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BAHAGIAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER ADDITIONAL MATHEMATICS www.cikgurohaiza.com

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BAHAGIAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

ADDITIONAL MATHEMATICS

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PAGE 1 2 3 4 5 6 7 8 ABOUT THIS MODULE WE LEARN EXAMINATION FORMAT ANALYSIS TABLE OF SPM ADD MATHS QUESTIONS LIST OF FORMULAE AND NORMAL TABLE ADDITIONAL MATHEMATICS NOTES PROBLEM SOLVING STRATEGY PARTITION I II II-III IV V-VII VIII-XVI XVII XVIII

This module is 1. specially planned for students who will be sitting for SPM. 2. to provide exposure and to familiarize students with the needs of the actual SPM exam questions. 3. to prepare students with adequate knowledge prior to the examination. 4. comprises challenging questions which incorporate a variety of questioning techniques and levels of difficulty and conforms to the current SPM farmat.

That which we persist in doing becomes easier not that the nature of the task has changed, but our ability to do has increased.

I

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Key towards achieving 1A Read question carefully Follow instructions Start with your favourite question Show your working clearly Choose the correct formula to be used +(Gunakannya dengan betul !!!) Final answer must be in the simplest form The end answer should be correct to 4 S.F. (or follow the instruction given in the question) 3.142

Kunci Mencapai kecemerlangan

Proper / Correct ways of writing mathematical notations Check answers! Proper allocation of time (for each question)

Paper 1 : 3 - 7 minutes for each question Paper 2 : Sec. A : 8 - 10 minutes for each question Sec. B : 15 minutes for each question Sec. C : 15 minutes for each question

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ANALYSIS TABLE OF SPM ADDITIONAL MATHEMATICS QUESTIONS 2004-2008AMaths (3472) SPM Chapter 1 2 3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 7 8 9 Functions Quadratic Equations Quadratic Functions Simultaneous Equations Indices and Logarithms Coordinate Geometry Statistics Circular Measure Differentiation Solution of Triangle Index Number Progressions Linear Law Integration Vectors Trigonometric Functions Permutations / Combinations Probability Probability Distributions Motion Along A Straight Line Linear Programming Total 04 3 1 2 05 3 2 1 Paper 1 06 2 1 1 07 3 1 2 08 3 1 2 1 2 2 3 1 1 1 2 1 2 3 1 1 1 3 2 2 1 1 2 2 2 1 1 2 1/2 1/2 2/3 1 1 1 1 1 1 1 1/2 1 1/3 1 1 1/3 1 1/3 1 1 4 1 1 2 1 1 1 1 3 1 1 2 1 1 1 1 2 1 2 2 1 1 1 1 3 1 1 2 1 1 1 1 3 1 1 2 1 1 1 1 1 1 1 1 1 1 1 25 25 25 25 25 6 6 6 6 6 5 5 5 5 5 4 1 1 4 1 1 4 1 1 4 1 1 4 1 1 1 1 1 1 1 1/3 1 1 1 1 1 1 1 1/2 1 1 2/3 1 1 1 2/3 1 1 2/3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 04 05 Section A 06 07 1 08 04 05 Paper 2 Section B 06 07 08 04 05 Section C 06 07 08

10

IV

SULIT

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The following formulae may be helpful in answering the questions. The symbols given are the ones commonly used. ALGEBRA 1 x=

b b 4ac 2a2

8

logab =

log c b log c a

2 am a n = a m + n 3 am an = a m - n

9 10 11 12

Tn = a + (n-1)d Sn =

4 (am) n = a nm 5 6 7 loga mn = log am + loga n loga

n [2a + (n 1)d ] 2

Tn = ar n-1 Sn =

m = log am - loga n n

log a mn = n log a m

a(r n 1) a(1 r n ) , (r 1) = r 1 1 r a , r 0 2 different (distinct) roots. b2- 4ac = 0 2 equal roots b2 - 4ac < 0 no real roots. b2 - 4ac 0 with real rootsyy y

_ _ +a

_ + _b

+ + +

4. INDICES & LOGARITHM (a) x = an Index Form loga x = n Logarithmic Form

(b) Laws of Indices

1. a n a m = a n+m 2. a n a m = a nm 3. (a n ) m = a nmLaws of Logarithm 1. logaxy = logax + logay x 2. loga = logax logay y 3. loga xn = n logax 4. loga a = 1 5. loga 1 = 0 log c b 6. loga b = log c a 1 7. loga b = log b a

0

x

0

x

0

b2 - 4ac > 0

b2 - 4ac = 0

b2 - 4ac < 0

x

(b) Completing the Squares y = a(x - p)2 + q a +ve minimum point (p, q) a ve maximum point (p, q) (c) Quadratic Inequalities (x a)(x b) 0 Range: x a, x b (x a)(x b) 0 Range: a x b

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X

5. COORDINATE GEOMETRY (a) Distance between A(x1, y1) and B(x2, y2) AB = ( x 2 x1 ) 2 + ( y 2 y1 ) 2 (b) Midpoint of AB x + x 2 y1 + y 2 M= 1 , 2 2 (c) P divides AB internally in the ratio m : nm : A(x 1 , y1 ) P n B(x2 , y2 )

6. STATISTICS Measure of Central Tendency (a) Mean x x= n for ungrouped datax=

fx f fx f

for ungrouped data with frequency.x=i

nx + mx 2 ny1 + my 2 P= 1 , n+m n+m (d) Gradient of AB y y1 m= 2 x 2 x1m=

for grouped data , xi = midpoint of each class interval (b) Median The centre value of a set of data after the data is arranged in the ascending or descending order. FormulanF C fm L = Lower boundary of the Median class n = Total frequency F = Cumulative frequency before the median class fm = Frequency of the median class C = Size of the class interval

y-intercept x-intercept

(e) Equation of a straight line (i) given m and A(x1, y1) y y1 = m(x x1) (ii) given A(x1, y1) and B(x2, y2) y y1 y 2 y1 = x x1 x 2 x1 (f) Area of polygon x 1 x1 x 2 x3 L= ......... 1 y1 2 y1 y 2 y 3 (g) Parallel lines m 1 = m2 (h) Perpendicular lines m1 m2 = -1

M=L+

1 2

From the OgiveCumulative Kekerapan Frequency longgokann

n __ 2

0

Median Sempadan atas

Median

Upper Boundary

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XI

(c) Mode Data with the highest frequency From the Histogram :KFrequency ekerapan

For ungrouped data

==

( x x)n

2

x

2

n For grouped data

x

2

=

f ( x x) f

2

=0 Mod Sempadan kelas Mode Class Boundary

fx f

2

x

2

Measure of Dispersion (a) Interquartile Range Formulae : 1 n F1 Q1 = L1 + 4 C f Q1 Q3 = L3 + Ogive:Cumulative FrequencyKekerapan longgokan3 4

7. INDEX NUMBERS (a) Price Index P I = 1 100 P0 where Po = price at the base time P1 = price at a specific time (b) Composite Index Iw I= w where I = price index or index number w = weightage 8. CIRCULAR MEASURE (a) Radian Degree 180 0 r= (b) Degree Radian rad 180 (c) Arc length s = j (d) Area of sector 1 1 L = j2 = js 2 2

n F3 C f Q3

3 __ n 4

1 __ n 4

0

Q

Interquartile Range = Q3 Q1 (b) Variance, Standard Deviation Variance = (Standard Deviation)2

1

Q 3 Sempadan atas Upper Boundary

o =

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XII

(e) Area of segment 1 L = j2( r sin o) 2 9. DIFFERENTIATION (a) Differentiation using the First Principal

