + 2.17 ×
8
1M+1A
10
(48.0591, 51.9409)
1M for 50 ± d 1A for 2.17
1A (5)
7.
(a)
(b)
P(a player is rewarded) =
1 2 ⋅
2 5 = 0 .3
+
1 1 ⋅
2 5 1A
P(both players are rewarded | one player is rewarded) = =
0.3 × 0.3 0.3 × 0.3 + 0.3 × 0.7 × 2 3 17
(c)
E(no. of players having drawn a blue ball from A ) = 60 × =
1 ×2 2 5
0.3 × 0.3
1M
OR
1A
OR 0.1765
1 − 0.7 × 0.7
1M
0.3
40
1A (5)
8.
(a)
P(a box contains more than 1 rotten eggs) = 1 − (0.96)
30
30
− C 1
(0.96) 29 (0.04)
1M+1M
1M for binomial prob 1M for correct cases
≈
0.338820302 ≈ 0.3388 (b)
(i)
st
P(the 1
1A
box containing more than 1 rotten egg is the 6 5
th
box inspected)
=
(1 − 0.338820302) (0.338820302)
1M
≈
0.0428
1A
PP-DSE-MATH-EP(M1)–5
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]
1 0
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Solution
Marks
(ii) E(no. of boxes inspected until a box containing more than 1 rotten egg is found) 1 =
0.338820302 ≈ 2.9514
1M 1A (7)
9.
(a)
P( A) = P( A ∩ B) + P ( A ∩ B ′) =
0.12 + k P( A ∩ B ′) P( A | B ′) = P( B ′) 0.6 =
1A
k
1 − P( B)
P( B ) = 1 −
5k
3 P( A ∪ B) = P( A) + P( B) − P( A ∩ B) =
(0.12 + k ) + 1 −
= 1−
(b)
5k
− 0.12
3
2k 3
1A
1M 1A
If A and B are independent, P ( A)P ( B) = P( A ∩ B) .
(0.12 + k )1 −
5k
= 0.12
1M
=0 3 k = 0.48 or 0 (rejected)
1A
0.8k −
3
5k 2
Alternative solution 1 If A and B are independent, P( A) = P( A | B ′) . 0.12 + k = 0.6 k = 0.48
1M 1A
Alternative solution 2 If A and B are independent, P ( A)P( B ′) = P( A ∩ B ′) .
5k = k 3
(0.12 + k )
1M
5k 2
− 0.8k = 0 3 k = 0.48 or 0 (rejected)
1A
Alternative solution 3 If A and B are independent, P ( A | B) = P( A | B ′) . ∴
P ( A ∩ B) P( B )
=
P( A | B ′)
0.12 = 0.6 5k 1− 3 k = 0.48
1M
1A (6)
PP-DSE-MATH-EP(M1)–6
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Remarks
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Solution 10. (a)
d x
Remarks
61t
=
dt
Marks
5
(t + 1) 2
Let u = t + 1 and hence du = dt . The amount of alloy produced by A 10 61t = dt 5
∫
=
∫
0
(t + 1) 2
11 61(u − 1)
=
∫
1
du
5
1
11
1A
u2
−3 61u 2
−1 = − 122u 2
du − 61u −5
2
+
122 3
−3
2
u
1M
11
1
1A
For primitive function
Alternative Solution 61t dt x = 5
∫
(t + 1) 2 =
∫
61(u − 1)
du
1A
−5 −3 61u 2 − 61u 2 du
1M
5
u2 =
∫
−1
= −122u
2 +
122 3
−3
u
−1
122
−3
(t + 1) 2 + C 3 The amount of alloy produced by A = −122(t + 1)
2 +
2 + C
−1 = − 122(10 + 1) 2
+
122 3
−3
(10 + 1)
1A
2 + C − − 122 +
122
≈ 45.6636
3
+ C
1A
OR =
(4)
(b)
The amount of alloy produced by B =
≈
∫
10 15 ln(t 2 + 100)
16
0
2 15 ⋅
2 16
dt
{ln(0 + 100) + ln(10 2 + 100) + 2[ln(2 2 + 100) 2
2
1M
2
+ ln(4 + 100) + ln(6 + 100) + ln(8 + 100)]} ≈ 45.6792
1A (2)
PP-DSE-MATH-EP(M1)–7
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244 3
−
3904 33 11
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Solution (c)
Marks
d d y d 15 ln(t 2 + 100) = dt dt dt 16 15t
=
1A
8(t 2 + 100)
d 2 d y 15 (t 2 + 100) − t ( 2t ) = ⋅ dt 2 dt 8 (t 2 + 100) 2 2
=
15(100 − t )
1A
2 2 8(t + 100)
2
d d y > 0 for 0 < t < 10 dt 2 dt Thus, 45.6792 is an over-estimate of the amount of alloy produced by B . 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. ∴
1A 1A (4)
a
11.
