2012 m1 marking eng

FOR TEACHERS' USE ONLY

m m ~ nt & ~ ~ foj HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY

2012 ~mmcp~x:~~nt HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2012

-~ m{lfJffB~ ~ - ( filHJl~W~~t >

MATHEMATICS Extended Part Module 1 (Calculus and Statistics)

MARKING SCHEME

*n@~~-*~~&~~~--~~*M~~~OO@ft8~Zm~@ ft~~~M~I~~~~*n~-~-~~ffft9~ffi~*M~$8M *~~~ &~mmw~trff~MRT~~M~~*~@~-~*~~$~~~00~ *~JJ:t~3tf!f oo @ft 1 *~silim!&Hi>lte*~ izg~~~OJn~:w~@~wiYt~.t~u~~~ .i:; ~m~~~~~ ~~~~mmMffi~~~ ~Rm~ftW~~ ~~~~ ~it~~ IID m n~ n w ji m tY: :i:75 z ' ~ JJ:t * FoU i1i ~jlj ~!Ml@ ffe41~!z8rlHm :;IJ !% f . ~ ~ J:: ~ ~fllJ 0

This marking scheme has been prepared by the Hong Kong Examinations and Assessment Authority for markers ' reference. The Authority has no objection to markers sharing it, after the completion of marking, wi th colleagues who are teaching the subject. However, under no circumstances should it be given to students because they are likely to regard it as a set of model answers. Markers/teachers should therefore finn ly resist students' requests for access to this document. Our examinations emphasise the testing of understanding, the practical application o f knowledge and the use of processing skills. Hence the use of model answers, or anything else which encourages rote memorisation, should be considered outmoded and pedagogically unsound. The Authority is counting on the co-operation of markers/teachers in this regard .

Hong Kong Examinations and Assessment Authority A ll Rights Reserved 2012

2012-DSE-MATH-EP(M I )-1

RPI~ ftili ~IHI FOR TEACHERS' USE ONLY


General Marking Instructions

I. It is very important that all markers should adhere as closely as possible to the marking scheme. In many cases, however, candidates will have obtained a correct answer by an alternative method not specified in the marking scheme. In general, a con ect alternative solution merits all the marks allocated to that part, unless a particular method has been specified in the question. Markers should be patient in marking alternative solutions not specified in the marking scheme.

2. For the conveni ence of markers, the marking scheme was written as detai led as possible. However, it is likely that candidates would not present their solution in the same explicit manner, e.g. some steps would either be omitted or stated implicitly. In such cases, markers should exercise their discretion in marking candidates' work. In general, marks for a certain step should be awarded if candidates' solution indicated that the relevant concept I technique had been used.

3. In marking candidates ' work, the benefi t of doubt should be given in the candidates' favour.

4. Unless the form of the answer is specified in the question, alternative simplified fonns of answers different from those in the marking scheme should be accepted if they are co1Tect.

5. Unless otherwise specified in the question, use of notations different from those in the marking scheme should not be penalised.

6. 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. Markers should follow through candidates' work in awarding 'M ' marks.) However, 'A' marks for the corresponding answers should NOT be awarded, unless otherwise specifi ed.

7. In the marking scheme, steps which can be skipped are enclosed by ;_~!~~c:J-~

FOR TEACHERS' USE ONLY Solution

I. (a) (I+ 3x)" = 1 + C{' (3x) + C~' (3x)2 + ... 911(11 - 1) '

= I + 3nx + x- + .. 2

(b) ,-''(I+ 3x)" = [1 + (- 2x) + H2~J' +][I + 3nx + 9"(';- I) x2 + .. J ? [ 911(11- I) ? ]

= (l- 2x+2.c + .. ) I +3nx+ 2

x- + ...

