Add Mth 1,2 Sabah

MOZ@C NAMA : _____________________ KELAS : _____________________ NO K.P : _____________________ A. GILIRAN : _____________________________________

SULIT 3472/1 ADDITIONAL MATHEMATICS PAPER 1 AUGUST 2008 2 HOURS

JABATAN PELAJARAN NEGERI SABAH SIJIL PELAJARAN MALAYSIA TAHUN 2008 EXCEL 2 ___________________________________________________________________________

ADDITIONAL MATHEMATICS PAPER 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 yang disediakan.

2.

Calon dikehendaki membaca arahan di halaman 2 .

___________________________________________________________________________ This question paper consists of 14 printed pages.

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MOZ@C 2 3472/1

(Kertas soalan ini terdiri daripada 14 halaman bercetak.) [Turn over (Lihat sebelah)

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 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.

SMS MUZAFFAR SYAH , MELAKA

MOZ@C 3

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.

a m  a n  a mn

9.

Tn  a  (n  1)d

3.

a m  a n  a m n

4.

(a m ) n  a mn

n 10. S n  [2a  (n  1)d ] 2

5.

log a mn  log a m  log a n

m  log a m  log a n n

6.

log a

7.

log a m n  n log a m

11. Tn  ar n 1 a (r n 1) a (1  r n ) 12. Sn   ,r 1 r 1 1 r 13. S 

a , r 1 1 r

CALCULUS 1.

y  uv,

dy dv du u v dx dx dx

4.

Area under a curve b

=  y dx or a

b

2.

du dv v u u dy y ,  dx 2 dx v dx v

3.

dy dy du   dx du dx

=

 x dy a

5.

Volume generated b

=   y 2 dx or a

SMS MUZAFFAR SYAH , MELAKA

MOZ@C 4 b

=   x 2 dy a

STATISTICS 1.

2.

3.

x

x N

7.

i i

W

 fx x f 

i

 (x  x )

x

2

N

 f (x  x ) f

4.



5.

1  N F m  L 2 fm  

6.

I 

W I



N

2



8.

n

9.

n

2

P

r



n!  n  r !

 x2

 fx f

2

 x2

  c  

n!

C   n  r !r ! r

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.   npq

Q I  1 100 Qo

14. Z 

X  

GEOMETRY 1.

Distance =

2.

4.

 x1  x2    y1  y2  2

1 ( x1 y2  x2 y3  x3 y1 )  ( x2 y1  x3 y2  x1 y3 ) 2

2

Midpoint x1  x2 y1  y2  ,  2   2

5.

r  x2  y2 

6.

rˆ  

 x, y    3.

A point dividing a segment of a line

Area of triangle =

xi  yj   x2  y 2

SMS MUZAFFAR SYAH , MELAKA

MOZ@C 5 nx1  mx2 ny1  my2  ,  mn   mn

 x, y   

TRIGONOMETRY 1.

Arc length, s  r

2.

Area of sector, A 

1 2 r 2

3.

sin 2 A  cos 2 A  1

4.

sec2 A  1  tan 2 A

5.

cosec 2 A  1  cot 2 A

6.

sin 2 A  2sin A cos A

7.

cos 2 A  cos 2 A  sin 2 A

 2 cos 2 A  1

8.

sin ( A  B )  sin A cos B  cos A sin B

9.

cos ( A  B)  cos Acos B  sin A sin B

10. tan ( A  B) 

11. tan 2 A 

12.

tan A  tan B 1  tan A tan B

2 tan A 1  tan 2 A

a b c   sin A sin B sin C

13. a 2  b 2  c 2  2bc cos A

 1  2sin 2 A 14. Area of triangle 

SMS MUZAFFAR SYAH , MELAKA

1 ab sin C 2

MOZ@C 6

For Examiner’s Use

Answer all questions. 1. In Diagram 1, set P is the domain and set Q is the codomain of a relation. 4

 4

3

 9

2

 16

2 

Set P

Set Q Diagram 1

(a) (b)

1

State the type of relation between set P and set Q . Using function notation, state the relation between set P and set Q. [2 marks]

Answer : (a) …………………… (b) …………………….. 2

2.

Given that f : x  7 x  4 . Find the value of p if f (2)  5 p  3 . [2 marks]

2

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MOZ@C 7

Answer : p = ……………………

3.

3 Given that g 1 : x  ax  2, and g ( x)  b  x. Find the values of a and b. 2 [3 marks]

Answer : a =………………………..

For Examiner’s Use

3

b =……………………….. 3

4.

