Q&a Add Maths P2 Trial Spm Phg 09

3472/2 Form Five Additional Mathematics Paper 2 September 2009 2 ½ hours Trial SPM Pahang

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009 ADDITIONAL MATHEMATICS Paper 2 Two hours and thirty minutes

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 Malaysia.

3.

Calon dikehendaki membaca maklumat di halaman belakang kertas soalan ini.

4.

Calon dikehendaki menceraikan halaman 18 dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan.

Kertas soalan ini mengandungi 19 halaman bercetak.

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The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used.

ALGEBRA

log c b log c a

1

− b ± b 2 − 4ac x= 2a

8

log a b =

2

am x an = a m + n

9

Tn = a + (n − 1)d

3

am ÷ an = a m – n

10. S n 

4

( am )n = a m n

5

log a mn  log a m  log a n

6

log a

7

log a mn = n log a m

n  2a  (n  1)d  2

n 1 11 Tn  a r

m  log a m  log a n n

12 S  n



  a  1 r  , r  1

a r n 1

n

r 1 1 r a , r 1 13 S  1 r

CALCULUS KALKULUS 4 1

y = uv ,

dy dv du =u +v dx dx dx

Area under a curve Luas di bawah lengkung b

=

b

 y dx or (atau)

 x dy

a

2

u dy y= , = v dx

v

du dv −u dx dx v2

5

Volume generated Isipadu janaan

b

3

dy dy du = × dx du dx

2 =   y dx a

a

b

or (atau )

 x

2

dy

a

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STATISTICS STATISTIK 1

x=

Σx N

7

2

x=

Σ fx Σf

8

n

Pr =

n! (n − r )!

9

n

Cr =

n! (n − r )!r !

Σ( x − x ) 3 σ= = N 2

Σx 2 −x2 N

Σ f ( x − x) = Σf 2

Σ fx 2 −x2 Σf

4

σ=

5

1   N −F  C m = L+  2  fm     

6

I=

Σ Wi I i ΣWi

10 P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )

n r nr , p  q 1 11 P ( X  r )  Cr p q

12 Mean / min,   np

Q I  1  100 Q0

13  

npq

14 Z 

x 

GEOMETRY GEOMETRI 1 =

Distance/jarak

( x1 − x 2 )

2

+ ( y1 − y 2 )

4 Area of a triangle/ Luas segitiga =

1 ( x1 y 2 + x 2 y 3 + x3 y1 ) − ( x 2 y1 + x3 y 2 + x1 y 3 ) 2

2

2 Mid point / Titik tengah

( x, y ) =  x1 + x2 , y1 + y 2  

3

2

2



A point dividing a segment of a line Titik yang membahagi suatu tembereng garis

( x , y ) =  nx1 + mx2 , ny1 + my 2  m+n   m+n

5

r = ~

^

6 r = ~

x2 + y2

x i+ y j ~

~

x + y2 2

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TRIGONOMETRY TRIGONOMETRI

1

2

8

Arc length, s = rθ Panjang lengkok, s= jθ

sin ( A ± B ) = sin A kos B ± ko s A sin B 1 2 r 2

Area of a sector, A 

9

4

5

10

tan  A  B  

11

tan 2 A 

sin 2 A  k os 2 A  1 sec 2 A 1  tan 2 A se k 2 A  1  tan 2 A co sec 2 A  1  cot 2 A

tan A  tan B 1 mtan A tanB

2 tan A 1  tan 2 A

2

12

a b c   sin A sin B sin C

sin 2A = 2 sin A cos A

13

a 2  b 2  c 2  2bc cos A

ko se k A  1  k ot A 2

6

cos  A  B   cos A cos B msin A sin B ko s  A  B   k os A k os B msin A sin B

1 2 j 2 sin 2 A  cos 2 A  1

Luas sektor, L = 3

sin ( A ± B ) = sin A cos B ± cos A sin B

a 2  b 2  c 2  2bc kos A

sin 2A = 2 sin A kos A 7 cos 2A = cos2 A – sin2 A

= 2 cos2A – 1 = 1 – 2 sin2 A kos 2A = kos2 A – sin2 A

= 2 kos2A – 1 = 1 – 2 sin2 A

14

Area of triangle/ Luas segitiga =

1 ab sin C 2

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THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1) 1

