K1 SARAWAK 2011

SULIT 3472/1 Matematik Tambahan Kertas 1 Sept 2011 2 Jam

1 Name : ………………..…………… Form : ………………………..……

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN PEPERIKSAAN PERCUBAAN SPM 2011

MATEMATIK TAMBAHAN

Kertas 1 Dua jam JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU 1

This question paper consists of 25 questions.

2. Answer all questions. 3. Give only one answer for each question. 4. Write your answers clearly in the spaces provided in the question paper. 5. Show your working. It may help you to get marks. 6. If you wish to change your answer, cross out the work that you have done. Then write down the new answer. 7. The diagrams in the questions provided are not drawn to scale unless stated. 8. The marks allocated for each question and sub-part of a question are shown in brackets. 9. A list of formulae is provided on pages 2 to 3. 10. A booklet of four-figure mathematical tables is provided. . 11 You may use a non-programmable scientific calculator. 12 This question paper must be handed in at the end of the examination .

For examiner’s use only Question

Total Marks

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

2 3 4 3 3 3 3 4 3 3 4 3 3 3 3 3 4 3 3 3 3 3 4 3 4

TOTAL

80

Kertas soalan ini mengandungi 16 halaman bercetak

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SULIT

Marks Obtained

2

SULIT

3472/1

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

−b ± b 2 − 4ac 2a

1

x=

2

a m × an = a m +

n

3

a m ÷ an = a m −

n

4

(am)n = a mn

5

log a mn = log a m + log a n m log a = log a m − log a n n log a mn = n log a m

6 7

log c b log c a

8

log a b =

9

Tn = a + (n − 1)d n [2a + (n − 1)d ] 2 Tn = ar n − 1 a (r n − 1) a (1 − r n ) Sn = = , (r ≠ 1) r −1 1− r a S∞ = , r <1 1− r Sn =

10 11 12 13

CALCULUS 1

y = uv , du

4 Area under a curve

dv

u dy v dx − u dx 2 y= , = , v2 v dx dy dv du =u +v dx dx dx

b

∫y

=

dx or

a b

=

∫ x dy a

3

dy dy du = × dx du dx

5 Volume generated b

= πy 2 dx or

∫ a b

=

∫ πx

2

dy

a

GEOMETRY

1 Distance =

( x1 − x 2 ) 2 + ( y1 − y 2 ) 2

2 Midpoint

y + y2   x1 + x 2 , 1  2   2

( x , y) = 

3

r = x2 + y2

4

rˆ =

3472/1

5

A point dividing a segment of a line  nx + mx 2 ny1 + my 2  (x, y) =  1 ,  m+n   m+n

6 Area of triangle 1 2

= ( x1 y2 + x2 y3 + x3 y1 ) − ( x2 y1 + x3 y2 + x1 y3 )

xi + yj x2 + y 2

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STATISTIC

1

x =

2

x =

3

4

5

6

σ =

σ =

∑x N

7

∑ fx ∑f

8

∑ ( x − x )2

∑ x2

=

N

∑ f ( x − x )2 ∑f

N

=

9 − x2

∑ fx 2 − x 2 ∑f

1  2 N −F C m = L+  fm   

∑ w1 I1 ∑ w1 n! n Pr = (n − r )! n! n Cr = (n − r )!r!

I=

10

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

11

P(X = r) = nCr p r q n − r , p + q = 1

12

Mean µ = np

13

σ = npq

14

z=

Q I = 1 × 100 Q0

x−µ

σ

TRIGONOMETRY

1 Arc length, s = r θ 2 Area of sector , A = 3 sin 2A + cos 2A = 1

9 sin (A ± B) = sinA cosB ± cosA sinB 1 2 rθ 2

10 cos (A ± B) = cosA cosB ∓ sinA sinB 11 tan (A ± B) =

tan A ± tan B 1 ∓ tan A tan B

4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A

12

a b c = = sin A sin B sin C

13

a2 = b2 + c2 − 2bc cosA

14

Area of triangle =

6 sin 2A = 2 sinA cosA 7 cos 2A = cos2A – sin2 A = 2 cos2A − 1 = 1 − 2 sin2A 8 tan 2A =