(a) Solve the following quadratic equation: 4x2  4x  3  0 (b) Given the quadratic equation x 2  px  9  0 has two equal roots. Find the values of p. [4 marks] 4

4

Answer : (a)….…………………… (b)….……………………

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5

MOZ@C 8 Find the range of the values of x for ( x  1) 2  (2 x  3)( x  1).

5.

[3 marks]

Answer : ……………………………

For Examiner’s Use

6.

The quadratic function f ( x)  3 x 2  6 x  5 can be expressed in the form 3( x  p ) 2  q , where p and q are constants. Find the values of p and q. [3 marks]

6

Answer : p = ….………….…… 3

q = ….………….……

7.

Given that log a p  2 and log a q 

1 p 2q , find the value of log a . 2 a

[3 marks]

7

3

Answer : ….………….…………

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MOZ@C 9

8.

2y 1 Solve the equation  y. y 1 16 4

[3 marks ]

Answer : ….………….………… 9.

An arithmetic progression has 15 as the second term and 3 as the common difference. List the first five terms of the progression. [2 marks]

For Examiner’s Use

9

Answer : ….………………………

2

10. The first three terms of an arithmetic progression are p  4, p  3, 2 p  2 . Find (a) the value of p, (b) the sum of the first 8 terms of the progression. [4 marks]

10

Answer : (a)….…………………… (b)….……………………

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4

MOZ@C 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 : ..………………………..

For Examiner’s Use

12. The variables x and y are related by the equation y 2  5 x  2 x 2 . y2 A straight line graph is obtained by plotting against x, as shown in Diagram 2. x y2 (3, p) x

(q, 4)

12

[3 marks] x

Diagram 2 3

Find the values of p and q .

Answer : p =.…………….…….……… q =…………………………… 13. Given a straight line 3 y  mx  1 is parallel to

x y   1. Find the value of m. 3 5 [3 marks]

13

3

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MOZ@C 11

Answer : m = …………………………. 14. Given that the points K (2, 1), L( 2,5h  1) and M ( 2h, 4) lie on a straight line, find the possible values of h . [3 marks]

Answer : h = …………………………...

15. In Diagram 3, PQRS is a parallelogram and the point H lies on the straight   line PT. Given that PS  12b and PQ  10 a . T is a midpoint of QR and  PH  3HT . Express PH , in terms of a and b . [3 marks] S R

12b

T

H

P

10a

For Examiner’s Use

Q

Diagram 3 15

3

 Answer : PH =……………………

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MOZ@C 12 16.

  Given STUV is a parallelogram, TV  2i  3 j and UV  2i  2 j . Find  (a) TU ,  (b) the unit vector in the direction of ST in terms of i and j . [4 marks]

Answer : (a) ………………………. (b).. ……….……………..

For Examiner’s Use

17. Solve the equation 3cos 2  8sin   5 for 0o    360o .

[ 4 marks ]

17

Answer : ….………….……………

4

dy p   2  1 , where p is a constant. dx x 1 The gradient of the curve at x  2 is . Find the value of p. [2 marks] 2

18. The curve y  f ( x) is such that

18

2

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MOZ@C 13 Answer : ….………….……………

19. Diagram 4 shows a circle with centre O and radius 4 cm . Given that the area of the minor sector AOB is 9 cm2, calculate the length, in cm, of the major arc AB. [Use  = 3.142] [4 marks] A 4 cm

O

B Diagram 4

Answer : ….………….………… 20. The curve y  2 x 2  8 x  3 has a maximum point at x = p , where p is a constant. Find the value of p . [3 marks]

For Examiner’s Use

20

Answer : ….……………………… 3 3

21. Given that

 g ( x) dx  6 , find 1

1

(a) the value of

 g ( x) dx, 3

3

(b) the value of p if  [ px  g ( x)]dx  18 .

[4 marks]

1

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21

4

MOZ@C 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) ...……………………… For Examiner’s Use

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]

23

4

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]

24

4

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MOZ@C 15

Answer : (a) ………….…………… (b) ……………………….

25. X is a random variable of a normal distribution with a mean of 75 and a standard deviation of 3 . (a) Find the Z-score if X is 70. (b) P (72  X  79) . [4 marks]

Answer : (a) ………….…………… (b) ……………………….

END OF QUESTION PAPER

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MOZ@C 16 PERATURAN PEMARKAHAN EXCEL 2 PAPER 1

NO. 1. 2.

3.

4.

