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SULIT 1 3472/23472/2 MatematikTambahanKertas 22 ½ jamOgos 2008

SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN PENGURUSAN

SEKOLAH BERASRAMA PENUH / KLUSTERKEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2008

MATEMATIK TAMBAHAN

Kertas 2

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

3. Give only one answer / solution to each question..

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

5. The diagram 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 3.

8. A booklet of four-figure mathematical tables is provided.

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

Kertas soalan ini mengandungi 13 halaman bercetak

3472/2 SULIThttp://tutormansor.wordpress.com/

SULIT 3472/2

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

ALGEBRA

1 x =a

acbb

2

42 −±−

2 am × an = a m + n

3 am ÷ an = a m - n

4 (am) n = a nm

5 loga mn = log am + loga n

6 loga n

m = log am - loga n

7 log a mn = n log a m

8 logab = a

b

c

c

log

log

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan −+

11 Tn = ar n-1

12 Sn = r

ra

r

ra nn

−−=

−−

1

)1(

1

)1( , (r ≠ 1)

13 r

aS

−=∞ 1

, r <1

CALCULUS

1 y = uv , dx

duv

dx

dvu

dx

dy +=

2 v

uy = ,

2

du dvv udy dx dx

dx v

−= ,

3 dx

du

du

dy

dx

dy ×=

4 Area under a curve

= ∫b

a

y dx or

= ∫b

a

x dy

5 Volume generated

= ∫b

a

y 2π dx or

= ∫b

a

x 2π dy

3472/2 SULIT

2

5 A point dividing a segment of a line

( x,y) = ,21

++

nm

mxnx

++

nm

myny 21

6. Area of triangle =

)()(2

1312312133221 1

yxyxyxyxyxyx ++−++

1 Distance = 221

221 )()( yyxx −+−

2 Midpoint

(x , y) = +

221 xx

, +

221 yy

3 22 yxr +=

4 2 2

xi yjr

x y

∧ +=+

GEOM ETRY

http://tutormansor.wordpress.com/

SULIT 3472/2

STATISTICS

[ Lihat sebelah3472/2 SULIT

TRIGONOMETRY

1 Arc length, s = r

2 Area of sector , A = 3 sin 2A + cos 2A = 1

4 sec2A = 1 + tan2A

5 cosec2 A = 1 + cot2 A

6 sin2A = 2 sinAcosA

7 cos 2A = cos2A – sin2 A = 2 cos2A-1 = 1- 2 sin2A

8 tan2A =

9 sin (AB) = sinAcosB cosAsinB

10 cos (AB) = cos AcosB sinAsinB

11 tan (AB) =

12

13 a2 = b2 +c2 - 2bc cosA 14 Area of triangle =

1 =

2 = 3 σ = =

4 σ = =

5 M =

6

7 8 9

10 P(AB)=P(A)+P(B)-P(AB)

11 p (X=r) = , p + q = 1

12 Mean , = np

13 14 z =

3


SECTION A

[40 marks]

Answer all questions in this section .

1. Solve the simultaneous equations h + 2k = 5 and k2 – 3h = 7. Give your answers correct to three decimal places. [5 marks]

2. Solution to this question by scale drawing will not be accepted.

Diagram 1 shows a straight line PRQ

The points P and Q intersects the y –axis and the x-axis. R is the midpoint of line PQ.The equation of line PR is 01223 =−+ yx

(a) Find

(i) the coordinates of R. [2 marks]

(ii) the area of triangle OPS if SR was extended to the origin and PS is parallel to the x-axis. [2 marks]

(b) A point X moves such that its distance from P is always 2

1 of its distance from Q

[3 marks]

4

y

x0

S

R

P

Q

Diagram 1


SULIT 3472/2

3 Diagram 2 shows a histogram representing the distribution of the heights of 30 students in a class.

Calculate

(a) the value of fx [3 marks]

(b) the value of 2fx [2 marks]

(c) the standard deviation of the number of patients visiting the clinic. [3 marks]

4. Diagram 3 shows the particles A and B are projected simultaneously towards each other from the opposite end of a straight tube, 9 m long.

