sulit 3472/1 3472/1 matematik tambahan kertas 1 ......matematik tambahan kertas 1 ogos 2013 2 jam...

SULIT 3472/1

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3472/1 MATEMATIK TAMBAHAN KERTAS 1 OGOS 2013 2 Jam

PEPERIKSAAN PRASPM SEKOLAH-SEKOLAH MENENGAH 2013

MATEMATIK TAMBAHAN Kertas 1 Dua Jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

NAMA : ________________________________ KELAS : ________________________________

Negeri Sembilan SPM 2013 http://edu.joshuatly.com/

SULIT 2 3472/1

INFORMATION FOR CANDIDATES MAKLUMAT UNTUK CALON

allsemua

onesatu

[Lihat sebelah]Negeri Sembilan SPM 2013 http://edu.joshuatly.com/

SULIT 3 3472/1

ALGEBRA

CALCULUS / KALKULUS


SULIT 4 3472/1 STATISTICS / STATISTIK

GEOMETRY / GEOMETRI

Negeri Sembilan SPM 2013

http://edu.joshuatly.com/

SULIT 5 3472/1 TRIGONOMETRY / TRIGONOMETRI

𝜃

𝑗 𝜃

∓

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SULIT 6 3472/1


SULIT 7 3472/1

allsemua

1

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SULIT 8 3472/1

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SULIT 9 3472/1

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SULIT 10 3472/1 6

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SULIT 11 3472/1

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SULIT 12 3472/1

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SULIT 13 3472/1

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SULIT 14 3472/1 13

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SULIT 15 3472/1 14

15 sin x = k , 90 ≤ ≤ 𝑠𝑖𝑛 𝑥 = 𝑘 , 90 ≤ ≤

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15


SULIT 16 3472/1 16 OA⃗ = 4i − 3j OB⃗ = −i + j

𝑂�⃗� = 4𝑖 − 3𝑗 𝑂𝐵 = −𝑖 + 𝑗

AB⃗

AB⃗

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SULIT 17 3472/1 17

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SULIT 18 3472/1

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SULIT 19 3472/1

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20

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SULIT 20 3472/1 21 ∫ f(x)dx = 3

∫ f(x)dx = 3

∫ 2f(x)dx

∫ (2x − f(x))dx

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SULIT 21 3472/1 22

P Q R S T 7 8 9

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SULIT 22 3472/1

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SULIT 23 3472/1 24

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SULIT 24 3472/1 25

END OF QUESTION PAPER KERTAS SOALAN TAMAT

25


IngredientPrice (RM) for the year

Price index based on the year

Weightage


Section A, Section BSection

all Section A four Section B twoSection C


KEBARANGKALIAN HUJUNG ATAS Q BAGI TABURAN NORMAL


Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah yang biasa digunakan.


No. Suggested solution and marking scheme Sub Marks

Total Marks

1. (a) 3 (b) 4

(c) many to many relation ( hubungan banyak kepada banyak ),

m – m , m m.

1

1

1

3

2. ( a) 6

( b) 32k

B1: k(6) + 3 = 7 or 6k = 4

1 2

3

3. 13121 xfg

B2: 5)32(6 x B1: 321 xg

3

3

4. 23p

B2: 23SOR and 2

)3(pPOR (Both) or equivalent.

B1: 23SOR or 2

)3(pPOR

3

3

5. ,2x 8x

B2: or

B1: 0)8)(2( xx or 0)8)(2( xx

3

3

6 (a) k = 8 (b) a = 2

(c) x = 1

1

1

1

3

7 23x or x = – 1.5

B2: 424 33 x or equivalent.

B1: Seen 243 x or x33 or 43 or equivalent.

OR using logarithms method: x = –1.5 (accept -1.499 1.501)

B2: (x+2)0.4771 +x (0.4771) = –1.908 or 1.9081x = – 2.8622

B1: log10 3x+2 + log10 27x = log10 811 ( (accept any base)

3 3

8x-2 8 x-2


8. 1x

B2: 98

xx or x + 8 = 9x or equivalent

B1: 28

log3 xx or 233 3log

8log

xx

3

3

9. 17 or 17 terms or n = 17

B1: 13 + (n – 1)(5) = 67

or by listing method: (Must list all the terms and correct )

13, 8, 3, 2, 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67

2

2

10. 2732a ,

23r or equivalent (Both correct )

B2 : 43ar and 2

276ar (Both )

B1 : 43ar or 216 13ar

3

3

11. h = 41 , k = 11

B2: S = 01.01

72.0 or S = 9972

3 or S = 118

3 or S = 1141

B1: r = 0.01 or 9972 or

118

3

3

12. h =100, k = 7 (both )

B3: h =100 or k = 7

B2: 2log10 h or 3)10(21k or 21

0102k

B1: hxy 10102110 logloglog accept 221 XY

4

4

13. ( a ) 1

)6(5yx

or equivalent

(b) 10x + 12y + 11 = 0

B2: 36122510 2222 yyxyxx or equivalent

B1: 22 2510 yxx or 361222 yyx

1

3

4

14. 41.81o , 138.19o or 41 49’, 138 11’