(a)
P′(t ) ln
= kte 20
P ′(t )
t
a
=
20
t
t + ln k
1A (1)
(b) t P′(t )
ln
ln
P ′(t ) t
1 22.83
2 43.43
3 61.97
4 78.60
3.13
3.08
3.03
2.98
1A
P ′(t ) t
3.3
3.2
3.1
3.0 1A 2.9
O
t
1
2
3
4
5
PP-DSE-MATH-EP(M1)–8
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Remarks
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Solution From the graph,
a
≈
Marks
2.98 − 3.13
1M
4 −1
20
Remarks
a ≈ −1 From the graph, ln k ≈ 3.18 k ≈ 24
1A
Either one
1A (5)
(c)
(i)
− t d 20 ′ P (t ) = 24te dt dt
d
− t
d
∴
dt
d dt
24e 20 1 −
=
t
1A
20
P′(t ) = 0 when t = 20
t
<
20
20
P′(t )
+ve
0
>
20 1M
-ve
Alternative Solution −t
d2
P ′(t )
dt 2
= 24e 20
=
6 5
d2
∴
dt 2
− t
e 20
− 1 t − 1 20 1 − 20 + 20
t − 2 20
1M
P ′(t ) < 0 when t = 20
Hence the rate of change of the population size is greatest when t = 20 .
(ii)
−t −t − t d te 20 = e 20 − 1 te 20 dt 20
1A
−t
−t
24te 20 =
480e 20 − 480
− t d te 20 dt
∫
− t
24te 20
− t
− t
dt = −9600e 20 −
480te 20 + C
−t
P(t )
1A
1M
− t
= C − 480te 20 − 9600e 20
1A
Since P (0) = 30 , we have 0
0
C − 480(0)e − 9600e = 30
1M
C = 9630 − t
∴
P(t )
− t
= 9630 − 480te 20 − 9600e 20
1A
− t − t 20 20 (iii) lim P(t ) = lim 9630 − 480te − 9600e t →∞ t →∞
= 9630 ∴ the population size after a very long time is estimated to be 9630 thousands.
1A
(9) PP-DSE-MATH-EP(M1)–9
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Solution 12. (a)
The estimate of the mean =
Marks
Remarks
0×6 +L+ 7× 4 100
= 3.21
1A (1)
(b)
(i)
The sample proportion of school days with less than 4 visits =
(ii)
An approximate 95% confidence interval for the proportion
= 0.57 − 1.96
0.57 × 0.43 100
, 0.57 + 1.96
57 100
0.57 × 0.43
100
1A
1M
= (0.4730 , 0.6670)
1A (3)
(c)
(i)
By (a), λ = 3.21 .
P(crowded on a day) = 1 − e −3.21 1 + 3.21 +
3.212 2!
+
3.213
3!
