add math p1 trial spm zon a 2009 - tutor mansor ... 3472/2 3472/2 zon a kuching 2009 sulit 6 section...

SULIT 1 3472/2

3472/2 ZON A KUCHING 2009 SULIT

3472/2 Matematik Tambahan Kertas 2 2 ½ jam 2009

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

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 11 halaman bercetak

http://mathsmozac.blogspot.com

http://tutormansor.wordpress.com/

SULIT 3472/2


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 mn 5 log a mn = log a m + log a n

6 log a n

m = log a m − log a n

7 log a mn = n log a m

8 log a b = 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

y2π dx or

= ∫b

a

x2π dy

5 A point dividing a segment of a line

(x, y) = ,21

++

nm

mxnx

++

nm

myny 21

6. Area of triangle =

1 2 2 3 3 1 2 1 3 2 1 3

1( ) ( )

2x y x y x y x y x y x y+ + − + +

1 Distance = 221

221 )()( yyxx −+−

2 Midpoint

(x, y) =

+2

21 xx ,

+2

21 yy

3 22 yxr +=

4 2 2

xi yjr

x y

∧ +=+

GEOM ETRY



SULIT 3472/2


3

STATISTICS

TRIGONOMETRY

7 1

11

w

IwI

∑

∑=

8 )!(

!

rn

nPr

n

−=

9 !)!(

!

rrn

nCr

n

−=

10 P(A∪ B) = P(A) + P(B) − P(A∩ B)

11 P(X = r) = rnr

rn qpC − , p + q = 1

12 Mean µ = np

13 npq=σ

14 z = σ

µ−x

1 x = N

x∑

2 x = ∑∑

f

fx

3 σ = 2( )x x

N

−∑ = 2

2xx

N−∑

4 σ = 2( )f x x

f

−∑∑

= 2

2fxx

f−∑

∑

5 m = Cf

FNL

m

−+ 2

1

6 1

0

100Q

IQ

= ×

9 sin (A± B) = sinA cosB ± cosA sinB

10 cos (A± B) = cosA cosB m sinA sinB

11 tan (A± B) = BA

BA

tantan1

tantan

m

±

12 C

c

B

b

A

a

sinsinsin==

13 a2 = b2 + c2 − 2bc cos A

14 Area of triangle = Cabsin2

1

1 Arc length, s = rθ

2 Area of sector , A = 21

2r θ

3 sin 2A + cos 2A = 1 4 sec2A = 1 + tan2A 5 cosec2 A = 1 + cot2 A

6 sin 2A = 2 sinA cosA 7 cos 2A = cos2A – sin2 A = 2 cos2A − 1 = 1 − 2 sin2A

8 tan 2A = A

A2tan1

tan2

−



SULIT 4 3472/2


SECTION A

[40 marks]

Answer all questions in this section .

1 Solve the simultaneous equations 2 6 0x y− + = and 2 20 0x xy+ − = . Give your answer correct to 3 decimal places.

[5 marks]

2 Diagram 1 shows a circle with centre O. PTQ is a tangent to the circle at T and PQ = OQ = 20 cm. Calculate (a) the length of the arc STR, [4 marks] (b) the area of the shaded region. [4 marks] 3 Table 1 shows the frequency distribution of scores of a group of players in a game.

Score 0 4− 95− 1410− 15 19− 20 24− 2925− 30 34−

Number of players 2 3 10 20 w 6 2

TABLE 1

It is given that the median of the distribution is 17. (a) Calculate the value of w. [3 marks] (b) Hence, calculate the variance of the distribution. [4 marks]

4

DIAGRAM 1

300°

O

R S

P Q T



SULIT 3472/2


5

4 (a) If the volume of a cube decreases from 125 3cm to 124.4 cm3, find the small change in the sides of the cube.

[3 marks]

(b) Given that 23 4

( )3

xf x

x

+=−

, find the value of f ′(2). [3 marks]

5 (a) Prove that sin 2x = 2 sin2 x cot x. [2 marks]

(b) Sketch the graph of y = x2sin2 for 0 ≤ x ≤ π. Hence, using the same axes,

sketch a suitable straight line to find the number of solutions of the equation x2sin2

= πx2

for 0 ≤ x ≤ π. State the number of solutions.

