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ADDITIONAL MATHEMATICS / PKertas 1

September 2011Dua jam

NAMA

TINGKATAN

ADDITIONAL MATHEMATICS

TINGKATAN 5KERTAS 1

2 JAM

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3472/2

ADDITIONAL

MATHEMATICS

Paper 2Aug/Sept 201121/2 hours


Tingkatan 5

Kertas 2

Dua jam tiga puluh minit

3472/2

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Kertas soalan ini adalah dalam dwibahasa.

2. Soalan dalam bahasa Inggeris mendahului soalan yang sepadan dalam

Bahasa Melayu.

3. Calon dikehendafd membaca maklumat di halaman belakang kertas soalan

ini.

Kertas soalan ini mengandungi 20 halaman bercetak.

©2011 Hak Cipta MPSM Pahang [Lihat halaman sebelahhttp://edu.joshuatly.com/

SULIT 3472/1(PP)

[Lihat halaman sebelah] 3472/1(PP) © 2011 Hak Cipta MPSM Cawangan Pahang SULIT

3472/1(PP) Additional Mathematics Kertas 1 Sept 2011 Peraturan Pemarkahan

MAJLIS PENGETUA SEKOLAH MENENGAH MALAYSIA CAWANGAN NEGERI PAHANG

PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2011


Paper 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 5 halaman bercetak


SULIT 2 3472/1(PP)

3472/1(PP) © 2011 Hak Cipta MPSM Cawangan Pahang SULIT

ADDITIONAL MATHEMATICS PAPER 1 MARKING SCHEME Question Solution and Marking Criteria Marks 1 (a) 2 1 3

(b) 2 1 (c) {4, 8} 1

2 2=k ,

434=m (both)

2=k or 434=m

( ) mkxxfg +-= 2141)(

3

B2

B1

3

3 2

1)( -=

xxg or 21

21)( -= xxg

xfxg 1)( -= or 21

=m

2

B1

2

4 2=p ( )[ ] ( )( ) 03144 2 =+-+- pp or equivalent

( ) ( ) 0341 2 =++-+ xpxp in general form or 1+= pa , 4--= pb and 3=c (all)

3 B2

B1

3

5 3,1 ³-£ xx ( )( ) 013 ³+- xx or 21,21 -£-³- xx (both)

0322 ³-- xx or 21 ³-x or 21 -£-x (Reject 21 ±³-x )

3 B2 B1

3

6 1.677

7lg2lg42lg2

-=x or equivalent

7lg2lg)24( xx -- or equivalent (log both sides of the equation)

3 B2

B1

3

7 (a) 1=x 1 3 (b) (2, -3) 1 (c) No (real) roots 1

8 2

22 qp -+ or equivalent

4log5log3log24log

2

222 -+ or equivalent

4log2.7log

2

2 or 5log36log 44 - or 10log72log 44 - or equivalent

3

B2

B1

3


SULIT 3 3472/1(PP)


9 (a) 3 )102()17()9()102( +--=+-+ pppp

2 B1

4

(b) 160

[ ] [ ])4(2)12(223)4(7)12(2

28

+-+ or 24+28+32+36+40

or [ ])4(4)24(225

+

*follow through the values of a and d

2 B1

10 (a) 28 1 3 (b)

271341

311

31128*

4

-

úúû

ù

êêë

é÷øö

çèæ-

or 333

28*33

28*328*28*

´´+

´++

*follow through the value of a from part (a)

2

B1

11 (a) 1

p-+

=-+

222

4513 or

p-+-

=--

421

2523 or

242

25 +

=+p

Or 542 -=-p or 524 -=-p

2

B1

4

(b) 11

[ ]2)2(1)1()4(3)5(2)2(1)2(4)1(5)3(221

-+-++-+-+-+

Or equivalent

2 B1

12 P (0, 2) R (10,7)

41)0(4)(12

++

=x or

41)2(4)(13

++

=y

Or )02((42 -+=x , )23(43 -+=y (both)

1 2

B1

3

13 2 02 =-h

2 B1

2

14 (a) ji 99 +-

)827(23 jiji ++--

2 B1

3

(b) ji -10 1 15 12=h , 3-=k (both)

12=h or 3-=k

421 2 -= x

xy

3 B2 B1

3


SULIT 4 3472/1(PP)


16 (a) p- 1 3 (b) 212 pp --

pq2-

2 B1

17 (a) (i) (ii)

3 4

1 1

4

(b) °10.24 , °91.155 , °10.204 , °91.335 °19.48

2 B1

18 (a) 1.031 rad 1 4 (b) 21.85 cm

Perimeter = )031.1(6661010 22 +-++ Arc AB = 6(*1.031) *follow through the values from part (a)