B3 : sin x = 32 , sin x = 4 ( both)

B2 : (3 sin x 2)(sin x + 4) = 0

B1 : 3( 1 – sin 2 x ) – 10sin x + 5 = 0

4

4


10 sin x

15. (a) k1

(b) 21 k

B1: xx sin180sincos180cos 00 or xcos

1

2

3

16. (a) ji 45 , Accept

45

B1: jiji 34 or equivalent

(b) 41 or 6.403

2

1

3

17. (a) baAC 46 (b) baBD

29

B2: )46(41

6 baa or )46(43

4 bab

B1: )46(41 ba or )46(

43 ba

1

3

4

18. (a)

34

or 31

1 accept 1.333

(b) 30

B1 : 34

921 2 or

34

621 2

1

2

3

19. 2)16(29 x

B2: 643 2u

dxdy or )6()16)(3(

41 2x

dxdy

B1: 243 u

dudy or 6

dxdu

3 3

20. 32p

B2: 6p(-1)2 – 12 (-1) = 8 or equivalent

B1 : xpxdxdy

126 2 or 8dxdy

3

3


21. (a) 6 (b) 12

B2 : 4

1

2

2x 3

B1 : 4

1

4

1)(2 dxxfxdx

1

3

4

22. (a) 40320 (b) 4320

B1 : 3 ! or 6 ! or equivalent

1

2

3

23 (a) 359

B1 : 53

73

(b) 3518

B1 : 74

53

73

52

2

2

4

24 3

24 ku

B2 : k = 24 – 3u

B1 : 2)3(5

120 u

3

3

25 (a) 1.267

B1 : 12

482.63

(b) k = – 0.39 B1 : 0.39

2

2

4


SULIT 3472/1

3472/1

MATEMATIK TAMBAHAN

KERTAS 1

OGOS 2013

2 Jam

PEPERIKSAAN PRASPM

SEKOLAH-SEKOLAH MENENGAH 2013

MATEMATIK TAMBAHAN / ADDITIONAL MATHEMATICS

KERTAS 1 / PAPER 1

SKEMA PERMARKAHAN / MARKING SCHEME


PEPERIKSAAN PERCUBAAN 2013 3472/2 ADDITIONAL MATHEMATICS Kertas 2 Ogos

221 jam. Duajamtigapuluhminit

SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2 PEPERIKSAAN PRASPM TINGKATAN 5, 2013

No Solution and mark scheme Sub

Marks Full Marks

1 y = 2x – 4 Or

(2x – 4)2 = 4(x+5)

0152 xx Solve the equation

x = )1(2

)1)(1(4)5()5( 2

x = 5.193 ,– 0.193 y= 6.386 ,–4.386

P1 K1 K1 N1 N1

5

2

(a) (i) 01272 xx

0)4)(3( xx x = 3, x = 4, since < , = 3 , = 4. (ii) 01272 xx

0)4)(3( xx 3 4

3 x 4

K1 N1 K1 N1

6

x =2

4y

y2= 5

24y

02822 yy

Solve the equation

y = )1(2

)28)(1(4)2()2( 2

y = 6.385 ,– 4.385 x=5.193, x = –0.193


No Solution and mark scheme Sub Marks

Full Marks

(b)

1 = 3 1 = 2 12 = 2(4) + 1 = 9.

0)9(2)92(2 xx 018112 xx

K1 N1

3

(a)

T1 = )80)(160(

21

= 6400 T2= 1600 T3=400

r = 41 or 0.25

Tn> 30 6400 ( 0.25 )n-1> 30 log 6400 ( 0.25 )n-1> log 30 log 6400 + log( 0.25 )n-1> log 30 log( 0.25 )n-1> log 30 log 6400

n 1 <25.0log

6400log30log

n < 4.868 n = 4 or listing method 6400,1600,400,100, 25 ..all correct

n = 4

P1 P1 K1 N1 K1K1 N1

6



Full Marks

(b)

25.01

6400S

= 8533.33

K1 N1

4

(a) LHS

xkosxx

sinsin 2

tan x

K1 N1

8

(b)

Shape as in the above diagram Amplitude is 4 Number of cycle is 1 Modulus

Equation of straight line y = 2– x

Straight line is drawn Number of solution 4

P1 P1 P1 K1 N1 N1

2

23

2

0 x

4

2

y =2 - x



Full Marks

5

a)

or

i) ii)

b)

Comparing ,

,

43k

K1 N1 N1 P1 K1 K1 N1 N1

8

6

(a)

109

90x

22 109

1050

3216 // 16.67

(b) (i)

1110

90 x

20x

P1 K1 N1 K1 N1



Full Marks

(ii)