10. INTEGRATION ax n +1 (a) ax n dx = +c n +1

(ax + b) n+1 (b) (ax + b) dx = +c a (n + 1)n

dy y = lim dx x0 x

(c)

f ( x) + g ( x) dxa

b

(b)

d (a) = 0, a = constant dx d n (x ) = nxn-1 (c) dx d (d) (axn) = anxn-1 dx

=b

a

b

f ( x) dx + g ( x) dxac c

b

(d) (e) (f)y

f ( x) dx + f ( x) dx = f ( x) dxa b a

af ( x) dx = a f ( x) dxa a

b

b

(e) Product Rule d dv du (uv) = u +v dx dx dx (f) Quotient Rule dv d u v du dx u dx = dx v v2 (g) Composite Function dy du d (ax+b)n = dx du dx = an(ax+b)n-1 dy =0 (h) Turning point dx Maximum point: d2y dy = 0 and 0 dx dx 2 (i) Rate of change dy dy dx = dt dx dt

f ( x) dx = f ( x) dxa b

b

a

(h) Area under the curve

A=

y dxa

b

0 y b

a

b

x

A=a

x dya

b

0

x

(i) Volume of revolutiony

0

a

b

x

V = y 2 dxa

b

(j) Small change : dy y . x dxXIII

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y b

V = x 2 dya

b

a 0

x

(a) s = 0 at the fixed point O (b) v = 0 stops momentarily maximum / minimum displacement (c) a = 0 v constant v maximum/ minimum 13. TRIGONOMETRIC FUNCTIONS (a) y y P(x, y) sin = r x cos = r r y y tan = x0 x x

11. PROGRESSIONS Arithmetic Progressions (a) Tn = a + (n - 1)d n (b) Sn = {2a + (n - 1)d} 2 n = (a + l) 2 (c) d = T2 - T1 Geometric Progressions (a) Tn = arn-1 a(1 r n ) (b) Sn = for r < 1 1 r a(r n 1) Sn = for r > 1 r 1 a (c) S = for -1 < r < 1 1 r and n T (d) r = 2 T1 General (a) S1 = T1 = a (b) Tn = Sn Sn-1 (c) Sum of terms from Ta to Tb = Sb Sa-1

(b) tan =

sin cos 1 sec = cos 1 sin

cosec = cot = (c)

1 cos = tan sin

Sin +ve

All Semua +ve Cos Kos +ve

Tan +ve

12. MOTION ALONG A STRAIGHT LINE ds dv dt dt s v a v dt a dtXIV

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(d) Special Angles 0o 30o 0 1 Sin 2 1 Cos 3 Tan 0

45 1

o

60

o

2 121

2 1 3

3 2 1 2 3

(f) sin2 + cos2 = 1 1 + tan2 = sec2 1 + cot2 = cosec2 (g) sin(A B) = sinA cos B cos Asin B cos(A B) = cosA cosB m sinA sinB tan (A B) tan A tan B = 1 m tan A tan B (h) sin2A = 2 sinA cosA cos2A = cos2A sin2A = 2 cos2A 1 = 1 2 sin2A 2 tan A tan 2A = 1 tan 2 A 14. VECTORS (a) Addition of Vectors 1. Triangle Lawb a +

Sin Cos Tan

90o 1 0

180o 270o 0 -1 -1 0 0

360o 0 1 0

(e) Trigonometric Graphs y = a sin bxy a

0

__ 90 b

180 __ b

270 __ b

__ 360 b

x

-a

y = a cos bxy aa

b

2. Parallelogram Law0__ 90 b 180 __ b 270 __ b __ 360 b

xb a +

-a

b

y = a tan bxya

3. Polygon Law0__ 90 b 180 __ b 270 __ b __ 360 b

xA

B C

uuu r uuu v uuu v uuu v D uuuv AE = AB + BC + CD + DEXV

E

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(b) Subtraction of vectorsa

b

-b

(b) Combination n Cn = 1 n! n Cr = (n r ) !r !

a

(c) Vectors in the Cartesan Coordinatesy P(x, y)

C r = nC n r (c) Binomial Distribution P(X = r) = n C r pr qn-r (d) Mean = = npn

Standard deviation = npq

(e) Converting Normal Distribution to Standard Normal X Distribution Z =

r yj

(f) Probability Distribution Graph

0

xi

x

r = xi + yj

r = x2 + y2 xi + y j r = = r r x2 + y 215. SOLUTIONS OF TRIANGLE (a) Sine Rule

0

a

Z

P(Z > a)

1. P(Z < a) = 1 P(Z > a) use P 2. P(Z < -a) = P(Z > a) use P 3. P(Z > -a) = 1 P(Z > a) use R4. 5. 6. P(a < Z < b) = P(Z>a) P(Z > b) P(-a < Z< b) = 1 P(Z >a) P(Z > b) P(-a < Z < -b) = P(b < Z < a)

a b c = = sin A sin B sin C(b) Cosine Rule a2 = b2 + c2 2bc cos A

Examples: a) P(Z> 0.1) b) P ( Z< 0.1) c) P ( -1.2 < Z < 0.4) Examples: d) P( Z > a ) = 0.3, find a e) P (Z > a) = 0.6, find a f) P (Z < a) = 0.1, find a g) P ( Z < a ) = 0.73, find a h) P(X > a) = 0.3, given = 45, =3

b2 + c2 a2 2bc (c) Area of Triangle 1 L = ab sin C 2cos A = 16. PROBABILITY DISTRIBUTIONS (a) Permutation n Pn = n! n! n Pr = (n r )!

XVI

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How to Solve a Problem

Understandthe Problem

Plan yourStrategy

Do - Carry outYour Strategy

Check yourAnswers

Which Topic / Subtopic ?

Choose suitablestrategy

Carry out thecalculations

Is the answerreasonable?

1.

PN ZABIDAH BINTI AWANG SM AGAMA PERSEKUTUAN, LABU. EN AMIRULFAIZAN BIN AHMAD SBP INTEGRASI SELANDAR, MELAKA. PN ROHANI MD NOR SEKOLAH SULTAN ALAM SHAH, PUTRAJAYA EN ZUZI BIN SHAFIE SM AGAMA PERSEKUTUAN, KAJANG. PN SARIPAH BINTI AHMAD SM SAINS MUZAFFAR SYAH, MELAKA.

2.

3.

4.

5.

XVII

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Master these questions

XVIII

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For examiners use only

Answer all questions. f 1. 3 -1 6 Diagram 1 g

Set A

Set B

Set C

In Diagram 1, the function f maps set A to set B and the function g maps set B to set C. Determine (a) f (3 ) (b) g(-1) (c) gf (3) [ 3 marks ] Answer : (a) .. (b) ... (c).................................... 2. Given function f : x 3 4x and function g : x x2 1, find (a) f 1, (b) the value of f 1g(3). [ 3 marks ] 13

2 Answer : (a) .. (b) ...3

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For examiners use only

3

Given the function f (x) = 4x, x 0 and the composite function f g(x) =

16 . Find x

(a) g(x), (b) the value of x when g(x) = 8. [3 marks]

33

Answer : .........

4

Solve the quadratic equation 2 x ( x 5) = (2 x )( x + 3) . Give your answer correct to four significant figures. [ 3 marks ]

43

Answer : .........

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5

(a) Given x =

4 y , find the range of x if y > 10. 2 (b) Find the range of x if x2 2x 3.

For examiners use only

[4 marks]

Answer : .................................___________________________________________________________________________ 6

Diagram below shows the graph of a quadratic function y = f ( x) . The straight line y = 9 is a tangent to the curve y = f ( x) .

y y = f ( x)

0

1

x 7 y = -9 Diagram 1

a) Write the equation of the axis of symmetry of the curve. b) Express f ( x) in form of ( x + p ) 2 + q , where p and q are constants. [ 3 marks ]

6

Answer : (a) ........................ (b) ........................3

3472/1

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For examiners use only 7

Solve the equation 324x = 48x + 6 [3 marks]

73

Answer : ..................................8.

Given log5 3 = 0.683 and log5 7 = 1.209. Calculate (i) log5 1.4, (ii) log7 75.

[ 4 marks]

84

Answer : ...................................

9.