3 B2

B1

19 (a) -2 0633 2 =-+ xx

2 B1

4

(b) 10=y

)2(*6)2(*23)2(* 23 ---+-

*follow through the values from part (a)

2 B1

20 3

76

15)2(915)4(9=

--

--

- kk or equivalent

)3(1)53( 1

-- -x

3 B2

B1

3

21 5 12103715 ´=+´ x or equivalent

2 B1

2

22 (a) 240 5! 2! or 5

52

2 PP ´ 2

B1 4

(b) 216 1

34

41

3 PPP ´´ or 3!43 ´´ 2

B1 23 (a)

94

1 3

(b) 41

121

321)1( --==XP or equivalent

2

B1

24 0.4 13.01.02.0 =+++ a

2 B1

4

0.8 )2(1 =- XP or 0.1 + 0.3 + *0.4

*follow through the values from part (a)

2 B1


SULIT 5 3472/1(PP)


25 30

6241 m-

=-

Z = -1

3 B2

B1

3

END OF MARKING SCHEME PERATURAN PEMARKAHAN TAMAT


SULIT 3472/2(PP)


3472/2(PP) Additional Mathematics Kertas 2 Sept 2011 Peraturan Pemarkahan

MAJLIS PENGETUA SEKOLAH MENENGAH MALAYSIA CAWANGAN NEGERI PAHANG

PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2011


Kertas 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 12 halaman bercetak


SULIT 2 3472/2(PP)


ADDITIONAL MATHEMATICS PAPER 2 MARKING SCHEME

SECTION A [40 MARKS] Question Marking Criteria Marks 1 82 =- yx ___①

8 12 8x y

- = ___② (any two equations : one linear, one non-linear)

82 += yx or 2

8-=

xy (x in terms of y or y in terms of x)

Substitute x or y into non-linear equation …

81282

8=-

+ yy or 8

28

128=

÷øö

çèæ -

=xx

0652 =++ yy or 0862 =+- xx 0)3)(2( =++ yy or 0)4)(2( =-- xx [factorisation or formula]

2-=y , 3-=y ; 2=x , 4=x 4=x , 2=x ; 3-=y , 2-=y

[ If answers given without factorisation or using formula: K1 N1 N1 OW – 1]

P1

P1

K1

K1 N1 N1

6

2 (a) Listing : 6x, 6x + 6, 6x + 12, … or a = 6x 6=d

[ ] 750)6(9)6(22

1010 =+= xS (Equating 750=nS )

8=x

P1 N1

K1 N1

7

(b) Area = 2)(2 x , 2)1(2 +x , 2)2(2 +x , ...=> 2(8)2, 2(9)2, 2(10)2, ... AP: x , 1+x , 2+x , ... => 8, 9, 10

)110(8*10 -+=T = 17 Area of largest rectangle = 2[*8 + (10 – 1)1]2 = 578 cm2 OR Perimeter : 6(8), 6(8) + 6, 6(8) + 12, …

yT 61026)110()8(610 ==-+= 17=y [any letter except x] ___K1

Area of largest rectangle = yy 2´ )17(217 ´ ___K1

578 cm2 ___N1

K1 K1 N1


SULIT 3 3472/2(PP)


3 § Shape of graph of y = sin x , 0 ≤ x ≤ 2π [must begin from origin, O(0, 0)]

§ Shape of graph of y = sin 2x , 0 ≤ x ≤ 2π [exactly 2 complete cycles]

§ Amplitude of graph = 3 § Reflection of graph y = 3 sin 2x at x – axis, 0 ≤ x ≤ 2π

[all correct]

xyp

= [Equation of straight line]

Draw any straight line with positive gradient passing through (0, 0) Number of solutions = 5

N1

N1

N1 N1

P1

K1 N1

7

4 (a) [ ])2(346 --=k = 12

P1 6

(b) )3()34)(1(12* 2 ---= -x

dxdy

[ ] )3()2(34)1(12* 2 ---= -

dxdy [substituting x = 2 in his

dxdy ]

= 9

K1

K1

N1

(c) y – (–6) = 9(x – 2) or equivalent [finding equation of tangent using his dy/dx ]

249 =- xy

K1

N1

x

y

O 2π π ½ π 3/2 π

+3

-3

2

y = - 3 sin 2x


SULIT 4 3472/2(PP)


5 (a) Marks (Markah)

0-19 20-39 40-59 60-79 80-99

Number of students (Bilangan Murid)