5

SOLUTION AND MARK SCHEME (a) many to one (b) f : x  x 2 or f ( x)  x 2 P =3 B1 : 7(2) + 4 = 5p + 3 2 a =  , b  3 [both] 3 2 3 1 2 B2 : 2  b or   , b 3 2 a a 2 2 x 2 B1 : g 1 ( x)  b  x or g ( x)   3 3 a a 1 3 (a) x  ,  (both) 2 2

4  42  4(4)(3) B1 : (2 x  1)(2 x  3)  0 or 2(4) (b) p = 6 B1 : p 2  4(1)(9)  0  4 < x < 1 or 1> x >  4



B2 :

+



6

7

8

4

B1 :



2

4( y 1)



2

2

3

3

2 4

2 3

3

3

3

3

3

3

3

+ 1

y  4( y  1)  2 y or

2y

TOTAL MARK 2

+

B1 : (x + 4 ) (x – 1 ) < 0 p = 1 , q = 2 (both) B2 : 3 ( x  1 ) 2 + 2 5 B1 : 3 ( x 2  2 x + ) 3 1 7 3 or or 3.5 2 2 1 B2 : 4 + 1 2 B1: log a p 2 + log a q – log a a y=4 B2 :

SUB MARK 1 1

3y = 4y – 4

1 or equivalent 22 y

SMS MUZAFFAR SYAH , MELAKA

MOZ@C 17 9

18, 15, 12, 9, 6

2

2

10

B1: a = 18 (a) p = 8

2

4

B1 : ( p  4)(2 p  2)  2( p  3) or equivalent 2

(b) 228 B1 : S 8 = 11

32 B2 :

B1 : 12

13

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

16 1 1 ( ) 2

r=

15

3

3

3

3

3

3

3

3

1 2

1 , p =  1 ( both) 2 B2 : 4 = 5  2q or p =  2 ( 3) + 5 y2 B1 : =  2x + 5 x 5 m 5 = 3 3

B1 : gradient = 14

3

or equivalent

q =

B2 :

3

m , 3

gradient = 

5 3

(both)

8 , h = 1 (both) 5 B2 : 5h 2 + 3h – 8 = 0

h= 

1 10h  2  8  2h  2  10h 2  2h  8 B1 : 2 4  (5h  1) 4 1 OR  or equivalent 2 h  2 2h  2 3 ( 5a+ 3 b ) 2  3 B2 : PH  (10 a  6b ) 4  B1 : PT  10 a  6b

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MOZ@C 18 16

 4 (a)   or 4 i + 5 j 5

2

4

 2   2 B1 :      or 2 i + 3 j  (  2 i  2 j )  3   2 1

(b)

17

18

2

1 1   or equivalent 2 1

( i + j ) or

B1 : 4  4 Or 8 Or 2 2 41.81, 138.19 2 B3 : sin  = , sin  =  2 ( both) 3 B2 : (3 sin   2) (4 sin  +2) = 0 B1 :3 ( 1 – 2 sin 2  ) = 8 sin   5 p=2 1 p  1 2 (2) 2 s = 20.64 cm B3 : s = 4 (5.159) rad

2

4

4

2

2

4

4

3

3

1

4

B1 : 19

20

9 B2 : major AOB = (2− ) rad or 5.159 rad 8 2 1 B1 : 9 = 2 ( 4)  p= 2

B2 : x – 2 = 0 B1 :  2 [ ( x  2 ) 2 21

 4p + 8 = 0 3 dy  (  2) 2 + ] or  4 x  8 2 dx

or

(a)  6 (b) p = 6

3 2

B2 : (

2

p(3) 1  )  6  18 or equivalent 2 2 3

B1 : 22

 px 2     6  18  2 1

1

(a) mean = 35 (b) 12 B1:

6845  (35) 2 or equivalent 5

SMS MUZAFFAR SYAH , MELAKA

2

3

MOZ@C 19 23

24

25

(a) 210 B1 : 7 C 3 x

2 6

4

C5

(b) 111 B1 : 7 C 7 x 6 C 1 + 7 C 6 x 6 C 2 (a) 0.1780 B1 : 6 C 6 ( 0.75) 6 (0.25) 0 or equivalent

2

2

(b) 0.9998 B1 : 1  P( x = 0) (a) Z =  1.667 70  75 B1 : or equivalent 3

2

(b) 0.7449 – 0.7501

2

B1 : 1 –P(x   1) – P( x  1.333) or equivalent

SMS MUZAFFAR SYAH , MELAKA

2

4

4

MOZ@C NAMA : _____________________ KELAS : _____________________ NO K.P : _____________________ A. GILIRAN : _____________________________________

SULIT 3472/2 ADDITIONAL MATHEMATICS PAPER 2 AUGUST 2008 2 ½HOURS

JABATAN PELAJARAN NEGERI SABAH SIJIL PELAJARAN MALAYSIA TAHUN 2008 EXCEL 2 ___________________________________________________________________________

ADDITIONAL MATHEMATICS PAPER 2 (KERTAS 2) TWO HOURS THIRTY MINUTES (DUA JAM TIGA PULUH MINIT) ___________________________________________________________________________ JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

1.