2

3

7

8

9

12

4 5 6 Minus / Tolak 16 20 24

0.4641

4

8

0.4247

4

8

28

32

36

12

16

20

24

28

32

0.3897

0.3859

4

36

8

12

15

19

23

27

31

35

0.3557

0.3520

0.3483

0.3192 0.2843

0.3156 0.2810

0.3121 0.2776

4

7

11

15

19

22

26

30

34

4 3

7 7

11 10

15 14

18 17

22 20

25 24

29 27

32 31

0.2546

0.2514

0.2483

0.2451

3

7

10

13

16

19

23

26

29

0.2266

0.2236

0.2206

0.1977

0.1949

0.1922

0.2177

0.2148

3

6

9

12

15

18

21

24

27

0.1894

0.1867

3

5

8

11

14

16

19

22

0.1736 0.1492

0.1711 0.1469

0.1685 0.1446

25

0.1660 0.1423

0.1635 0.1401

0.1611 0.1379

3 2

5 5

8 7

10 9

13 12

15 14

18 16

20 19

23 21

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

0.1170

2

4

6

8

10

12

14

16

18

0.1112

0.1093

0.1075

0.0934

0.0918

0.0901

0.1056

0.1038

0.1020

0.1003

0.0985

2

4

6

7

9

11

13

15

17

0.0885

0.0869

0.0853

0.0838

0.0823

2

3

5

6

8

10

11

13

14

0.0793 0.0655

0.0778 0.0643

0.0764 0.0630

0.0749 0.0618

0.0735 0.0606

0.0721 0.0594

0.0708 0.0582

0.0694 0.0571

0.0681 0.0559

1 1

3 2

4 4

6 5

7 6

8 7

10 8

11 10

13 11

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

0..0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

1.7

0.0446

0.0436

1.8

0.0359

0.0351

0.0427

0.0418

0.0409

0.0401

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

0.0344

0.0336

0.0329

0.0322

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

1.9 2.0

0.0287 0.0228

0.0281 0.0222

0.0274 0.0217

0.0268 0.0212

0.0262 0.0207

0.0256 0.0202

0.0250 0.0197

0.0244 0.0192

0.0239 0.0188

0.0233 0.0183

1 0

1 1

2 1

2 2

3 2

4 3

4 3

5 4

5 4

2.1

0.0179

0.0174

0.0170

0.0166

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

2

2

3

3

4

2.2

0.0139

0.0136

0.0132

0.0129

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

0

1

1

1

2

2

2

3

3

2.3

0.0107

0.0104

0.0102

0

1

1

1

1

2

2

2

2

z

0

1

2

3

4

5

6

7

8

9

0.0

0.5000

0.4960

0.4920

0.4880

0.4840

0.4801

0.4761

0.4721

0.4681

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.4 0.5

0.3446 0.3085

0.3409 0.3050

0.3372 0.3015

0.3336 0.2981

0.3300 0.2946

0.3264 0.2912

0.3228 0.2877

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.9 1.0

0.1841 0.1587

0.1814 0.1562

0.1788 0.1539

0.1762 0.1515

1.1

0.1357

0.1335

0.1314

1.2

0.1151

0.1131

1.3

0.0968

0.0951

1.4 1.5

0.0808 0.0668

1.6

0.00990

0.00964

0.00939

0.00914

3

5

8

10

13

15

18

20

23

0.00889

0.00866

0.00842

2 2

5 4

7 6

9 8

12 11

14 13

16 15

16 17

21 19

2.4

0.00820

0.00798

0.00776

0.00755

0.00734

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00714 0.00539

0.00695 0.00523

0.00676 0.00508

0.00657 0.00494

0.00639 0.00480

2 2

4 3

6 5

7 6

9 8

11 9

13 11

15 12

17 14

2.6

0.00466

0.00453

0.00440

0.00427

0.00415

0.00402

0.00391

0.00379

0.00368

0.00357

1

2

3

5

6

7

9

9

10

2.7

0.00347

0.00336

0.00326

0.00317

0.00307

0.00298

0.00289

0.00280

0.00272

0.00264

1

2

3

4

5

6

7

8

9

2.8

0.00256

0.00248

0.00240

0.00233

0.00226

0.00219

0.00212

0.00205

0.00199

0.00193

1

1

2

3

4

4

5

6

6

2.9 3.0

0.00187 0.00135

0.00181 0.00131

0.00175 0.00126

0.00169 0.00122

0.00164 0.00118

0.00159 0.00114

0.00154 0.00111

0.00149 0.00107

0.00144 0.00104

0.00139 0.00100

0 0

1 1

1 1

2 2

2 2

3 2

3 3

4 3

4 4

f (z)

Example / Contoh:

 1  exp  z 2  2π  2  1

f ( z) 

If X ~ N(0, 1),

then Jika X ~ N(0, 1), maka

Q(z)

∞

Q( z ) = ∫ f ( z ) dz

P(X > k) = Q(k)

k

P(X Q(2.1) = 0.0179

k O Section A Bahagian A

z

>

2.1)

[40 marks] CONFIDENTIAL

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6 [40 markah] Answer all questions. Jawab semua soalan

1.

Solve the following simultaneous equations , give your answers correct to three decimal places. Selesaikan persamaan serentak berikut dengan memberi jawapan anda tepat kepada tiga tempat perpuluhan : 2 1 + = x + 3y = 5 x y

[6 marks] [6 markah]

2.

In Diagram 1, ABCD is a quadrilateral. BFC and DEF are straight lines. Dalam Rajah 1, ABCD ialah sebuah sisi empat. BFC dan DEF adalah garis lurus. D A

•E

Diagram 1 Rajah 1 B

C

F

1 2 BC and DE = DF , 4 5 1 2 Diberi BA = 24 x , BF = 10 y , CD = 30 x − 30 y , BF = BC dan DE = DF , 4 5 Given that BA = 24 x , BF = 10 y , CD = 30 x − 30 y , BF =

(a) express in terms of x and/or y , ungkapkan dalam sebutan x dan/atau y , (i)

AC

(ii)

[3 marks] [3 markah]

DF

(b) show that the points A, E and C are collinear. marks] tunjukkan bahawa titik A, E dan C adalah segaris markah] 1 + cos x + cos 2 x = cot x . 3.(a) Prove that sin 2 x + sin x marks]

[3 .

