2012 maths practice paper(m1 & m2) marking scheme

只限教師參閱 FOR TEACHERS’ USE ONLY

© 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


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.




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.


Solution Marks Remarks



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





(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





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





(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


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


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)





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





(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

′





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


−+

−

−=′

−

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





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





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


© 香港考試及評核局保留版權

Hong Kong Examinations and Assessment Authority

All Rights Reserved 2012



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.




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.





1. The general term of 9)2( x− is rrr xC )(299 −− 1M

rrrr xC )1(299 −= − 1A


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





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=


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 ++





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)





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


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





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





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


)()( kjiji ++×+−=×OCAB 1M

kji 2−+=

( )OCABMN ×//

211

1

−=

−=

−+∴

babba 1M

Solving, we get 2

1=a and

3

1=b . 1A+1A

For both





(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


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





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





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


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





(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)

2012 maths practice paper(m1 & m2) marking scheme

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