1M
For Poisson probability
≈ 0.399705729 ≈ 0.3997
(ii)
1A
P(crowded on alternate days | crowded on at least 2 days) =
(0.399705729) 3 (1 − 0.399705729) 2 + (1 − 0.399705729) 3 (0.399705729) 2 1 − (1 − 0.399705729) 5 − 5(1 − 0.399705729) 4 (0.399705729)
≈ 0.0869
1M for numerator 1M+1M+1M 1M for denominator 1M for binomial probability 1A (6)
PP-DSE-MATH-EP(M1)–10
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Solution
Marks
Remarks
13. Let X r minutes and X e minutes be the waiting times for a customer in the regular and express counter respectively. 6 − 6.6 (a) P( X r > 6) = P Z > 1.2
1M
= P( Z > −0.5) ≈ 0.6915
1A (2)
(b)
(i)
P(more than 10 from 12 customers with X r > 6 ) 12
= C 11 (0.6915)
11
(1 − 0.6915) + (0.6915)12
1M+1M
≈ 0.0759
1A
(ii) Let Y minutes be the average waiting time of the 12 customers
Y ~ N 6.6 ,
1.2 2
= N(6.6 , 0.12)
1A
12
6 − 6.6
0.12
P(Y > 6) = P Z >
≈ P ( Z > −1.73)
OR P ( Z > −1.732)
≈ 0.9582
1A
OR 0.9584
(5)
(c)
(i)
P( X r < k ) = 0.2119
P Z <
k − 6.6
= 0.2119
1M
1.2
k − 6.6
= −0.8 1.2 k = 5.64 P( X e > k ) = 0.0359
P Z >
5.64 − µ
5.64 − µ 0.8 µ = 4.2
(ii)
1A
0.8
= 0.0359
1M
= 1.8
1A
P( X r > µ ) = P Z >
4.2 − 6.6 1.2
≈ 0.9772
1A
P(1 customer pays at regular counter | 2 customers wait more than µ min) ≈
2(0.88)(0.9772)(0.12)(0.5) [(0.88)(0.9772) + (0.12)(0.5)]2
≈ 0.1219
1M+1M
1M for numerator 1M for denominator
1A (8)
PP-DSE-MATH-EP(M1)–11
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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)
PROVISIONAL MARKING SCHEME
This marking scheme has been prepared by the Hong Kong E xaminations 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.
© Hong Kong Examinations and Assessment Authority All Rights Reserved 2012
PP-DSE-MATH-EP(M2)–1
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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 t he 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 – ‘A’ marks – Marks without ‘M’ or ‘A’ –
awarded for applying correct methods awarded for the accuracy of the answers 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) (b)
Unless otherwise specified in the question, numerical answers not given in exact values should not be accepted. 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.
PP-DSE-MATH-EP(M2)–2
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Solution 1.
Marks
The general term of ( 2 − x) 9 is C r 9 29−r (− x ) r 9
= C r 2
9−r
Remarks
1M
(−1) r x r
1A
Alternative Solution (2 − x ) 9 = 2 9 − C 19 28 x + C 29 2 7 x 2 − C 39 2 6 x 3 +C 49 25 x 4 − C 59 2 4 x 5 + L
1M+1A
Hence the coefficient of x 5 is − C 59 2 4
1M
= −2016
1A (4)
2.
If the system of homogeneous equations has non-trivial solutions, then 1 −7 7 1
− k
2
1
3 =0
1M+1A
k
2
− k + 7 − 42 + 14k + 7 k − 3 = 0
1M
2
k − 21k + 38 = 0 k = 19 or 2
1A (4)
3.
For n = 1 , 41 + 15(1) − 1 = 18 which is divisible by 9 . ∴ the statement is true for n = 1 .
1
Assume 4 k + 15k − 1 is divisible by 9 , where k is a positive integer.
1 Withdraw the last mark if “ N is an integer” was omitted
i.e. let 4 k + 15k − 1 = 9 N , where N is an integer. ∴
4 k = 9 N − 15k + 1
4 k +1 + 15( k + 1) − 1 = 4(9 N − 15k + 1) + 15k + 15 − 1
(by induction assumption)
1
= 36 N − 45k + 18 = 9( 4 N − 5k + 2)
which is divisible by 9
1
Hence the statement is true for n = k + 1 . By the principle of mathematical induction, the statement is true for all positive integers n .