[6 marks]

6 (a) A piece of wire is cut into 15 parts which are bent to form circles as shown in

Diagram 2. The radius of each circle increases by 3 cm consecutively. Calculate

(i) the radius of the last circle, [2 marks]

(ii) the area of the last circle. [1 mark] (b) Diagram 3 shows a rectangular geometric pattern. Diagram 2 The first rectangle is ABCD and followed by MBNP and so on. The length and width

of the next rectangle is half of the length and width of the previous rectangle. Given that AB = 30 cm and BC = 20 cm. Find the perimeter of the seventh rectangle.

[3 marks]

5 cm 2 cm 8 cm

A

D

B

C

M

P N

DIAGRAM 3

DIAGRAM 2

. . .



SULIT 3472/2


6

SECTION B

[40 marks]

Answer four questions from this section. 7 Use graph paper to answer this question.

Table 2 shows the values of two variables x and y which are related by 2+= xpqy , where p and q are constants.

(a) Convert 2+= xpqy to a linear form of cmXY += . [2 marks]

(b) Plot y10log against )2( +x by using a scale of 2 cm to 1 unit on the Y-axis and 2 cm to 1 unit on the X-axis. Hence, draw the line of best fit. [4 marks]

(c) From the graph in (b), find the value of p and of q. [4 marks]

8 Diagram 4 shows a triangle OPQ. The point R lies on OP and the point S lies on PQ. The

straight line QR intersects the straight line OS at point T.

Given OP : OR = 4 : 3, PQ : PS = 2 : 1, 12OP x=→

% and 9OQ y=

→

%

.

(a) Express, in terms of x and / or y,

(i) QR→

,

(ii) OS→

. [3 marks]

(b) If OT hOS=→ →

and QT k QR=→ →

, where h and k are constants, find the values of h and k.

[5 marks]

(c) Given that 3x =%

units, 5y =%

units and ∠POQ = 90o, find PQ→

. [2 marks]

x 1 2 3 4 5 6 y 25.6 125.9 640 3163 15849 63096

TABLE 2

O R P

T

S

Q

DIAGRAM 4



SULIT 3472/2


7

9 (a) In a certain area, 30% of the trees are rubber trees.

(i) If 8 trees in the area are chosen at random, find the probability that at least two of the trees are rubber trees.

(ii) If the variance of the rubber trees is 315, find the number of rubber trees in the

area. [5 marks]

(b) The masses of the children in the Primary One in the school have a normal

distribution with mean 33.5 kg and variance 25 kg2. 150 of the children have masses between 30 kg and 36.5 kg. Calculate the total number of children in Primary One in that school.

[5 marks]

10 Solution by scale drawing is not accepted. In Diagram 5, point T lies on the perpendicular bisector of AB. (a) Find the equation of straight line AB. [2 marks] (b) A point P moves such that 2PA AB= . Find the equation of locus of P.

[3 marks]

(c) Locus of P intersects the x-axis at points Y and Z. State the coordinates of Y and Z. [3 marks]

(d) Find the x-intercept of CD. [2 marks]

B(3, 2)

T

C O x

y

A (1, 4)

D

DIAGRAM 5



SULIT 3472/2


8

11 (a) Given that a curve has a gradient function xpx +2 such that p is a constant. xy 26−= is the equation of tangent to the curve at the point (2, q). Find the value of

p and of q. [3 marks]

(b) Diagram 6 shows the curve y = (x − 3)2 and the straight line y = 2x + 2 intersect at

point (1, 4). Calculate (i) the area of the shaded region, [4 marks] (ii) the volume of revolution, in terms of π , when the region bounded by

the curve, the x-axis, the y-axis and the straight line x = 2 is revolved through 360° about the x-axis. [3 marks]

y

y = (x − 3)2

x O

y = 2x + 2

• (1, 4)

DIAGRAM 6



SULIT 3472/2


9

SECTION C

[20 marks]

Answer two questions from this section. 12 A particle moves along a straight line and passes through a fixed point O. Its velocity, v

ms−1, is given by v = t2 − 6t + 5, where t is the time, in seconds, after passing through O. (Assume motion to the right is positive).

Find (a) the initial velocity, in ms−1, [1 mark] (b) the minimum velocity, in ms−1, [3 marks] (c) the range of values of t at which the particles moves to the left, [2 marks] (d) the total distance, in m, travelled by the particle in the first 5 seconds. [4 marks] 13 In Diagram 7, ABC is a triangle. BMC and AM are straight lines.