This question paper consists of three sections: Section A, Section B and Section C.

2.

Answer all questions in Section A, four questions from Section B and two questions from Section C.

3.

Give only one answer / solution for each question.

4.

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

5.

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

6.

The marks allocated for each question and sub-part of a question are shown in brackets.

7.

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

8.

A booklet of four-figure mathematical tables is provided.

9.

You may use a non-programmable scientific calculator.

___________________________________________________________________________ This question paper consists of 13 printed pages. (Kertas soalan ini terdiri daripada 13 halaman bercetak.) [Turn over (Lihat sebelah)

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

3472/2

SMS MUZAFFAR SYAH , MELAKA

2 MOZ@C

CONFIDENTIAL

3472/2

ALGEBRA log c b log c a

1.

b  b 2  4ac x 2a

8.

log a b 

2.

a m  a n  a mn

9.

Tn  a  (n  1)d

3.

a m  a n  a m n

10.

4.

(a m ) n  a mn

n S n  [2a  (n  1)d ] 2

11.

Tn  ar n 1

12.

Sn 

a(r n 1) a(1  r n )  ,r 1 r 1 1 r

13.

S 

a , r 1 1 r

5.

log a mn  log a m  log a n

m  log a m  log a n n

6.

log a

7.

log a m  n log a m n

CALCULUS 1.

y  uv,

dy dv du u v dx dx dx

4.

Area under a curve b

=  y dx or a

b

2.

du dv v u u dy y ,  dx 2 dx v dx v

=

 x dy a

5.

Volume generated b

3.

=   y 2 dx or

dy dy du   dx du dx

a

b

=   x 2 dy a

STATISTICS

3472/2

SMS MUZAFFAR SYAH , MELAKA

CONFIDENTIAL

3 MOZ@C

CONFIDENTIAL

x

1.

3472/2

x N

7.

I 

W I

i i

W

 fx x f

2.

 (x  x )



3.

f ( x  x )2

f





1  2N F m  L fm  

5.

I

6.

x

2

N



4.  

i

N

 f

fx 2

n!  n  r !



n!  n  r !r !

n

9.

n

10.

P  A  B   P  A  P  B   P  A  B 

11.

P  X  r   n Cr p r q n  r , p  q  1

12.

Mean, μ = np

13.

  npq

14.

Z

2

x



8.

P

r

2

 x2

  c  

Q1 100 Qo

C

r

x 

GEOMETRY 1.

Distance =

2.

4.

 x1  x2    y1  y2  2

1 ( x1 y2  x2 y3  x3 y1 )  ( x2 y1  x3 y2  x1 y3 ) 2

2

Midpoint x1  x2 y1  y2  ,  2   2

5.

r  x2  y2 

6.

rˆ  

 x, y    3.

Area of triangle =

A point dividing a segment of a line nx  mx2 ny1  my2  ,  x, y    1  mn   mn

xi  yj   x2  y 2

TRIGONOMETRY 1.

Arc length, s  r

3472/2

8.

sin ( A  B )  sin A cos B  cos A sin B

9.

cos ( A  B)  cos Acos B  sin A sin B

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4 MOZ@C

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2.

Area of sector, A 

3.

sin 2 A  cos 2 A  1

4.

sec2 A  1  tan 2 A

5.

cosec 2 A  1  cot 2 A

1 2 r 2

6.

sin 2 A  2sin A cos A

7.

cos 2 A  cos 2 A  sin 2 A

3472/2 tan A  tan B 1  tan A tan B

10.

tan ( A  B) 

11.

tan 2 A 

12.

a b c   sin A sin B sin C

13.

a 2  b 2  c 2  2bc cos A

14.

Area of triangle 

2 tan A 1  tan 2 A

 2 cos 2 A  1  1  2sin 2 A

3472/2

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1 ab sin C 2

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5 MOZ@C

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3472/2

Section A [40 marks] Answer all questions.

1

Solve the simultaneous equations 4 x  y  x 2  x  y  3 .

2

Diagram 1 shows a straight line CD which meets a straight line AB at point D. The

[5 marks]

point C lies on the y-axis.

Diagram 1

3

(a)

State the equation of AB in the intercept form.

