ADDITIONAL MATHEMATICS - Penang Free School papers Add Maths... · program didik cemerlang akademik spm model spm questions ( paper 1 ) organised by: jabatan pelajaran negeri pulau

PROGRAM DIDIK CEMERLANG AKADEMIK

SPM

MODEL SPM QUE

ORGA

JABATAN PELAJARA

ADDITMATHEM

MODU

http://maths

http://sahatmozac.blogspot.com

IONALATICS

STIONS ( PAPER 1 )

NISED BY:

N NEGERI PULAU PINANG

LE 19

mozac.blogspot.com

Additional Mathematics ( Paper 1) SPM

Jabatan Pelajaran Pulau Pinang 2

3472/1 NO. KAD PENGENALANAdditionalMathematicsPaper 1 ANGKA GILIRAN

2 Hours

JABATAN PELAJARAN NEGERI PULAU PINANG

ADDITIONAL MATHEMATICS

Paper 1

Two Hours

JANGAN BUKA KERTAS SOALAN INISEHINGGA DIBERITAHU

1. Tuliskan angka giliran dan nombor kadpengenalan anda pada ruang yangdisediakan.

2. Calon dikehendaki membaca arahandi halaman 2.

Kod PemeriksaQuestions Marks Actual Marks

1 22 33 34 35 36 37 38 39 310 211 412 413 414 215 216 317 418 419 420 421 322 423 324 325 4

Total

http://mathsmozac.blogspot.com




INFORMATION FOR CANDIDATES

1. This question paper consists of 25 questions.

2. Answer all questions.

3. Give only one answer for each question.

4. Write your answer 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-figure mathematical tables is provided .

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

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 examination.





Answer all questions

1 . Diagram 1, the function h maps x to y and the function k maps y to z.

DIAGRAM 1

Determine

(a) hk –1 (– 4),

(b) kh –1 (– 2). [2 marks]

Answer : (a) ___________

(b) ___________

2. The function p–1 is defined as p–1 (x) =x

x

4

3, x ≠ 4.

Find

(a) p(x),

(b) p(5). [3 marks]

Answer : (a) ___________

(b) ___________

x y z

4

3

2

hk




Jabatan Pelajaran Pulau Pina

3. The following information refers to the functions h and g.

Find f (x).

4. The straight line y = t

Find the range of values

5. Solve the quadratic equ

places .


g (x) = 4 – 3 x

ng 5

[3 marks]

Answer : ___________

intercept with the curve y = 3 + 5x – 4x2 at the two points A and B .

of t . [3 marks]

Jawapan : ___________

ation5

2

3

12 2

xxGive your answer correct to three decimal

[3 marks]

Answer : x = ___________

fg (x) = 2 x + 5




6. Diagram 2 shows the graph of a quadratic functions g(x) = 4 – 2(x + h)2 where h is constant.

DIAGRAM 2

The curve y = g(x) has maximum point (3, k) , where k is a constant .State

(a) the value of h,

(b) the value of k,.

(c) the equation of the axis of symmetry .[3 marks]

Answer : (a) h = ___________

(b) k = ___________

(c) ___________

7. Solve the equation .)4)(8(

164

1

3

xx

x [3 marks]

Answer : x = ___________

y = g(x)●

(3,k)

x

y

O





8. Solve the equation log2 (x – 1) + 2 = 2 log2 x [3 marks]

Answer : x = ___________

9. Given that log2 x = p and log2 y = r, express

2

3

2

32log

y

xin term p and r.

[3 marks]

Answer : ___________

10. The sum of the first n term of an arithmetric progression is given by Sn= 5n – 3. Find the

fifth term . [2 marks]

Answer : ___________





11. The first three terms of an arithmetic progression are –2

1, 1, –2 , …

Find

(a) the common ratio ,

(b) the sum of the first 10 terms after the 5th term .

[ 4 marks]

Answer : (a) ___________

(b) ___________

12. The sum of the first n terms of an arithmetric progression 47, 44, 41, … is –1325. Find

(a) the common difference of the progression ,

(b) the value of n. [4 marks]

Answer : (a) ______________

(b) n = ___________




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13. Diagram 3 shows a straight line graph of log10 y againts log10 x.

The variables x and y are related by equation y = p x q, where p and q are contants .

DIAGRAM 3

(a) Calculate the value of p and q,

(b) Find the value of y if x = 10. [4 marks]

Answer : (a) p = ___________

q = ___________

(b) y =___________

14. The following information refers to the equations of two straight lines , AB and CD , which

parallel to each other.

Express p

log10 y

log10 x

2

4 O


AB : 2y = p x + q

CD : 3y = (q + 1) x + 2

n Pulau Pinang 9

in terms q. [2 marks]

Answer : p =___________

Where p and q are constants




15. Given that B point is (–8, 5) and O is origin.

(a) ExpressOB in terms of

~i and

~

j .