DIAGRAM 7 (a) Calculate (i) ∠AMB, (ii) the area, in cm2, of triangle ABC. [7 marks]

(b) A new triangle A′B′M′ is formed with A′B′ = AB, B′M′ = BM and ∠B′A′M′ = ∠BAM,

find the length of A′M′. [3 marks]

A

B C M



SULIT 3472/2


10

14 Use the graph paper provided to answer this question. A factory produces two types of school bags M and N using two types of machines A and

B. Given that machine A requires 20 minutes to produce a bag M and 30 minutes to produce a bag N while machine B requires 25 minutes to produce a bag M and 40 minutes to produce a bag N. The machines produce x units of M and y units of N in a particular day according to the following conditions.

I : Machine A is operated for not more than 8 hours. II : Machine B is operated for at least 4 hours. III : The number of units of bag M produced is not more than twice the number of

units of bag N. (a) Write the three inequalities for the above conditions. [3 marks] (b) Using a scale of 2 cm to 2 units for both axes, construct and shade the region R

which satisfies all the above conditions. [3 marks] (c) Use the graph constructed in 14 (b), to find (i) the maximum number of units of bag M that can be produced if the factory

produces 12 units of bag N . (ii) the maximum profit obtained if the profit from one unit of bag M and bag N

are RM 18 and RM 20 respectively. [4 marks]



SULIT 3472/2


11

15 Table 3 shows the price indices and percentage of usage of four components, P, Q, R and S,

which are the number of parts in the making of an electronic device.

Item Price index for the year 2000 based on the year 1997

Percentage of usage (%)

P 125 20 Q 140 10 R x 30 S 110 40

TABLE 3

(a) Calculate (i) the price of Q in the year 1997 if its price in the year 2000 is RM 50.40, (ii) the price index of P in the year 2000 based on the year 1994 if its price index in

the year 1997 based on the year 1994 is 120. [5 marks] (b) The composite index number for the cost of production in the year 2000 based on

the year 1997 is 122. Calculate (i) the value of x, (ii) the price of an electronic device in the year 1997 if the corresponding price in

the year 2000 was RM 288. [5 marks]

END OF QUESTION PAPER



3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2009

SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

MATEMATIK TAMBAHAN

Kertas 2

Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

Skema Pemarkahan ini mengandungi 14 halaman bercetak

MARKING SCHEME



2

ADDITIONAL MATHEMATICS MARKING SCHEME

TRIAL SBP 2009 – PAPER 2

QUESTION

NO. SOLUTION MARKS

1

2

2

2 6

(2 6) (2 6) 20 0

3 15 8 0

x y

y y y

y y

= −

− + − − =

− + =

0.6070, 4.393

@

4.786, 2.786

y y

x x

= =

= − =

5

2

(a)

17.3205 cmOT =

17.32053STRsπ= ×

18.1380 cmSTRs =

4

Solve the quadratic equation by using the factorization @ quadratic formula @ completing the square must be shown

Eliminate orx y

3π

Use the formula s = rθ

Note : OW-1 if the working of solving quadratic equation is not shown.

5

P1

K1

K1

N1

N1

K1

K1

N1

K1



3

QUESTION NO.

SOLUTION MARKS

(b)

Area OPQ = @ 173.2051cm2

Area OSTR = 2117.3205

2 3π× × @ 157.0795 cm2

1

20 20 sin 602

× × × o − 2117.3205

2 3π× × @

173.2051 – 157.0795 16.1256 cm2

4

17 = 14.5 + )5(20

152

43

−+ w

w = 7

3

3

(a)

(b)

Score Frequency,

f M id- point,

x

fx fx 2

0-4 2 2 4 8 5-9 3 7 21 147

10-14 10 12 120 1440 15-19 20 17 340 5780 20-24 7 22 154 3388 25-29 6 27 162 4374 30-34 2 32 64 2048

∑ = 50f ∑ = 865fx 171852 =∑ fx

Variance = 2

50

865

50

17185

−

= 44.41

4

120 20 sin 60

2× × × o

P1

Lower boundary OR

)5(20

152

43

−+ w

7

8

K1

K1

K1

N1

K1

N1

K1

P1

N1

OR

P1



4

QUESTION NO.