[1 mark]

(b)

Given that 2AD = DB, find the coordinates of D.

[3 marks]

(c)

Given that CD is perpendicular to AB, find the y-intercept of CD. [3 marks]

(a)

Sketch the graph of y  3sin 2 x for 0  x  2 .

(b)

Hence, using the same axes, sketch a suitable straight line to find the number of solutions for the equation 3sin 2 x 

x =1 for 0  x  2 . State the number 

of solutions.

4

[4 marks]

[3 marks]

Given that the gradient of the tangent to the curve y  2 x3  6 x 2  9 x  1 at point P is 3, find

3472/2

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6 MOZ@C

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5

3472/2

(a)

the coordinates of P,

[2 marks]

(b)

the equation of the tangent and normal to the curve at P.

[4 marks]

Table 1 shows the distribution of the ages of 100 teachers in a secondary school. Age (years) Number of teachers

<30

<35

<40

<45

<50

<55

<60

8

22

42

68

88

98

100

Table 1 (a)

Based on Table 1, copy and complete Table 2. Age (years)

25 - 29

Frequency Table 2 [2 marks] (b)

Without drawing an ogive, calculate the interquartile range of the distribution. [5 marks]

6

The first three terms of a geometric progression are also the first, ninth and eleventh terms, respectively of an arithmetic progression. (a)

Given that all the term of the geometric progressions are different, find the common ratio.

(b)

3472/2

[4 marks]

If the sum to infinity of the geometric progression is 8, find (i)

the first term,

(ii)

the common difference of the arithmetic progression.

SMS MUZAFFAR SYAH , MELAKA

[4 marks]

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7 MOZ@C

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3472/2

Section B [40 marks] Answer four questions.

7

Use graph paper to answer this question. Table 3 shows the values of two variables, x and y, obtained from an experiment. Variables x and y are related by the equation y  ab  x , where a and b are constants. x

1

2

3

4

5

6

y

41.7

34.7

28.9

27.5

20.1

16.7

Table 3 (a)

Plot log10 y against x by using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.2 unit on the log10 y -axis. Hence, draw the line of best fit.

(b)

[4 marks]

Use your graph from (a) to find (i)

the value of y which was wrongly recorded, and estimate a more accurate value of it,

8

(ii)

the value of a and of b,

(iii)

the value of y when x = 3.5.

[6 marks]

  Diagram 2 shows a trapezium PQRS. U is the midpoint of PQ and PU  2SV . PV and TU are two straight lines intersecting at W where TW : WU = 1 : 3 and PW = WV.

S T P

V

R

W U

Q

Diagram 2    It is given that PQ  12a, PS  18b and QR  18b  5a .     (a) Express in terms of a and/or b ,  

3472/2

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8 MOZ@C

CONFIDENTIAL (i) (ii) (iii)

(b)

 SR ,  PV ,  PW .

[5 marks]

 Using PT : TS = h : 1, where h is a constant, express PW in terms of h, a and/or b and find the value of h.  

9

3472/2

[5 marks]

Diagram 3 shows a circle with centre C and of radius r cm inscribed in a sector OAB of a circle with centre O and of radius 42 cm. [Use  = 3.142]

Diagram 3 Given that AOB 

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 rad , find 3

(a)

the value of r,

[2 marks]

(b)

the perimeter, in cm, of the shaded region,

[4 marks]

(c)

the area, in cm2, of the shaded region.

[4 marks]

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9 MOZ@C

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10

3472/2

Diagram 4 shows part of the curve y  x  1 . y A R

y  x 1 y=k

S x

O Diagram 4

The curve intersects the straight line y = k at point A, where k is a constant. The gradient of the curve at the point A is (a)

Find the value of k.

(b)

Hence, calculate

1 . 4

[3 marks]

(i)

area of the shaded region R : area of the shaded region S.

(ii)

the volume generated, in terms of π, when the region R which is bounded by the curve, the x-axis and the y-axis, is revolved through 360o about the y-axis.

11

(a)

[7 marks]

A committee of three people is to be chosen from four married couples. Find how many ways this committee can be chosen (i)

if the committee must consist of one woman and two men,

(ii)

if all are equally eligible except that a husband and wife cannot both serve on the committee.

(b)

[5 marks]

The mass of mango fruits from a farm is normally distributed with a mean of 820 g and standard deviation of 100 g. (i)

Find the probability that a mango fruit chosen randomly has a minimum mass of 700 g.

(ii)

Find the expected number of mango fruits from a basket containing 200 fruits that have a mass of less than 700 g.

