peperiksaan percubaan spm 2009 3472/1 · 8 ± = ± sin( ) sin cos cos sin a b a b a b ... the...

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…

…

_________________________________________________________________________________

Kertas soalan ini mengandungi 19 halaman bercetak.

Nama : ………….……..…………………………………

Tingkatan : ………….………………………………………..

JABATAN PELAJARAN

WILAYAH PERSEKUTUAN KUALA LUMPUR

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tuliskan nama dan tingkatan anda

pada ruang yang disediakan.

2. Kertas soalan ini adalah

dalam dwibahasa.

3. Soalan dalam bahasa Inggeris

mendahului soalan yang sepadan

dalam bahasa Melayu.

4. Calon dibenarkan menjawab

keseluruhan atau sebahagian soalan

sama ada dalam bahasa Inggeris atau

bahasa Melayu.

5. Calon dikehendaki membaca maklumat

di halaman belakang kertas soalan ini.

Untuk Kegunaan Pemeriksa

Kod Pemeriksa:

Soalan Markah

Penuh

Markah

Diperoleh

1 2

2 4

3 4

4 3

5 3

6 3

7 3

8 3

9 3

10 3

11 3

12 4

13 3

14 3

15 2

16 4

17 3

18 3

19 3

20 4

21 3

22 3

23 4

24 3

25 4

Jumlah 80

PEPERIKSAAN PERCUBAAN SPM 2009 3472/1 ADDITIONAL MATHEMATICS

Kertas 1

September

2 jam Dua jam

SHARED BY PN. DING HONG ENG OF SMK ALAM SHAH, KL

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SULIT 3472/1

3472/1 @ 2009 Hak Cipta JPWPKL [Lihat Sebelah SULIT

2

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

ones commonly used.

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

yang biasa digunakan.

ALGEBRA

1 a

acbbx

2

42 −±−= 8

a

bb

c

ca

log

loglog =

9 Tn = a + (n – 1)d

10 [ ]dnan

Sn )1(22

−+=

2 a m x a

n = a

m + n

3 a m

÷ a n = a

m – n

4 (a m )

n = a

m n

5 nmmn aaa logloglog += 11 Tn = ar n – 1

6 nmn

maaa logloglog −= 12

( ) ( )1,

1

1

1

1≠

−−

=−−

= rr

ra

r

raS

nn

n

7 log a m n = n log a m 13 1,

1<

−=∞ r

r

aS

CALCULUS / KALKULUS

1 y = uv , dx

duv

dx

dvu

dx

dy+=

4 Area under the curve

Luas di bawah lengkung

∫=b

adxy or (atau)

2 2

,v

dx

dvu

dx

duv

dx

dy

v

uy

−==

∫=b

adyx

3 dx

du

du

dy

dx

dy×=

5 Volume generated

Isipadu janaan

∫

∫

=

=

b

a

b

a

dyx

ataudxy

2

2 )(or

π

π


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SULIT 3472/1


3

STATISTICS / STATISTIK

1 N

xx

Σ= 7

i

ii

W

IWI

ΣΣ

=

2 f

fxx

ΣΣ

= 8 ( )!

!

rn

nPr

n

−=

3 ( ) 2

22

xN

x

N

xx−

Σ=

−Σ=σ 9

( ) !!

!

rrn

nCr

n

−=

4 ( ) 2

22

xf

fx

f

xxf−

ΣΣ

=Σ−Σ

=σ

5 Cf

FNLm

m

−+= 2

1

10 ( ) ( ) ( ) ( )BAPBPAPBAP ∩−+=∪

11 ( ) 1, =+== − qpqpCrXP rnrr

n

12 Mean / Min , µ = np

13 npq=σ

6 1000

1 ×=Q

QI 14

σµ−

=X

Z

GEOMETRY / GEOMETRI

5 22 yxr += 1 Distance / Jarak

= ( ) ( )2212

21 yyxx −+−

2 Midpoint / Titik tengah

( )

++=

2,

2, 2121 yyxx

yx

6 22

ˆ

yx

jyixr

+

+=

3 A point dividing a segment of a line

Titik yang membahagi suatu tembereng garis

( )

