146873446-Add-Maths-Paper-1-SBP-Form-4-Akhir-06 PDF

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SULIT 3472/1 3472/1 Form Four Name : ………………..…………… Additional Mathematics Paper 1 Form : ………………………..…… Oct 2006 2 hours SEKTOR SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2006 ADDITIONAL MATHEMATICS For examiner’s use only Form Four Total Marks Question...

Description

SULIT

3472/1

3472/1 Form Four Additional Mathematics Paper 1 Oct 2006 2 hours

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

SEKTOR SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2006 ADDITIONAL MATHEMATICS Form Four Paper 1 2 hours

DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO

1

Write your name and class clearly in the space provided on this page.

2

Read the information on page 2 carefully.

3

This question paper must be handed in at the end of the examination.

For examiner’s use only Question 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

Total Marks 2 3 4 2 3 3 3 4 3 3 4 4 3 4 3 3 3 3 4 3 4 3 3 3 3

Marks Obtained

TOTAL

This question paper consists of 13 printed pages

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INFORMATION FOR CANDIDATES 1.

This question paper consists of 25 questions.

2.

Answer all questions in this question paper.

3.

Give only one answer/solution to 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 are shown in brackets.

7.

A list of formulae is provided on pages 2 to 3.

9.

You may use a non-programmable scientific calculator.

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

ALGEBRA

 b  b 2  4ac 2a

5

log a mn  log a m  log a n

am x an = a m + n

6

log a

3

am  an = a m – n

7

log a mn = n log a m

4

( am )n = a m n

8

1

x

2

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m  log a m  log a n n

log a b 

log c b log c a

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GEOMETRY

1

Distance =

x1  x 2 2   y1  y 2 2

3

A point dividing a segment of a line

x , y    nx1  mx2 , ny1  my2  

2 Mid point

x, y    x1  x2 , y1  y 2  

2

2



mn

mn



4 Area of a triangle =

1 x1 y 2  x 2 y 3  x3 y1   x 2 y1  x3 y 2  x1 y 3  2 STATISTICS

1

x

x N

2

x

 fx f

4.  

x  x  3   N 2

x 2 x2 N

 f x  x   f 2

 fx 2 x2 f

1   N F  C 5. m  L   2  fm     

TRIGONOMETRY 1

Arc length, s =r

2

Area of a sector, A 

1 2 r 2

CALCULUS

1

dy dv du y = uv , u v dx dx dx

3

dy dy du   dx du dx

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2

y

u dy ,  v dx

v

du dv u dx dx 2 v

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For examiner’s use only

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Answer all questions.

1

Diagram 1 represents the relation between set R and set S. S 10 8 6 4 2 1

2

R

3

DIAGRAM 1 State (a)

the range,

(b)

the type of relation. [2 marks]

1 Answer : (a) …………………….. 2

(b) ……………………...

2

Diagram 2 illustrates the mapping of the function f and g – 1. y

x f (x)

2

z g–1 (y) 3

5 DIAGRAM 2 Given that f 1 ( x)  2 x  k , find (a) the value of k, (b) the image of f 1 g (3) .

[3 marks] 2

Answer : (a) …………………….. 3

(b) ……………………...

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Given the function f ( x)  1  3 x and the composite function fg ( x)  4  3 x 2 , find (a) g (x) , (b) the value of x when f

1

For examiner’s use only

( x)  5 . [4 marks]

Answer : (a) ...………………………......

3

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

4

Express the quadratic equation

4

x3 x  in general form. 6 2x  4

[2 marks]

4 Answer : .........…………………

5

2

One of the roots of the quadratic equation 2 x 2  nx  24  0 is three times the value of the other root. Find the value of n , if n > 0 . [3 marks]

5 Answer : ……………………..

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6

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The quadratic equation x 2  px  q  0 has roots  3 and

1 where p and q are 2

constants . Find the value of p and q. [3 marks]

Answer : p = ……........................

6

q =....……………….... 3

7

Find the range of values of x if y  x 2  4 and 2 y  x  11 . [3 marks]

7 Answer :

3

8

………………………...

Diagram 3 shows the graph for the function f ( x)  p( x  q) 2  r , where p, q and r f(x) are constants. f ( x)  p( x  q)2  r

4 2 0

1

5

x

DIAGRAM 3 Find the values of (a) q and r, (b) p

[4 marks]

8 Answer : (a) q = …….. r = ........... 4

(b) p =....………………..