.. I 911(l1-I) + (- 2)(311) + 2 I = 62 2

911 2 -2 111 - 120 = 0 :--- ~g-- ---- - ----:

11 = 5 i or J (rejected) i ~------- --- ---- - --- J

2. Let u = 4t + I . du= 4dt

... j,

When t = 0 , 11 = I ; when t = 2 , 11 = 9 . The change in the value of the flat

f2 I = Jo J41 + I dt = f9 _1_, 11- I ~

J1 ..r;; 4 4 = _!__ f9(j;; -_I )du

16 J1 ..r;;

1[2 3 1]9

= 16 3 11 2 - 2112 I 5

=

6 .i

Hence the percentage change = - 6 x I 00% 3

(a)

(b)

kl -

P = ae 40 - 5 k ln(P +5) = -t +Ina 40

I 2 p 2.36

ln(P + 5) 2.00

7 =27 - %

9

4 2.81 2.06

From the graph on the next page, In a ~ 1. 96 a~7 k 2.2 1- 1.96 -~---40 10-0 k ~I

2012-DSE-MATH-EP(M 1)-3

6 8 3.23 3.55 2.11 2.15

10 4.01 2.20

Marks

IA

IA

JM

IA

(4)

IM

IM

IA

IA

IA (5)

IA

IM

Remarks

(-2x) 2 For l+(-2x)+ - - +"

2!

I 2 - -[

3 '] For - -11 2 - 2u2 16 3

OR 27.7778%

IM ~,--__,,

Either one I

IA For both a and k


RPl~Bili~tm FOR TEACHERS' USE ONLY Solution Marks

ln(P + 5)

I I 2.25 2.20 / / . 2. 15 v I / ~-v 2.10 i

--b / IA v I I 2.05 >-- --2.00

v Vi 1.95 ~ I I ~ $ ~ 7 I ~ I ' g I '

(5)

4. (a) y=V3x - 1 x-2 I I ln y = - ln(3x- 1)-- ln(x-2) 3 3

IA

I dy I I .. - - = - - - IA

y dx 3x - I 3(x - 2)

(b) By(a), dy =[-I__ I J v3x- I IA d\ 3x- I 3(x - 2) x - 2

:;' = ![3x1- I - 3(x~2)]v~yx_-; + [ 3x1- J - 3(x~2)]![ v~x_-;)

[ lf [ ]2f -3 I 3x- I I I 3x - I = + - -+ - -- -- by(a) (3x- J) 2 3(x- 2)2 x-2 3x- I 3(x-2) x - 2 2 { [ ]2)~ d y -3 I I I 33-1 When x = 3 - = + + - -- 3--

, dx 2 (33 - 1)2 3(3-2)2 33 - 1 3(3 - 2) 3-2

Alternative Solution dy -5

When x = 3 , y = 2 and so - = - . dx 12

I d2 y I dy dy -3 I By(a) --- - - -= +

' y dx2 y 2 dY dx (3x- 1)2 3(x - 2)2

I d2y I - 5 -5 - 3 I When x=3 ----- - = + , 2 d 2 22 12 12 (33-1)2 3(3-2)2

20 12-DSE-MATH-EP(M I )--4

JM

JM

IA

JM

IM

IA

(6)

Remarks

For using (a)

For both y and dy d Y

For chain rule

OR 0.6597


.R ~R ~ Bili ~Im FOR TEACHERS' USE ONLY Solution Marks Remarks

5. (a) dy 2 r -=e. d\' y=_!_e2x +C

2 IA

)~I\ I

Since A(O, I) lies on S, we have I= _!_ e2 + C . 2

JM

i.e. I C=- x= I 2

v Hence the equation of S is I ?, I IA y = -e- + -2 2

I ?, I (b) At A(O I) dy = e 2 =I Y"2'i i1(0, I) , ' dx

Hence the equation of l is y - l = l(x-0). IM y- x +I __ x i.e. y=x+I IA / 0

/

(c) The area of the region bounded by S, l and the line x =I

= J~[(+e2x ++)-(x+ l)]d\' JM IM for A= f~(y1 - y 2 )dx =[ e2x - ~x2 -+xl

e2

- 5 4

6. (a) Let X be the weight of a student. The sample mean X - N( 67, 1;; l P(X>70) = P[Z >

7067 l = P(Z > 1.2) "" 0.1151

9 (b) The sample proportion is - = 0.25 . 36

An approximate 95% confidence interval for the proportion

,,,,(0.25 _ 1.96 x /0.25x0.75 ,0.25 +1.%x 0.25x0.75l 36 36

"" (0.1085 , 0.39 I 5)

- .{ 7. (a) .:.___ = 0.1653

O! A. = - In 0.1653

""1.8

e- 1.8 e-1.8 (1 8) e- 1.8 (1 8)2 (b) P(no. of goals in a match< 3) = --+ + O! I! 2!

