Modul Perfect Score Sbp 2009

BAHAGIAN SEKOLAH BERASRAMA PENUHDAN SEKOLAH KLUSTER

ADDITIONAL MATHEMATICS

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ABOUT THIS MODULE

I

2

WE LEARN

II

3

EXAMINATION FORMAT

II-III

4

ANALYSIS TABLE OF SPM ADD MATHS QUESTIONS

IV

5

LIST OF FORMULAE AND NORMAL TABLE

V-VII

6

ADDITIONAL MATHEMATICS NOTES

VIII-XVI

7

PROBLEM SOLVING STRATEGY

XVII

8

PARTITION

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


III

Key towards achieving 1A …

Read question carefullyFollow instructionsStart with your favourite questionShow your working clearlyChoose the correct formula to be used +(Gunakannya dengan betul !!!)Final answer must be in the simplest formThe end answer should be correct to 4 S.F.(or follow the instruction given in the question)

π ≅ 3.142

Kunci Mencapai kecemerlangan

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

Paper 1 : 3 - 7 minutes for each questionPaper 2 :

Sec. A : 8 - 10 minutes for each questionSec. B : 15 minutes for each questionSec. C : 15 minutes for each question


ANALYSIS TABLE OF SPM ADDITIONAL MATHEMATICS QUESTIONS 2004-2008 AMaths (3472)

SPM Paper 1 Paper 2

Chapter Section A Section B Section C 04 05 06 07 08 04 05 06 07 08 04 05 06 07 08 04 05 06 07 08

1 Functions 3 3 2 3 3 1 2 Quadratic

Equations 1 2 1 1 1

3 Quadratic Functions 2 1 1 2 2 1

4 Simultaneous Equations 1 1 1 1 1

5 Indices and Logarithms 2 3 3 2 2

6 Coordinate Geometry 2 1 1 2 2 1 1 1 1

7 Statistics 1 1 1 1 1 1 1 1 1 8 Circular

Measure 1 1 1 1 1 1 1 1 1 1

9 Differentiation 2 2 3 2 2 1/2 1/2 2/3 1/2 1/3 1/3 1/3 10 Solution of

Triangle 1 1 1 1 1

11 Index Number 1 1 1 1 1 1 Progressions 4 3 2 3 3 1 1 1 1 1

2 Linear Law 1 1 1 1 1 1 1 1 1 1 3 Integration 1 1 2 1 1 ½ ½ 1/3 1/2 2/3 1 2/3 2/3 4 Vectors 2 2 2 2 2 1 1 1 1 1 5 Trigonometric