++

++

=nm

myny

nm

mxnxyx 2121 ,,

4 Area of triangle / Luas segitiga

( ) ( )1 2 2 3 3 1 2 1 3 2 1 3

1

2x y x y x y x y x y x y= + + − + +


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3472/1 @ 2009 Hak Cipta JPWPKL [Lihat Sebelah SULIT

4

TRIGONOMETRY / TRIGONOMETRI

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

8 BABABA sincoscossin)sin( ±=±

BABABA sinkoskossin)sin( ±=±

9 BABABA sinsincoscos)cos( m=±

BABABA sinsinkoskos)(kos m=±

2 Area of sector, θ= 2

2

1rA

Luas sektor, θ= 2

2

1jL

3 sin2 A + cos

2 A = 1

sin2 A + kos

2 A = 1

10 BA

BABA

tantan1

tantan)tan(

m

±=±

4 AA 22 tan1sec +=

AA 22 tan1sek +=

11 A

AA

2tan1

tan22tan

−=

5 AA 22 cot1cosec +=

AA 22 kot1kosek += 12

C

c

B

b

A

a

sinsinsin==

6 AAA cossin22sin =

AAA kossin22sin =

13 a2 = b

2 + c

2 – 2bc cos A

a2 = b

2 + c

2 – 2bc kos A

7 AAA 22 sincos2cos −=

A

A

2

2

sin21

1cos2

−=

−=

A

A

AAA

2

2

22

sin21

1kos2

sinkos2kos

−=

−=

−=

14 Area of triangle / Luas segitiga

= Cab sin2

1


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5 For

Examiner’s

Use Answer all the questions.

Jawab semua soalan.

1 Diagram 1 shows the relation of set A to set B.

Rajah 1 menunjukkan hubungan set A kepada set B.

Diagram 1

Rajah 1

State

Nyatakan

(a) the image of 2,

imej bagi 2,

(b) the object that maps onto itself.

objek yang memeta kepada dirinya sendiri.

[2 marks]

[2 markah]

Answer / Jawapan: (a) ….....…………………….

(b) ……..………….………...

2

1

–1 0 1 2 3

2

1

–1

–2

•

• •

•

• Set A

Set B


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6 For

Examiner’s

Use

2 Given that f : x → 7 – 3x and kxxg −→ 2: .

Diberi f : x → 7 – 3x dan kxxg −→ 2: .

Find

Cari

(a) f 2 (x),

(b) the possible values of k, if g(1) = 4.

nilai-nilai yang mungkin bagi k, jika g(1) = 4.

[4 marks]

[4 markah]

Answer / Jawapan: (a) ……..……………………..

(b) k = ….………….………...

__________________________________________________________________________

3 Given the function g(x) = 12

5

−x, x ≠

2

1 and hg(x) = 2 – 3x.

Diberi fungsi g(x) = 12

5

−x, x ≠

2

1 dan hg(x) = 2 – 3x.

(a) Find g –1(x).

Cari g–1

(x).

(b) Hence, find h(x). [4 marks]

Seterusnya, cari h(x). [4 markah]

Answer / Jawapan: (a) .………..…….….………

(b) …………….……………

4

2

4

3


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7 For

Examiner’s

Use 4 The sum of the roots of a quadratic equation 2(x

+ p)

2 = 1 is 7. Find the value of p.

[3 marks]

Hasiltambah punca-punca bagi persamaan kuadratik 2(x + p)2 = 1 ialah 7. Carikan

nilai bagi p. [3 markah]

Answer / Jawapan: p = ….…..………….……...

__________________________________________________________________________

5 Find the range of the values of x for x( x – 2) > 5x – 6. [3 marks]

Cari julat nilai x bagi x( x – 2) > 5x – 6. [3 markah]

Answer / Jawapan: ……….…..………………...

__________________________________________________________________________

6 A quadratic function f(x) = 3( x + h)2 – 5 has a minimum point at (–4, k).

Suatu fungsi kuadratik f(x) = 3( x + h)2 – 5 mempunyai titik minimum di (–4, k).

Find the value of

Cari nilai bagi

(a) h,

(b) k,

(c) the y-intercept. [3 marks]

pintasan-y. [3 markah]

Answer / Jawapan: (a) h = ……………………..

(b) k = ….………………….

(c) ………………………….