[3 [3

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Buktikan bahawa

7

1 + cos x + cos 2 x = cot x . sin 2 x + sin x

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[3 markah]

(b)(i) Sketch the graph of the trigonometric function y = 3 cos x + 1 for the domain 0 ≤ x ≤ 2π . Lakar graf bagi fungsi trigonometri y = 3 cos x + 1 untuk domain 0 ≤ x ≤ 2π . (ii)

On the same axes, sketch the graph of a suitable straight line that can be used to solve the equation 3π cos x = 3 x − π . State the number of solutions to the equation 3π cos x = 3 x − π for 0 ≤ x ≤ 2π .

Pada paksi yang sama, lakar graf bagi satu garis lurus yang sesuai digunakan untuk menyelesaikan persamaan 3π cos x = 3 x − π . Nyatakan bilangan penyelesaian bagi persamaan 3π cos x = 3 x − π untuk 0 ≤ x ≤ 2π . [5 marks] [5 markah] 4.

Table 1 shows the frequency distribution of the Additional Mathematics marks of a group of students. Jadual 1 menunjukkan taburan kekerapan markah Matematik Tambahan bagi sekumpulan pelajar. Marks 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60

Number of students 2 3 5 10 K 2 Table 1 Jadual 1

(a) Given that the median mark is 34.5, Diberi markah median adalah 34.5, (i)

calculate the value of k, hitungkan nilai k,

(ii)

find the median mark if the mark of each student is increased by 8. cari markah median jika markah setiap pelajar ditambahkan sebanyak 8. [4 marks] [4 markah]

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(b) Given that k = 4, draw a histogram to represent the frequency distribution of the mark by using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 1 student on the vertical axis. Diberi k = 4, lukis sebuah histogram untuk mewakili taburan kekerapan markah dengan menggunakan skala 2 cm kepada 10 markah pada paksi ufuk dan 2 cm kepada 1 pelajar pada paksi tegak. Hence, find the modal mark.

[3

marks] Seterusnya, cari markah mod.

5. (a)

[3 markah]

Find the equation of the normal to the curve y = 3 x +

1 at (1 , 4). x

[3

marks] Cari persamaan normal kepada lengkung y = 3 x +

1 pada (1 , 4). x

[3

markah] (b)

Diagram 2 shows a leaking hemispherical container with a radius of r cm. Rajah 2 menunjukkan sebuah bekas bocor yang berbentuk hemisfera dengan jejari r cm.

r cm

Given that the radius of the water Diagram 2 surface is decreasing at the rate of 0.1 cm s–1, find Rajah 2 in terms of π, the rate of change of the volume of water in the container at the instant the radius of the water surface is 20 cm.

[3

marks]

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Diberi jejari permukaan air menyusut dengan kadar 0.1 cm s–1, cari dalam sebutan

π, kadar perubahan isipadu air dalam bekas itu pada ketika jejari permukaan air adalah 20 cm.

6.

[3 markah]

Diagram 3 shows a few circles. The first circle is the largest circle with a radius of 2 R cm. The second circle has a radius of R cm . The third circle has a radius which is 3 2 of the radius of second circle and this process is continued indefinitely. 3 Rajah 3 menunjukkan beberapa bulatan. Bulatan pertama adalah bulatan terbesar dan 2 mempunyai jejari R cm. Bulatan kedua mempunyai jejari R cm. Bulatan ketiga 3 2 mempunyai jejari yang merupakan daripada jejari bulatan kedua dan proses ini 3 diteruskan sehingga ketakhinggaan.

R cm

Diagram 3 Rajah 3 (a) Show that the perimeters of the circles form a geometric progression with common 2 ratio by using first three circles. [2 marks] 3 Tunjukkan bahawa perimeter bulatan-bulatan itu membentuk satu janjang geometri 2 dengan nisbah sepunya dengan menggunakan tiga bulatan pertama. [2 3 marks] (b) Given that the area of the largest circle is 900π cm2, find in terms of π, Diberi bahawa luas bulatan yang terbesar ialah 900π cm2, cari dalam sebutan π, (i)

the circumference of the tenth circle, Ukurlilit bagi bulatan yang kesepuluh,

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(ii) the sum of the circumference of infinite number of circles formed. [5 marks] jumlah ukurlilit bagi semua bulatan yang dapat dibentuk sehingga ketakterhinggaan [5 markah]

Section B Bahagian B [40 marks] [ 40 markah] Answer any four questions from this section. Jawab mana-mana empat soalan daripada bahagian ini. 7.

Use graph paper to answer this question. Gunakan kertas graf untuk menjawab soalan ini. Table 2 shows the values of two variables, x and y , obtained from an experiment. p x+2 Variables x and y are related by the equation y = , where p and q are constants. q One of the values of y is incorrectly recorded. Jadual 2 menunjukkan nilai-nilai bagi dua pembolehubah, x dan y , yang diperoleh daripada satu eksperimen. Pembolehubah x dan y dihubungkan oleh persamaan p x+2 y= , dengan keadaan p dan q adalah pemalar. Satu daripada nilai y telah salah q direkodkan. x –1 0 1 2 3 4 y 8.4 10.1 12.1 13.2 17.4 20.9

Table 2 Jadual 2 (a) Plot log10 y against ( x + 2) , using a scale of 2 cm to 1 unit on the ( x + 2) - axis and 2 cm to 0.05 unit on the log10 y -axis. [Start the log10 y -axis with the value 0.8]. Hence, draw the line of best fit. [4 marks]

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Plot log10 y melawan ( x + 2) , dengan menggunakan skala 2 cm kepada 1 unit pada paksi- ( x + 2) dan 2 cm kepada 0.05 unit pada paksi- log10 y . [Mulakan paksi- log10 y dengan nilai 0.8] Seterusnya, lukis garis lurus penyuaian terbaik. [4 markah] (b) Use your graph from 7(a), find Gunakan graf anda di 7(a), cari (i)

8.

the correct value of y that is wrongly recoded. nilai yang betul bagi nilai y yang salah direkodkan.