3472

1 absin C 2

2 tan A 1 − tan 2 A

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SULIT

3472/1

THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1) 7

8

9

24

28

32

36

24

28

32

36

19

23

27

31

35

19

22

26

30

34

15

18

22

25

29

32

14

17

20

24

27

31

10

13

16

19

23

26

29

9

12

15

18

21

24

27

5

8

11

14

16

19

22

25

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

19

21

0.1170

2

4

6

8

10

12

14

16

18

0.0985

2

4

6

7

9

11

13

15

17

0.0838

0.0823

2

3

5

6

8

10

11

13

14

0.0694

0.0681

1

3

4

6

7

8

10

11

13

0.0582

0.0571

0.0559

1

2

4

5

6

7

8

10

11

0..0475

0.0465

0.0455

1

2

3

4

5

6

7

8

9

0.0392

0.0384

0.0375

0.0367

1

2

3

4

4

5

6

7

8

0.0322

0.0314

0.0307

0.0301

0.0294

1

1

2

3

4

4

5

6

6

0.0256

0.0250

0.0244

0.0239

0.0233

1

1

2

2

3

4

4

5

5

0.0207

0.0202

0.0197

0.0192

0.0188

0.0183

0

1

1

2

2

3

3

4

4

0.0162

0.0158

0.0154

0.0150

0.0146

0.0143

0

1

1

2

2

2

3

3

4

0.0125

0.0122

0.0119

0.0116

0.0113

0.0110

0

1

1

1

2

2

2

3

3

0

1

1

1

1

2

2

2

2

3

5

8

10

13

15

18

20

23

2

5

7

9

12

14

16

16

21

1

2

3

4

5

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

0.4641

4

8

12

16

20

0.1

0.4602

0.4562

0.4522

0.4483

0.4443

0.4404

0.4364

0.4325

0.4286

0.4247

4

8

12

16

20

0.2

0.4207

0.4168

0.4129

0.4090

0.4052

0.4013

0.3974

0.3936

0.3897

0.3859

4

8

12

15

0.3

0.3821

0.3783

0.3745

0.3707

0.3669

0.3632

0.3594

0.3557

0.3520

0.3483

4

7

11

15

0.4

0.3446

0.3409

0.3372

0.3336

0.3300

0.3264

0.3228

0.3192

0.3156

0.3121

4

7

11

0.5

0.3085

0.3050

0.3015

0.2981

0.2946

0.2912

0.2877

0.2843

0.2810

0.2776

3

7

10

0.6

0.2743

0.2709

0.2676

0.2643

0.2611

0.2578

0.2546

0.2514

0.2483

0.2451

3

7

0.7

0.2420

0.2389

0.2358

0.2327

0.2296

0.2266

0.2236

0.2206

0.2177

0.2148

3

6

0.8

0.2119

0.2090

0.2061

0.2033

0.2005

0.1977

0.1949

0.1922

0.1894

0.1867

3

0.9

0.1841

0.1814

0.1788

0.1762

0.1736

0.1711

0.1685

0.1660

0.1635

0.1611

1.0

0.1587

0.1562

0.1539

0.1515

0.1492

0.1469

0.1446

0.1423

0.1401

0.1379

1.1

0.1357

0.1335

0.1314

0.1292

0.1271

0.1251

0.1230

0.1210

0.1190

1.2

0.1151

0.1131

0.1112

0.1093

0.1075

0.1056

0.1038

0.1020

0.1003

1.3

0.0968

0.0951

0.0934

0.0918

0.0901

0.0885

0.0869

0.0853

1.4

0.0808

0.0793

0.0778

0.0764

0.0749

0.0735

0.0721

0.0708

1.5

0.0668

0.0655

0.0643

0.0630

0.0618

0.0606

0.0594

1.6

0.0548

0.0537

0.0526

0.0516

0.0505

0.0495

0.0485

1.7

0.0446

0.0436

0.0427

0.0418

0.0409

0.0401

1.8

0.0359

0.0351

0.0344

0.0336

0.0329

1.9

0.0287

0.0281

0.0274

0.0268

0.0262

2.0

0.0228

0.0222

0.0217

0.0212

2.1

0.0179

0.0174

0.0170

0.0166

2.2

0.0139

0.0136

0.0132

0.0129

2.3

0.0107

0.0104

0.0102 0.00990

2.4

0.00820

0.00798

0.00776

0.00755

0.00964

0.00939

0.00914 0.00889

0.00866

0.00842

0.00734

6

Minus / Tolak

2

4

6

8

11

13

15

17

19

0.00714

0.00695

0.00676

0.00657

0.00639

2

4

6

7

9

11

13

15

17

2.5

0.00621

0.00604

0.00587

0.00570

0.00554

0.00539

0.00523

0.00508