3

4

3

5

3

6


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8 For

Examiner’s

Use

7 The straight line y = 2x – p intersects the quadratic function y – 2 = 3x2 – x + 5p at

two different points. Find the range of values of p. [3 marks]

Garis lurus y = 2x – p menyilangi fungsi kuadratik y – 2 = 3x2 – x + 5p di dua titik

yang berbeza. Cari julat bagi nilai p. [3 marks]

Answer / Jawapan : ……………………………...

__________________________________________________________________________

8 Solve the equation: Selesaikan persamaan:

2x − 2

(3 x) = 4 [3 marks]

[3 markah]

Answer / Jawapan: ………………………………...

__________________________________________________________________________

9 Given log 3 7 = m and log 9 2 = n. Find, in terms of m and n, log 3 14 . [3 marks]

Diberi log 3 7 = m dan log 9 2 = n. Cari, dalam sebutan m dan n, log 3 14. [3 markah]

Answer / Jawapan : ……….………..……………

3

7

3

8

3

9


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9 For

Examiner’s

Use

10 An arithmetic progression 3, 2m, 13, … has a common difference of 5.

Suatu janjang aritmetik 3, 2m, 13, … mempunyai beza sepunya 5.

Find

Cari

(a) the value of m,

nilai bagi m,

(b) the term with value 103. [3 marks]

sebutan yang mempunyai nilai 103. [3 markah]

Answer / Jawapan : (a) …...……..………………

(b) ......................................... _______________________________________________________________________________________________________________

11 In a geometric progression, the sum to infinity is –8 and the sum of the first five terms

is 4

31− . Find the value of the common ratio. [3 marks]

Dalam suatu janjang geometri, hasil tambah ketakterhinggaannya ialah –8 dan

hasiltambah lima sebutan yang pertama ialah4

31− . Cari nilai nisbah sepunyanya.

[3 markah]

Answer / Jawapan : ……………..……………

3

10

3

11


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10 For

Examiner’s

Use

12 The variables x and y are related by the equation y = p x 2 k where p and k are constants.

Diagram 2 shows the straight line graph obtained by plotting log10 y against log10 x.

Pembolehubah x dan y dihubungkan oleh persamaan y = p x 2 k

dengan keadaan p dan

k ialah pemalar. Rajah 2 menunjukkan graf garis lurus yang diperoleh dengan

memplot log10 y melawan log10 x.

Diagram 2

Rajah 2

(a) Express the equation y = p x 2

k in its linear form used to obtain the straight line

graph shown in Diagram 2.

Ungkapkan persamaan y = p x 2

k dalam bentuk linear yang digunakan untuk

memperoleh graf pada Rajah 2.

(b) Hence, find the values of p and k.

Seterusnya, cari nilai-nilai bagi p dan k.

[4 marks]

[4 markah]

Answer / Jawapan : (a) ……………..……...………..……

(b) p = .............................................

k = …………………………….

log10 y

log10 x

2 •

• (5, 6)

4

12

O


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11 For

Examiner’s

Use 13 Digaram 3 shows a triangle ABC with the vertices A(1, 5), B(4, 0) and C (h, –1),

where h is a constant.

Rajah 3 menunjukkan segitiga ABC dengan titik-titik A(1, 5), B(4, 0) dan C(h, –1),

di mana h ialah pemalar.

Diagram 3

Rajah 3

If the area of the triangle is 29 unit2, find the value of h. [3 marks]

Jika luas segitiga itu ialah 29 unit2, cari nilai bagi h. [3 markah]

Answer / Jawapan : h = ………..……………

_________________________________________________________________________

14 A point R (3, 1) divides the straight line AB in the ratio of 1 : 3. Given that the

coordinates of A are (0, –1), find the coordinates of B. [3 marks]

Satu titik R (3, 1) membahagi garis lurus AB dalam nisbah 1 : 3. Diberi koordinat A

ialah (0, –1), cari koordinat B. [3 markah]

Answer / Jawapan : ……….……..……………

3

14

3

13

B (4, 0)

• A (1, 5)

x

y

C(h, –1) O


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12 For

Examiner’s

Use

15 Diagram 4 in the answer space, shows vectors mOM = , nON = drawn on a

grid of equal squares. Draw, on the grid, vector OR where OR = m + n.