SOLUTION MARKS

δV = −0.6 OR dV

dx = 3x2 OR x = 5

20.6 3(5) xδ− ≈ × 0.008xδ ≈ −

3

4

(a)

(b)

[ ]

2

2 2

2

22

(3 )(3) (3 4)( 2 )( )

(3 )

3 (2) (3) (3(2) 4)( 2(2))(2)

3 (2)

37

x x xf x

x

f

− − + −′ =−

− − + − ′ = −

3

5 (a)

2 sin2 x cos

sin

x

x

= 2 sin x cos x = sin 2x

2

(b)

πx

y2=

Number of solutions = 4

6

K1

N1

K1

N1

K1

6

K1

N1

P1

8

π

xy 2sin2=

x

πx

y2=

y

O

2 1

K1

Sine curve Period Amplitude Modulus

P1

P1

P1

P1

Sketch straight line correctly

N1



5

QUESTION NO.

SOLUTION MARKS

6 (a)

(i) 15T = 2 + 14(3) = 44 cm (ii)Area of the fifteenth circle = 1936π 2cm

3

(b)

100, 50 , 25 , ………

a= 100 AND r = 2

1

7T = 1.563

3

K1

N1

N1

K1

K1

N1

6



6

2 3 4 5

1

×

4

−2

−1

0

2

3

×

x + 1

log10 y

×

x + 2 3 4 5 6 7 8

log 10 y 1.408 2.100 2.806 3.500 4.200 4.800

(a) Each set of values correct (log10 y must be at least

2 decimal places) N1, N1 Y = mX + c log 10 y = (x + 2) log 10 q + log 10 p K1 where Y = log 10 y, X = (x + 2), m = log 10 q and c = log 10 p (c) log 10 q = gradient

log 10 q = ( )

07

6.02.4

−−−

K1

q = 4.85 N1 log p = Y-intercept log p = −0.6 K1 p = 0⋅251 N1

Correct both axes (Uniform scale) K1 All points are plotted correctly N1 Line of best fit N1

1 6 7 8

×

× 5

×

10



7

QUESTION NO.

SOLUTION MARKS

8

(a) (i)

(ii)

9 9QR x y= −uuur

% %

( )19 9 12

2OS y y x= + − +uuur

%% %

9

62

x y= +% %

3

OT hOS=uuur uuur

9

62

hx hy= +% %

OR QT kQR=uuur uuur

9 9kx ky= −

% %

QT QO OT= +uuur uuur uuur

9

9 62

y hx hy= − + +%% %

9

6 92

hx h y = + − + % %

Comparing QTuuur

,

9 6k h= ⇒ 2

3k h= --------------------------(1)

OR 9

9 92

k h− = − + ⇒ 1

12

k h= − --------------(2)

Solving the simultaneous equations

6 4,

7 7h k= =

5

(b)

(d)

12 9PQ x y= − +uuur

% %

( ) ( )2 212(3) 9(5)PQ = +

uuur

= 57.63 units

2

10

K1

K1

K1

K1

N1

N1

N1

N1

K1

K1



8

QUESTION NO.

SOLUTION MARKS

9

(a) (i)

(ii)

p = 0.3, q = 0.7, n = 8 1 − (0.7)8 − 8C1(0.3)(0.7)7 = 0.7447 n(0.3)(0.7) =315 n = 1500

5

(b) (i)

(ii)

P30 33.5 36.5 33.5 150

5 5Z

n

− − < < =

OR

P [ ] 1500.7 0.6Z

n− < < =

1 − P[Z > 0.7] − P[Z > 0.6] = 150

n

1 − 0.2420 − 0.2743 = 150

n

n = 310.11 Total number = 310

5

N1

K1

K1

N1

K1

N1

10

K1

K1

K1

K1



9

QUESTION NO.

SOLUTION MARKS

( )

1

4 1 1

5 0

Get m

y x

y x

= −

− = − −

+ − =

3

( ) ( ) ( ) ( )( )

2 2 2 2

2 2

2 2

1 4 2 4 2 1 3

2 1 8 16 4 8

2 8 15 0

x y

x x y y

x y x y

− + − = − + −

− + + − + =

+ − − − =

2

( ) ( )

( ) ( )

2 2 15 0

5 3 0

5, 3

Get both

/ 5,0 and / 3,0

x x

x x

x x

Y Z Z Y

− − =− + =

= = −

= = −

3

10 (a)

(b)

(c)

(d) ( )Get 2, 3 ,

intercept 1

T

x

=

− = −

2

10

N1

K1

K1

K1

K1

K1

N1

N1

K1

N1



10

QUESTION NO.