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7

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Diagram 4 shows the graph of the function f ( x)  4 x 2  (4k  12) x  15  5k in relation to the x–axis . Find the range of values of k. [3 marks] f(x)

x DIAGRAM 4

9 Answer : k =……..……...……….....

10

Solve the equation

125

x 1



3

5 x 3 . 25 [3 marks]

10 Answer : ……………...……….....

11

3

Given log 5 2  m and log 5 7  p , express log 5 3.43 in terms of m and p. [4 marks]

11 Answer : …...…………..………... 4

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8

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Solve the equation log 2 ( x  2)  1  2 log 4 (4  x) [4 marks]

12 Answer : ……………...………..... 4

13

Given points P(5, 6), Q(10, 3), R(7, n) and S(6, 1 ) are the vertices of a kite. Find the value of n. [3 marks]

13 Answer : n =…………...……….....

3

14

Diagram 5 shows points A, B and C on a Cartesian plane. Given that point A lies on the y-axis and the area of the triangle formed by the three points is 10 unit2. y A k

 B (4, 1) x

O

 C (2, 3) DIAGRAM 5 Calculate the value of k.

[4 marks]

14 Answer : k = ………..……….......... 4

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9

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Find the equation of the locus of the moving point M such that its distances from the

For examiner’s use only

points A (2 , 3) and B (5 , 2) are in the ratio of 1 : 2 . [3 marks]

15 Answer : ...........................................

16

3

A set of five numbers has a mean of 9. When q is added to the data the new mean is 10. Find the value of q. [3 marks]

16 Answer : q =...................................... 3

17

Table 1 shows the frequency distribution of the masses of 40 students. Mass (kg) Frequency

46-50 7

51-55 10

56-60 12

61-65 8

66-70 3

TABLE 1

Without drawing an ogive, find the median.

[3 marks]

17 Answer : …...…………..……..…... 3

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10

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For a set of four numbers, it is given that the sum of all the numbers,  x , is 28 and the standard deviation is 2 3 . Find the value of  x 2 . [3 marks]

18 Answer : …...…………..……..…... 3

19

In Diagram 6 , PQR is a semicircle with centre O.

Q 4 cm P

O DIAGRAM 6

R

Given that the radius is 5 cm, find the length of the arc PQ. (Use  = 3.142) [4 marks]

19 4

Answer : …...…………..……..…...

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11

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Diagram 7 shows sector OAB with centre O and radius of 5p. A

0.5 rad O

C DIAGRAM 7

B

Point C is on OB such that OC : CB = 3 : 2. Given that the perimeter of the shaded region is 34 cm, calculate the value of p. (Use  = 3.142) [3 marks]

20 Answer : …...…………..……..…... 3

21

Diagram 8 shows sector OPQ with centre O. O

20 cm

P

50

Q

DIAGRAM 8 Given that the radius is 20 cm and OPQ  50 . Find the area of the shaded region. (use   3.142) [4 marks] 21 Answer : …...…………..……..…...

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12

Determine the value of

lim n 1

n 2 n2 . n 1

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[3 marks]

22 Answer : …...…………..……..…... 3

23

Given that f ( x) 

2x2 1 , find f ' (1) . 3x  4 [3 marks]

23 Answer : …...…………..……..…... 3

24

Find the equation of the tangent to the curve y = 3x 2 – 2x + 1 that passes through point (1, 2) . [3 marks]

24 3

Answer : …...…………..……..…...

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13

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Given that y  (3 x  2) 4 , and x decreases at the rate of 4 unit per second, find the rate at which y changes when x =

For examiner’s use only

1 . 3

[3 marks]

Answer : …...…………..……..…...