Particle A travels 47 cm in the 1st second, 45 cm in nd2 second, 43 cm in the rd3 second, etc. Particle B travels 25 cm in the 1st second, 24 cm in nd2 second, 23 cm in the rd3 second, etc. Find how long it takes for both particles to meet.

[6 marks]


2

10

8

6

4

145.5 160.5155.5150.5140.5 170.5165.5

Height (cm)

Num

ber

of s

tude

nts

Diagram 2

Diagram 3A B

• •→ ←

5


5. (a) Find the equation of the normal to the curve 3 22y x x at the point (1, -1).

[3 marks]

(b) A cylindrical tank with a circular base of radius 0.5 m is filled with h m of turpentine. If the turpentine is evaporating at a uniform rate of 0.001 m3 s-1, find the rate of change in the level of turpentine. Leave your answer in terms of .

[3 marks]

6. (a) Prove the identity cos 4 θ - sin 4 θ = cos 2θ. [3 marks]

(b) By sketching the graph of y = 2sin4 θ - 2cos4 θ and a suitable straight line on the same axis for 0 ≤ θ ≤ π, state the number of solutions for the equation

2sin4 θ - 2cos4 θ = 1x

. [5 marks]

3472/2 SULIT


SULIT 3472/2

SECTION B

[40 marks]

Answer four questions from this section.

7. Use the graph paper provided to answer this question.

Table 1 shows the values of two variables, x and y, obtained from an experiment. Variables x and y are related by an equation 1xy pk , where p and k are constants.

Table 1

(a) Plot log 10 y against (x – 1) , by using the scale of 2 cm to 1 unit on the x-axis and 2 cm to 0.2 unit on the y-axis. Hence, draw the line of best fit.

(b) Use the graph in 7 (a) to find the value of

(i) p,

(ii) k

[5 marks]

[5 marks]


x 2.0 3.0 4.0 5.0 6.0 7.0

y 8.2 11.8 16.7 23.9 34.8 50.2

7


8. In the Diagram 4, OP = 8 p , OQ = 10 q and PS = 4 q .

(a) Express each of the following vectors in terms of p and/or q .

(i) OS

(ii) QP [4 marks]

(b) Given that OT = aOS and QT = bOP , express OT in terms of

(i) a, p and q

(ii) b, p and q [3 marks]

(c) Hence, find the values of a and b. [3 marks]

3472/2 SULIT

Q

10

O

ST

P8

4

Diagram 4


SULIT 3472/2

9. Diagram 5 shows sector AOB and sector OED with centre O and E respectively. OCE is a right angle triangle. (Use = 3.142).

Given that ∠ AOB = 500 , OA = 10 cm , OE = 8 cm and OB : BC = 2 : 1. Calculate

(a) θ in radian, [2 marks]

(b) perimeter of the shaded region in cm, [4 marks]

(c) area of the shaded region in cm2. [4 marks]


C B

A

O

E

Dθ

Diagram 5

9


10. (a) Given y = (t – 2)(t + 1) and x = 2t + 1, find dy

dx in terms of x. [2 marks]

(b) Diagram 6 shows the curve y = (x − 3)(x − 5) intersects with a straight line

2y x p at the point (1, 8).

Find,

(i) the value of p,

(ii) the area of the shaded region. [5 marks]

(iii) the volume generated when the region bounded by the straight line x = 1, y = 2x + p, the x and y-axes is revolved 360o on the x-axis.

[3 marks]

11. (a) In one housing area, 20% of the residents are senior citizens.

i) If a sample of seven persons is chosen at random, calculate the probability that at least two of them are senior citizens.