3 10 14 8 5

N1

6

(b) Percentage of students scoring A+ = 100

405*

´

= 12.5%

K1

N1 (c) L = 39.5

)20(14

13)40(21

5.39-

+=m

= 49.5

P1

K1 N1

6 (a) 𝑃𝑅��⃗ = −6𝑥 + 4𝑦

𝑃𝑆��⃗ =34 𝑃𝑅��⃗

yx 329

+=

N1

K1

N1

8

(b) 𝑂𝑆��⃗ = 𝑂𝑃��⃗ + 𝑃𝑆��⃗ 𝑜𝑟 𝑂𝑅��⃗ + 𝑅𝑆��⃗

yx 323

+=

𝑃𝑄��⃗ = 2𝑥 + 4𝑦

𝑃𝑄��⃗ = ÷øö

çèæ + yx 4

23

34 And PQ is parallel to OS

OR

𝑂𝑆��⃗ = ( )yx 4243

+ and OS is parallel to PQ

N1

N1

N1

(c) OP = 12, OR = 20 and QR = 16 [Using 2=x , 5=y to find the length] Area of OPQR = 280 units2

K1

N1


SULIT 5 3472/2(PP)


SECTION B [40 MARKS]

Question Marking Criteria Marks 7 (a) 2x 1.00 4.00 9.00 12.25 16.00 30.25

xy 4.50 8.90 16.20 21.53 27.00 48.40 Graph of xy against x2 : Please refer to Appendix 1 (End of Mark Scheme) Correct axes with uniform scale and a point All points correctly plotted Line of best fit

N1 N1

K1 N1 N1

10

(b) (i) hkhxxy += 22

From Graph, gradient, 50.12 »= hm 75.0»h [0.70 – 0.80]

P1

N1

(ii) xy -intercept, c = 3.0

0.375.0

»k

25.2»k [accept 2.1 – 2.4]

N1

(iii) 30=xy From graph, x2 = 18

24.4»x

N1

8 (a) 10)0)4(0()000(

21

=+-+-++ OB or 10421

=´OB

OB = 5 B (5, 9)

K1

N1

10

(b) (i) Let C be (h, k) From area ratio, AO : OC = 10 : 20

021

)3(2)(1=

+-+h or 0

21)4(2)(1

=+

-+k

OR Equation of OC

xy34

= ___K1

K1

K1

Equating 2h with its gradient OR

Equating hk with its xy-intercept

OR Construction of the line xy = 30 on the graph

K1


SULIT 6 3472/2(PP)


20)000()05*0(21

=++-++ k __K1

OR Area of ΔOBC

20)5(*21

=´ k ___K1

hk34

= ___K1

C (6, 8)

N1 (ii) PC = 2PA

2222 )]4([)]3([2)8*()6*( --+--=-+- yxyx [ ]2222 )4()3(4)8()6( +++=-+- yxyx 0161222 =+++ yxyx

At x-axis, y = 0

0)12(0122

=+=+

xxxx

0=x or 12-=x (0, 0), (-12, 0)

K1 N1

N1

K1

N1 9 (a) °= 37.346.0 rad

106.0tan QRrad = or

1037.34tan QR

=°

84.6»QR cm

K1 N1

10

(b) (i) °=Ð 55.63or rad 971.0RQO OR °=Ð 111.26or rad 942.1PQR

OR)90(34.37or rad )

26.0(

21

major °+°+=ÐpPQR

OR °=Ð 248.74or rad 342.4majorPQR

Arc PSR = 342.484.6 ´ = 29.70 cm

K1

N1 (ii)

Area of OPQR = 284.61021

´´´

Area of sector OPR = 2.11021 2 ´´ or 210

36074.68

´´°° p

Area of shaded region = 68.4 – 60 or 68.4 – 59.99 = 8.4 cm2 or 8.41 cm2

K1

K1

K1 N1


SULIT 7 3472/2(PP)


10 (a) (i) 75.0=p , 25.0=q

433

7 )25.0()75.0()3( CXP == or 43

37

41

43

÷øö

çèæ

÷øö

çèæC

= 0.05768 or 16384

945

K1

N1

10

(ii) )7()6(1)5( =+=-=£ XPXPXP 716

67 )75.0()25.0()75.0(1 +-= C

= 0.5551

K1 K1 N1

(b) (i) 16

162175 -=z

= 0.75

N1 (ii)

6.016163

-=-x

cm 4.153=x

N1 (iii) )75.01()175147( ££-=££ ZPXP

)75.0()1(1 ³-³-= ZPZP = 1 – 0.1587 – 0.2266 = 0.6147

K1 K1 N1

11 (a) (i) At P, 0123)( 21 =-=- xxxf 0)4(3 =-xx

4 ,0 == xx P(4, 0)

