Add. Maths Paper 1 SBP Form 4 Akhir 06

SULIT

3472/1

3472/1 Form Four Additional Mathematics 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 INSTRUCTED TO DO SO

1

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

For examiner’s use only

1

Total Marks 2

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2

5

3

6

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7

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8

4

9

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10

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Question

2

Read the information on page 2 carefully.

11

4

3

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

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15

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16

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25

3 TOTAL

This question paper consists of 13 printed pages

Marks Obtained

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3472/1

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. working. It may help you to get marks.

5.

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

6.

The marks allocated allocated for each question question are shown in brackets. 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

1

 x 

2

am

x

3

am

 an = a m – n

4

( a m )n = a

5

log a mn  log a m  log a n

6

log a

7

log a mn = n log a m

2a an = a

mn

m+n

8

m n

 log a m  log a n

log a b 

log c b log c a

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GEOMETRY

1

Dist Distan ancce =

 x1   x 2    y1  y 2  2

2

3

A point point divi dividin ding g a seg segme ment nt of a line line

nx1  mx2

 x , y    

  m  n

2 Mid Mid point point

 x1   x 2  y1  y 2   ,  2     2

 x, y    

ny1  my 2  



,

m  n  

4 Area Area of of a triangl trianglee =

1 2

 x1 y 2   x 2 y 3   x3 y1    x 2 y1   x3 y 2   x1 y 3 

STATISTICS

1

 x 

2

 x 

x

  f   x  x    fx 2   x 2 4.      f     f   2

  fx   f    x  x 

2

3

  

 N 



 x 2  N 

 x 2

  1     N   F   2  C  5. m   L     f  m       

TRIGONOMETRY

1

2

Arc Arc leng length th,, s =r    

Area Area of a sect sector or,,  A



1 2

r 2 

CALCULUS

1

3

y = uv ,

dy dx



dy du

dy dx



u

du dx

dv dx

v

du dx

2

 y 

u v

,

dy dx

v



du dx

u v

2

dv dx

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 For  examiner’s use only

3472/1

 Answer  all questions.

1

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

2

 R

3

DIAGRA DIAGRAM M 1 State ( a)

the range,

(b) (b)

the the type type of rela relati tion on.. [2 marks] marks]

1

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

(b) ……………………...

2

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

2

z  g  –  (y)

3

5 DIAGRAM 2 Given that  f 1 ( x)  2 x  k  , find (a) (a) the the valu valuee of k  of k , (b) (b) the the imag imagee of   f 1 g ( 3) .

[3 marks] marks] 2

 Answer :  Answer  : (a) …………………….. …………………….. 3

(b) ……………………...

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3472/1

Given the function  f  ( x)  1  3x and the composite composite function function  fg ( x)  4  3 x 2 , find find

 For  examiner’s use only

(a)  g ( x ) , (b) the value of  x when  f  1 ( x )  5 . [4 marks] marks]

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

3

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

4

Express the quadratic equation

 x  3 6



 x 2 x  4

4

in general form. [2 marks] marks]

4

 Answer :  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 other root. Find Find the value value of  n , if  n > 0 . [3 marks] marks]

5

 Answer :  Answer  : ………… ……………… ………… …….. ..

3

 For  SULIT examiner’s use only

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The quadratic equation  x 2   px  q  0 has roots  3 and

1 2

where p and q are

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

 Answer :  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] marks]

7

 Answer :  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)  x) are constants.  f ( x )  p ( x  q) 2  r 

4 2 0

1

x

5 DIAGRAM 3

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

[4 marks] marks]

8

 Answer :  Answer  : (a) 4

q = …….. r = r  = ...........

(b) p =....………………..

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3472/1

 For  examiner’s use only

Diag Diagra ram m 4 show showss the the grap graph h of the the func functio tion n  f ( x )  4 x 2  ( 4k  12) x  15  5 k  in relation relation to the x –axis . Find the range of values of  k . [3 marks] marks]  x

 x DIAGRAM 4

9

 Answer :  Answer  : k =……..……...………..... k  =……..……...……….....

10

Solve the equation

 x 1

125



5 x

3

3

25

. [3 marks] marks]

10

 Answer :  Answer  : ……………...……….....

11

3

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

11

 Answer :  Answer  : …...…………..… …...…………..………... ……... 4

 For  examiner’s use only

SULIT 12

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3472/1

Solve the equation log 2 ( x  2)  1  2 log 4 ( 4  x) [4 marks] marks]

12

 Answer :  Answer  : ……………...………..... 4

13

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

13

 Answer :  Answer  : n =…………...……….....

3

14

Diagram 5 shows points A, B and C  on a Cartesi Cartesian an plane. plane. Given Given that point point A lies on the y-axis and the area of the triangle formed by the three points is 10 unit 2.  y A k  

  B (4, 1)  x

O

 C (2, C  (2, 3) DIAGRAM 5 Calculate Calculate the value of k  of  k .

[4 marks] marks]

14

 Answer :  Answer  : k = k  = ………..……….......... 4

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

 For  examiner’s use only

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

15

 Answer :  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 value of q of  q. [3 marks] marks]

16

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

17

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

46-50 7

51-55

56-60

10

12

61-65

66-70

8

3

TABLE 1

Without drawing an ogive, find the median.

[3 marks] marks]

17

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... .. 3

 For  SULIT examiner’s use only

18

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3472/1

For a set of four numbers, it is given that the sum of all the numbers,  x , is 28 and

the standar standard d devia deviation tion is 2 3 . Find Find the value value of   x 2 . [3 marks] marks]

18

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... .. 3

19

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

Q 4 cm  P 

O

R

DIAGRAM 6 Given that the radius is 5 cm, find the length of the arc PQ. PQ. (Use   = 3.142) [4 marks] marks]

19 4

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... ..