0.00494

0.00480

2

3

5

6

8

9

11

12

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

0.00187

0.00181

0.00175

0.00169

0.00164

0.00159

0.00154

0.00149

0.00144

0.00139

0

1

1

2

2

3

3

4

4

3.0

0.00135

0.00131

0.00126

0.00122

0.00118

0.00114

0.00111

0.00107

0.00104

0.00100

0

1

1

2

2

2

3

3

4

Example / Contoh:

 1  exp − z 2  2π  2  1

f ( z) =

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

∞

Q ( z ) = ∫ f ( z ) dz

P(X > k) = Q(k)

Q(z)

k

P(X > 2.1) = Q(2.1) = 0.0179

O

3472/1

k

z

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SULIT

3472/1 For examiner’s use only

Answer all questions. 2

Diagram 1 shows the graph of the function f(x) = (x − 1) .

1.

f(x)

f(x) = (x − 1)

0

2

x

k Diagram 1

State (a) (b)

the type of relation, the value of k. [2 marks]

Answer : (a)

(b)

1 2

2.

The function f −1 is defined by f −1 ( x ) = (a) (b)

3 ,x ≠ k. x−2

State the value of k. Find the function f . [3 marks]

Answer : (a) (b)

2 3

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3472/1

3. Given the function f : x → 2 x − 3 and composite function fg : x → 6 x 2 − 4 x + 1 . Find (a) (b)

g(x), the value of gf(−1). [4 marks]

Answer : (a)

(b)

3 4

4.

Given the equation x 2 + 2 x = − k has two distinct roots, find the range of values of k. [3 marks] Answer :

4 3

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5.

3472/1 2

Diagram 5 shows the graph of function y = (x − 2) + q, where q is a constant. Given that the line y = 3 is the tangent to the curve.

For examiner’s use only

y 2

k

y = (x − 2) + q

y=3 O

x Diagram 5

(a)

State the equation of axis of symmetry.

(b)

State the value of q.

(c)

Find the value of k. [3 marks]

Answer : (a) (b) (c)

5 3

___________________________________________________________________________ 6.

Find the range of values of x which satisfies 4x − 5x2 ≤ −1

[3 marks]

Answer :

6 3

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SULIT

For examiner’s use only

Solve the equation 64 x+3 = 8 x 4 x+1 .

7.

3472/1 [3 marks]

Answer :

7 3

 m  8. Given that log 2 m = r and log 2 n = t , express log 8  3  in terms of r and / or t.  16n  [4 marks] Answer :

8 4

9.

If 3, x, y and 15 are consecutive terms of an arithmetic progression, find the value of x and y. [3 marks] Answer :

9 3

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10.

3472/1

The third and sixth terms of a geometric progression are 1 and 8 respectively. Find the [3 marks] first term and common ratio of the progression.

For examiner’s use only

Answer :

10 3

11.

Express 0.363636... in the form of 2.363636... as a single fraction.

p where p and q are positive integers. Hence express q [4 marks]

Answer :

11 4

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SULIT For examiner’s use only

12.

3472/1

The variables x and y are related by the equation y = 7 x − 2 x 2 . A straight line graph y is obtained by plotting against x, as shown in Diagram 12. x y x (k, 1)

x

O

(7, h) Diagram 12 Find the value of h and of k.

[3 marks]

Answer :

12

3

13.

Diagram 13 shows a quadrilateral PQRS. y Q(4a, 3a) P(−5, 4)

O

x R(6, −1)

S(−2, −4) Diagram 13 Given the area of the quadrilateral is 80 unit2, find the value of a.