(ii) the values of p and q . nilai p dan nilai q. Solutions to this question by scale drawing will not be accepted. Penyelesaian secara lukisan berskala tidak diterima.

[6 marks] [6 markah]

Diagram 4 shows the triangle OAB where O is the origin. Point C lies on the straight line AB. Rajah 4 menunjukkan segitiga OAB dengan O ialah titik asalan. Titik C terletak pada garis lurus AB. y A(-4 , 2) x

O

C• Diagram 4 Rajah 4

B(6 , -8)

(a)

Calculate the area, in unit2, of triangle OAB. Hitungkan luas, dalam unit2, bagi segitiga OAB. markah]

[2 marks] [2

(b)

[3 marks]

Find the equation of the perpendicular bisector of line segment AB. Cari persamaan pembahagi dua sama serenjang bagi tembereng garis AB. markah] (c)

[3

4 of the distance of point B from the perpendicular 5 bisector of the line segment AB, find the coordinates of point C. [2 marks] Given that the length BC is

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4 daripada jarak titik B dari pembahagi dua sama 5 serenjang bagi tembereng garis AB, cari koordinat bagi titik C. [2 markah] Diberi panjang BC ialah

(d)

9.

A point P moves such that its distance from point B is always twice its distance from point C. Find the equation of the locus of P. [3 markah] Satu titik P bergerak dengan keadaan jaraknya dari titik B adalah sentiasa dua kali jaraknya dari titik C. Cari persamaan lokus bagi P. [3 markah]

Diagram 5 shows the straight line y = 2x which passes through the maximum point of a quadratic curve y = −( x − α )( x − β ) , where α and β are constants. Rajah 5 menunjukkan garis lurus y = 2x yang melalui titik maksimum suatu garis lengkung kuadratik, y = −( x − α )( x − β ) , dengan keadaan α and β ialah pemalar.

y

y = 2x

Q P

O

4

x

Diagram 5 Rajah 5 (a)

State Nyatakan (i)

the coordinates of the maximum point,

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koordinat titik maksimum itu, (ii)

the equation of the quadratic curve.

[2 marks]

persamaan garis lengkung kuadratik itu. (b)

[2 markah]

Calculate the area of the shaded region P.

[4 marks]

Hitungkan luas rantau berlorek P. (c)

[4 markah]

Find the volume of the solid generated, in terms of π , when the region Q is revolved through 360o about the x-axis.

[4 marks]

Cari isipadu pepejal yang dijanakan, dalam sebutan π , apabila rantau Q dikisarkan melalui 360o pada paksi-x.

[4 markah]

10. (a) In a house to house check carried out in Taman Maju, aedes mosquitoes were found in 2 out of every 5 houses. If 8 houses in Taman Maju are chosen at random, calculate the probability that Dalam suatu pemeriksaan dari rumah ke rumah di Taman Maju, nyamuk aedes telah dijumpai dalam 2 daripada setiap 5 buah rumah. Jika 8 buah rumah di Taman Maju dipilih secara rawak, hitung kebarangkalian bahawa (i)

exactly 3 houses are infested with aedes mosquitoes, tepat 3 buah rumah akan dijumpai dengan nyamuk aedes,

(ii)

more than 2 houses are infested with aedes mosquitoes. lebih daripada 2 buah rumah akan dijumpai dengan nyamuk aedes. [5 marks] [5 markah]

(b) A study on the body mass of a group of students is conducted and it is found that the mass of a student is normally distributed with a mean of 50 kg and a variance of 256 kg2.

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Satu kajian jisim badan dijalankan ke atas sekumpulan pelajar dan didapati jisim seorang pelajar adalah mengikut taburan normal dengan min 50 kg dan varians 256 kg2. (i)

If a student is selected randomly, calculate the probability that his mass is more than 60 kg. Jika seorang pelajar dipilih secara rawak, hitungkan kebarangkalian bahawa jisimnya adalah lebih daripada 60 kg.

(ii) Given that 28% of the students weigh less than m kg, calculate the value of m. Diberi bahawa 28% daripada pelajar itu mempunyai jisim kurang daripada m kg, cari nilai m. [5 marks] [5 markah] 11. Use π = 3.142 in this question. Gunakan π = 3.142 dalam soalan ini. Diagram 6 shows a circular sector OAC with a radius of 5 cm and ∠AOC is 1.2 radian. BC is an arc of the circle with centre A. Rajah 6 menunjukkan satu sektor bulatan OAC dengan jejari 5 cm dan ∠AOC adalah 1.2 radian. BC adalah lengkok bulatan dengan pusat A. B

A

5 cm

Diagram 6 Rajah 6

1.2 rad O

C CONFIDENTIAL

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(a)

(b)

15

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Find Cari (i)

the length, in cm, of the arc AC. panjang, dalam cm, lengkok AC.