[5 marks]

Section C

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10 MOZ@C

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3472/2

[20 marks] Answer two questions. 12

A particle moves along a straight line and passes through a fixed point O. Its velocity, v m s–1, is given by v  pt 2  qt  16 , where t is the time, in seconds, after passing through O, p and q are constants. The particle stops momentarily at a point 64 m to the left of O when t = 4. [Assume motion to the right is positive.] Find

13

(a)

the initial velocity of the particle,

[1 mark]

(b)

the value of p and of q,

[4 marks]

(c)

the acceleration of the particle when it stops momentarily,

[2 marks]

(d)

the total distance traveled in the third second.

[3 marks]

Table 4 shows the prices of four types of book in a bookstore for three successive

years. Price in year (RM) Book 2000 2001 2002

Price index in 2001

Price index in 2002

based on 2000

based on 2000

Weightage

P

w

20

30

150

225

6

Q

50

x

65

115

130

5

R

40

50

56

125

140

3

S

80

z

150

y

y

2

Table 4

(a)

Find the values of w, x, y and z.

[4 marks]

(b)

Calculate the composite index for the year 2002 based on the year 2001. [4 marks]

(c)

A school spent RM4, 865 to buy books for the library in the year 2002. Find the expected total expenditure of the books in the year 2003 if the composite index for the year 2003 based on the year 2002 is the same as for the year 2002 based on the year 2001. [2 marks]

3472/2

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11 MOZ@C

CONFIDENTIAL

14

3472/2

Use graph paper to answer this question. A farmer wants to plant x-acres of vegetables and y-acres of tapioca on his farm. Table 5 shows the cost of planting one acre and the number of days needed to plant one acre of vegetable and one acre of tapioca.

Cost of planting per acre Number of days needed per acre

Vegetables

Tapioca

RM100

RM 90

4

2

Table 5 The planting of the vegetables and tapioca is based on the following constraints: I

The farmer has a capital of RM1800.

II

The total number of days available for planting is 60.

III

The area of his farm is 20 acres.

(a)

Write down three inequalities, other than x  0 and y  0 , which satisfy all the above constraints.

(b)

By using a scale of 2 cm to 4 acres on both axes, construct and shade the region R that satisfies all the above constraints.

(c)

[3 marks]

[3 marks]

By using your graph from (b), find (i)

the maximum area of tapioca planted if the area of vegetables planted is 10 acres,

(ii)

the maximum profit that the farmer can get if the profit for one acre of vegetables and one acre of tapioca planted are RM60 and RM20 respectively.

3472/2

SMS MUZAFFAR SYAH , MELAKA

[4 marks]

CONFIDENTIAL

12 MOZ@C

CONFIDENTIAL 15

3472/2

Diagram 5 shows a quadrilateral ABCD such that ABC is acute.

Diagram 5 (a)

(b)

Calculate (i)

ABC ,

(ii)

ADC ,

(iii)

the area, in cm2, of quadrilateral ABCD.

[8 marks]

A triangle AB’C has the same measurement as triangle ABC, that is, AC = 15 cm, CB’ = 9 cm and B ' AC  30 , but is different in shape to triangle ABC. (i)

Sketch the triangle ABC .

(ii)

State the size of AB ' C .

[2 marks]

END OF QUESTION PAPER

3472/2

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CONFIDENTIAL

13 MOZ@C

CONFIDENTIAL

3472/2

NO. KAD PENGENALAN

ANGKA GILIRAN

Arahan Kepada Calon 1

Tulis nombor kad pengenalan dan angka giliran anda pada ruang yang disediakan.

2

Tandakan (√ ) untuk soalan yang dijawab.

3

Ceraikan helaian ini dan ikat sebagai muka hadapan bersama-sama dengan buku jawapan. Kod Pemeriksa Bahagian

A

B

C

Soalan

Soalan Dijawab

Markah Penuh

1

5

2

6

3

5

4

9

5

7

6

8

7

10

8

10

9

10

10

10

11

10

12

10

13

10

14

10

15

10

Markah Diperoleh (Untuk Kegunaan Pemeriksa)

Jumlah EXCEL 2 PAPER 2 MARKING SCHEME

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No.

3472/2

Sub Total Marks Marks

Solution and Mark Scheme

1

K1

y  4 x  3

or equivalent x 

3  y 4

Eliminate x or y

3  y   3  y  or      y  3  4   4  2

x 2  x  (4 x  3)  3 K1 Solve the quadratic equation x2  5x  6  0 K1 ( x  2)( x  3)  0 x  3, 2 y  9,5

2

N1

y 2  14 y  45  0 ( y  9 )( y  5 )  0

for both values of x.

