AddMath1 Q&A(Sbh)

8/14/2019 AddMath1 Q&A(Sbh)

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SULIT

3472/1

ADDITIONAL

MATHEMATICS

PAPER 1

AUGUST 20082 HOURS

JABATAN PELAJARAN NEGERI SABAH

SIJIL PELAJARAN MALAYSIA TAHUN 2008

EXCEL 2

___________________________________________________________________________

ADDITIONAL MATHEMATICSPAPER 1 (KERTAS 1)

TWO HOURS (DUA JAM)

___________________________________________________________________________

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1. Tuliskan angka giliran dan nombor kad

pengenalan anda pada ruang yangdisediakan.

2. Calon dikehendaki membaca arahan di

halaman 2 .

___________________________________________________________________________This question paper consists of 14 printed pages.

NAMA : _____________________

KELAS : _____________________

NO K.P : _____________________

A. GILIRAN : _________________-


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2(Kertas soalan ini terdiri daripada 14 halaman bercetak.)

3472/1 [Turn over (Lihat sebelah)

INFORMATION FOR CANDIDATES

1. This question paper consists of 25 questions.2. Answerall questions.3. Give only one answer for each question.4. Write your answers clearly in the space 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 are shown in brackets.9. A list of formulae is provided on pages 3 to 5.10.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.


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3

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

ones commonly used.

ALGEBRA

1.2 4

2

b b acx

a

=

2. m n m na a a + =

3. m n m na a a =

4. ( )m n mna a=

5. log log loga a a

mn m n= +

6. log log loga a am

m nn

=

7. log logna am n m=

8.log

loglog

ca

c

bb

a=

9. ( 1)nT a n d = +

10. [2 ( 1) ]2

n

nS a n d = +

11. 1nn

T ar =

12.( 1) (1 )

, 11 1

n n

n

a r a r S r

r r

= =

13. , 11

aS r

r = + [3 marks]

Answer:

6. The quadratic function 2( ) 3 6 5 can be expressed in thef x x x= + 2form 3( )x p q + , wherep and q are constants. Find the values ofp and q.

[3 marks]

Answer:p = ..

q = ..

7. Given that log 2a p = and1

log2

a q = , find the value of2

logap q

a.

[3 marks]

Answer: ..

For

ExaminersUse

7

3

6

3


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9

8. Solve the equation1

2 1

16 4

y

y y= . [3 marks ]

Answer: ..

9. An arithmetic progression has 15 as the second term and 3

as the commondifference. List the first five terms of the progression. [2 marks]

Answer: .

10. The first three terms of an arithmetic progression are 4, 3,2 2p p p + + .

Find

(a) the value ofp,

(b) the sum of the first 8 terms of the progression. [4 marks]

Answer: (a).

(b).

9

2

For

Examiners

Use

10

4


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10

11. The first term and the fourth term of a geometric progression are 16 and 2

respectively. Calculate the sum to infinity of the geometric progression.

[3 marks]

Answer: ....

12. The variablesx andy are related by the equation 2 25 2 .y x x=

A straight line graph is obtained by plotting2y

xagainstx, as shown in Diagram 2.

[3 marks]

Find the values ofp and q .

Answer:p =...

q =

13. Given a straight line 3 1y mx= + is parallel to 1.3 5

x y

+ = Find the value ofm.[3 marks]

For

Examiners

Use

12

3

13

3

2y

x (3,p)

(q, 4)

xDiagram 2


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11

Answer: m = .

14. Given that the points (2, 1), ( 2,5 1) and ( 2 ,4)K L h M h lie on a straight line,

find the possible values ofh . [3 marks]

Answer: h = ...

15. In Diagram 3, PQRS is a parallelogram and the pointHlies on the straight

line PT. Given that 12PS b=uuur

and 10PQ a=uuur

. Tis a midpoint ofQR and

3PH HT = . Express PHuuur

, in terms of anda b . [3 marks]

S R

Answer: PHuuur

=

For

Examiners

Use

15

3

P Q

TH

10a

12b

Diagram 3


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12

16. Given STUVis a parallelogram, 2 3 and 2 2TV i j UV i j= + = uuur uuur

. Find

(a) TUuuur

,

(b) the unit vector in the direction of STuuur

in terms of andi j .

[4 marks]

Answer: (a) .

(b).. ...

17. Solve the equation 3cos 2 8sin 5 for 0 360o o = . [ 4 marks ]

Answer: ..

18. The curve ( )y f x= is such that2

1dy p

dx x= + , wherep is a constant.

The gradient of the curve at 2x = is1

2. Find the value ofp. [2 marks]

For

Examiners

Use

17

4

18

2


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13

Answer: ..

19. Diagram 4 shows a circle with centre O and radius 4 cm . Given that the area of theminor sectorAOB is 9 cm

2, calculate the length, in cm, of the major arcAB.

[Use = 3.142] [4 marks]

Diagram 4

Answer: ..

20. The curve 22 8 3y x x= + has a maximum point atx =p , wherep is a

constant. Find the value ofp . [3 marks]

Answer: .

21. Given that

3

1

( ) 6g x dx = , find

(a) the value of1

3

( ) ,g x dx

(b) the value ofp if

3

1

[ ( )] 18px g x dx = . [4 marks]

For

Examiners

Use

21

4

20

3

O

A

B

4 cm


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14

Answer: (a) ...

(b) ...