1
Follow through
(5) 4.
(a)
2 x 1+ x
2
=
=
2 tan θ 1 + tan 2 θ 2 tan θ
1M
sec 2 θ sin θ 2 =2 ⋅ cos θ cos θ = sin 2θ
(b)
(1 + x) 2 1 + x 2
=
1
1 + x 2 + 2 x 1 + x 2 2 x
= 1+
1 + x 2 Since x is real, we can let x = tan θ for some θ . ∴
(1 + x ) 2 1 + x 2
= 1 + sin 2θ
by (a)
1M
1M
PP-DSE-MATH-EP(M2)–3
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Solution
Marks
Since the maximum value of sin 2θ is 1 , the maximum value of
(1 + x) 2 1 + x 2
is 2 .
Remarks
1A (5)
5.
(a)
cos( x + 1) + cos( x − 1) = 2 cos
x + 1 + x − 1
2
cos
x + 1 − x + 1
1M
2
OR cos x cos1 − sin x sin 1 + cos x cos1 + sin x sin 1
= 2 cos1 cos x
(b)
Alternative Solution cos(0 + 1) + cos(0 − 1) = k cos 0
1M
i.e. k = 2 cos1
1A
cos1
cos 2
cos 3
cos1 + cos 3
cos 2
cos 3
cos 4
cos 5
cos 6 = cos 4 + cos 6
cos 5
cos 6
cos 7
cos 8
cos 9
cos 7 + cos 9
cos 8
cos 9
2 cos1cos 2 cos 2
1M
For column (or row) operations
1M
For using (a) or sum-toproduct formula of cosine
cos 3
= 2 cos1 cos 5
cos 5
cos 6
2 cos1cos 8
cos 8
cos 9
cos 2 cos 2
cos 3
= 2 cos 1 cos 5
cos 5
cos 6
cos 8
cos 8
cos 9
=0
by (a)
1M 1A (6)
1 6.
−
1
d 1 = lim x + h x d x x h→0 h = lim
h→0
= lim
h→0
=
1M+1A
x − x − h h( x + h) x −1
1A
( x + h) x
−1
1A
x 2
(4) 7.
(a)
f ( x ) = e x (sin x + cos x ) f ′( x ) = e x (sin x + cos x ) + e x (cos x − sin x ) x
= 2e cos x
1A
f ′′( x ) = 2e x cos x − 2e x sin x x
= 2e (cos x − sin x )
(b)
1A
f ′′( x ) − f ′( x) + f ( x ) = 0 2e x (cos x − sin x) − 2e x cos x + e x (sin x + cos x) = 0
1M
e x (cos x − sin x) = 0
sin x = cos x or e x = 0 (rejected) tan x = 1 x =
π 4
for 0 ≤ x ≤ π
1A 1A (5)
PP-DSE-MATH-EP(M2)–4
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Solution 8.
(a)
Marks
Let x = 2 sin θ . d x = 2 cos θ dθ
d x
∫
=
4− x
2
1M
OR x = 2 cos θ
2cosθ
∫ 4 4 sin θ dθ ∫ 1 dθ 2
−
=
Remarks
1A
= θ + C = sin
∫ ln x d x
(b)
−1
x
+ C
2
1A
∫ 1 x ∫ x d x
= x ln x − x d ln x = x ln x −
1M
⋅
= x ln x − x + C
1A (5)
9.
x 2 − xy − 2 y 2 − 1 = 0 --------------------------------------------- (*)
2 x − x
d y d x
d y
− y − 4 y
d x
=0
1A
For the tangents parallel to y = 2 x + 1 , ∴
d y d x
=2 .