[2 marks]

Rajah 4 pada ruang jawapan, menunjukkan vektor-vektor mOM = , nON = yang

dilukis pada garisan bergrid segiempat sama. Lukis, pada grid itu, vektor OR dengan

keadaan OR = m + n. [2 markah]

Answer / Jawapan:

Diagram 4

Rajah 4

_______________________________________________________________________

16 Given that

=1

7u and

−

=3

2v .

Diberi bahawa

=1

7u dan

−

=3

2v .

Find

Cari

(a) vu + ,

(b) the unit vector in the direction of u + v. [4 marks]

vector unit dalam arah u + v. [4 markah]

Answer / Jawapan : (a) …….....……..………………

(b) ………………………………

4

16

O

M m

n

•

N •

2

15


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13 For

Examiner’s

Use

17 Given an equation of a curve 162 23 +−= xxy . Find the value of x when y is

maximum. [3 marks]

Diberi satu persamaan lengkung 162 23 +−= xxy . Cari nilai x apabila y adalah

maksimum. [3 markah]

Answer / Jawapan : ………....……..……………

__________________________________________________________________________

18 The area of a circle increases at the rate of 16π cm2 s

–1. Find the rate of change of the

radius when the radius is 4 cm. [3 marks]

Luas suatu bulatan bertambah pada kadar 16π cm2

s–1

. Cari kadar perubahan jejari

ketika jejari ialah 4 cm. [3 markah]

Answer / Jawapan : ……….………..……………

3

17

3

18


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14 For

Examiner’s

Use

19 Given that ∫ =4

17dx)x(g and ∫ ∫ =++

k

kdxxgdxxg

1

4

29)](5[2)(2

where 1 ≤ k ≤ 4 and g(x) > 0. Find the value of k. [3 marks]

Diberi bahawa ∫ =4

17dx)x(g dan ∫ ∫ =++

k

kdx)]x(g[dx)x(g

1

429522

dengan keadaan 1 ≤ k ≤ 4 dan g(x) > 0. Cari nilai bagi k. [3 markah]

Answer / Jawapan : k = .………..……..…………

__________________________________________________________________________

20 Solve 2cos 2x = cos x + 1 for 0° ≤ x ≤ 360°. [4 marks]

Selesaikan 2kos 2x = kos x + 1 untuk 0° ≤ x ≤ 360°. [4 markah]

Answer / Jawapan : …….……..……..………………….

3

19

4

20


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15 For

Examiner’s

Use 21 Diagram 5 shows two sectors of circles with center O. The radii of the quadrant OAB

and the sector ORS are 8 cm and 12 cm respectively. Given ∠SOR = θ.

Rajah 5 menunjukkan dua sektor bulatan berpusat O. Jejari bagi sukuan bulatan OAB

dan sektor ORS masing-masing ialah 8 cm dan 12 cm. Diberi ∠SOR = θ.

Diagram 5

Rajah 5

If the area of ORS is equal to the area of OAB, find the value of θ in radian.

( Use π = 3.142 ) [3 marks]

Jika luas ORS adalah sama dengan luas OAB, cari nilai θ dalam radian.

( Guna π = 3.142 ) [3 markah]

Answer / Jawapan : …….……..……..………….

3

21

S

O

8 cm

12 cm

A

B R

θ


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16 For

Examiner’s

Use

22 Table 1 shows the number of goals obtained by a football team in several games.

Jadual 1 menunjukkan jumlah gol yang diperoleh satu pasukan bola sepak dalam

beberapa perlawanan.

Goals

Gol 0 1 2 4

Number of games

Bilangan perlawanan 1 x 3 2

Table 1

Jadual 1

(a) Find the possible values of x if the mode is 2.

Cari nilai-nilai yang mungkin bagi x jika mod ialah 2.

(b) If x = 2 and the mean = 2, find the standard deviation for the number of goals.

Jika x = 2 dan min = 2, cari sisihan piawai bagi bilangan gol.

[3 marks]

[3 markah]

Answer / Jawapan : (a) ……..……..……………

(b) …………....……………

3

22


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17 For

Examiner’s

Use

23 Five letters from the word S U P E R B are chosen to be arranged in a row.

Lima huruf daripada perkataan S U P E R B dipilih untuk disusun sebaris.

(a) Find the number of arrangements that can be formed.

Cari bilangan susunan yang boleh dibuat.

(b) If the vowels must be chosen, find the number of arrangements that can be formed

so that the vowels are next to each other.