"" 0.7306

2012-DSE-MATH-EP(M I )-5

IA

(7)

JM

IA

IA

IM

IA

(5)

IA

IM

IA

OR 0.5973


(c)

8. (a)

(b)


The number of goals scored in two matches by the team - Po (3.6) . .. P(no. of goals in two matches< 3)

e-3.6 e-3.6(3.6) e-3.6(3.6)2 =--+ +

O! l! 2!

Alternative Solution P(no. of goals in two matches < 3) = P(O , O)+ P(O, I)+ P(I, O)+ P(l, I) + P(O, 2)+ P(2 ,0)

= ( ~! r + 2( ,~,I .-,~18) H ,-,~18) r + { ~, 1 .-;: 8)' i ~ 0.3027

P(X =I)+ P(X =3)+ .. + P(X = 13) =I 0.1 +a+0.25+0.15+b+0.05 = I a+ b = 0.45 --------------------------------------------------------------- (I) E(X) = 5.5 I xO.l + 3a+4x 0.25 + 6x O. l 5+9b+ 13x0.05 = 5.5 a+ 3b = 0.95 -------------------------------------------------------------- (2) Solving (I) and (2), we get a= 0.2 and b = 0.25 .

(i) P(F nG) = 0.25+0. 15 = 0.4

(ii) P(F) x P(G) = (0.25 + 0.15 + 0.25 + 0.05)(0.1+0.2 + 0.25 + 0.15) = 0.49 :t P(F nG)

Alternative Solution I P(FjG)= P(F n G)

P(G) 0.4

= - - - - - --

0.1+0.2 +0.25 + 0.15 ~ 0.571428571

P(F) = 0.25 + 0.15 + 0.25 + 0 .05 = 0.7 :tP(F jG)

Alternative Solution 2 P(GjF)= P(FnG)

P(F) 0.4

=-------

0.25 + 0 .15 + 0.25 + 0.05 ~ o.571428571

P(G) = 0.1+0.2+0.25 + 0.15 =0.7 :t P(G IF)

Hence, F and G are not independent.

Marks Remarks

JM

IM

IA

(5)

IM

IM

IA For both

IA

IA

IA

IA

(6) 2012-DSE-MATH-EP(M 1 )-6


FOR TEACHERS' USE ONLY Solution Marks Remarks

9. (a) Let X be the score of a student who had revised. P(X ~ 43) = p( z ~ 431~59)

= P(Z ~- 1. 6) ~ 0.9452 IA

Let Y be the score of a student who had not revised. I P(Y ~ 43) = P( Z ~ 43 ~;5 2 ) Either one

I = P(Z ~ 0.65) ~ 0.2578 :. P(pass the test) ~ 0.73 x 0.9452 + 0.27 x 0 .2578 JM

= 0.759602 IA OR 0.7596

(b) P(a student had not revised for the test I he passed the test) 0.27 x 0.2578 IM =

0.759602 ~ 0.091634829 ~ 0.091 6 IA

(c) P( 4 students had not revised for the test among I 0 passed students) ~ C~o (0.09 1634829)4 (I - 0.091634829)6 JM ~ 0.0083 IA

(7)

IO. (a) (i) 4 I ::!.. I = Ii fie 2 dt 1 4 - 1 [ 1 ::! I _., ( l ::!c' 1 -2 1 -25

=1 -6- Jle2 + J4e2 +2 Jl5e 2 + ..fi.e2 +Ne 2

1 - 3 1 -3 5) l + .fje2 +Ne 2 IM+IA

~ 0.692913377 ~ 0.6929 IA

( - I - ) -3 _, - I - d - - - I - - - -I -(ii) - t 2e2 =- t 2e2 + 1 2 -e2 IM+IA dt 2 2

I - - --r -J =~e2 1 2+1 2

d- - - - I - -3 - -I - - I - - -T' - J [ - ( _, -3 J l 3 - J l d,Tt2e2 =Te2 2 , 2+2 / 2 +Te2 1 2+ 1 2 JM =e2 312 + 212 +12

_, ( _, -3 _, l >0 for I :;; t 5 4 .

Hence the estimation in (i) is an over-est imate. I

(7)

2012-DSE-MATH-EP(M I )-7 FOR TEACHERS' USE ONLY


(b) Let I = x2 JM dt = 2xdx }1A When I= I , x =I ; when 1=4' X= 2 .