22. A set of data consists of five numbers. The sum of the numbers is 175 and the

sum of the squares of the numbers is 6845. Find, for the five numbers

(a) the mean,

(b) the standard deviation. [3 marks]

Answer: (a) ...

(b) ...

23. A badminton team that consists of 8 students is to be chosen from a group of 7

male students and 6 female students . Calculate the number of different teams

that can be formed if each team must consist of

(a) exactly 3 male students,

(b) not more than 2 female students. [4 marks]

Answer: (a) .

(b) .

24. In a shooting competition, the probability that Lim will strike the target is 0.75.If Lim fires 6 shots, calculate the probability that

(a) all the shots hit the target,

(b) at least one of the shots hits the target. [4 marks]

For

Examiners

Use

23

4

24

4


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15

Answer: (a) .

(b) .

25. Xis a random variable of a normal distribution with a mean of 75 and a standard

deviation of 3 .

(a) Find theZ-score ifXis 70.

(b) (72 79)P X . [4 marks]

Answer: (a) .

(b) .

END OF QUESTION PAPER


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PERATURAN PEMARKAHAN EXCEL 2 PAPER 1

NO. SOLUTION AND MARK SCHEME SUB

MARK

TOTAL

MARK

1. (a) many to one

(b) 2 2: or ( )f x x f x x =

1

1

2

2. P = 3

B1 : 7(2) + 4 = 5p + 3

2 2

3.a =

2, 3 [both]

3b =

B2 :2 3 1 2

2 or ,3 2

b ba a

= = =

B1 : 12 2 2

( ) or ( )3 3

xg x b x g x

a a

= =

3 3

4.(a)

1 3,2 2x = (both)

B1 :24 4 4(4)( 3)

(2 1)(2 3) 0 or 2(4)

x x

+ =

(b) p = 6

B1 : 2 4(1)(9) 0p =

2

2

4

5 4 x > 4

B2 : + +

+

4 1

B1 : (x + 4 ) (x 1 ) < 0

3 3

6 p = 1 , q = 2 (both)

B2 : 3 (x 1 ) 2 + 2

B1 : 3 (x 2 2x +3

5)

3 3

73

2

1or

2

7or 3.5

B2 : 4 +2

1 1

B1: log ap2 + log a q log a a

3 3

8 y = 4

B2 : 4( 1) 2y y y = or 3y = 4y 4

B1 :4( 1) 2

2 1

2 2

y

y y= or equivalent

3 3


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17

9 18, 15, 12, 9, 6

B1: a = 18

2 2

10 (a) p = 8

B1 : ( 4)(2 2) 2( 3)p p p + = + or equivalent

(b) 228

B1 : S8

=2

8[ 2 (4) + 7 ( 7) ] or equivalent

2

2

4

11 32

B2 :

)2

1(1

16

or equivalent

B1 : r= 2

1

3 3

12q =

2

1, p = 1 ( both)

B2 : 4 = 5 2q orp = 2 ( 3) + 5

B1 :x

y 2= 2x + 5

3 3

13 5

B2 :3

m=

3

5

B1 : gradient =3

m, gradient =

3

5(both)

3 3

14h =

5

8, h = 1 (both)

B2 : 5h 2 + 3h 8 = 0

B1 :

21 10 2 8 2 2 10 2 82

4 (5 1) 4 1OR or equivalent

2 2 2 2

h h h h

h

h h

+ +

=

+

3 3

15

2

3( 5 a + 3 b )

B2 :3

(10 6 )4

PH a b= +uuur

B1 : 10 6PT a b= +uuur

3 3


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18

16(a)

5

4or 4 i + 5 j

B1 :

3

2

2

2or 2 i + 3 j ( 2 i 2 j )

(b)2

1( i + j ) or

2

1

1

1or equivalent

B1 : 44 + Or 8 Or 2 2

2

2

4

17 41.81, 138.19

B3 : sin =3

2, sin = 2 ( both)

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

B1 :3 ( 1 2 sin 2 ) = 8 sin 5

4 4

18 p = 2

B1 :2

12

1( 2)

p= +

2 2

19 s = 20.64 cm

B3 : s = 4 (5.159) rad

B2 : majorAOB = (28

9) rad or 5.159 rad

B1 : 9 =21 ( 4) 2

4 4

20 p = 2

B2 : x 2 = 0 or 4p + 8 = 0

B1 : 2 [ (x 2 ) 2 ( 2) 2 +2

3] or 4 8

dyx

dx= +

3 3

21 (a) 6

(b) p = 6

B2 :2 2(3) 1

( ) 6 18 or equivalent2 2

p =

B1 :

32

1

6 18

2

px =

1

3

4

22 (a) mean = 35

(b) 12

B1: 26845

(35)5

or equivalent

1

2

3


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19

23 (a) 210

B1 : 7 C 3 x6 C 5

(b) 111

B1 : 7 C 7 x6 C 1 +

7 C 6 x6 C 2

2

2

4

24 (a) 0.1780B1 : 6 C 6 ( 0.75)

6 (0.25) 0 or equivalent

(b) 0.9998

B1 : 1 P(x = 0)

2

2

4

25 (a) Z= 1.667

B1 :3

7570 or equivalent

(b) 0.7449 0.7501

B1 : 1 P(x 1) P(x 1.333) or equivalent

2

2

4

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AddMath1 Q&A(Sbh)