Jika huruf vokal mesti dipilih, cari bilangan susunan yang boleh dibuat supaya

huruf-huruf vokal berada sebelah menyebelah.

[4 marks]

[4 markah]

Answer / Jawapan : (a) ……..……..……………

(b) …………....……………

_________________________________________________________________________

24 The probabilities of Aminah and Mei Mei solving a crossword puzzle in less than one

hour, are 3

2and

5

1respectively. Find the probability that, within one hour,

Kebarangkalian Aminah dan Mei Mei menyelesaikan satu teka silangkata dalam masa

kurang daripada satu jam, masing-masing ialah 3

2dan

5

1. Cari kebarangkalian,

dalam masa satu jam,

(a) none of them can solve the crossword puzzle,

tiada seorang pun yang dapat menyelesaikan teka silangkata itu,

(b) at least one of them can solve the crossword puzzle.

sekurang-kurangnya seorang daripada mereka dapat menyelesaikan teka

silangkata itu.

[3 marks]

[3 markah]

Answer / Jawapan : (a) .....................................

(b) .....................................

4

23

3

24


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18 For

Examiner’s

Use

25 Given 85% of the students in a class passed the Calculus test. If 10 students are chosen

randomly from the class, find

Diberi 85% pelajar dalam satu kelas telah lulus ujian Kalkulus. Jika 10 pelajar dipilih

secara rawak daripada kelas tersebut, cari

(a) the probability that exactly 6 students pass the test,

kebarangkalian bahawa tepat 6 pelajar lulus ujian itu,

(b) the number of students in the class, if the variance of students who pass the test is

5.355.

bilangan pelajar dalam kelas itu, jika varians bagi pelajar yang lulus ujian ialah

5.355.

[4 marks]

[4 markah]

Answer / Jawapan : (a) .....................................

(b) .....................................

END OF QUESTION PAPER

4

25


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19 For

Examiner’s

Use


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20 For

Examiner’s

Use


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INFORMATION FOR CANDIDATES

MAKLUMAT UNTUK CALON

1. This question paper consists of 25 questions.

Kertas soalan ini mengandungi 25 soalan

2. Answer all questions.

Jawab semua soalan.

3. Write your answers in the spaces provided in the question paper.

Jawapan hendaklah ditulis dalam ruangan yang disediakan dalam kertas soalan.

4. Show your working. It may help you to get marks.

Tunjukkan langkah-langkah penting dalam kerja mengira anda. Ini boleh

membantu anda untuk mendapatkan markah.

5. If you wish to change your answer, cross out the work that you had done.

Then write down the new answer.

Sekiranya anda hendak menukar jawapan, batalkan jalan kerja yang telah dibuat.

Kemudian tulis jawapan yang baru.

6. The diagrams in the questions provided are not drawn to scale unless stated.

Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.

7. The marks allocated for each question are shown in brackets.

Markah yang diperuntukkan bagi setiap soalan ditunjukkan dalam kurungan.

8. A list of formulae is provided on pages 2 to 4.

Satu senarai rumus disediakan di halaman 2 hingga 4.

9. You may use a non-programmable scientific calculator.

Anda dibenarkan menggunakan kalkulator saintifik yang tidak boleh diprogramkan.

10. Hand in this question paper to the invigilator in at the end of the examination.

Serahkan kertas soalan ini kepada pengawas peperiksaan pada akhir peperiksaan.


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SULIT PP 3472/1

PP 3472/1 @ 2009 Hak Cipta JPWPKL SULIT

…

…

PERATURAN PEMARKAHAN

JABATAN PELAJARAN

WILAYAH PERSEKUTUAN KUALA LUMPUR

PEPERIKSAAN PERCUBAAN SPM 2009 3472/1

ADDITIONAL MATHEMATICS

Kertas 1

September

2 jam Dua jam

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2

Question Solutions Sub Mark Total Mark

(a) 1, –2 1 1

(b) –1 1

2

(a) f(7 – 3x) = 7 – 3(7 – 3x)

= 9x – 14

1

1

2

(b) 2x – k = 4 or 2x – k = –4

2 – k = 4 or 2 – k = – 4

k = –2 , 6 (both)