Solve the equation log

x

16 log x 2 = 3.

[3 marks]

93

Answer : ......................................

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10.

The first terms of the series are 2, x , 8. Find the value of x such that the series is a (a) an arithmetic progression, (b) a geometric progression. [2 marks ]

For examiners use only

102

Answer : .........

11. The sum of the first n terms of an arithmetic progression is given by S n = 3n 2 + 13n. Find

(a) the ninth term, (b) the sum of the next 20 terms after the 9th terms. [3 marks]

11

Answer: a)............ b) ....................................

4

3472/1

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For examiners use only

12.

Given that

1 = 0.166666666..... p = 0.1 + a + b + ............ [ 3 marks ]

Find the values of a and b. Hence, find the value of p.

12

Answer: a =....4

b =.......

p = ........................___________________________________________________________________________ 13.

Diagram 2 shows a linear graph of

y against x2 x

y x

(4,1)

x2

(1,-5)

DIAGRAM 2

Given that

y = hx2 + k, where k and h are contants. x

Calculate the value of h and k.

[3 marks]

Answer :133

h = ... k = ........

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For examiners use only

14.

The equation of a straight line PQ is

x y + = 1. Find the equation of a straight line 3 2 that is parallel to PQ and passes through the point (6 , 3). [3 marks]

143 Answer : .

15

7 Given u = 9 dan v = case:

p 1 3 , find the possible values of p for each of the following [2 marks] [2 marks]

(a) u and v are parallel, (b) u = v .

154 Answer : a).. b)

3472/1

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For examiners use only

16

P

R Qr O

Ss

The diagram above shows OR = r, OS = s, OP and PQ are drawn in the square grid. Express in terms of r and s. (i) (ii)

OP uuu r PQ .

[ 3 marks ]

163

uuu r Answer: a) OP = ....

uuu r b) PQ =..... ___________________________________________________________________________17.

Solve the equation 3 cos2 + sin 2 = 0 for 0 0 360 0 .

[ 4 marks ]

174

Answer: ............www.cikgurohaiza.com

18.

P

For examiners use only

O Q

Diagram above shows a length of wire in the form of sector OPQ, centre O. The length of the wire is 100 cm. Given the arc length PQ is 20 cm, find (a) the angle in radian, (b) area of the sector OPQ. [2 marks] [2 marks]

18

Answer: a) b) ___________________________________________________________________________19.

4

Find the equation of the tangent to the curve y =

5 at the point (3, 4). ( x 5) 3 [2 marks]192

Answer:

3472/1

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For examiners use only 20.

A roll of wire of length 60 cm is bent into the shape of a circle. When above the wire is heated, its length increases at a rate of 0.1 cms1. (Use = 3.142) (i) Calculate the rate of change of radius of the circle. (ii) Hence, calculate the radius of the circle after 4 seconds. [2 marks] [2 marks]

205

Answer: ............ ___________________________________________________________________________21.

Given

4 0

f ( x) dx = 5 and

3 1

g ( x) dx = 6.

Find the value (a)

4 0

2 f ( x) dx + g ( x)dx ,3

1

[1 marks] [2 marks]

(b) k if

3 1

[ g ( x) k x] dx =14.

213

Answer: a) .. k =...

223

22. A chess club has 10 members of whom 6 are men and 4 are women. A team of 4 members is selected to play in a match. Find the number of different ways of selecting the team if (a) all the players are to be of the same gender,

(b) there must be an equal number of men and women. [3 marks]

Answer: p = . .3472/1www.cikgurohaiza.com

11

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For examiners use only

23. (a) Given that the mean for four positive integer is 9. When a number y is added to the four positive integer, the mean becomes 10. Find the value of y. [2 marks] (b) Find the standard deviation for the set of numbers 5, 6, 6, 4, 7. [3 marks]

23

Answer: a)............ b) ............................... ___________________________________________________________________________ 24. Hanif , Zaki and Fauzi will be taking a driving test. The probabilities that Hanif , 1 1 1 respectively. Calculate the , and Zaki and Fauzi will pass the test are 2 3 4 probability that (a) only Hanif will pass the test (b) at least one of them will pass the test. [ 3 marks ]

5

243

Answer:

3472/1

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For examiners use only

25.

Diagram below shows a standard normal distribution graph.

252

-k

k

z

Given that the area of shaded region in the diagram is 0.7828 , calculate the value of k. [ 2 marks ]

Answer: ............

END OF QUESTION PAPER

3472/1www.cikgurohaiza.com

13 JAWAPAN

3472/1

1 2

(a) 1 (a) f1

(b) 6 =3 x 4

(c) 6(b) 5 4

1314

h = 2 , k = 73y = 2x 3

3

(a) g(x) =

4 ,x0 x

15

1 (b) x = 2

(a)

10 3

(b) 10, 12

4

3.562 , -0.5616

16

(b)(i) 3r + 2s

(ii) r 3s

5. 6

(a) x < 3 a) x = 4

(b) 1 x 3

17

90, 123 41, 270, 303 41 1 2

b) f ( x) = ( x 4) 92

18 19 20 21

(a)

(b) 400

7 8 9 10 11 12

x=3 ( i) 0.209 (ii) 2.219 x = 4 a) 5 (a) 64 b) 4 ( b) 2540

15x + 16y 109 = 0 ( i) 0.01591 cms1 (a) 4 14 553 (a ) 14 (b) 1.020 (a) 9/35 k = 1.234 ( b) 5/6 (b) k = 2 (ii) 9.612

22 23 24 25

a = 0.06 , b = 0.006 , p = 6

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4THE UPPER TAIL PROBABILITY Q(z) FOR THE NORMAL DISTRIBUTION N(0, 1) KEBARANGKALIAN HUJUNG ATAS Q(z) BAGI TABURAN NORMAL N(0, 1)z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 0 0.5000 0.4602 0.4207 0.3821 0.3446 0.3085 0.2743 0.2420 0.2119 0.1841 0.1587 0.1357 0.1151 0.0968 0.0808 0.0668 0.0548 0.0446 0.0359 0.0287 0.0228 0.0179 0.0139 0.0107 1 0.4960 0.4562 0.4168 0.3783 0.3409 0.3050 0.2709 0.2389 0.2090 0.1814 0.1562 0.1335 0.1131 0.0951 0.0793 0.0655 0.0537 0.0436 0.0351 0.0281 0.0222 0.0174 0.0136 0.0104 2 0.4920 0.4522 0.4129 0.3745 0.3372 0.3015 0.2676 0.2358 0.2061 0.1788 0.1539 0.1314 0.1112 0.0934 0.0778 0.0643 0.0526 0.0427 0.0344 0.0274 0.0217 0.0170 0.0132 0.0102 0.00990 2.4 2.5 2.6 2.7 2.8 2.9 3.0 0.00820 0.00621 0.00466 0.00347 0.00256 0.00187 0.00135 0.00798 0.00604 0.00453 0.00336 0.00248 0.00181 0.00131 0.00776 0.00587 0.00440 0.00326 0.00240 0.00175 0.00126 0.00755 0.00570 0.00427 0.00317 0.00233 0.00169 0.00122 0.00964 0.00734 0.00714 0.00554 0.00415 0.00307 0.00226 0.00164 0.00118 0.00539 0.00402 0.00298 0.00219 0.00159 0.00114 0.00695 0.00523 0.00391 0.00289 0.00212 0.00154 0.00111 0.00676 0.00508 0.00379 0.00280 0.00205 0.00149 0.00107 0.00657 0.00494 0.00368 0.00272 0.00199 0.00144 0.00104 0.00639 0.00480 0.00357 0.00264 0.00193 0.00139 0.00100 0.00939 0.00914 0.00889 0.00866 0.00842 3 0.4880 0.4483 0.4090 0.3707 0.3336 0.2981 0.2643 0.2327 0.2033 0.1762 0.1515 0.1292 0.1093 0.0918 0.0764 0.0630 0.0516 0.0418 0.0336 0.0268 0.0212 0.0166 0.0129 4 0.4840 0.4443 0.4052 0.3669 0.3300 0.2946 0.2611 0.2296 0.2005 0.1736 0.1492 0.1271 0.1075 0.0901 0.0749 0.0618 0.0505 0.0409 0.0329 0.0262 0.0207 0.0162 0.0125 5 0.4801 0.4404 0.4013 0.3632 0.3264 0.2912 0.2578 0.2266 0.1977 0.1711 0.1469 0.1251 0.1056 0.0885 0.0735 0.0606 0.0495 0.0401 0.0322 0.0256 0.0202 0.0158 0.0122 6 0.4761 0.4364 0.3974 0.3594 0.3228 0.2877 0.2546 0.2236 0.1949 0.1685 0.1446 0.1230 0.1038 0.0869 0.0721 0.0594 0.0485 0.0392 0.0314 0.0250 0.0197 0.0154 0.0119 7 0.4721 0.4325 0.3936 0.3557 0.3192 0.2843 0.2514 0.2206 0.1922 0.1660 0.1423 0.1210 0.1020 0.0853 0.0708 0.0582 0..0475 0.0384 0.0307 0.0244 0.0192 0.0150 0.0116 8 0.4681 0.4286 0.3897 0.3520 0.3156 0.2810 0.2483 0.2177 0.1894 0.1635 0.1401 0.1190 0.1003 0.0838 0.0694 0.0571 0.0465 0.0375 0.0301 0.0239 0.0188 0.0146 0.0113 9 0.4641 0.4247 0.3859 0.3483 0.3121 0.2776 0.2451 0.2148 0.1867 0.1611 0.1379 0.1170 0.0985 0.0823 0.0681 0.0559 0.0455 0.0367 0.0294 0.0233 0.0183 0.0143 0.0110 1 2 3 4 5 6 7 8 9