(b) Find the unit vector in the direction ofOB . [2 marks]

Answer : (a) ___________

(b) ___________

16. Given thatOA = –2

~i + 3

~

j ,OB = 10

~i + 6

~

j and R is a point on AB such that AR :

AB = 2 : 3. Find

(a)AB ,

(b)OR . [3 marks]

Answer : (a) ___________

(b) ___________

17. Solved the equation 3cot

1 2

2 xsek

xfor 0 ≤ x ≤ 360.

[4 marks]

Answer : x = ___________http://mathsmozac.blogspot.com



Jabatan Pelajaran P

18. Diagram 4 shows the sector OPQ, centre O with a radius of 5 cm . Line PR is perpendicular tothe line OQ and QR = 1 cm.

[4 marks]

Find ( a ) P

( b ) the

19. Given that (f

Q

P O

R


ulau Pinang 11

OR in radians ,

perimeter of the shaded region .

Answer : ( a ) ………………….

( b ) ………………….

2)25(4) xxx , find )(" xf [4 marks]

Answer : ___________

RAJAH 4




20. Given that x = t + 3t 2 and y = 2t – 1

(a) finddy

dxin term of t.

(b) If y decreases from 4.0 to 3.98, Find the corresponding small change in t .

[4 marks]

Answer : (a) ___________

(b) ___________

21. Given that23

21

x

xy

and )(4 xf

dx

dy with )(xf is a function in x .

Calculate the value of dxxf )(32

2 . [3 marks]

Answer : ___________





22. A chess team consists that of 5 students . The team will be chosen from a group of 6 boys

and 4 girls . Calculate the number of teams that can be formed such that each team consists

of

(a) 4 boys,

(b) Not more than 2 girls.[4 marks]

Answer : (a) ___________

(b) ___________

23. The mean of five numbers is m. The sum of the squares of the numbers is 720 and the

standard deviation is 9h2. Express m in term of h.

[3 marks]

Answer: m = ___________





24. Ali has 6 accessories of the item which consists of P, Q, R, S, T and U. He wants to arrange

4 of the items in a row . Find the probability that

(a) the arrangment is TUPS ,

(b) the arrangment does not incule P item

[3 marks]

Answer : (a) ___________

(b) ___________

25. X is a random variable of normal distribution with a mean of 10 and a variance 9 , find the

value of r such that P ( X < r ) = 0 . 9 7 5.

[4 marks]

Answer : ___________

END OF QUESTION PAPER





Answers

1. ( a ) h k – 1 (– 4) = – 2

( b ) k h – 1 (– 2) = – 4

2 . ( a ) p – 1 ( x ) =x

x

4

3

p – 1 ( y ) = y

y

4

3

x =y

y

4

3

x ( 4 – y ) = 3 y

4 x – x y = 3 y

y ( x + 3 ) = 4 x

y =3

4

x

x

p ( x ) =3

4

x

x, x ≠ –3,

( b ) 35

)5(4)5(

p

=2

5

3 . g ( x ) = 4 – 3 x

y = 4 – 3 x

3 x = 4 – y

x =3

4 y

g – 1(x) =3

4 x

f ( x ) = f g g – 1 ( x ) = f g

3

4 x

= 2 53

4

x

=3

223 x

4 . y = t , y = 3 + 5 x – 4 x 2

3 + 5 x – 4 x 2 = t





4 x 2 – 5 x + t – 3 = 0

Intercept at two points , use b 2 – 4 a c > 0

( – 5 ) 2 – 4 ( 4 ) ( t – 3 ) > 0

1 6 t < 7 3

t <16

73

5 . 5 ( 2 x – 1 ) = 3 ( x 2 – 2 )

3 x 2 – 10 x – 1 = 0

Use x =a

acbb

2

42

x =)3(2

)1)(3(4)10()10( 2

x =6

11210

x = 3.43, x = – 0. 0972

6 . g (x) = 4 – 2 ( x + h ) 2

x + h = 0 , x = – h ; y = 4

( – h , 4 ) = ( 3 , k )

(a) h = – 3

(b) k = 4

(c) symmetry Axis : x = 3

7 . 2 6 ( x+3 ) x ½ = 2 – 3x . 2 – 2 ( x + 1 )

3 ( x + 3 ) = – 3 x + [ – 2 x – 2 ]