1

1

4

(a) 12

5

−=

xy using the correct method

0,2

5)(1 ≠

+=− x

x

xxg

1

1

3

(b) h(x) =

+−

x

x

2

532

= 0,2

15≠

−x

x

x

1

1

4

4 2 ( x2 + 2px + p

2 ) – 1 = 0 or

2x2 + 4px + 2p

2 – 1 = 0

SOR: 72

4=−

p

p = 2

7−

1

1

1

3

5 x2 – 7x + 6 > 0

(x – 1)(x – 6) > 0

x < 1 , x > 6 (both)

1

1

1

3

6 (a) h = 4

(b) k = – 5

(c) y-intercept = 43

1

1

1

3

(either)

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3


7 (2x – p) – 2 = 3x2 – x + 5p

3x2 – 3x + 6p + 2 = 0

(-3)2 – 4(3)(6p + 2) > 0

9 – 72p – 24 > 0

P < 24

5−

1

1

1

3

8

22

2 x 3x = 4 OR (x – 2)log 2 + x log 3 = log 4

2x 3x = 16

(6)x = 16

x log10 6 = log10 16 OR x(lg2 + lg3)= lg 4 + 2lg 2

x = 1.547

1

1

1

3

9

3log

2log2logor

9log

2log2log

9

9

3

3

3

9 ==

log3 14 = log3 7 + log3 2

= m + 2n

1

1

1

3

(a) 2m – 3 = 5

m = 4

1

10

(b) 103 = 3 + (n – 1)(5)

103 = 3 + 5n – 5

21 = n

1

1

3

11 8

1−=

−=∞

r

aS , or

4

31

1

)1( 5

5 −=−

−=

r

raS

4

31)1(8 5 −=−− r

r 5 =

32

1 � r =

2

1

1

1

1

3

(a) log10 y = log10 p + 2k log10 x

1

12

(b) 2k = 05

26

−

−

k = 52

log10 p = 2 � p = 100

1

1

1

4

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4


13 A =

2

1

0150

414

−h

= 29

294)1()(5)1(0)0()1(1)5)(4(2

1=−−−−+−+ hh

5h = 23 – 58 or 5h = 23 + 58

h = –7 , 5

81

� h = –7

1

1

1

3

14

31

)1(31,

31

)0(33

+

+−=

+

+=

yx

B( 12, 7)

1, 1

1

3

15

Without arrow ( 1 mark)

2

2

(a) u + v =

−

+

3

2

1

7

=

− 29

22 )2(9 −+=+ vu

= 85 = 9.220

1

1

1

16

(b) 85

29 ji −

1

4

O

M

m

n

•

N

R •

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5


17 xxdx

dy126 2 −= = 0

6x(x – 2) = 0

x = 0 , x = 2

12122

2

−= xdx

yd

When x = 0, 2

2

dx

yd= –12 < 0

� y max when x = 0

1

1

1

3

18 π16=

dt

dA , r = 4 A = πr2

rdr

dAπ2=

dtdr

drdA

dtdA ×=

16π = 2πr × dtdr

)(dt

dr

42

16

π

π= = 2 cm s

–1

1

1

1

3

19 ∫∫ =+

44

12910)(2

kdxdxxg

2(7) + 29104=

kx

10(4) – 10k = 15

k = 2.5

1

1

1

3

20 2cos 2x – cos x – 1 = 0

2( 2cos2 x – 1) – cos x – 1 = 0

4 cos2 x – cos x – 3 = 0

(4cosx + 3)(cos x – 1) = 0

cos x = – , cos x = 1

x = 0°, 138.59° , 221.41°, 360°

1

1

1

1

4

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6


21 AORS = AOAB

(12)2 θ = (8)2 (2π )

θ = 0.6982 rad

1, 1

1 3

(a) 0, 1 , 2 1 22

(b) 22222

)2(8

)2(4)3(2)2(1)1(0−

+++=σ

= 1.323

1

1

3

(a) 6P5 = 720 ways 1,1 23

(b) 4 × ( 2! × 4P3 )

= 192

1

1

4

(a) 5

4

3

1×

= 15

4

1

24

(b)

×+

×+

×

5

1

3

2

5

1

3

1

5

4

3

2 or 1 –

15

4

= 15

11

1

1

3

(a) P(X=6) = 466

10 )15.0()85.0(C

= 0.04010

1

1

25

(b) variance = n p q

5.355 = n (0.85)(0.15)

n = 42

1

1

4

END OF MARK SCHEME

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