5

5

x = −3, −2

N1

1

(a)

x y  1 6 3

(b)

AD : DB  1: 2

P1

P1

 1(6)  2(0) 1(0)  2( 3)  ,   3 3  

(c)

y = 5, 9

 2, 2 

N1

mCD  2

P1

K1 3

y  (2)  2( x  2) y  2 x  2

K1

y  intercept  2

N1

3

7

3 (a)

y y

3 O –3 3472/2

2π

x 1  x

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15 MOZ@C

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3472/2

P1 Shape of sin x P1 Maximum = 3, minimum = –3 P1

2 periods for 0  x  2

4

P1 Inverted sin x (b)

y

N1

x  1 or equivalent 

Draw the straight line y 

x 1 

K1

No. of solutions = 5

4

(a)

dy  6 x 2  12 x  9  3 dx

N1

3

7

K1

x2  2x  1  0 ( x  1)( x  1)  0 x 1 y  2(1)3  6(1) 2  9(1)  1

4 P (1, 4)

(b)

N1

Equation of tangent: y  4  3( x  1) y  3x  1 Equation of normal: 1 y  4   ( x  1) 3 3 y   x  13

3472/2

2

K1 N1

K1 N1

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4

6

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16 MOZ@C

CONFIDENTIAL

5

3472/2

(a) Age (years)

Frequency, f

25 – 29

8

30 – 34

14

35 – 39

20

40 – 44

26

45 – 49

20

50 – 54

10

55 – 59

2

N1

(b)

N1

L1  34.5, FQ1  22

2

P1

or L1 =34.5 , f Q1  20 L3  44.5, FQ3  68

P1

or L3  44.5, f Q3  20  1 N-FQ1  Use Q1  L1   4 C  f Q1 

or

K1

 34 N-FQ3  Q3  L 3   C  f Q3 

Interquartile Range = 46.25 – 35.25 = 11 6

(a)

5

K1

7

N1

GP : T1 = a, T2 = ar, T3 = ar2 AP : T1 = a, T9= a + 8d, T11 = a + 10d ar = a + 8d or ar2 = a + 10d

a(r2 – 1) = 10d or a(r−1) = 8d or ar(r−1)=2d r 2  1 10  r 1 8

3472/2

P1 K1

K1

SMS MUZAFFAR SYAH , MELAKA

4

CONFIDENTIAL

17 MOZ@C

CONFIDENTIAL r

(b)

1 4

N1

a

8

(i)

3472/2

K1

1

1

4

a=6

1 1 6( )  6  8d or 6( ) 2  6  10d 4 4 9 d  N1 16

(ii)

7

N1

4

K1

8

(a) x

1

log10 y

2

3

4

5

6

1.620 1.540 1.461 1.439 1.303 1.223 N1

Plot log10 y against x (Correct axes and correct scales)

K1

6 points plotted correctly

N1

Draw line of best fit

N1

(b)

(i)

(ii)

log 10 y  (log10 b) x  log10 a

4

N1

y = 27.5 should be y = 24.0

P1

(iii)

a = 50

N1

b = 1.2

N1

log 10 y  1.42 y = 26.3

K1 N1

3472/2

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6

10

CONFIDENTIAL

18 MOZ@C

CONFIDENTIAL

8 (a)

(i)

  Use Triangle Law to find SR or PV

 SR  7 a

(ii)

(iii)

(b)

3472/2

K1

N1

 SV  3a   PV  3a  18b   1 PW   3a  18b  2  

N1 N1 5

P1

    3  PW  PU  UW  PU  UT or equivalent 4

K1

   h UT  UP  PT  6a  (18b)  h 1 

K1

 3 18h PW  6a  (6a  b)  4  (h  1) 

N1

 3 27 h PW  a  b 2  2(h  1)  Comparing (1) & (2)

(2) K1 9

27h 2(h  1)

N1 5

10

h=2

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19 MOZ@C

CONFIDENTIAL

9

(a)

(b)

r  sin 30 42  r

K1

r  14

N1

14 

 2 or 14  3 3

3472/2

2

K1

282  142

K1

Perimeter = 24.249 + 24.249 + 29.325

K1

= 77.823 (accept 77.82) (c)

1   14 2  2 3

N1

K1

1  14  588 2

K1

Area = 2 ( 169.741 – 102.639) = 134.204

K1

N1 4

Accept 134.2 10 (a)

4

dy 1  dx 2 x  1

P1

1 1  2 x 1 4

K1

10

x = 5, N1

k =2

(b)

(i)