[3 marks]

Answer :

13 3

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SULIT 14.

3472/1 For examiner’s use only

Given A(−5, k), B(−1, 6), C(1, −5). Find the possible values of k if AB = 2BC. [3 marks] Answer :

14 3

15.

 Diagram 15 shows vector OA drawn on a Cartesian plane. y 6

A

4

2

2

0

(a) (b)

4

6

8

10

12

x

Diagram 15  x Express OA in the form   .  y  Find the unit vector in the direction of OA . [3 marks]

Answer : (a)

(b)

15 3

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3472/1

16 . Given that a = (2k − 1) i + 3 j and b = 4 i + 5 j . Find the value of k if 2a + 3b is parallel to y-axis.

[3 marks]

Answer :

16 3

. ___________________________________________________________________________

17.

5 and θ is an acute angle. 12 Find the value of each of the following It is given that tan θ =

(a) (b)

tan ( −θ ) ,

secθ + sin θ . [4 marks]

Answer : (a)

(b)

17 4

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3472/1

P

18.

O

For examiner’s use only

θ Q Diagram 18

Diagram 18 above shows a sector POQ with centre O. The perimeter of sector POQ is 40 cm. Given that the radius of the sector is 15 cm, find the value of θ , in radians. [3 marks] Answer :

18 3

19. Given that y = 6 x 2 − 4 x , find the small approximate change in y when x increases from 1 to 1.05. [3 marks] Answer :

19 3

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SULIT 20.

Given

14 5

∫ 2 f ( x) dx = 6 and −∫

2 0

f ( x) dx = 2. Find

3472/1

∫5 f ( x ) dx . 0

[3 marks]

Answer :

20 3

___________________________________________________________________________

21.

2

Diagram 21 shows the graph of y = (x − 3) and x = 5. y 2 y =x−3

O

3

5 Diagram 21

x

Find the volume generated when the shaded region is rotated through 360° about x-axis. [3 marks] Answer :

21 3

22.

Given that the mean and the standard deviation of a set of numbers are 7 and 2. If each of the numbers is multiplied by 3, find (a) the mean, (b) the variance of the new set of numbers.

22 3

[3 marks]

Answer :

3472/1

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23. The number of ways in which a group of 4 men and 3 women can be seated in a row of (a) (b)

For examiner’s use only

8 chairs, 8 chairs if the first two chairs in the row are occupied by the men. [4 marks]

Answer :

23 4

___________________________________________________________________________

24.

A box contains 40 marbles. The colours of the marbles are yellow and blue. If a marble 2 is drawn from the box, the probability that a yellow marble drawn is . 5 Find the number of blue marbles that have to be added to the box such that the 5 probability of obtaining a blue marble becomes . [3 marks] 7 Answer :

24 3

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For examiner’s use only

SULIT 25.

16

3472/1

The continuous random variable X is distributed normally with mean µ and variance 25. Given that P (X < 20) = 0.7881 , find the value of µ . [4 marks]

Answer :

25 4

END OF QUESTION PAPER

3472/1

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3472/1 Matematik Tambahan Kertas 1 2 jam Sept 2011

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2011

MATEMATIK TAMBAHAN Kertas 1 Dua jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

MARKING SCHEME

Skema Pemarkahan ini mengandungi 7 halaman bercetak

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MARKING SCHEME FOR PAPER 1 -2011 ZON A No 1.

Solution and marking scheme

Sub Marks

Total Marks

(a) many to one relation

1

2

(b) 1

1

(a) k = 2

1

2.

(b) f ( x)

=

3 + 2x , x ≠ 0. x

2

3 x−2

B1

(a) g ( x) = 3 x 2 − 2 x + 2

2

y=

3

3.

2 g ( x) − 3 = 6 x 2 − 4 x + 1 @ f −1(fg(x)) = f −1(6x2 −4x + 1) (b) 87

4

B1

2

f (−1) = −5

B1

4.

k>1 −4k > −4

3

or 4 < 4k

(2) 2 − 4(1)(−k ) > 0

B2 B1

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3

3 No 5.

Solution and marking scheme

Sub Marks

Total Marks 3

(a)

x=2

1

(b)

q=3

1

(c)

k=7

1

6. 3

1 x ≤ − , x ≥1 5

7.