8 x = – 11

x =8

11

8 . log 2 ( x – 1 ) + 2 = 2 log 2 x 2 = log 2 22 = log 2 4

log 2 4 ( x – 1 ) = log 2 x 2

4 x – 4 = x 2

x 2 – 4 x + 4 = 0

( x – 2 ) 2 = 0

x = 2





9 . 22

3222

3

2 loglog32log32

log yxy

x

= yx 225

2 log2log32log

= 5 + 3 p – 2 r

10 . S n = 5 n – 3

T n = S n – S n – 1

T 5 = S 5 – S 4

= [ 5 ( 5 ) – 3 ] – [ 5 ( 4 ) – 3 ]

= 22 – 17

= 5

11 . (a) r =1

2= – 2

(b) T 1 , T 2 , ……… T 5 , T 6 , ……….. T 15

the sum of = S 15 – S 5

S 15 =

1)2(

1)2(2

1 15

= – 5461.5 ;

S 5 =

1)2(

1)2(2

1 5

= – 5.5

S 15 – S 5 = – 5456

12 . 47, 44, 41, …………

( a ) d = 4 4 – 4 7 = – 3

( b ) S n = ])1(2[2

dnan

)]3)(1()47(2[2

nn

= –1325

3 n 2 – 9 7 n – 2 6 5 0 = 0

( 3 n + 5 3 ) ( n – 5 0 ) = 0

n = 50

13 . y = p x q

( a ) log 10 y = log 10 p + q log 10 x

log 10 p = 2

p = 10 2 = 100





q =2

1

40

02

( b ) log 10 y = 2 + ½ log 10 x

x = 10 , log 10 y = 2 + ½ log 10 10

log 10 y = 2.5

y = 10 2.5

y = 316 . 2

Alternatif method

y = p x q

= 100 x ( 10 ) ½

= 316 . 2

14 . y =22

qx

p

y =3

2

3

1

x

q

3

1

2

qp

p = )1(3

2q

15 . ( a )OB =

~~58 ji

( b ) the unit vector in the direction ofOB =

89

58 ji

16 . ( a )AB =

AO +

OB

= –

3

12

6

10

3

2

( b )OR =

12

)(1)(2

OAOB

OR =

3

21

6

102

12

1

OR =

5

6





17.x2cot

1. + sek 2 x = 3

tan 2 x + sek 2 x = 3

tan 2 x + 1 + tan 2 x = 3

2 tan 2 x = 2

tan 2 x = 1

tan x = ± 1

x = 45 o , 135 o , 225 o , 315 o

18 . ( a ) Cos POR =5

4= 0 . 8

POR = 0 . 6435 r

( b ) s PQ = j r

= 5 x 0.6435 = 3 . 2175

RQ = 1 cm , PR = 3 cm ( Pythagoras Teorem )

The perimeter of the shaded region = 3 + 1 + 3. 2175

= 7 . 2175

19 . f ( x ) = 2)25(4 xx

u = 4x v = ( 5 – 2 x ) 2

4dx

du)2)(25)(2( x

dx

dv

dx

duv

dx

dvu

dx

dy

)4()25()]25(4[4)(' 2xxxxf

= 4 8 x2 – 160 x + 100

16096)('' xxf

20 . (a) x = t + 3 t 2 y = 2 t – 1

tdt

dx61 2

dt

dy

)61(2

1

2

1)61(

t

t

dy

dt

dt

dx

dy

dx

(b) 498.3 y

= -0.02





2dt

dy

t

y

dt

dy

dt

dy

yt

01.0

2

02.0

21 . dxxfdy )(4

dxxfdy )(34

3

2

22

2

23

21

4

3)(3

x

xdxxf

=

12

41

12

41

4

3

=2

1

22 . ( a ) 62024

36 CC

= 120

( b ) Total of player boys and girls = 6 + 4 = 10

The number of ways of selecting from 5 player = 252510 C

The number of ways of selecting player consisting of 1 boy and 4 girls

= 61644

16 CC

The number of team is less than 2 boys = 2 5 2 – 6 = 2 4 6

23 . mx ; 7202x ; 22 9h

22

2 xn

x

9 h 2 = 2

5

720m

m 2 = 144 – 9h2 m =

29144 h

24 . ( a ) The probability of arrangement is TUPS





=360

1

3

1

4

1

5

1

6

1

( b ) The probability does not included the P accessory

=4

6

45

C

C

=3

1

25 . variance = 2 = 9

Standard deviation = = 3

mean = = 10

P ( X < r ) = 0 . 975

Z =

X

P

3

10

3

10 rX= 0 . 975

P

3

10rZ = 0 . 975

1 – P

3

10rZ = 0 . 975

P

3

10rZ = 1 – 0. 975

= 0 . 025

3

10r= 1 . 9 6

r = 1 5 . 8 8



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ADDITIONAL MATHEMATICS - Penang Free School papers Add Maths... · program didik cemerlang akademik spm model spm questions ( paper 1 ) organised by: jabatan pelajaran negeri pulau