2 x − x (2) − y − 4 y (2) = 0
1M
y = 0
1A
By (*), x 2 − 1 = 0 x = ±1 Hence the tangents are y − 0 = 2[ x − ( ±1)]
1M
i.e. y = 2 x + 2 and y = 2 x − 2
1A
1M For both
(6)
10. (a)
∫
xe
− x
∫
2
=
−1
2 (b)
2
d x = e − x e
1 2
−x
d x 2
1M
+ C
1A
2
OR
∫
e
−x
2
−1
2
d(− x 2 )
The volume of the solid
∫
2
∫
2
= 2π
x 2 2
x
1
= 2π
1
−e
−x
2
d x
1M+1A
∫
1M for V = 2π xy dx
x 3 2 − xe − x d x 2
x 4 = 2π 8
2
1
1M
15 −4 −1 + e − e π 4
1A
+
1 2
e
− x
2
=
For using (a)
(6)
PP-DSE-MATH-EP(M2)–5
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Solution 11.
(a)
α + β
A = 2
− αβ α + β
0 1
1 (α + β ) 2 − αβ = α + β
Marks
Remarks
− αβ
0
− αβ (α + β ) − αβ
α + β
1A
1 0 − αβ 0 1 0 1 (α + β ) 2 − αβ − αβ (α + β ) = α β αβ + −
(α + β ) A − αβ I = (α + β )
− αβ
Either one
i.e. A 2 = (α + β ) A − αβ I
1 (2)
(b)
( A − α I ) 2 = A 2 − 2α A + α 2 I 2
= (α + β ) A − αβ I − 2α A + α I
by (a)
1M
2
= ( β − α ) A + (α − αβ ) I = ( β − α )( A − α I )
1
Alternative Solution 2
α + β − αβ 1 0 − α ( A − α I ) = 0 1 1 0 β − αβ β − αβ = 1 − α 1 − α β 2 − αβ α 2 β − αβ 2 = 2 β − α − α αβ α + β − αβ 1 0 − α ( β − α )( A − α I ) = ( β − α ) 1 0 0 1 β − αβ = ( β − α ) 1 − α 2 2 β − αβ α β − αβ 2 = 2 β − α α αβ − 2
1A
i.e. ( A − α I ) 2 = ( β − α )( A − α I )
1
By interchanging α and β , we have ( A − β I ) 2 = (α − β )( A − β I ) .
1
Alternative Solution 1
( A − β I ) 2 = A 2 − 2 β A + β 2 I 2
= (α + β ) A − αβ I − 2 β A + β I
by (a)
2
= (α − β ) A + ( β − αβ ) I = (α − β )( A − β I )
1
PP-DSE-MATH-EP(M2)–6
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Solution
Marks
Remarks
Alternative Solution 2 2
α + β − αβ 1 0 − β ( A − β I ) = 0 1 1 0 α − αβ α − αβ = 1 − β 1 − β α 2 − αβ αβ 2 − α 2 β = 2 α − β β αβ − α + β − αβ 1 0 (α − β )( A − β I ) = (α − β ) − β 0 1 1 0 α − αβ = (α − β ) 1 − β α 2 − αβ αβ 2 − α 2 β = 2 α − β − β αβ 2
i.e. ( A − β I ) 2 = (α − β )( A − β I )
1 (3)
(c)
(i)
A = X + Y
α + β 1
− αβ
β − αβ α − αβ = s + t 0 1 − α 1 − β s β + t α − αβ ( s + t ) = s + t − − s t α β s β + t α = α + β Comparing the entries, we have s + t = 1 sα + t β = 0 Solving, s =
(ii)
1M
β α and t = β − α α − β
1A
β n α n ( A − α I ) and Y n = ( A − β I ) ”. β − α α − β β α When n = 1 , X = ( A − α I ) and Y = ( A − β I ) are true by (c)(i). β − α α − β
For both
Consider the statement “ X n =
1
β k α k k Assume X = ( A − α I ) and Y = ( A − β I ) , where k is a β − α α − β k
positive integer. + X k 1 =
=
=
Y k +1 = =
β k β ( A − α I ) ( A − α I ) β − α β − α β k +1 ( β − α ) 2
( β − α )( A − α I )
by the assumption
by (b)
1
β k +1 ( A − α I ) β − α
Either one
k
α α ( A − β I ) ( A − β I ) α − β α − β α k +1 (α − β ) 2
(α − β )( A − β I )
by the assumption
by (b)
α k +1 = ( A − β I ) α − β
1
PP-DSE-MATH-EP(M2)–7
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Solution
Marks
Hence the statement is true for n = k + 1 . By the principle of mathematical induction, the statement is true for all positive integers n .