8.

+

+ 1

−

1 − 5

(5 x + 1)( x − 1) ≥ 0

3

B2 B1

x = −16

3

6 x + 18 = 5 x + 2

B2

26(x+3) = 23x 22x+2

B1

r − 4 − 3t 3 r − log 2 24 − 3log 2 n 3

4

3

4

B3

log 2 m − log 2 16 − log 2 n3 3

B2

 m  log 2    16n3  log 2 23

B1

9. x = 7, y = 11

3

Solving equation x − 3 = y − x or

or

x = 7 @ y = 11

x – 3 = 15 – y

or

d=4

B2

B1

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3

4

No

Solution and marking scheme

Sub Marks

Total Marks

10. r =2,a= r=2

@

3

1 4

a=

1 4

ar2 = 1 ------------ (1)

3

B2 or

ar5= 8 -------------(2)

B1

11. 4

26 11

2.363636… = 2+ 0.363636… = 2 + 4/11

S∞ =

0.36 4 @ 1 − 0.01 11

a = 0.36 and r = 0.0036/0.36 = 0.01

12.

h = −7 ,

k=3

h = −7 @

k=3

y y = −2 x + 7 = 7 − 2 x or x x

4

B3

B2 B1

3

3

B2

B1

13. a=2 1 (53a + 54) = 80 2

B2

1 [ −5(−4) + (−2)(−1) + 6 × 3a + 4a × 4 − ( −2)4 − 6( −4) − 4a(−1) − (−5)(3a) ] = 80 2

B1

3

3

14. k = −16, 28

3

(k − 6) 2 = 484 @ k − 6 = ±22 @ (k + 16)(k − 28) = 0

(−5 − (−1)) 2 + (k − 6)2 = 2 ( −1 − 1)2 + (6 − ( −5)) 2

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B2 B1

3

5 Solution and marking scheme

No 15.

Sub Marks

Total Marks

.




  8  (a) OA =   or 8 i + 4 j  4 8i + 4 j 1 8 (b)   80  4  @ 80


 OA = 82 + 42 = 4 5 @

1

3

2

80

B1

16. k=−

17.

3

5 2

4k + 10 = 0

B2

2a + 3b = (4k + 10)i + 21 j

B1

3

(a) tan ( −θ ) = − tan θ 5 12

1

229 73 or 156 156

3

=−

(b) 1

sec θ =

13 13 5 or sec θ + sin θ = + 12 12 13

B2

sin θ =

5 12 @ cos θ = 13 13

B1

4

18. 3

2 θ = rad 3 10 = 15θ

B2

sPQ = 10 cm

B1

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3

No 19.

6 Solution and marking scheme ∂y ≈ 0.4

Sub Marks 1

∂y ≈ (12(1) − 4)(0.05)

B2

dy = 12 x − 4 or dx

B1

∂x = 1.05 − 1 = 0.05

Total Marks 3

20.

−4 2

3

3

2

f ( x ) dx − ∫ f ( x )dx

∫ 0

B2

5 2

∫

5

f ( x )dx + ∫ f ( x )dx

0

B1

2

21.

2

3

  52   32 − 3 × 5   −  − 3 × 3   5   5 

3

B2

5

 x2   − 3 x   2 3

22.

B1

variance = 36, mean = 21

3

Variance = 36 @ Mean = 21

B2

Variance = 32 × 22 @ Mean = 3 × 7 @ SD = 3×2 @ σ = 6

B1

3

23.

(a) 40320

2

8

B1

P7

(b) 8640 4

P2 × 6 P5

2 B1

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4

7 Solution and marking scheme

No 24.

16

3

168 + 7 x = 200 + 5 x

@

24 + x 5 = 40 + x 7

n(B) = 24 + x and n(S) = 40 + x 25.

µ = 16

20 − µ ) = 0.2119 5

P(X ≥ 20) = 0.2119

Total Marks 3

B2

B1

4

20 − µ = 0.8 (from table) 5 P(Z ≥

Sub Marks

B3

B2

B1

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4