(ii)

the length, in cm, of radius AB. marks] panjang, dalam cm, jejari AB. markah]

(i)

(ii)

[2 marks] [2 markah] [2 [2

Show that ∠BAC = 2.171 radian. Tunjukkan ∠BAC = 2.171 radian. Hence, calculate the area, in cm2, of the shaded region. Seterusnya, hitungkan luas, dalam cm2, rantau yang berlorek. [6 marks] [6 markah]

Section C Bahagian C [20 marks] [20 markah] Answer two questions from this section. Jawab dua soalan daripada bahagian ini. 12. Table 3 shows the price indices and the percentage of usage of 5 different ingredients A, B, C, D and E needed to make a cake. The composite index number for the cost of making the cake in the year 2007 based on the year 2005 is 132. Jadual 3 menunjukkan indeks harga dan peratus penggunaan lima jenis bahan A, B, C, D dan E yang diperlukan untuk membuat sejenis kek. Nombor indeks gubahan kos membuat kek itu pada tahun 2007 berasaskan tahun 2005 ialah 132. Ingredients Jenis bahan A B C D E

Price index for the year 2007 based on the year 2005 Indeks harga pada tahun 2007 berasaskan tahun 2005 140 x 110 104 120 Table 3 Jadual 3

Percentage of ingredient Peratus bahan (%) 30 20 15 10 25

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(a) Calculate Hitungkan (i)

the price of A in the year 2005 if its price in the year 2007 is RM7. harga A pada tahun 2005 jika harganya pada tahun 2007 ialah RM7. [2 marks] [2 markah]

(ii) the value of x. marks] nilai x.

[2 [2 markah]

(b) The cost of the cake increased 10% from the year 2007 to the year 2009. Find the price of the cake in the year 2009 if its price in the year 2005 is RM40. Kos penghasilan kek itu bertambah 10 % dari tahun 2007 ke tahun 2009. Cari harga kek itu pada tahun 2009 jika harganya pada tahun 2005 ialah RM40. [3 marks] [3 markah] (c) Find the price index of D in the year 2007 based on the year 2003 if its price index in the year 2005 based on the year 2003 is 125. [3 marks] Carikan indeks harga bagi D pada tahun 2007 berasaskan tahun 2003 jika indeks harganya pada tahun 2005 berasaskan tahun 2003 ialah 125. [3 markah] 13. Diagram 7 shows a quadrilateral PQRS where the sides PQ and RS are parallel. Rajah 7 menunjukkan sebuah sisiempat PQRS dengan keadaan sisi PQ dan sisi RS adalah selari. P 110o 10 cm S 50o R

Q

Diagram 7 Rajah 7 Given that PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 and SR : PQ = 2 : 5, calculate Diberi PQ = 10 cm, ∠RPQ = 1100 , ∠PQR = 500 dan SR : PQ = 2 : 5, hitungkan (a) the length, in cm, of PR and QR. marks] panjang, dalam cm, PR dan QR. markah]

[3 [3

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(b) the length, in cm, of diagonal QS. marks] panjang, dalam cm, perpenjuru QS. (c) the area, in cm2, of PQRS. marks] luas, dalam cm2, PQRS. markah]

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14. A particle moves along a straight line and passes through a fixed point O with a velocity of 3 ms-1. Its acceleration, a ms-2, is given by a = 2 – 2t, where t is the time, in seconds, after passing through O. The particle stops momentarily at time, t = k s. Suatu zarah bergerak di sepanjang suatu garis lurus dan melalui satu titik tetap O dengan halaju 3 ms-1. Pecutannya, a ms-2, diberi oleh a = 2 - 2t, dengan keadaan t ialah masa, dalam saat, selepas melalui O. Zarah itu berhenti seketika pada masa, t = k s. Find Cari (a) the maximum velocity of the particle, marks ] halaju maksimum zarah itu,

[3 markah]

(b) the value of k, nilai k,

[2 marks] [2 markah]

(c) the distance travelled in the third second, marks] jarak yang dilalui dalam saat ketiga.

[3

[2 [2 markah]

(d) the value of t , correct to two decimal places, when the particle passes O again. [3 marks] nilai t, betul kepada dua tempat perpuluhan, apabila zarah itu melalui titik O semula. [3 markah] 15. Use the graph paper provided to answer this question Gunakan kertas graf yang disediakan untuk menjawab soalan ini. A factory produces two types of school bags, type P and type Q. In a day, it can produce x bag of type P and y bag of type Q. The time taken to produce a bag of type P is 40 minutes and a bag of type Q is 50 minutes. Sebuah kilang menghasilkan dua jenis beg sekolah, jenis P dan jenis Q. Dalam satu hari, kilang itu boleh menghasilkan x beg jenis P dan y beg jenis Q. Masa yang diambil untuk menghasilkan satu beg jenis P ialah 40 minit dan satu beg jenis Q ialah 50 minit. The production of the bags per day is based on the following constraints : Pengeluaran beg dalam satu hari adalah berdasarkan kepada kekangan berikut : CONFIDENTIAL

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I

:

The total number of bags produced is not more than 160. Jumlah bilangan beg yang dihasilkan tidak melebihi 160.

II

: The time taken to make bag P is not more than twice the time taken to make bag Q. Masa yang diambil untuk membuat beg P tidak melebihi dua kali ganda masa yang diambil untuk membuat beg Q.