K1

N1

10

(ii) ( )cxx

xdxxxf

+-=

-= ò23

2

6

123)(

32=c cxxxf +-= 23 6)(

K1 N1 N1

(b) R(6, 32) Area of rectangle = 326´ OR )326*(32* 23 +-- xx = 192 ___① = 326 xx - Area under curve Area of shaded region from x = 0 to x = 6

( )ò +-=6

0

23 326* dxxx ( )ò -=6

0

326 dxxx

6

0

34

3224 ú

û

ùêë

é+-= xxx

6

0

43

436

úû

ùêë

é-=

xx

= 84 ___② 04)6(

3)6(6 43

-úû

ùêë

é-=

P1 K1

K1

K1


SULIT 8 3472/2(PP)


Area of shaded region = ① – ② = 192 – 84 = 108 cm2

N1

SECTION C [20 MARKS]

Question Marking Criteria Marks

12 (a) ttts 156 23 --=

15123 2 --== ttdtdsv

-1ms 15 ,0 -== vt

K1

N1

10

(b) For minimum velocity, 0126 =-= t

dtdv

2=t 15)2(12)2(3 2

min --=v = - 27 ms-1

K1

N1

N1 (c) At P, v = 0: 0)5)(1(3 =-+ tt

5=t 12)5(6 -=a

= 18 ms2

K1 N1

N1

(d) 0<v 0)5)(1(3 <-+ tt

50 << t

K1 N1


SULIT 9 3472/2(PP)


13 (a) (i) AB = 20 cm P1 10 (ii) 2 2 2 020 8.9 2(20)(8.9)cos 54AC = + -

AC = 16.43 cm K1 N1

(b) ACBÐ-+= cos)43.16)(*9.8(243.16*9.820 222 1739.0cos -=ÐACB

°±°=Ð 01.0 100ACB OR Using sine rule,

43.16*54sin

9.8sin °

=ÐCAB

OR

43.16*54sin

20sin °

=ÐACB

___K1

°=Ð 99.25CAB °-°-°=Ð 99.2554180ACB

= 100.01o °=Ð 100ACB ___N1

K1

N1

(c) (i) °÷

øö

çèæ=D 54sin)20)(9.8(

21 of Area ABC

= 72.00 cm2

ABCADC D´=D of Area 53 of Area

OR °99.25sin)43.16)(12(

21

= 43.20 cm2 = 43.20 cm2 OR

2.7722021

==>=´´ hh 2.712

21 of Area ´´=DADC

___K1 = 43.20 cm2 ___N1

K1

N1

K1

N1

(ii) Let h cm be the distance between AB and DC

722021

=´´ h

cm 2.7=h

N1

14 (a) 20>x 10³y

80£+ yx

xyyx41or 4 ³£

N1 N1 N1

N1

10


SULIT 10 3472/2(PP)


(b) § Axes correct and one *straight line correct § Draw correctly all 4 straight lines § Region R is correctly shaded and labelled

(Please refer to appendix 2)

K1 N1 N1

(c) (i) 6020 £< x N1 (ii) 200 (21) + 300 (59)

The maximum fees collected is RM 21 900 K1 N1

15 (a) Item P : 110100

52009 =´

P

5.50 RM2009 =P

K1

N1

10

(b) (i) Q2008 = RM 6.00 N1 (ii) Q2010 = RM 9.00 N1

(c) Item I2010 based on I2008 P 121 Q 150 R 120 S 104.5

[one mark each for every TWO correct answers]

N1

N1

(d) Item I2010 based on I2008 W WI P 121 80 9680 Q 150 150 22500 R 120 90 10800 S 104.5 40 4180 360=åW å = 47160WI

Weightage for item S = 40o or 4 [as seen in his formula]

40(104.5) + 90(120) + 150(150) + 80(121) =ii IWå OR 8(121) + 15(150) + 9(120) + 4(104.5)

360)5.104(40)120(90)150(150)121(80 +++

=I

OR 36

)5.104(4)120(9)150(15)121(8 +++

= 131

P1

K1

K1

N1

END OF MARKING SCHEME SKEMA PEMARKAHAN TAMAT


SULIT 11 3472/2(PP)


5 10 0 15 20 25 30 35 40

40

25

20

15

35

30

10

5

xy

X

X

X

X

2x

50

45

(0, 3.0)

X

18

(28, 45) X

Gradient,

45 328 0

1.50

m -=

-

=

x

Appendix 1 – Graph for Question 7


SULIT 12 3472/2(PP)


10 20 0 30 40 50 60 70 80

80

50

40

30

70

60

20

10

xy

R

x

100

90

(21, 59)

x+ y = 80

X

x = 4y

x = 20

y = 10

y Appendix 2 – Graph for Question 14


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