Minus / Tolak 4 4 4 4 4 3 3 3 3 3 2 2 2 2 1 1 1 1 1 1 0 0 0 0 3 2 2 2 2 1 1 1 0 0 8 8 8 7 7 7 7 6 5 5 5 4 4 3 3 2 2 2 1 1 1 1 1 1 5 5 4 4 3 2 2 1 1 1 12 12 12 11 11 10 10 9 8 8 7 6 6 5 4 4 3 3 2 2 1 1 1 1 8 7 6 6 5 3 3 2 1 1 16 16 15 15 15 14 13 12 11 10 9 8 7 6 6 5 4 4 3 2 2 2 1 1 10 9 8 7 6 5 4 3 2 2 20 20 19 19 18 17 16 15 14 13 12 10 9 8 7 6 5 4 4 3 2 2 2 1 13 12 11 9 8 6 5 4 2 2 24 24 23 22 22 20 19 18 16 15 14 12 11 10 8 7 6 5 4 4 3 2 2 2 15 14 13 11 9 7 6 4 3 2 28 28 27 26 25 24 23 21 19 18 16 14 13 11 10 8 7 6 5 4 3 3 2 2 18 16 15 13 11 9 7 5 3 3 32 32 31 30 29 27 26 24 22 20 19 16 15 13 11 10 8 7 6 5 4 3 3 2 20 16 17 15 12 9 8 6 4 3 36 36 35 34 32 31 29 27 25 23 21 18 17 14 13 11 9 8 6 5 4 4 3 2 23 21 19 17 14 10 9 6 4 4

f

Example / Contoh: If X ~ N(0, 1), then Jika X ~ N(0, 1), maka

f ( z) =

1 exp z 2 2 2 1Q(z)

Q ( z ) = f ( z ) dzk

P(X > k) = Q(k) P(X > 2.1) = Q(2.1) = 0.0179

O

k

z

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SECTION A

[40 marks] [40 markah]Answer all questions in this section .

1. Solve the equations x2 y + y2 = 2x + 2y = 10. [5 marks][ Answer x = 2, y = 3; x =

5 5 ,y= ] 2 2

2 Given kx2 x is the gradient function for a curve such that k is a constant. y 5x + 7 = 0 is the equation of tangent at the point (1, 2) to the curve. Find, (a) the value of k, (b) the equation of the curve. [2 marks] [3 marks] [ Answer k = 6 ] x2 7 [ y = 2x3 ] 2 2

3

Diagram 3 Diagram 3 shows a string of length 125 cm is cut and made into ten circle as shown above . The diameter of each subsequent circles are difrent by 1 cm from its previous. Calculate, (i) the diameter of the smallest circle , (ii) the number or a circle if the length of a circle is 400 [6 marks] Answer : (b)(i) 8 (ii) 21 4 Table 2 shows the frequency distribution of the marks of a group of form 4 students in a test. Mark 20 29 30 39 40 49 50 59 60 69 70 - 79 Number of students 2 10 36 55kwww.cikgurohaiza.com

5

(a) (b)

It is given that the first quartile score it 44.5. Find the value of k. [ 3 marks ] [ Use the graph paper to answer this question] Using the scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 10 students on the vertical axis, draw a histogram based on the given data. Hence, estimate the mode mark [ 3marks ] Calculate the mean marks. [2 marks ] [ Answer k = 12, mode = 52.25 mean = 55.81 1 - sec = - sin tan sec [3 marks]

(c)

5

a)

Prove that

(b)

Sketch y = 1 sin 2 x for 0 x . Hence using the same axes , draw a suitable straight line to find the number of solutions of the equation sin 2 x x = 0 . State the number of solutions [ line y = x

+ 1] , 4 number of solution ] [5 marks]

6 A 4y D

5x B P

C x In the diagram above, AB = 5x, AD = 4y and DC = x. (i) Express, (a) AC (b) BD in terms of x and y. (ii) Given AP = h AC and BP = k BD . State AP (a) in terms of h, x and y, (b) in terms of k, x and y. Hence, or otherwise, prove that h = k.

[2 marks]

[5 marks]

Answer (i)(a) 4y + x, (b) 5x + 4y3472/2

(ii)(a) h(4y + x) (b) ( (5 5k)x + 4ky; k =SULIT

5 6

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SECTION B [ 40 Marks ] Answer four equations from this section.

7 Table 7 shows the values of two variables x and y ,obtained from an experiment. Variable x and y are related by the equations y = ab x , where a and b are constants. One of the value of y is wrongly recorded. 1 2 3 4 5 x y 41.7 38.7 28.9 27.5 20.1 (a) Plot log 10 y against x. (b) By using your graph find, (i) the value of y which is wrongly recorded and determine the correct value (ii) the value of a and the value of b (iii) the value of y when x = 2.5 . 8

y

y=a

Q

y = x2 + 1

1 3

O

1 3

x

(a) Refer to the diagram above, answer the following question: (i) Calculate the area of the shaded region. (ii) Q is a solid obtained when the region bounded by the curve y = x2 + 1 and the line y = a is 1 revolved through 180 at the y - axis. If the volume of Q is unit2 Find the value a. 2 [6 marks] (b) Find the equation of tangent to the curve y = 2x2 + r at point x = k. If the tangent passes through the point (2, 0), find r in terms of k. [4 marks ] [Answer 16. (a)(i) 56 (ii) a = 2 81 (b) y (2k2 + r) = 4k(x k); r = 2k2 8k ]

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Solutions to this question by scale drawing will not be accepted. 9. y U(5, 6) T(0, 4)

x OV(p, q)

W Diagram above shows the vertices of a rectangle TUVW in a Cartesian plane. (a) Find the equation which relates p and q by using the gradient of UV.