1
Remarks
Follow through
(iii) XY = s( A − α I )t ( A − β I ) 2
= st [ A − (α + β ) A + αβ I ]
by (a)
0 0 0 0
=
YX = t ( A − β I ) s ( A − α I ) 2
= st [ A − (α + β ) A + αβ I ]
For both
by (a)
0 0 0 0
=
1
A n = ( X + Y ) n n
n
= X + Y
by the note given
n
1M
n
=
β α ( A − α I ) + ( A − β I ) β − α α − β
=
α n − β n αβ n − α n β A + I α − β α − β
by (ii) 1A (9)
12.
(a)
(i)
OM = (1 − a )i + a j
1A
ON = b(i + j + k ) ∴ MN = b(i + j + k ) − [(1 − a )i + a j ] = ( a + b − 1)i + (b − a ) j + bk
(ii)
1
AB = j − i MN ⋅ AB = 0 [( a + b − 1)i + (b − a ) j + bk ] ⋅ ( j − i ) = 0
1M
−a − b + 1 + b − a = 0
a=
1
1A
2
MN ⋅ OC = 0 [(a + b − 1)i + (b − a) j + bk ] ⋅ (i + j + k ) = 0 a + b −1 + b − a + b = 0 1 b= 3
1M
1A
Alternative Solution AB × OC = (−i + j) × (i + j + k )
1M
= i + j − 2k
MN // AB × OC ∴
a + b −1
1
=
b−a
=
1
Solving, we get a =
1 2
b
1M
−2
and b =
1 3
.
1A+1A
PP-DSE-MATH-EP(M2)–8
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Solution
Marks
Remarks
−1
1 1 i − j + k 6 6 3 The shortest distance between the lines AB and OC
(iii) MN =
= MN
=
=
− 1 6
2
− 1 + 6
2
1 + 3
2
1M
6
1A
6
(8)
(b)
(i)
AB × AC = (−i + j ) × ( j + k ) = i + j − k
(ii)
1A
z
C
Let the intersecting point of the two lines OG and MN be P . N
Since P lies on MN , let MP = λ MN .
1M
OP = OM + MP =
=
O x
A
− 1 1 1 i + j + λ i − j + k 2 2 6 3 6
1
M
1
3 − λ 6
i+
3 − λ λ j + k 6 3
G
1A
Since P lies on OG , OP // AB × AC . ∴
3 − λ
=−
6 λ = −3
P
λ
1M
3
Alternative Solution Since P lies on OG , OP // AB × AC . Let OP = t (i + j − k )
1M
1 1 i + j 2 2
∴ MP = t (i + j − k ) −
2t − 1 j − t k 2 2 Since P lies on MN , MP // MN . 2t − 1 2 = − t ∴ −1 1 =
6
2t − 1
i+
1A
1M
3
t = 1 Hence the coordinates of P are (1 , 1 , − 1) .
1A (5)
PP-DSE-MATH-EP(M2)–9
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y
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Solution 13.
(a)
Marks
Let u = x − p . ∴
1M
du = d x
When x = 0 , u = − p ; when x = 2 p , u = p .