III

:

The number of bag Q exceed the number of bag P by at most 80. Bilangan beg Q melebihi bilangan beg P selebih-lebihnya 80.

(a) Write down three inequalities, other than x ≥ 0 and y ≥ 0 which satisfy all the above constraints. [3 marks ] Tulis tiga ketaksamaan, selain x ≥ 0 dan y 0, yang memenuhi semua kekangan di atas. [3 markah] (b) By using a scale of 2 cm to 20 bags on both axes, construct and shade the region R that satisfies all the above constraints. [3 marks] Menggunakan skala 2 cm kepada 20 beg pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua kekangan di atas. [3 markah] (c) Use your graph in 15 (b) to answer the following : Gunakan graf anda di 15 (b) untuk menjawab yang berikut : (i)

Find the range of the number of bag Q that can be produced if the number of bag P is 50. Cari julat bilangan beg Q yang boleh dihasilkan jika bilangan beg P ialah 50.

(ii) If the profit of selling bag P is RM20 and bag Q is RM30, find the maximum profit that can be obtained. Jika untung jualan bagi beg P ialah RM20 dan beg Q ialah RM30, cari keuntungan maksimum yang boleh diperolehi. [4 marks] [4 markah] END OF QUESTION PAPER KERTAS SOALAN TAMAT NAMA: KELAS : NO. KAD PENGENALAN: CONFIDENTIAL

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ANGKA GILIRAN Arahan Kepada calon Tulis nama, kelas, nombor kad pengenalan dan angka giliran anda pada ruang yang INFORMATION FOR CANDIDATES disediakan. MAKLUMAT UNTUK CALON Tandakan ( √ ) untuk soalan yang dijawab. helaianpaper ini dan ikat sebagai hadapan bersama-sama kertas jawapan. 1 Ceraikan This question consists of threemuka sections : Section A, Sectiondengan B and Section C. Kod Kertas Pemeriksa soalan ini mengandungi tiga bahagian: Bahagian A, Bahagian B dan Bahagian C. Bahagian 2

3

4

Soalan

Soalan Markah Markah Diperoleh Dijawab Penuh (Untuk Kegunaan Pemeriksa) Answer all questions in Section A, four questions from Section B and two questions from 1 6 Section C. 2 6 Jawab semua soalan dalam Bahagian A, empat solan daripada Bahagian B dan dua A 3 8 soalan daripada Bahagian C. 4 7 5 6 Show your working. It may help you to get marks. 6 7 Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh membantu anda 7 10 untuk mendapat markah. 8 10 B 9 the questions provided are 10 not drawn to scale unless stated. The diagrams in 10 10 mengikut skala kecuali dinyatakan. Rajah yang mengiringi soalan tidak dilukis 11

5

12 10sub –part of a question are shown in brackets. The marks allocated for each question and C 13 10 dan ceraian soalan ditunjukkan dalam Markah yang diperuntukkan bagi setiap solan kurungan.

6

10

14

10

15

10

A list of formulae is provided on page 2 to 4. Satu senarai rumus disediakan di halaman 2 hingga 4

7

You may use a non-programmable scientific calculator. Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogram

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SULIT 3472/2[PP] Additional Mathematics Paper 2 September 2009

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SEKTOR PENGURUSAN AKADEMIK JABATAN PELAJARAN PAHANG

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009

ADDITIONAL MATHEMATICS Paper 2

MARKING SCHEME This marking scheme consists of 12 printed pages

Question 1.

Working/Solution 2 1 + = 5 and x + 3 y = 5 x y 2 1 or any correct pairs from + = x + 3 y = 5 x y 5− x x = 5 − 3y y= or 3

Marks

Total

P1

P1

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5− x into the non-linear 3 equation to obtain a quadratic equation in terms of x or y 15 y 2 − 26 y + 5 = 0 or 5 x 2 − 24 x + 10 = 0 Substitute x = 5 − 3 y or y =

Solve quadratic equation using formula or completing the squares

K1

y = 1.513 ; y = 0.220 or x = 4.339 ; x = 0.461

N1

x = 0.461 ; x = 4.340 or y = 0.220 ; y = 1.513 2(a)

N1 K1

Use triangle law correctly to find AC or DF (i) AC = AB + BC AC = −24 x + 40 y

2(b)

K1

6

(ii) DF = DC + CF DF = −30 x

N1 N1

Find vector CE or EC or AE or EA by correct triangle law.

K1

CE = CF + FE 3 = CF + FD 5 3 = CF − DF 5 = 6(3 x − 5 y ) 1 CE 6 Find AC in term of CE or vice versa or any equation that can conclude A, C, E are collinear. AC = −24 x + 40 y 3x − 5 y =

K1

AC = −8(3 x − 5 y ) 4 AC = − CE 3 4 ∴A, C, E are collinear because AC = − CE 3 Question 3(a)

Working/Solution Use cos 2x or sin 2x and factorize

N1

6

Marks K1

Total

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3(b)(i)

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1 + cos x + cos 2 x 1 + cos x + (2 cos 2 x − 1) = sin 2 x + sin x 2 sin x cos x + sin x cos x + 2 cos 2 x = sin x (2 cos x + 1) cos x(2 cos x + 1) = sin x( 2 cos x + 1) cos x = sin x = cot x Correct shape of cosine function with amplitude of 3 units or Correct shape of cosine function and translated 1 unit OR Correct shape of cosine function with amplitude of 3 units and translated 1 unit. OR Correct shape of cosine function with amplitude of 3 units π and translated 1 unit and passing through (0 , 4), ( , 1), 2 3π (π , –2), ( , 1) and (2π , 4). 2 3x y y= π 4 y = 3 cos x + 1