[3 marks] ii) If the variance of the senior citizens is 128, find the number of residents in

the housing area. [2 marks]

b) In a field study, it is found that the mass of a student is normally distributed with a mean of 50 kg and standard deviation 15 kg. i) If a student is selected randomly, calculate the probability that his mass is

less than 41 kg. [2

marks] ii) Given that 12% of the students have a mass greater than m kg, find the value

of m.

3472/2 SULIT

Diagram 6

(1, 8)

y = (x - 3)(x - 5)

x

y

O

y = 2x + p


SULIT 3472/2

[3 marks]

SECTION C

[20 marks]

Answer two questions from this section.

12. A particle moves along a straight line from a fixed point O. Its velocity, V ms-1, is given by V = 15t – 3t2, where t is the time, in seconds, after leaving the point O. (Assume motion to the right is positive) Find

a) the maximum velocity of the particle, [3 marks] b) the distance travelled during the fourth second, [3 marks]

c) the value of t when the particle passes the point O again, [2 marks]

d) the time between leaving O and when the particle reverses its direction of motion.

[2 marks]

13

Food Price Index, I Weightage, w

Fish 110 3

Chicken m 2

Rice 130 5

Meat 105 n

Prawn 115 1

Table 2 shows the price indices and weightage of 5 types of food consumed in the year 2007 using 2006 as the base year. The composite index of these 5 items in the year 2007

using 2006 as the base year is 117 and ∑w = 13.

a) Calculate the values of m and n. [4 marks]

b) Find the price of a kilogram of rice in the year 2007 if its price in the year 2006 is RM 12.50. [2 marks]


Table 2

11


c) Given that the projected rate of change in the prices of all the foods from 2007 to 2008 is the same as that from 2006 to 2007. Find

(i) the composite index number of these foods in the year 2008, using the year 2006 as the base year.

(ii) the amount to be paid for these foods in the year 2008 if the amount paid for these items in 2006 was RM650.

[4 marks]

14. An institution offers two types of Mathematics courses, Calculus and Statistic . The number of students taking Calculus course is x and the number of students taking Statistic course is y. The number of students taking the Mathematics courses is based on the following constraints:

I : The ratio of the number of students taking Calculus course to the number of students taking Statistic course is not more than 80 : 20.

II : The total number of students taking Mathematics courses is less than or equal to 80.

III : The number of students taking Statistic course is at least 10

IV: The number of students taking Calculus course is more than 20.

(a) Write four inequalities, other than x ≥0 and y ≥ 0,which satisfy all the above constraints.

[4 marks]

(b) Using a scale of 2 cm to 10 students on both axes, construct and shade the region R which satisfies all of the above constraints.

[3 marks]

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

(i) the range of the number of students for Calculus course if the number of students for Statistics course is 20.

(ii) the maximum examination fee that can be collected if the examination fees for Calculus and Statistics courses are RM 200 and RM 400 respectively.

[3 marks]

3472/2 SULIT


SULIT 3472/2

15.

Diagram 7 shows the triangle ABC, where D is a midpoint of the line AC and ABC is an obtuse angle. Triangle CDE is an isosceles triangle such that CD = DE. Given that the length of AC = 10 cm, EC = 6 cm , AB = 5 cm and 27ACB o.

a) Calculate the ABC [3 marks]

b) Find the area of triangle ABC [3 marks]

c) If the line CB is extended to point F, find the length of the shortest distance from point A to line CF. [2 marks]

d) Calculate the CDE [2 marks]

END OF QUESTION PAPER


A

E

B C

5 cm6 cm

27o

D

Diagram 7

13


3472/2 MatematikTambahanKertas 22 ½ jam Ogos 2008

SEKOLAH BERASRAMA PENUHBAHAGIAN PENGURUSAN

SEKOLAH BERASRAMA PENUH / KLUSTERKEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2008

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

Skema Pemarkahan ini mengandungi 13 halaman bercetak

MARKING SCHEME


QUESTION NO.