4 l :!.. I = Ii Jf e 2 dt

, 2 l -x-

= i -e 2 2xd\" I X

, 2 -x-

=21e2c:Jx I

(3)

, ' -x-

(c) 2 re 2 dx < 0.692913377 IM ,

') - x-2& r &e2 dx


(c) RI - RI = lni!_ /=40 /=4 1 50

? 2 61 - ln[(40 -35)- +b]+C-{- ln[(41 - 35) +b]+C} = ln-

50 61

- ln(25 +b) + ln(36 + b) = ln-50

In 36 +b = In~ 25 + b 50

b = 25 .. R = -ln[(t -35)2 + 25]+ C Ri,=35 = 6

- ln[(35 - 35) 2 + 25] + C = 6 C= 6 + 1n25 1.e. R = - ln[(t - 35)2 + 25] + 6 +In 25

(d) dR = 2(30- t) + 10 dt (I - 35)2 + 25

70-21 = -----

12- 701+1250 d 2 R (1 2 - 701+ 1250)(-2) - (70- 2t)(2t - 70) dt 2 = (1 2 - 701+1 250)2

212 - 1401 +2400 =

(1 2 -701+ 1250)2

d2R When the rate of change of the radiation intensity attains its greatest value, - = 0 .

d/2

212 - 140/ + 2400 = 0 I = 30 ::~i:~f (~~l~~!~~)J

I 0 ~ I < 30 I = 30 30 < I~ 35

d/2 +ve 0 -ve

Hence, the rate of change of the radiation intensity would attain its greatest value

Marks

IM

IA

IM

IA

(4)

lM+ lA

JM

when I = 30 . I A

(4)

20 I 2-DSE-MATH-EP(M I )-9 FOR TEACHERS' USE ONLY

Remarks


12. (a) (i) The sample mean = 56+ + 50 16

= 5 1.5625 IA A 90% confidence interval

"' ( 51.5625 - 1.645 x k , 5 1.5625 + 1.645 x k) IM = (47.86125, 55.26375) IA OR (47.86 13, 55.2638)

(ii) Let n be the sample size.

.. 2(1.645},; ) 24.354225 IA Hence, the least samp le size is 25 . IA

(7)

(b) (i) P(a tourist waits for more than 65 minutes) = r( 2 > 65 -

951.5) JM

= P(Z > 1.5) "'0.0668 IA P(Jess than 2 coupons are sent to the first I 0 tourists interviewed) "' (I - 0.0668) 10 + c:0 (I - 0.0668)9 (0.0668) JM "'0.8594 IA

(ii) P(the 5th coupon is sent to the 20th tourist interviewed) "'c~9 o-0.0668) 15 (0.0668)4 . 0.0668 JM "'0.0018 IA

(6)

2012-DS E-MATH-EP(M I)-I 0

.R PN ~ ftili ~Im FOR TEACHERS' USE ONLY


13. (a) P(at least 2 drunk drivers are prosecuted) = 1-e-2.3 -e-2.3(2.3) IA ""0.6691458 15 ""0.669 1 lA

(2)

(b) P( ;


SUPPLEMENTARY NOTES ON MARKING Cl) I. Marking Scheme

x = ..f2 I IA Sample 1 Sample 2

IA x = 1.414 x OA

= 1.414

2. Marking Scheme (a) JM x = ..f2 IA

(b) Substitute: x = ..f2 IM

y=4 IA

Sample 3 Sample 4

(a) ; ../ JM (a) ; ./ JM x=l.414 x OA

(b) Substitute: x = 1.414 ../ IM :

x = J3 x OA

(b) Substitute: x = J3 ../ JM y=4 ../ IA

y = 4 x OA

3. Marking Scheme (a) IM

x=2 IA (b) : IM

y=4 IA

Sample 5 Sample 6

(a) ..... . x OM ../ JM ...... x OA x=2 ../ IA

(b) ; ../ IM ../ IM x=2 x OA y = 4 ../ IA

: ../ JM

y=4 ../ IA

4. Final Answer

Like terms were not collected Answers were not simplified

xcos(x 2 + I) h

12x - 3y+2 =!Ox+ I dy - =

dx x2 IA x = 2111r + 1T - !!_ :>- OA X= ~, x =J8 ~(pp- I if 2 4 applicable) y - 6 sin (J = (cos B)(x - 2 tan B) 8

20x - 30y = 0 I J

20 I 2-DSE-MATH- EP


2012 m1 marking eng

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