SOLUTION MARKS

11 (a)

2

2

2

(2) 2 2

1, 2

px x

p

p q

+ = −

+ = −

= − =

3

(b) (i)

(ii)

Area of the shaded region

= 1 1

2

0 0( 6 9) (2 2)x x dx x dx− + − +∫ ∫

= 1

2

0( 8 7)x x dx− +∫

= 13 2

0

87

3 2

x xx

− +

= 1 8

7 03 2 − + −

= 1

33

unit 2 .

Volume of revolution

= 2 2

0y dxπ ∫

= 2 40

( 3)x dxπ −∫

= 5 2

0

( 3)

5

xπ −

= 5 5(1 3) (0 3)

5 5π − −−

= 2

485

π unit3 .

7

N1

K1

K1

10

K1

N1

K1

K1

K1

K1

N1



11

QUESTION NO.

SOLUTION MARKS

v = 5 ms−1

1

a = 2t − 6 and a = 0 or dv

dt = 0

t = 3 vmin = −4 ms−1

3

(t − 1)(t − 5) < 0 1 < t < 5

2

12 (a)

(b)

(c)

(d)

323 5

3t

s t t= − +

| s1 − s0 | + | s5 − s1 | OR 1 5

0 1v dt v dt+∫ ∫

7 25 7

03 3 3

− + − −

OR 3 3 3

2 2 21 5 13(1) 5(1) 0 3(5) 5(5) 3(1) 5(1)

3 3 3

− + − + − + − − +

13 m

4

10

K1

K1

K1

K1

N1

N1

P1

K1

N1

K1



12

QUESTION

NO. SOLUTION MARKS

)8)(5(2

1285cos

222 −+=∠AMB

∠AMB = 133.43˚ @ 133˚26'

2

4

57.46sin

5

sin °=∠ACM

∠ACM = 65.20˚ @ 65˚24' ∠MAC = 180˚ – 46.57˚ - 65.20˚ = 68.23˚ Area of ∆ACM

= 2

1(5)(4)sin 68.23˚ OR

= 9.287 cm2

Area of ∆ABC = 14.52 + 9.287

= 23.807 cm2

5

13 (a) (i)

(ii)

(b)

Get ∠B 'M 'M = 46.57˚ @ ∠B 'MM ' = 46.57˚ @ ∠M 'B 'M = 86.86˚

°=

° 57.46sin

8

86.86sin

'MM

MM ' = 10.999 cm @ 11 cm A'M '= 5 + 10.999 = 15.999 cm @ 16 cm

3

10

K1

K1

K1

N1

K1

N1

K1

Area of ∆AMB

= 2

1(5)(8)sin 133.43˚

= 14.52 cm2

K1

K1

N1



13

Answer for question 14

2 4 6 8 10 12 14 0 16

2

4

14

12

10

16

8

6

(12, 8)

(a) I. 2 3 48x y+ ≤

II. 5 8 48x y+ ≥ III. 2 y x≥

(b) Refer to the graph, 1 or 2 graph(s) correct 3 graphs correct Correct area (c) i) 6 ii) max point (12, 8) k = 18x + 20y Maximum Profit = RM 18(12) + RM 20(8) = RM 376

10

N1

N1

N1

N1

N1

N1

K1

N1

K1

N1

R



14

QUESTION

NO. SOLUTION MARKS

97

RM 50.40140 100

Q= ×

Q97 = RM 36

00,9400,97

97,94

100I

II

= ×

00,94125 100120

I= ×

I00, 94 = 150

5

15 (a) (i)

(ii)

(b) (i)

(ii)

125 20 140 10 30 110 40

122100

x× + × + + × =

30x + 8300 = 12200 x = 130

1997

RM 288122 100

Q= ×

1997 RM 236.07Q =

5

10

N1

N1

N1

K1

K1

K1

N1

K1

K1

K1



add math p1 trial spm zon a 2009 - tutor mansor ... 3472/2 3472/2 zon a kuching 2009 sulit 6 section...

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