SOLUTION MARKS

1

2

2

2

2

2

h + 2k = 5 ------(i)

k - 3h = 7 -----(ii)

From (i) h = 5 - 2k ------(iii)

k - 3(5 - 2k) = 7

k + 6k - 22 = 0

-b b - 4acUse

2a

-6 6 - 4(1)(-22) k =

2(1)

= 2.568 , -8.568

h = -0.136

, 22.136

2 a i)

ii)

b)

P( 0 , 6 ) , Q (4 , 0 )

R = ( 2 ,3 )

OPS = 01

02

6

0

6

4

0

0 or 64

2

1 ××

= 12

XQPX2

1=

= 2 2 2 21( 0) ( 6) ( 4) ( 0)

2x y x y− + − = − + −

2 23 3 8 48 128 0x y x y⇒ + + − + =

2

2

3

2

N1

K1 Eliminate h or k

Solve the quadratic equation by using the quadratic formula or completing squareK1

N1

N1

N1

N1

K1Find coord. P,Q and use midpoint formula

K1

N1

P1P1 P1P1P1P1

N1

P1

P1

P1P1 P1

K1

5

7


3 a)

b)

c)

See 143, 148, 153, 158, 163 and 168

4(143) + 5(148) + 8(153) + 9(158) +3(163) + 1(168)

4615

4(1432) + 5(1482) + 8(1532) + 9(1582) + 3(1632) + 1(1682)

711195

σ = 2

2

xf

fx−

∑∑

2711195153.83

30

23706.5 23664.69

6.466

3

2

3

4 2,47 −== AA da or 25=Ba , 1−=Bd

[ ])2)(1()47(22

−−+ nn

or [ ])1)(1()25(22

−−+ nn

[ ])2)(1()47(22

−−+ nn

+ [ ])1)(1()25(22

−−+ nn

= 900

018001473 2 =+− nn

Factorize/solve 24=n

5 a) Find

dy

dxcorrectly and subtitude x=1

Use 1 2 1m m = −

y = x – 2

3

3

K1

N1

K1

N1

K1

K1

N1Use s.d formula

Find the value of min

P1

K1

K1

K1

N1

N1

K1

K1

N1

8

6

P1

Accept listing method


b)

0.25

0.25

0.001 (0.25 )

1

250

V h

dv

dhdv dv dh

dt dh dtdh

dtdh

dt

3

6 a)

b)

4 4 2 2 2 2

2 2

cos sin (cos sin )(cos sin )

(cos sin )(1)

cos 2

3

5

4

N1

K1

K1

6

K1

K1

N11 2 – 1 – 2

y

x

4

4

2

2

34

34

1

1

2

2

– 1

– 1

– 2

– 2

P1P1N1 P1P1

P1

P1

P1

P1

Graph of cosMin/max = -2 & 2

1 period for 0 x

K1K1 K1

K1

N1

Draw a straight line of

1x

y

Graph of cos

Number of solutions = 2

8


7 a)

b)

log10 y = log10 p + (x – 1) log10 k

Refer to the graph

5

P1

N1x - 1 1 2 3 4 5 6log y 0.91 1.07 1.22 1.38 1.54 1.70

p = 5.623

Plot all the points correctly

Draw the best line fit

y-intercept = log10 p

t = 4

gradient = log10 k

k =1.440 1.1.445

N1

K1

N1

N1

N1

K1

K1

N1

10



6

Answer for question 7

1.4

1.6

x

x

yLog10

0 1 2 3 4 5 6 7 8

0.2

0.4

0.6

0.8

1.0

1.2

x

x

x

x

x

(a)

x - 1123456log y0.911.071.221.381.541.70

Acce


8 a i)

ii)

b i)

ii)

c)