∫
∴
2 p
∫
f ( x − p ) d x = 0
p
f (u ) du
=0
∫
∴
2 p 0
1A
− p
since f is an odd function
1
[f ( x − p) + q] d x = 0 + [qx]20 p = 2 pq
1A (4)
π 6 π 3 − tan x − 6
3+
3 + tan x − (b)
1 3
tan x − 1 3
1+
=
tan x
3−
1 3
1+
1M
1 3
tan x −
tan x
3 + 3 tan x + 3 tan x − 1
=
3 + 3 tan x − 3 tan x + 1 =
1 + 3 tan x
1
2
(2)
(c)
∫
π 3
ln(1 + 3 tan x) d x =
0
∫
=
Consider f ( x ) = ln
f (− x) = ln
= ln
3 ln 0 π 3 ln 0 π
∫
3 + tan x
π 6 ⋅ 2 d x π 3 − tan x − 6
3 + tan x −
by (b)
π 6 + ln 2 d x π 3 − tan x − 6 3 + tan x −
1M
.
3 − tan x
3 + tan(− x) 3 − tan(− x) 3 − tan x 3 + tan x
3 + tan x = ln 3 − tan x = − ln
−1
1M
3 + tan x 3 − tan x
= −f ( x) ∴
∴
f ( x) is an odd function
∫
π 3
ln(1 + 3 tan x ) d x =
0
1A
∫
=
2×
π 6
0
π f x − 6 + ln 2 d x
π ln 2 3
by (a)
1A (4)
PP-DSE-MATH-EP(M2)–10
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Remarks
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Solution 14. (a)
Marks
Remarks
The volume of the solid of revolution = π
∫
h
( 25 − y 2 ) d y
1M
0 h
3 y = π 25 y − 3 0
3 h
= 25h −
π
1
3
(2)
(b)
(i)
h 3
By (a), V = 25h −
dV dt
= 25
dh dt
−h
2
π for 0 ≤ h ≤ 4
3
dh
π
1A
dt
2
When h = 3 , 8 = ( 25 − 3 )π
dh dt
=
dh dt
1 2π
i.e. the rate of increase of the depth of coffee is (ii)
1 2π
cm s
−1
.
1A
Let x , l , r and h be the lengths as shown in the figure. x 2 + 4 2 = 25 x = 3
6 1A r
By similar triangles,
x
=
l
6
8
1M
8+l
24 + 3l = 6l l =8
x
By similar triangles, r = ∴
r h−4+l
=
6
4
8+l
5
3(h + 4) 2 ( 4) 3 π 3(h + 4) π (h + 4) − (3) 2 (8) V = 25(4) − π + 3 3 8 3
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 y − 0 8 − 0 =
x − 3 6 − 3 3 x = ( y + 8) 8
236π
+
3 164π 3
1M
y
(6, 8)
1M 1A
∴ V = 25( 4) −
i.e. V =
l
1A
8
=
h
+
(4) 3
∫
π + π 3
9π ( y + 8) 3
64 3π 64
3
h−4
0
9 64
( y + 8) 2 d y
1M O
h− 4
0
(h + 4) 3
1
PP-DSE-MATH-EP(M2)–11
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3
x
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Solution (iii) After 15 seconds,
3π 64
164π 3
+
3π 64
Marks
(12 + 4) 3 − 2 × 15 =
164π 3
+
3π 64
( h + 4) 3
1M
(h + 4) 3 = 192π − 30 1
64π − 10 3 h + 4 = 4
π
1A
h ≈ 11.73 > 4 164π 3π + ( h + 4) 3 V = 3 64 dV 9π dh (h + 4) 2 = dt 64 dt
1A 2
1 9π 64π − 10 3 dh After 15 seconds, − 2 = 4 64 π dt
dh dt
=
−8 1
2
9π 3 (64π − 10) 3 ≈ −0.0183
i.e. the rate of decrease of the depth of coffee is 0.0183 cm s −1 .
1A (11)
PP-DSE-MATH-EP(M2)–12
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Remarks