25 END OF THE QUESTION PAPER 3

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[See overleaf SULIT

SULIT 3472/1 Additional Mathematics Paper 1 October 2006

SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2006

ADDITIONAL MATHEMATICS Paper 1 MARKING SCHEME

This marking scheme consists of 7 printed pages

3472/1

© 2006 Copyright SBP Sector, Ministry of Education, Malaysia

SULIT

Sub Marks

Full Marks

(a) { 2, 4, 6, 10 }

1

2

(b) many to many

1

(a) k = 1

2

Number 1

2

Solution and marking scheme

2(2) + k = 5

3

B1

(b) 5

1

(a) g(x) = x 2 + 1

3

( 4  3 x 2 ) 1 3 x 1 f – 1 (x) = 3

4

4

B2 B1

(b) x = 16

1

x2 – 8x + 6 = 0

2

(x-3)(2x -4) = 6x

5

3

2

B1

n = 16

3

SOR = 4 = 

n 2

and

POR = 3

2

 12

B2

SOR = 4 = 

n 2

or

POR = 3

2

 12

B1

2

3

Number 6

7

8

Solution and marking scheme 3 2 5 p=  2

q= 

SOR =  3 

1 2

x  1 , x 

3 2

1

3

POR = 3 

or

1 2

B1

3

(2 x  3)( x  1)  0

B2

2(x2 + 4)  x + 11

B1

(a) q = - 3

(b) p =

1

1 2

4

2 or

4 = p(5 – 3) 2 + 2

-2 < k < 3

B1

3

(k + 2)(k – 3) < 0

B2

(12 – 4k) 2 – 4(4)(15 – 5k) < 0

B1

x = 1

3

2 + (3x – 3) =

x3 2

B2

x 3

5 3( x 1)

3

1

4 = p(1 – 3) 2 + 2

10

Full Marks

2

r=2

9

Sub Marks

5 2  2 5

or

5 3( x 1) ( 5 2 )  5

3

x 3 2

B1

3

3

Number 11

12

Solution and marking scheme 3p – 2 – 2m

4

4

B3

log 5 343 – log 5 100

B2

log 5

343 100

x =

10 3

log 2

B1

4

4

B3

x2 = 1 4 x

B2

 log 2(4  x)  log 2 ( x – 2 ) = 1 + 2    log 2 4  n = -2

B1

3

 3  1  n  6      1  10  6  7  5   3 1  m QS =    10  6 

14

Full Marks

3 log 5 7 - 2 log 5 5 – 2log 5 2

x2  2 4x

13

Sub Marks

or

3

B2

 n6  m PR =    7 5 

k=3

B1

4

14 + 2k = - 20

and

14 + 2k = 20

B3

14 + 2k = - 20

or

14 + 2k = 20

B2

1 14  2k  10 2

B1

4

4

Number 15

Solution and marking scheme 3x2 + 3y2 – 6x + 28y + 23 = 0 2 ( x  2) 2  ( y  3) 2  ( x  2) 2  ( y  3) 2

16

17

( x  5) 2  ( y  2) 2 ( x  5) 2  ( y  2) 2

or

q = 15

3

3

B2 B1

45  q  10 6

B2

 x  45

B1

Median = 56.75

3

3

B1

 x 2  244

3

 x2  (7) 2  2 3 4

3

B2



B1

x = 7 S PQ = 11.71 cm

4

S PQ = 5 ( 2.342 )

B3

 POQ  3.142  0.8  2.342 rad

B2

 QOR 

3

B2

L = 55.5

19

Full Marks

3

1   2 (40)  17  55.5   5 12    

18

Sub Marks

B1

4 radian 5

5

4

Number 20

21

22

Solution and marking scheme p = 4

Full Marks

3

3

2.5p + 2p + 4p = 34

B2

S AB = 5p (0.5)

B1

82.24

to

82.33 cm2

4

1  20  2 2

  3.142   80   180   

1  20  2 2

  3.142   80   180   

and

1  20  2 2

  3.142   80   180   

or

-

1  20  2 sin 80  2 1  20  2 sin 80  2 1  20  2 sin 80  2

3

4

B3

B2

B1

3

lim ( n  2 )

3

B2

n 1

 n  2  n 1 23

Sub Marks

B1

7

3

4   1 3   1  4   3  2  1 2 1 [ 3   1  4 ] 2

B2

4 x  3x  4  3  2 x 2  1 [3x  4 ] 2

B1

6

3

Number 24

Solution and marking scheme y = 4x – 2 2 = 4 (1) + c dy = 6x – 2 dx

25

or

or

y2 4 x 1

Sub Marks

Full Marks

3

3

B2 B1

dy 4 dx

3

dy  48 dt dy  1   12  3    2  3  4 dt  3 

B2

dy = 4 (3x – 2) 3 (3) dx

B1

7

3...