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20

11

3472/1  For  examiner’s use only

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

0.5 rad O

C 

 B

DIAGRAM 7 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] marks]

20

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... .. 3

21

Diagram Diagram 8 shows shows sector OPQ 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] marks] 21

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... ..

4

 For  examiner’s use only

SULIT

22

12

Determine Determine the value of 

lim n 1

n 2 n2 n 1

3472/1

.

[3 marks] marks]

22

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... .. 3

23

Given that  f ( x) 

2 x 2  1 3 x  4

, find  f  ' ( 1) . [3 marks] marks]

23

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... .. 3

24

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

24 3

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... ..

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3472/1

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

1 3

 For  examiner’s use only

.

[3 marks] marks]

 Answer :  Answer  : …...…………..… …...…………..……..…. …..…... ..

25 END OF THE QUESTION PAPER  3

SULIT 3472/1 Additional Mathematics Paper 1 October 2006

SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN 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 many

1

( a) k = 1

2

 Number 1

2

Solution and marking scheme sch eme

2(2) + k = 5

B1

(b) 5

3

3

1

2

(a) g(x) = x + 1

3

( 4  3 x 2 )  1

4

B2

3

f 

–1

(x) =

 x  1

B1

3

(b) x = 16

4

1

2

x  – 8x + 6 = 0

2

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

5

2

B1

n = 16

3

SOR = 4  = 

n

SOR = 4  = 

n

2

2

and

POR = 3  2  12

B2

or

POR = 3  2  12

B1

2

3

Sub Marks

Full Marks

3

1

3

2 5

2

 Number 6

Solution and marking scheme sch eme

q=   p = 

2

SOR =  3 

7

 x   1 , x 

1

POR = 3 

or

2

1

B1

2

3

3

2

(2 x  3)( x  1)  0

B2

2(x + 4)  x + 11

B1

2

8

( a) q = - 3

1

r= 2 (b) p =

1 2

2 2

2

4 = p(5 – 3) + 2

or

-2 < k < 3

B1

3

(k + 2)(k – 3) < 0 2

B1

x = 1

3

2 + (3x (3x – 3) =

3 B2

2

 x  3

5 3( x 1) 

5 5

2 2

3

B2

(12 – 4k)  – 4(4)(15 – 5k) < 0

10

4

1

4 = p(1 – 3) + 2

9

3

or

5

3( x 1)

 x  3

(5 )  5 2

3

2

B1

3

 Number 11

Solution and marking scheme sch eme 3 p – 2 – 2m

4

log 5 343 – log 5 100

B2

x =

log 2

B1

343 100

4

10

4   x

 2  x  2 4  x

B3

B2

= 1

 log 2 (4  )    log 2 4 

B1

n = -2

3

 3 1   n  6      1  10  6   7  5   3 1    10  6 

m QS = 

or

 n6    7 5 

m PR  = 

B1

4

14 + 2k = - 20

and

14 + 2k = 20

B3

14 + 2k = - 20

or

14 + 2k = 20

B2

2

3

B2

k=3

1

4

3

log 2 ( x – 2 ) = 1 + 2 

14

4

B3

 x  2

13

Full Marks

3 log 5 7 - 2 log 5 5 – 2log 5 2

log 5

12

Sub Marks

14  2k   10

B1

4

4

 Number 15

Solution and marking scheme sch eme 2

2

3x + 3y  – 6x + 28y + 23 23 = 0 2

( x  2) 2  ( y  3)2  ( x  2) 2  ( y  3) 2

16

( x  5) 2  ( y  2) 2 ( x  5)2  ( y  2) 2

or

q = 15 45  q 6

B1

3

B1

3

 (7) 2  2 3

3

B2

B1

= 7 S PQ = 11.71 cm

4

S PQ = 5 ( 2.3 2.342 42 )

B3

 POQ  3.142  0.8  2.342 rad 

B2

 QOR 

3

B2



19

B2

3

 x 2  244

4

3

B1

Median = 56.75

 x 2

3

B2

 10

1  (40)  17   2 (40 55.5   5 12     L = 55.5

18

Full Marks

3

 x  45

17

Sub Marks

4 5

B1

radian

5

4

 Number 20

21

Solution and marking scheme sch eme p = 4

3

3

B2

S AB = 5p (0.5) (0.5)

B1

82.24

to

1



3.142 



 180 



3.142 

 20  2  80   2

1

2

82.33 cm

 20  2  80   2





 180 



3.142 

 20  2  80   2 



 180  

4

-

1 2

80   20  2 sin 80

1

and

2 1

or

2

3

80   20  2 sin 80

80   20  2 sin 80

4

B3

B2

B1

3

lim ( n  2 )

3

B2

n 1

 n  2   n  1 23

Full Marks

2.5p 2.5p + 2p 2p + 4p = 34

1

22

Sub Marks

B1

7

3

4   1  3   1  4   3  2   1 2  1 [ 3   1  4 ] 2

B2

4 x  3 x  4  3  2 x 2  1 [ 3 x  4 ] 2

B1

6

3

 Number 24

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

25

dy dt 

= 6x – 2

 y  2

or

or

 x  1 dy dx

Sub Marks

Full Marks

3

3

B2

4

B1

4

3

 48

 1   12  3    2  3  4 dt   3 

B2

dy

B1

dy

dx

3

= 4 (3x (3x – 2) (3)

7

3