[3maks] 5 (b) Shows that the area of the TUV can be expressed as p q + 10. [2marks] 2 (c) Hence, calculate the coordinates of V given the area of the rectangle TUVW is 5 unit2. [3marks] (d) Find the equation of the straight line TW in the intercept form. [2marks] 10

Diagram above shows a sector MJKL of a circle centre M and two sectors, PJM and QML, with centre P and Q respectively. Given the angle of the major sector JML is 3.6 radian. Find, (a) the radius of the sector MJKL, [2 marks] (b) perimeter of the shaded region, [2 marks] (c) the area of sector PJM, [2 marks] (d) the area of the shaded region. [4 marks] [ Answer 11. (a) 4.795 (b) 27.24 (c) 25 ] 2

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SULIT

11 (a) In a centre of chicken eggs incubation, 30% of the eggs hatched are male chickens. If 10 newly born chickens are chosen at random, find the probability (correct to four decimal places) that (i) 4 eggs hatched are male chicken, (ii) at least 9 eggs hatched are female chickens. [4 marks] (b) The mass of guava fruits produced in a farm shows a normal distribution with mean 420 g and standard deviation 12 g. Guava fruits with mass between 406 g and 441 g are sold in market, whereas those with mass 406 g or less are sent to the factory to be processed as drinks. Calculate, (i) the probability (correct to four decimal places) that a guava fruit chosen randomly from the farm will be sold in the market, (ii) the number of guava fruits that has been sent to the factory and also not sold in the market, if the farm produced 2 500 guava fruits. [ Answer (a)(i) 0.2001 (ii) 0.1493

[6 marks]

(b)(i) 0.8383 (ii) 100 ]

Sections C Answer two questions from this section.

12 . A P 8mB Q

In the diagram above, P and Q are two fixed points on a straight line such that PQ = 8 m. At a certain instant, particle A passes the point P with a velocity VA = 2t 6, whereas particle B passes the point Q with a velocity VB = 5 t where t is time in second after the particle A and the particle B have passed the point P and the point Q. [Assume direction P to Q is the positive.] (a) Find the distance between the particle A and particle B at the instant when particle A stopped momentarily. (d) [3marks ] Find the time, t1, when the distance between the particle A and particle B is maximum before the two particles meet.. [ 2 marks ]

(c) For how long the two particles A and B are moving in the same direction? (d)(i) Find the time, t2, when the particles A and B meets. (ii) Hence, find the distance from the point P when the two particles meet..

[3 marks ]

1 [Answer (a) 27 m 2

(b)

11 s 3

(c) 2 s (d)(i) 8 s (ii) 16 m ]

13 A small factory produces a certain goods of A model and B model. In a day, the factory produces x units of A model and y units of B model where x 0 and y 0. Time taken to produce one unit A model and one unit B model is 5 minutes and minutes respectively. The production of these goods in a certain day iswww.cikgurohaiza.com

restricted by the following conditions: I. The number of units of A model is not more than 60, II. The number of units of B model is more than two times the number of units of A model by 10 units or less. III. The total time for the production of A model and B model is not more than 400 minutes. Write an inequality for each of the above condition.. Hence draw the graphs for the three inequalities. Marks and shades the region R which satisfy the above conditions. Use your graph to answer the following questions: (a) Find the range of the number of units of A model which can be produced if 40 units of B model are produced. (b) Find the total maximum profits which can be obtained if the profit gained from one unit of A model and one unit of B model is RM 6 and RM 3 respectively. (c) If the factory intends to produce the same number of units of A model and B model, find the maximum number of units which can be produced for each o A model and B model. Answer x 60, y 2x 10, 5x + 4y 400 (a) 15 x 48 (b) RM435 14 . Diagram 6 shows a quadrilateral ABCD such that ABC is acute. D9.8 cm

(c) 44

5.2 cm C12.3 cm

A

40.5

9.5 cm B

DIAGRAM 6 (a) Calculate, (i) ABC, (ii) ADC, (iii) area, in cm2, of quadrilateral ABCD. [8 marks] (b) A triagle ABC has the same measurements as those given for triangle ABC, that is, AC = 12.3 cm, CB = 9.5 cm and BAC = 40.5, but which is different in shape to triangle ABC. (i) Sketch the triangle ABC. (ii) State the size of ABC. [2 marks] Answer . (a)(i) 57.21 - 57.25 (ii) 106.07 - 106.08 (iii) 82.37 - 82.39 (b)(i) C (ii) 122.75 - 122.79 A B

10. Table 2 shows the price indices and percentage usage of four items, P, Q, R, and S, which are the main ingredients of a type biscuits.3472/2www.cikgurohaiza.com

SULIT

Item

Price index for the year 1995 based on the year 1993

Percentage of usage (%)

P Q R S

135 x 105 130

40 30 10 20

(a) Calculate, (i) the price of S in the year 1993 if its price in the year 1995 is RM37.70 (ii) the price index of P in the year 1995 based on the year 1991 if its price index in the year 1993 based in the year 1991 is 120. [5 marks] (b) The composite index number of the cost of biscuits production for the year 1995 based on the year 1993 is 128. Calculate, (i) the value of x, (ii) the price of a box of biscuit in the year1993 if the corresponding price in the year 1995 is RM 32. [5 marks] [ Answer (a)(i) RM 29 (ii) 162 (b)(i) 125 (ii) RM 25 ]

Section C Alternative

Answer two questions from this section. 12. Diagram 6 shows STQ such that ST = 12.1 cm and TQ = 9.5 cm.

T

SDiagram 6

Q

The area of the triangle is 45 cm2 and STQ is obtuse. (a) Find (i) STQ [ STQ = 128.47 or 12828' ] (ii) the length, in cm, of SQ [19.49 cm] (iii) the shortest distance, in cm, from T to SQ. [4.613] [ 5 marks]

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T

13 cm

5 cm

R P

Q 5 cm Diagram 7

(b) Diagram 7 shows a pyramid TPQR with a horizontal triangular base PQR. T is vertically above Q. Given that PQ = QT = 5 cm, TR = 13 cm and PRQ = 15 .Calculate two possible values of PQR [PQR = 126.60o and 23.40o] (c) Using the acute PQR in (i), calculate ( i) the length of PR [7.673] (ii) the value of PTR [29.420] (iii) the surface area of the plane TPR [22.58] [ 5marks]

13. shows the bar chart for the monthly sales of five essential items sold at a sundry shop. Table 3 shows their price in the year 2000 and 2006, and the corresponding price index for the year 2006 taking 2000 as the base year.Cooking Oil Rice Salt Sugar Flour 10 20 30 40 50 60 70 80 90 100 Diagram 2 units

Items Cooking Oil Rice Salt Sugar Flour

Price in the Price in the year 2000 year 2006 x RM1.60 RM0.40 RM0.80 RM2.00 RM2.50 RM2.00 RM0.55 RM1.20 z TABLE 4

Price Index for the year 2006 based on the year 2000 125 125 y 150 120

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SULIT

(a)

Find the values of (i) x, (ii) y (iii) z.

[x=2.00,y=137.5,z=2.40]

[3 marks]

(b) (c)

Find the composite price index for cooking oil, rice, salt, sugar and flour in the year 2006 based on the year 2000. [131.17] [2 marks] Calculate the total monthly sales for those essential items in the year 2006, given that the total monthly sales in the year 2000 was RM 150.[3 marks] [120] monthly sales of those

(d) the composite index for the year 2008 based on the year 2000 if the essential items increased by 20% from the year 2006 to the year 2008. [157.40] [3 marks] 14. Use the graph paper provided to answer this question. The Mathematics Society of a school is selling x souvenirs of type A and y souvenirs of type B in a charity project based on the following constraints : I II III : : : The total number of souvenirs sold must not exceed 75.

The number of souvenirs of type A sold must not exceed twice the number of souvenirs of type B sold. The profit gained from the selling of a souvenir of type A is RM9 while the profit gained from the selling of a souvenir of type B is RM2. The total profit must not be less than RM200. [

(a)

Write down three inequilities other than x 0 dan y 0 which satisfy the above constraints. Answer x + y 75 , x 2y and 9x + 2y 200] [3 marks]

(b)

Hence, by using a scale of 2 cm to 10 souvenirs on both axes, construct and shade the region R which satisfies all the above constraints. [ 3 marks] By using your graph from (b), find (i) the range of number of souvenirs of type A sold if 30 souvenirs of type B are sold. [ 16 number of A type souvenirs sold 45] (ii) the maksimum which may be gained. [Answer RM 500] [4 marks]

(c)

15.