3

Area of R

or

2

=  ( y 2  1)dy 0

3472/2

K1

Area of S



5

1

x  1 dx

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20 MOZ@C

CONFIDENTIAL

3472/2 5

3  2 x  1    =  3   2 1 or 5 1 3

2

y  =   y 3 0 3

=4

2 3

N1

Area of S

or Area of R 2 1 = 25  4 = 25  5 K1 3 3 1 =5 =4 2 3 3 Area of R : Area of S = 7 : 8

N1

2

V    ( y 2 1) 2 dy

(ii)

K1

0

2

 y5 2  V     y3  y  5 3 0 11 V  13  15

K1 N1 7

11

(a)

4

(i)

C1 4C2

10

K1 N1

= 24

(ii) If 4C3 4 C0 or 4C2 2C1 or 4C1 3C2 or 4 C0 4C3 is shown K1 4

C3 4 C0 + 4C2 2C1 + 4C1 3C2 + 4 C0 4C3

= 32

or

(b)

(i)

N1

8  6  4 or 3! 8 6 4 3!  32

700  820 100 P( X  1.2)

3472/2

K1 5

K1 K1 N1

K1

K1

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CONFIDENTIAL

21 MOZ@C

CONFIDENTIAL

3472/2

=1 – 0.1151 = 0.8849

(ii)

12

(a)

N1

200 x 0.1151

K1

= 23

N1

–16 m s–1

5

10

N1 1

(b)

Integrate pt 2  qt  16 with respect to t p q s  t 3  t 2  16t K1 3 2 t = 4, v = 0 16p + 4q = 16 or p=3

N1

q = –8

N1

64 p  8q  0 3

K1

4

(c)

(d)

a = 6t – 8

K1

t = 4, a = 16

N1

s  t 3  4t 2  16t

Find



3

2

v dt or

St 3  St  2

Substitute t  2 or t  3 into s

2

K1 K1

d = |[ s  33  4(32 )  16(3) ] – [ s  23  4(22 )  16(2) ]| d = 17 m

3472/2

N1

SMS MUZAFFAR SYAH , MELAKA

3

10

CONFIDENTIAL

22 MOZ@C

CONFIDENTIAL

13

(a)

(b)

w = 13.33

N1

x = 57.50

N1

y = 187.5

N1

z = 150

N1

3472/2

4 P1

I2002 / 2001 : 150 , 113.04, 112, 100 Use I 

W I W

i i

K1

i

150  6  113.04  5  112  3  100  2 I 6 5 3 2 2001.2  K1 16

N1

= 125.08

(c)

14

125.08  4865 100

K1

=6085.14

N1

4

2

100 x  90 y  1800 or equivalent

(a)

10

N1

4 x  2 y  60 or equivalent

N1

x  y  20 or equivalent

N1 3

(b)

K1

Draw correctly at least one straight line Draw correctly all the three straight lines

N1

N1

Region R shaded correctly

3

(c)

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(i)

y = 8.0 – 9.0

(ii)

maximum point (15, 0)

N1 N1

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23 MOZ@C

CONFIDENTIAL

RM15  60 + RM20  0

3472/2

K1

= RM900

N1 4

15 (a)

sin ABC 

(i)

sin 30o  15 9

K1

ABC  56.44o or 56o 27'

N1

152  102  82  2(10)(8) cos ADC

(ii)

ADC  112.41 or 112 25 '

K1

N1

1 area of ACD  10  8  sin112.41 2

(iii)

10

K1

1 area of ABC  15  9  sin(180  56.44  30 ) 2

K1

area of quadrilateral ABCD = 36.98 + 67.37

K1

= 104.35

N1

C

N1

8

B

(b)

(i)

A AB ' C must be obtuse

(ii)

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123.56 or 123 33’

N1

SMS MUZAFFAR SYAH , MELAKA

2

10

CONFIDENTIAL

24 MOZ@C

CONFIDENTIAL

3472/2

GRAPH FOR QUESTION 7

log10 y

1.8

×

1.6

× × 1.4

× × ×

1.2

1.0

0.8

0.6

0.4

0.2

GRAPH FOR QUESTION 14 0

3472/2

1

2

3

4

5

SMS MUZAFFAR SYAH , MELAKA

6

CONFIDENTIAL

x

25 MOZ@C

CONFIDENTIAL

3472/2

y

32

28 2x+ y = 30

24

20

16

12

8 R x+ y = 20 4

0

4

3472/2

8

12

16

x

20 10x+9 y = 180

SMS MUZAFFAR SYAH , MELAKA

CONFIDENTIAL