N1

N1

P1 OR P2 OR P3

1 O -2 3(b)(ii)

Question 4.(a)

π 2

π

3π 2

2π

x

Get the correct linear equation and draw a straight line. 3π cos x = 3 x − π 3π cos x + π = 3x π (3 cos x + 1) = 3 x 3x 3 cos x + 1 = π 3x The equation of straight line : y = π Straight line drawn correctly and number of solutions = 1

K1

N1

8

Working/Solution (i) Use median formula with at least two of the L, N, F, f

Marks K1

Total

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and c correctly substituted.  22 + k  − 10   (10) 34.5 = 30.5 +  2 10       k =6 4(b)

5(a)

5(b)

Question 6.

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(ii) median mark = 42.5 Correct axes and uniform scales with all the lower and upper boundaries correctly labeled and the Height of at least three bars are proportional to the frequency

N1 N1 P1 K1

Correct way of finding the value of mode.

K1

Modal mark = 35.0 1 y = 3x + x dy 1 = 3− 2 dx x

N1

Find gradient of normal and use y − y1 = m( x − x1 ) dy =2 at (1 , 4), dx 1 Gradient of normal = − 2 Equation of normal : 1 y − 4 = − ( x − 1) 2 x + 2 y − 9 = 0 or equivalent dV Get the expression for V and find to determine the dr dV value of at r = 20 cm. dr 2 V = π r3 3 dV = 2π r 2 dr dV dV dr = × Use dt dr dt dV = 2π (20) 2 (−0.1) dt = −80π cm 3 s −1

Working/Solution (a) Perimeters of the first three circles :

7

P1 K1

N1 K1

K1

N1

6

Marks

Total

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4 8 πR , πR 3 9 T3 T2 Check the ratios and T1 T2

2πR ,

K1

Make a correct conclusion : T2 T3 2 = = , the perimeters of the first three Since T1 T2 3 circles form a geometric progression with common 2 ratio . 3

N1

(b)

πR 2 = 900π R = 30

P1

(i) Use the formula Tn = ar n −1 :

K1

9

2 T10 = (60π )  3 = 1.561π cm

N1

a : 1− r 60π S∞ = 2 1− 3 = 180π cm

(ii) Use the formula S ∞ =

7(a)

K1

N1

All values of log10 y correct. x+2 log10 y

1 0.92

2 1.00

3 1.08

4 1.12

5 1.24

6 1.32

N1

Plot log10 y against ( x + 2 ) with correct axes, uniform scales and at least one point plotted correctly.

K1 N1

6 points plotted correctly. Line of best fit, ( passes through as many points as possible and balance in terms of numbers point appear above and below the line, if any .) 7(b)

(i) Recognize the wrong recorded value of y and use graph to find the should be value of y.

N1

K1

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7

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log 10 y = 1.16

N1

y = 14..45

P1

(ii) log10 y = ( x + 2) log10 p − log10 q K1

8(a)

8(b)

8(c)

8(d)

Use − log10 q = c or Use log10 p = m log10 q = −0.84 log10 p = 0.08 p = 0.1445 ; p = 1.202 1 x1 x 2 x3 x1 Use correctly to find the area of triangle 2 y1 y 2 y 3 y1

N1 N1 10 K1

Area = 10 unit2 Either midpoint of AB or gradient of AB correct. Mid point of AB = (1 , –3) Gradient of AB = –1

N1 P1

Use y − y1 = m( x − x1 ) with his midpoint of AB and his gradient of normal. y + 3 = 1(x -1) y=x–4 Use Ratio Theorem follow his midpoint  4(1) + 1(6) 4( −3) + 1(−8)  C = ,  5 5   C = (2 , − 4) Use of distance formula correctly for PB or PC PB = ( x − 6) 2 + ( y + 8) 2 ; PC = ( x − 2) 2 + ( y + 4) 2

K1 N1 K1 N1 K1

Use BP = 2 PC ( x − 6) 2 + ( y + 8) 2 = 2 ( x − 2) 2 + ( y + 4) 2 2

9(a) 9(b)

K1 N1

2

3 x + 3 y − 4 x + 16 y − 20 = 0 (i) (2 , 4) y = − x( x − 4) or eqivalent (ii) Correct method of finding area under curve or area of ∆ 4

P1 P1 K1

2

 2x 2   4x 2 x 3  1 −  or × 2 × 4   or  2 3 2  2  2 0 Find integration value using correct limits  2(2) 2 2(0) 3   4(4) 2 4 3   4(2) 2 2 3  − − − − or       3  2 3 2   2  2

9(c)

10

K1

Find the sum of two area 28 unit2 or equivalent 3

K1

Correct method of finding volume of revolution or volume

K1

N1

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of cone.