OS

= 8p+4q

QP QO

= 8p-10q

OT (8 4 ) 8 4

QT

OT 8 10

8 8

=

4 = 10

4 = 10

OP OS

OP

a p q ap aq

OT OQ bOP

bp q

bp ap

b a

aq q

a

uuur uuur uuur

% %

uuur uuur uuur

% %

uuur

% %

uuur uuur uuur uuur

uuur

% %

% %

5 5 = ,

2 2

a b

4

3

3

9 a)

b)

tan -1 8

15

61.93 o = 1.081 rad

ABS = 10 ( π×180

50) = 8.728

Perimeter of OAB = 10 + 8.728 or EC2 = 152 + 82 = 17

Perimeter of shaded region = 10 + 8.728 + 5 + 9 + 1.081x 8 = 41.376

2

4

7

K1

N1

K1

N1

P1

K1

K1

N1

P1

K1

K1

K1

N1

K1

K1

N1

Comparison

Solve

Both correct

10

N1


c)

2

2

110 0.8728

243.64

18 1.081 34.59

2

18 15

260

43.64 60 34.592

69.048

AreaOAB

AreaOED

Area

area shaded region

4

8

K1

N1

K1

K1

N1

K1

Area OAB or Area OED

10


10 a)

b i)

ii)

iii)

dy

dt= 2t – 1 or

dx

dt = 2 or

1

2

12

2

1 xxy

dy

dx=

dy

dtx

dt

dx

=2 1

2

t

=

2

12

12

x

= 2

2

x

p = 6

32

1

332

1

Area of shaded region

1(6 8)(1) ( 8 15)

2

7 4 153

213

3

x x dx

xx x

12

0

12

0

32

1

0

(2 6)

(4 24 36)

412 36

3

1

349

Volume x dx

x x dx

xx x

2

5

3

9

K1 K1

K1

N1

K1

K1

N1

K1

Area of triangle

Integration

L1 + L

2

K1

K1

N1

Use

1

0

P1

10

Integrate


11 a i)

ii)

b i)

ii)

3

2

2

3

12 a)

b)

c)

d)

a = 15 – 6t

15 – 6t = 0, t = 15

6s

V max = 3

184

ms-1 // 18.75 // 4

75

S = 2 315t t

2

Distance travelled = S4 – S 3

= 56 - 1

402

S4-S3

= 1

152

m

2 315t t

2 = 0, S=0

t = 15

s2

V=015 t – 3 t2 = 0

Time between = 5 s

3

3

2

2

10

K1

K1

N1

K1

N1

K1

N1

K1

K1

N1 10

K1

K1

N1

a = 0

Substitute t into v

K1

K1

N1

K1

N1

K1

N1 10



13 a)

b)

c i)

ii)

07

06

07

07

08 07 08

06 06 06

08

08

n = 2

110(3)+m(2)+130(5)+105(2)+115(1)117

13m = 108

Qx 100 = 130

Q

12.50 x 130Q =

100Q = 16.25

I I I = X

I I I

= 1.17 x 1.17 x 100

= 136.89

Q X 100 = 136.89

650 Q

= 889.79

3

2

3

2

14 Refer to the graph

11

P1

K1

N1

K1

N1

K1

K1

N1

K1

N1

10


12Answer for question 14

(a) I.

II.

III.IV

(b) Refer to the graph,

1 graph correct3 graphs correct

Correct area

(c) max point ( 50,120 )

i)

ii) k = 200x + 400y

Max fees = 200(21) + 400(59)

= RM 27,800

10

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

10

20

70

60

50

80

90

40

30

y

(21,59)

10 20 30 40 50 60 700 80

x


15 a)

b)

c)

d)

'

'

2 2 2

5 10

sin 27 sinsin 0.9080

65.23 65 14

180 65.23

114.77 114 46

180 27 114.77

38.23

1(10)(5)sin 38.23

215.47

sin 27104.54

6 5 5 2(5)(5)cos

cos

ABCABC

ABC or

ABC

or

BAC

Area of triangle ABC

x

x

CDE

CDE

'73.74 73 44or

3

3

2

2

13

K1

K1

N1

P1

K1

N1 *

K1

N1

K1

N1

10


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