An object, P, moves along a straight line which passes through a fixed point O.www.cikgurohaiza.com

Figure 8 shows the object passes the point O in its motion. t seconds after leaving the point O , the velocity of P, v m s1 is given by v = 3t2 18t + 24. The object P stops momentarily for the first time at the point B. P O Figure 8 (Assume right-is-positive) Find: (a) the velocity of P when its acceleration is 12 ms 2 , [9 ms 1 ] [3 marks] (b) the distance OB in meters, [20 m] [4 marks] (c) the total distance travelled during the first 5 seconds. [28 m] [ 3 marks] B

12. (a)

(i)

Use area formula 1 (12.1)(9.5) sin STQ = 45 2 STQ = 128.47 or 12828' Using cosine Rule SQ 2 = 12.12 + 9.5 2 2(12.1)(9.5) cos STQ SQ = 19.49cm TQS = 29.05 h sin 29.05 = or equivalent 9.5 = 4.613 cm

(ii)

(iii)

(b) 5 12 = sin 15 sin p 12 sin p = sin 15 = 0.6212 5 QPR = 3824' ,14136' @ 38.40, 141.60 PQR = 180o 15o 38.40o @ PQR = 180o 15o 141.60o PQR = 126.60o and 23.40o (c) (i) PR 5 = sin 23.4 sin 15 5 PR = sin 23.4 sin 15 = 7.672 cm PR = 7.673 cmwww.cikgurohaiza.com

3472/2

SULIT

(ii)

Use Cosine Rule13 2 + ( 50 ) 2 (7.672) 2 2(13)( 50 ) = 0.8710

cos PTR =

PTR = 29.420(iii) Area PVR =13.

1 (13)( 50 ) sin 29.42 2

= 22.58 cm2

(i) x = 2.00 (ii) y = 137.5 (iii) z = 2.40 (b) Use composite index formula 125(80) + 125(100) + 137.5(50) + 150(60) + 120(30) I= 80 + 100 + 50 + 60 + 30 = 131.17 P2006 100 = 131.17 (c) 150 P2006 = RM 196.762008 I 2006 = 120 2008 (d) I 2000 =

(a)

120 131.17 100 = 157.40

14. (a) (b) (c)

The three inequalities are x + y 75 , x 2y and 9x + 2y 200 refer by graph (i) 16 number of A type souvenirs sold 45 (ii)Maximum profit = RM [ 9(50) + 2(25) ] = RM500.

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y

100

90 9x + 2y = 200 80

70

60

50 x = 2y 40 y = 30 ( 50 , 25 ) 20 x + y = 75 10

R

30

O

x 10 16 20 30 40 45 50 60 70 80

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SULIT

For examiners use only

Answer all questions. 1. Function f is defined by 2 x,x 3 f ( x) = 11 3 x, x 3 2 2

Find the range corresponding to the domain 0 x 4

[3 marks]

1 Answer : ..

3

2.

x+2 mx + n and fg: x , 5 5 where m and n are constants , find the value of m and of n,Given the function f: x 2x + 5 , g : x

[2 marks ]

2

Answer : m =. n =..............................2

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[ Lihat sebelah SULIT

For examiners use only

Diagram 1 shows part of the mapping of x to z by the function 12 f : x ax + b followed by the function g : y , y c . Calculate the values of a, b, c yc and d.

3.

12 4 6 1 d Diagram 1 [ 4 marks] 3

Answer: a=b=c=d=..

4.

If the x-axis is a tangent to the curve x 2 + 3 px = p 3 , find the values of p. [3 marks ]

4

3 Answer : p =.........

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[ Lihat sebelah SULIT

5.

Given and are the roots of 2 x 2 4 x + 1 = 0 . Form the quadratic equation with roots 2 and 2 . [ 4 marks ]

For examiners use only

5

4 Answer : .................................___________________________________________________________________________ 6.

Given the quadratic function of f(x) = 6x 1 3x2. a) Express the quadratic function f(x) in the form k + m(x + n)2, where k, m and n are constants. b) write the equation of the axis of symmetry

[ 3 marks ]

6

3 Answer : (a) ......................... (b) ........................

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For examiners use only

7.

Find the range of values of x if

f ( x ) = 3 x 2 + 2 x 5 always positive. [3 marks]

7

3

Answer : .................................. Simplify and state your answer in the simplest form 5 3n +1 + 5 3n 2 125 n 1 . [2 marks]

8.

8

Answer : ................................... 39.

Solve the equation 9

y +1

= 24 + 9 y .[3 marks]

9

3

Answer : ....................................

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10.

Given 2 + log 3 k = log 9 (m + 3) , express k in terms of m. [4 marks]

For examiners use only

10

Answer : .........11. Solve the equation log 3 x = log 9 (2 x + 3)

4

[3 marks]

11

Answer: ............

3

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For examiners use only

12.

Given that the n th term , Tn = 20 4n for an arithmetic progression. Find the sum of the first 12 terms of the progression. [3 marks]

12

4 Answer: ............. ___________________________________________________________________________13. Given the sum of the first 3 terms of a geometric progression is 567 and the sum of the next three terms of the progression is 168. Find the sum to infinity of the progression. [4 marks]

13

Answer : 4

.

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14.

Given that the sum of the first three terms of a geometric progression is 13 times the third term of the progression. If the common ratio, r > 0, find the common ratio. [ 2 marks ]

For examiners use only

14

Answer : .

2

15.

Diagram 2 shows the graph of log2 y against log2 x. Values of x and values x2n , where n and k are constants. of y are related by the equation y = k Find the value of n and the value of k. log2 y

*(5, 6)

0

(2, 0)

log2 x

Diagram 2[4 marks]

Answer : n= ..k=.

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For examiners use only

16.

Diagram 3 shows a semicircle KLMN, of diameter KLM , with centre L. y

K

L

N (x,y)

0Diagram 3

M

x

x y + = 1 and point N( x , y ) lies 4 3 on the circumference of a circle KLMN , find the equation of the locus of the moving point N. [ 3 marks ] Given that the equation of the straight line KLM is

16

3 Answer: 17. ......

If a = 2i + ( p + 1) j and b = 3i + 6 j , find the value of p if a + b is parallel to the x-axis. [3 marks]

17

4

Answer:

......

___________________________________________________________________________

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18. Given that sin 20 0 = a and cos 30 0 = b , express sin 50 0 in terms of a and

For examiners use only

[3 marks]

18

Answer: ...

3

___________________________________________________________________________19.Diagram below shows two sectors , OAB and OCD with centre O.

E

D

C

A

B

OGiven that COD = 0.92 rad, BC = 5 cm and perimeter of sector OAB is 20.44 cm. Using = 3.142 , find the area of the shaded region of ABCED. [ 3 marks ]

19

3 Answer:Perfect Score 2009www.cikgurohaiza.com

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SULIT For examiners use only 20. Given that y=1 1

3472/1

2x 1 dy and = 2 g(x) where g(x) is a function in x . 2 x dx

Find g ( x)dx .

[3 marks]

20

3

Answer: ............ ___________________________________________________________________________21.

The gradient of the curve y = hx +and the value of k.

k at the point x2

7 1, is 2. Find the value of h 2 [3 marks]

21

Answer: ..3

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SULIT

SULIT

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For examiners use only

22.