π ∫ (16 x 2 − 8 x 3 + x 4 )dx or π 2

0

10(a)

∫

2

0

4 x 2 dx or

1 π (4) 2 (2) 3

K1

Find integration value using correct limits  16(2) 3 8(2) 4 2 5    4( 2) 3  π  − +  − 0 or π  − 0 4 5    3   3 Find the difference of two volume 32 π cm3 or equivalent 5 2 3 (i) Both p = and q = or equivalent 5 5 3 5  2  3 8 C 3     or equivalent  5 5

K1 N1 P1 K1

0.2787 or other more accurate answers. 0

8

1

7

N1 2

 2 3  2  3  2  3 (ii) 1 - C 0     + 8C1     + 8C 2     5 5  5 5  5 5

6

8

3

5

4

4

10

8

 2  3  2  3  2  3 or C 3     + 8C 4     + ...+ 8C8      5 5  5 5  5 5 All terms must be correct and completed in full

K1 0

8

10(b)

11(a)

11(b)

0.6846 or other more accurate answers. 60 − 50    (i) P Z ≥ 16   0.2660 m − 50  m − 50     = 0.28 or P Z ≥ −  = 0.28 (ii) P Z < 16  16    or equivalent m − 50 − = 0.583 16 40.672 kg (i) SAC = 5(1.2) 6 cm (ii) Correct method of finding AB 1 AC 1.2 or equivalent OR cosine rule 2 = sin rad 5 2 AB = AC = 5.646 cm (5.64578271...) π + 1.2 (i) ∠BAC = 1.2 + or equivalent 2 2.171 rad or equivalent

N1 K1 N1 K1

K1 N1 K1 N1

10

K1

N1 K1 N1

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1 (5.646)2(2.171) 2 1 2 Area of segment AC = (5) (1.2 − sin 1.2rad ) 2 or equivalent

(ii) Area of sector BAC =

Find difference of the two area 1 1 (5.646) 2 (2.171) − (5) 2 (1.2 − sin 1.2rad ) 2 2 Area of shaded region = 31.25 cm2 (31.25261028...) 12(a)

7 × 100 = 140 or equivalent (i) A05 RM5 (ii)

12(b)

12(c)

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140(30) + x(20) + 110(15) + 104(10) + 120(25) = 132 30 + 20 + 15 + 10 + 25 165.5

Q07 × 100 = 132 40 Q07 = RM 52.80 Q09 = RM 58.08 Q05 Q 104 × 100 = 125 and 07 = Q03 Q05 100 Q07 Q 125 × 100 = 104 and 05 = Or Q05 Q03 100 Q I 07,03 = 07 × 100 Q03 Q Q 125 104 I 07,03 = 07 × 05 × 100 or × 104 or 125 × or Q05 Q03 100 100

K1 K1

K1

N1

10

K1 N1 K1 N1 K1 N1 N1

P1

K1

125 × 104 100 = 130

13(a)

Using correct rules to find PR and QR. PR 10 QR 10 = = o o o or sin 110 sin 20 o sin 50 sin 20

N1

10

K1

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13(b) 13(c)

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PR = 22.40 cm (22.39764114...) QR = 27.47 cm (27.47477419...) Use cosine rule to find QS. QS2 = 42 + 27.472 – 2(4)(27.47)cos (110 + 20)o QS = 30.20 cm Use formula correctly to find area of triangle PQR or PRS. 1 Area ∆PQR = (27.47)(10) sin 50 o 2 1 o or Area ∆PRS = (22.40)(4) sin 110 2 Use Area PQRS = sum of two area

14(b)

14(c)

N1

K1

Area PQRS = 147.32 cm2 14(a)

N1 N1 K1 N1 N1 K1

N1 K1

Find expression for v. 2t 2 v = 2t − +c 2 Find t for maximum v and substitute in his v. a = 2 - 2t = 0 2(1) 2 and v max = 2 − +3 2 vmax = 4 ms-1 Try solving v = 0 2k - k2 + 3 = 0 (3 - k)(1 + k) = 0 or equivalent k=3s Correct method in finding distance traveled.

10

K1

N1 K1 N1 K1

3

14(d)

15(a)

 2t 2 t 3  1 s= − + 3t  or s = t 2 − t 3 + 3t and s3 − s 2 3 3  2 2 5 s= m 3 Find expression for s. t3 2 s = t − + 3t 3 Solve his s = 0 t3 2 t − + 3t = 0 3 3t2 - t3 + 9t = 0 t2 - 3t -9 = 0 t = 4.854 s

N1 K1

K1

N1

x + y ≤ 160 note: ss-1 for all answers in terms of P and Q 40 x ≤ 100 y or equivalent y ≤ x + 80 or equivalent

10

N1 N1 N1

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15(c)

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Axes correct and one correct straight line. All three straight lines are correct. The shaded region of R is correct. Please refer to attached graph. (i) Drawing of the straight line, x = 50, completely across the shaded region R. 20 ≤ y ≤ 110

Graph For Question 4(b) (ii) Construction of the line 20x + 30y = k at (40, 120) Number of students or any clear indication of point (40, 120) only or 30(120) + 20(40) = k

K1 K1 N1 K1 N1 K1

N1

maximum profit = RM4400

10

9

8

7

6

5

4

3

2

1

0.5

10.5

20.5

30.5 40.5 Modal mark = 35.0

50.5

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10

log10 y

No.7(a) x

1.30

1.25

x

1.20

1.15 x

1.10 x

1.05

x

1.00

0.95 x

0.90

0.85

0.80

0

1

2

3

4

5

7 6 CONFIDENTIAL

x+2

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(6 , 1.32)

1.16

1.32 − 0.92 6 −1 m = 0.08 c = 0.84 m=

(1 , 0.92)

Qn. 15

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