222

1110

20x

= 24 // 4.90

K1 N1

7

(a) x 3 4 6 8 10 11

log 10 y 0.62 0.74 0.92 1.21 1.46 1.62 (b) Plot y10log against x (Correct axes and uniform scales)

6 points are correctly plotted. Note: 5 or 4 points correctly plotted Award 1 mark

Line of best fit

(c) (i) sxty 101010 log)(loglog

Use sc 10log 26.0log10 s

s = 02.082.1 (ii) Use tm 10log

115.0log10 t 02.030.1t

N1 K1 N2 N1 P1 K1 N1 K1 N1

10



Full Marks

8 a) 4xdxdy ,

y – 9 = 4 (x – 4)

y = 4x – 7

b)

44221Area

dxxcurveunderArea4

0

2

12

= 4

0

3

6xx

= 14.67 // 3214 //

344

Area = 67.144

= 67.183218

356 oror

c) dyyVolume9

122

= 912 2yy

= 31881 = 96

K1 K1 N1 K1 N1 K1 N1 K1 K1 N1

10

9 (a) kos �� =

3015

SOT = 60 SOT = 1.047 rad (b)

Lengkok ST = (1.047)(15) = 15.705 ST=TR PT=TQ PT = 15302 = 25.981

K1 N1 K1 K1



Full Marks

Perimeter kawasan berlorek =ST + TR + RQ + TQ +SP + PT = 15.705 + 15.705+15+15+ 25.981 + 25.981 = 113.372

(c)

Luassektor OST = )047.1()15(

21 2

= 117.7875 Luas SOT = luas OTR

Luassegitiga OPQ = 15)981.25(221

= 389.715 Luasakawasanberlorek = luas OPQ 2(luas OST) = 389.715 2 ( 117.7875 ) = 154.14

K1 N1 K1 K1 K1 N1

10

(a) or

= 28 unit2 (b)

= (-1 , 3)

(c)

or

K1 N1 K1 N1 P1 K1 N1

10



Full Marks

(d) Use PA = 5

5)(y3)-(x 22 = 5

0910622 yxyx

P1 K1 N1

11 (a) (i0 p = 101 // 0.1 , q =

109 // 0.9

733

10 9.01.0)3( CXP = 0.0574

(ii) )]2()1()0([1)2( XPXPXPXP

1 8221091

110100

010 9.01.09.01.09.01.0 CCC

or 1937.03874.03487.01

= 0.0702

(b) (i) 5.0

0.36.3zP or 2.1zP

0.88493 // 0.8849

(ii) Seen -0.385

385.05.0

0.3m

m = 2.8075

K1 N1 P1 K1 N1 K1 N1 P1 K1 N1

10



Full Marks

12

(a) a = t49 = 0

t=49

2

492

499makV

13.10881

8110 ororVmak

(b) ctts32

32

29

t = 2, 3212s //

338

// 12.67

ort = 3, 2122s //

245 // 22.5

23 ssd

659 // 9.83 m

(c) 032

29 32 tts

27–4 t = 0

427t // 6.75 s

(d) 029 2ttv

29t // 4.5 s

K1 K1 N1 K1 K1 N1 K1 N1 K1 N1



Full Marks

13

(a)

Guna I = 1000

1

QQ

p= 1.60 , q = 1.10 , r = 160

(b)

I = 250150200400

)250(160)150(110)200(125)400(112

= 126.3

(c)

3.12610025

13Q

Q13 = 31.575 (d)

10010085

1003.126

= 107.36

P1P1P1 K1K1 N1 K1 N1 K1 N1



Full Marks

14 (a)

//

(b) Draw a line correctly Draw all the lines correctly Correct region R

(c) (i) 350 (ii) Profit P = 12x + 15y Maximum point ( 0 , 450 ) Maximum profit = 12 ( 0 )+15( 450 ) = RM 6750

N1 N1 N1 K1 N1 N1 N1 N1 K1 N1

15

(i) 0222 70)4.18)(8(24.188 kosPR = 17.37

(ii) 0107sin37.17

sin5.6PRQ

097.20PRQ

(b) Q P’ R

000 97.20107180QPR = 52.030

00' )03.52(2180PQP

K1 N1 K1 N1 N1 K1 K1

10



Full Marks

= 75.940

000' 97.12703.53180RQP

00 97.20sin5.6

97.127sinQR

QR = 14.32

Luas QRP ' = 006.31sin32.145.621

= 24.01

K1 K1 K1 N1


Graf Q7 0 2 4 6 8 10 12 x

log10 y

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

1.8

Plot lg y against x (Correct axes and uniform scales) P1 6 points are correctly plotted N2 or 5 or 4 points correctly plotted N1 Line of best fit N1


Q14

50 100 150 200 250 300 350 400 450

500 450 400 350 300 250 200 150 100 50

x

y

R


sulit 3472/1 3472/1 matematik tambahan kertas 1 ......matematik tambahan kertas 1 ogos 2013 2 jam...

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