A coach wish to choose a player from two bowlers to represent the nation in a tournament. The following data show the number of pins scored by the two players in six sucessive bowls: Player A: 8, 9, 8, 9, 8, 6 Player B: 7, 8, 8, 9, 7, 9 By using the values of mean and standard deviation, determine the player which qualified to be choosen because the score is consistent. [3 marks]

22

Answer: ............___________________________________________________________________________23. In a debate competition, the probability of team A, team B and team C will qualify for 1 1 1 the final are , , respectively. Find the probability that at least 2 teams will qualify 3 4 5 for the final. [3 marks]

3

23

Answer: 3

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SULIT

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3472/1

24. The letters of the word G R O U P S are arranged in a row. Find the probability that an arrangement chosen at random (a) begins with the letter P, (b) begins with the letter P and ends with a vowels. [3 marks]

24

Answer: ( a)............3 ( b )...................................

25. The life span of certain computer chip has a normal distribution with a mean of 1500 days and a standard deviation of 40 days. a) Calculate the probability that a computer chip chosen at random has a life span of more than 1540 days b) Given that 6% of the computer chips have a life span of more than n days, find the value of n. [4 marks]

25

Answer : (a)..........4 (b)..........................................

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Paper 2 Time: Two hours and thirty minutes Instruction : This question paper consists of three sections: Section A, Section B and Section C. Answer all questions in Section A, four questions from Section B and two questions from Section C. Give only one answer/ solution for each question. All the working steps must be written clearly. Scientific calculator that are non-programmable are allowed. Section A [40 marks] 1. Given that (-1, 2k) is a solution for the simultaneous equation x2 + py 29 = 4 = px xy where k and p are constants. Find the value of k and of p. [5 marks]

Jawapan: k = 4, p = 4; k = 2, p = 8 2. Given function f : x 4 3x. (a) Find, (i) f 2(x), (ii) (f 2)1(x). (b) Hence, or otherwise, find (f 1)2(x) and show (f 2)1(x) = (f 1)2(x).

[3 marks] [3 marks]

(c) Sketch the graph of f 2(x) for the domain 0 x 2 and find its corresponding range. [2 marks]

Jawapan: (a)(i) 9x 8 (ii)

x+8 9

(c) y10 -------------------------8

0 y 10

0

8/9

x

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3. Diagram 3 shows five semicircles.

DIAGRAM 3 The area of the semicircles form a geometric progression. Given that area of the 1 smallest semicircle is of the area of the largest semicircle. If the total area of the 16 semicircles is 30 cm 2 , find (a) area of the smallest semicircle (b) area of the largest semicircle [5 marks] Jawapan: (a) 10 (b) 160

4. Given that tan( x y ) = 1 and tan y =

3 1 , show that tan x = . 4 7 0 0 Sketch the graph of y = tan x for 0 x 360 .

Hence, using the same axes , draw a suitable straight line and find the number of solutions for the equation 3 tan x + x = 6 [6 marks]

Jawapan: Number of solutions = 3

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5. Diagram 5 shows a parallelogram OABC. O

A

P D C DIAGRAM 5 Given that APD, OPC and DCB are straight lines. Given that OA = 6a, OC = 12c and OP : PC = 3 : 1. (i) (ii) Express AP in terms of a and/or c. Given the area of the ADB = 32 unit 2 and the perpendicular distance from A to DB is 4 units, find a. [5 marks]

B

Jawapan: (a) 6a + 9c % % (b)2

6.Cumulative frequency (20.5, 74) x x(15.5, 58) (10.5, 26) x x (5.5, 6) x (25.5, 80)

O

0.5 DIAGRAM 6

Length of fish in cm

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Diagram 5 shows an ogive for the distribution of 80 fishes in a tank when the cumulative frequency is plotted against upper boundaries for a certain classes. O is the origin. (a) Construct a frequency table with a uniform class interval from the information given in the ogive. [2 marks] (b) Draw a histogram and determine the mode. [3 marks] (c) From the frequency table, find (i) the variance, (ii) the median for the length of fish in the tank. [4 marks]

7. Use the graph paper provided to answer this question. An experiment which involves samples of red blood cell used to trace the percentage, P, of the red blood cell which experience creanation when it is added by drops of sodium chloride solution with different concentration, K mol dm3. Table below shows the results of the above experiment. Sodium chloride concentration (K) Percentage of red blood cells which experience creanation (P) 0.50 0.4 0.75 5.0 TABLE 7 Variables P and K are related by the equation P =constants. (a) Draw the graph ofP against K.

1.00 14.5

1.25 27.6

1.50 46.2

1.75 68.9

4

2

(K + A)2 where and A are

[5 marks] (b) From your graph, find the value of and the value of A. [4 marks] (c) Find the value of P when K = 1.4? [1 mark]

Jawapan: (b) = 0.33, A = 0.40 (c)37.21 38.44

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8.

y y = x(x 1)(x + 3)

O

x

(a) Diagram above shows the curve y = x(x 1)(x + 3). Calculate the area bounded by the curve, x-axis, line x = 2 and line x = 1.

[6 marks]

(b) Diagram below shows the shaded region bounded by the curve y = 2 x + 1 , line x = 1 and line x = k. When the region is revolved 360 at the x-axis, the volume generated is 18 unit3. Find the value of k. [4 marks] [ answer (a) 9. y P(0, ) Jawapan: (a) 2 + 2 = 81 2 (b)(ii ) ,5.85 3 (c)0.35 47 12 (b) k = 4 ]

Wall

R(x, y)

O Floor

Q(, 0)

x

Diagram 9 shows the x-axis and the y-axis which represent the floor and the wall. The end of a piece of wood PQ with length 9 m touches the wall and the floor at the point P(0, ) and point Q(, 0). (a) Write the equation which relates and . [1 mark] (b) Given R is a point on the piece of wood such that PR : RQ = 1 : 2. (i) Show that the locus of the point R when the ends of the wood is slipping along the x-axis and the y-axis is 4x2 + y2 = 38. (ii) Find the coordinates of R when = 2. (iii) Find the value of tan ORQ when = 2. [9 marks]

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10.L J P Q rad O R

K

T

DIAGRAM 10Diagram 10 shows a circle PQRT, centre O and radius 5 cm. JQK is a tangent to the circle at Q. The straight lines, JO and KO, intersect the circle at P and R respectively. OPQR is a rhombus. JLK is an arc of a circle, centre O. Calculate [ 2 marks] (a) the angle , in terms of (b) the length, in cm, of the arc JKL [ 4 marks] (c) the area, in cm2, of the shaded region. [4 marks]

Jawapan: 2 (a) 3 20 (b) 3 (c)61.50

11. (a) A study on post graduate students, revealed that 70% out of them obtained jobs six months after graduating. (i) If 15 post graduates were chosen at random, find the probability of not more than 2 students not getting jobs after six months. (ii) It is expected that 280 students will succeed in obtaining jobs after six months. Find the total number of students involved in the study. [5 marks] (b) The mass of 5000 students in a college is normally distributed with a mean of 58kg and variance of 25 kg2. Find (i) the number of students with the mass of more than 90 kg. (ii) the value of w if 10% of the students in the colleges are less than w kg. [5 marks] Jawapan:(a) (i) 0.1268 (ii)400 (b) (i) 82or 83 (ii) 38.77 or 38.79

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12. Diagram 12 shows the position and direction of motion for two objects, P and Q, which move along a straight line and passes through two fixed points, A and B respectively. At the instant when P passes through the fixed point A, Q passes through the fixed point B. Distance AB is 28 m. P A 28 m DIAGRAM 12 The velocity of P, vp ms , is given by vp = 6 + 4t 2t 2, where t is the time in seconds, after passing through A, whereas Q moves with a constant velocity of 2 ms1. Object P stops instantaneously at the point C. (Assume towards the right is positive.) Find, [3 marks] (a) the maximum velocity, in ms1, for P, (b) the distance, in m, C from A, [4 marks] (c) the distance, in m, between P and Q at the instant when P is at the point C. [3 marks]1

Q C B

Jawapan: (a) 8 m/s (b) 18 (c) 4

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13. . Diagram 13 shows a quadrilateral ABCD such that ABC is acute. 9.8 cm40.50 D

5.2 cmC

A

12.3 cm

9.5 cm

B

DIAGRAM 13 (a) Calculate, (i) ABC, (ii) ADC, [8 marks] (iii) area, in cm2, of quadrilateral ABCD. (b) A triangle ABC has the same measurements as those given for triangle ABC, that is, AC = 12.3 cm, CB = 9.5 cm and BAC = 40.5, but which is different in shape to triangle ABC. (i) Sketch the triangle ABC. [2 marks] (ii) State the size of ABC.

Jawapan: (a) (i) 57.21-57.25 (ii) 106.07-106.08 (iii)82.37-82.39 (b) (ii) 122.75-122.79

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14. Use the graph paper provided to answer this question.Cloth T-shirt Slack Preparation time (minutes) Sewing time (minutes)

45 30

50 70

A tailor shop received payment only for sewing T-shirt and slack. Preparation time and sewing time for each T-shirt and slack are shown in the table above. The maximum preparation time used is10 hours and the sewing time must be at least 5 hours 50 minutes. The ratio of the number of T-shirt to slack is not more than 4 : 5. In a certain time, the shop is able to complete x pieces of T-shirt and y pieces of slack. (a) Write three inequalities, other than x 0 and y 0, which satisfy the above conditions. [3 marks] (b) By using a scale of 2 cm to I unit on the x-axis and 2 cm to 2 units on the y-axis, draw the graphs for the three inequalities. Hence, shades the region R which satisfies the above conditions. [3 marks] (c) Based on your graph, find (i) the minimum number of slacks which can be sewn in that time if 3 pieces of of T-shirt has been sewn.. (ii) maximum total profit received in that time if the profit gained from each piece of T-shirt and slack are RM16 and RM 10 respectively. [4 marks]

Jawapan:(a)45 x + 3 y 600 50 x + 70 y 350 5x 4 y (c)4 (6,11), RM 206

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15. (a) Index number, Ii Weightage, Wi 105 5x 94 x 120 4 [4 marks]

The composite index number for the data in the table above is 108. Find the value of x.

(b) (i) In the year 1995, price and price index for one kilogram of certain grade of rice is RM2.40 and 160 respectively. Based on the year 1990, calculate the price per kilogram of rice in the year 1990. [2 marks] Item Timber Cement Iron Steel Price index in the year 1994 180 116 140 124 Change of price index from the year 1994 to the year 1996 Increased 10 % Decreased 5 % No change No change Weightage 5 4 2 1

(ii) Table above shows the price index in the year 1994 based on the year 1992, the change in price index from the year 1994 to the year 1996 and the weightage respectively. Calculate the composite price index in the year 1996. . [4 marks] Jawapan: (a) x=3 (b) (i) 1.50 (ii) RM152.90

End of question paper

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-1-

FUNCTIONS 1. Given that f : x 4 x + m and f1

: x nx +

3 , find the values of m and n. 4Answer:- m = 3 ; n =

1 4 2. Given that f : x 2 x 1 , g : x 4 x and fg : x ax + b , find the values of a and b . Answer:- a = 8 ; b = 12 2 3. Given that f : x x + 3 , g : x a + bx and gf : x 6 x + 36 x + 56 , find the values of a and b . Answer:- a = 2 ; b = 6

1 4. Given that g : x m + 3 x and g : x 2kx

4 , find the values of m and k. 3Answer:- k =

1 ;m=4 6

5. Given the inverse function f

1

( x) =

2x 3 , find 211 1 (b) 2 2

(a) the value of f(4), (b) the value of k if f 1 (2k) = k 3 .Answer:-(a)

6. Given the function f : x 2 x 1 and g : x (a) f 1 (x) , (b) f 1 g(x) , (c) h(x) such that hg(x) = 6x 3 . Answer:-(a)

x 2 , find 3

x +1 2

(b)

1 1 x 6 2

(c) 18x + 33

7. Diagram 1 shows the function g : x

p + 3x , x 2 , where p is a constant. x2

x 7

g

p + 3x x25

Diagram 1 Find the value of p.

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-2Answer:- p = 4 8. x 4 2 y 4 2 z 4 2

0

0

01

2 2 Diagram 2 Diagram 2 shows the mapping of y to x by the function g : y ay + b and mapping 6 b to z by the function h : y , y . Find the, 2 2y b (a) value of a and value of b, (b) the function which maps x to y, (c) the function which maps x to z. 10 y 18 Answer:- (a)a= 6, b=10 (b) (c) 6 y 20 9. In the Diagram 3, function h mapped x to y and function g mapped y to z.

2

x

h

y

g

z8

5 2 Diagram 3 Determine the values of, (a) h1(5), (b) gh(2)

Answer:- (a)2 (b)8 10. Given function f : x 2 x and function g : x kx + n. If composite function gf is given as gf : 3x2 12x + 8, find (a) the value of k and value of n, (b) the value of g2(0). Answer:-(a) k = 3 ,n = 4 (b)442

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-311. The following information refers to the functions f and g.g (x) = 4 3 x fg (x) = 2 x + 5

Find f (x). 12. (a) Function f, g and h are given as f : x 2x 3 g:x ,x2 x2 h : x 6x2 2.

Answer:-

23 2 x 3

(i) Determine the function fh(x). At the same axis, sketch the graphs of y = g(x) and y = fh(x). Hence, determine the number of solutions for g(x) = fh(x). (ii) Find the value of g1(2). (b) Function m is defined as m : x 5 3x. If p is another function and mp is defined as mp : x 1 3x2, determine function p. Answer:-(a)(i)12x2 4 (b) p ( x ) = 2 + x 2

13. Given function f : x 4 3x. (a) Find (i) f 2(x), (ii) (f 2)1(x). (b) Hence, or otherwise, find (f 1)2(x) and show (f 2)1(x) = (f 1)2(x).

(c) Sketch the graph of f 2(x) for the domain 0 x 2 and find its corresponding x +8 Answer:-(a)9x 8 (b) range. 9 14. A function f is defined as f : x are constants. (a) Determine the value h. (b) Given value 2 is mapped to itself by the function f. Find the (i) value p, (ii) another value of x which is mapped to itself, (iii) value of f 1(1). 3 Answer:-(a) h = (b)(i)p =12(ii)x = 3 (iii)5 2p+x , for all values of x except x = h and p 3 + 2x

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-4QUADRATIC EQUATIONS 1. One of the roots of the quadratic equation Find the possible values of p. is twice the other root.Answer ; p = 5, 7

2. If one of the roots of the quadratic equation . find an expression that relates

is two times the other root,Answer : 2b 2 = 9ac

3. Find the possible value of m , if the quadratic equation roots.

has two equal

Answer ;

4. Straight line y = mx + 1 is tangent to the curve x2 + y2 2x + 4y = 0. Find the possible values of m.

Answer : and are roots of the equation k x(x 1) = 2m x. 2 2 If + = 6 and = 3, find the value of k and of m.

1 or 2 2

5. Given

Answer : k =

1 3 ,m= 2 16

6. Find the values of such that the equation (3 )x2 2( + 1)x + + 1 = 0 has equal roots. Hence, find the roots of the equation base on the values of obtained.

Answer : = 1; roots: = 1, x = 1; = 1, x = 0

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-5QUADRATIC FUNCTIONS 1. Diagram 1 shows the graph of the function y = 2 ( x p ) + 5 , where p is constant. y2

0 ( 0, 3 ) ( 4 , 3 )

x

Find, Diagram 1 (a) the value of p , the equation of the axis of symmetry, (b) (c) the coordinate of the maximum point.

Answer:- (a) p = 2 (b) x = 2 (c) ( 2, 5 )2.

y

0

x

( 0, 2 )

f ( x ) = ( p 1) x 2 + 2 x + q

Diagram 2 Diagram 2 shows the graph of the function f ( x ) = ( p 1) x 2 + 2 x + q . (a) State the value of q . (b) Find the range of values of p .

Answer:-(a) q = 2 (b) p