Functions 1 1 1 1 1 1 1 1 1 1

6 Permutations / Combinations 1 1 1 1 1

7 Probability 1 1 1 1 1 8 Probability

Distributions 1 1 1 1 1 1 1 1 1 1

9 Motion Along A Straight Line

1 1 1 1 1

10 Linear Programming 1 1 1 1 1

Total 25 25 25 25 25 6 6 6 6 6 5 5 5 5 5 4 4 4 4 4

IV


SULIT 3472/2

[ Lihat sebelah 3472/2 SULIT

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

ALGEBRA

1 x =a

acbb2

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 nm

= log am - loga n

7 log a mn = n log a m

8 logab = ab

c

c

loglog

9 Tn = a + (n-1)d

10 Sn = ])1(2[2

dnan−+

11 Tn = ar n-1

12 Sn = rra

rra nn

−−

=−−

1)1(

1)1(

, (r 1) ≠

13 r

aS−

=∞ 1 , r <1

CALCULUS

1 y = uv , dxduv

dxdvu

dxdy

+=

2 vuy = , 2

du dvv udy dx dxdx v

−= ,

3 dxdu

dudy

dxdy

×=

V

4 Area under a curve

= dx or ∫b

a

y

= dy ∫b

a

x

5 Volume generated

= dx or ∫b

a

y 2π

= dy ∫b

a

x 2π

5 A point dividing a segment of a line

( x,y) = ,21⎜⎝⎛

++

nmmxnx

⎟⎠⎞

++

nmmyny 21

6. Area of triangle =

)()(21

312312133221 1yxyxyxyxyxyx ++−++

1 Distance = 221

221 )()( yyxx −+−

2 Midpoint

(x , y) = ⎜⎝⎛ +

221 xx

, ⎟⎠⎞+

221 yy

3 22 yxr +=

4 2 2

xi yjrx y

∧ +=

+

GEOM ETRY


[ Lihat sebelah 3472/2 SULIT

STATISTICS

TRIGONOMETRY

1 Arc length, s = rθ

2 Area of sector , A = 212

r θ

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 = A

A2tan1

tan2−

9 sin (A± B) = sinAcosB cosAsinB ±

10 cos (A

± B) = cos AcosB m sinAsinB

11 tan (A± B) = BABA

tantan1tantan

m

±

12 Cc

Bb

Aa

sinsinsin==

13 a2 = b2 +c2 - 2bc cosA

14 Area of triangle = Cabsin21

1 x = N

x∑

2 x = ∑∑

ffx

3 σ = N

xx∑ − 2)( =

2_2

xN

x−∑

4 σ = ∑

∑ −

fxxf 2)(

= 22

xf

fx−

∑∑

5 M = Cf

FNL

m⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡ −+ 2

1

6 1000

1 ×=PPI

7 1

11

wIwI

∑∑

=

8 )!(

!rn

nPrn

−=

9 !)!(

!rrn

nCrn

−=

10 P(A∪B)=P(A)+P(B)-P(A∩B)

11 p (X=r) = rnrr

n qpC − , p + q = 1

12 Mean , μ = np 13 npq=σ

14 z = σμ−x

VI


3472/2 SULIT

[ Lihat sebelah

VII


TO EXCEL in

You need to…

• set a TARGET

• familiar with FORMAT of EXAM PAPERS • analyse the EXAM QUESTIONS

• master the TECHNIQUES OF ANSWERING QUESTIONS

• do EXERCISES

VIII


ADDITIONAL MATHEMATICS NOTES

(c) Absolute Value Function 1 FUNCTIONS (a)

i. Domain = {a,b,c} ii Codomain = {1,2,3,4}

iii. Range = {1,2,3} iv. Objects of 1 are a and b v. Images of b are 1,2 and 3.

(b) Types of Relations i. One-to-one

ii. Many-to-one

iii. One-to-many

iv. Many-to-many

The corresponding range of values of f(x) is 0 ≤ f(x) ≤ 5 The corresponding range of values of f(x) means the range from the smallest value of y to the largest value of y, based on the given domain. (d) Composite Functions fg(x) = f[g(x)] In general, gf (x) ≠ fg(x) f 2 = ff, f 3 = fff or ff 2

(e) Determining one of the functions in a given composite function i. Given f and fg , find g. - Substitute g into f(x)

ii. Given f and gf , find g. - Let y= f(x) (f) To find the Inverse Function : - Let y = f(x), then x = f -1(y).

y

a 1 b 2

- 4 - 3 -2 - 1 0 1 2

5

4

3

2 1

a 1 b 2

a 1 b 2

3 c

c

a 1 b 2

3 c

3 c

4

x

f(x) x g[f(x)] = gf(x)

f g

gf

IX

a 1 2 3 b


2. QUADRATIC EQUATIONS

(a) ax2 + bx + c = 0

x = a

acbb2

42 −±−

Sum of roots:

α + β = ab

−

Product of roots

αβ = ac

(b) Form quadratic equation from

2 given roots: x2 - (sum of two roots)x + product of two roots = 0

3. QUADRATIC FUNCTIONS

(a) Types of roots b2 - 4ac > 0 → 2 different (distinct) roots. b2- 4ac = 0 → 2 equal roots b2 - 4ac < 0 →no real roots. b2 - 4ac ≥ 0 → with real roots

y

x0

y

x0

y

x0b2 - 4ac > 0 b2 - 4ac = 0 b2 - 4ac < 0 (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

y

x0

a b

+

++

++

__

_

_

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

1.2.3. ( )

n m n m

n m n

n m nm

a a aa a aa a

+

−

× =

÷ =

=

m

Laws of Logarithm 1. logaxy = logax + logay

2. logayx

= logax – logay

3. loga xn = n logax 4. loga a = 1 5. loga 1 = 0

6. loga b = ab

c

c

loglog

7. loga b = ablog

1

a b

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5. COORDINATE GEOMETRY

(a) Distance between A(x1, y1) and B(x2, y2) AB = 2

122

12 )()( yyxx −+− (b) Midpoint of AB

M = ⎟⎠⎞

⎜⎝⎛ ++

2,

22121 yyxx

(c) P divides AB internally in the ratio m : n

m : n

A(x , y ) B(x , y )P1 1 2 2

P= ⎟⎠⎞

⎜⎝⎛

++

++

mnmyny

mnmxnx 2121 ,

(d) Gradient of AB

m = 12

12

xxyy

−−

m = y-interceptx-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)

12

12

1

1

xxyy

xxyy

−−

=−−

(f) Area of polygon

L = 1

1

3

3

2

2

1

1 .........21

yx

yx

yx

yx

(g) Parallel lines m1 = m2 (h) Perpendicular lines m1 × m2 = -1

6. STATISTICS Measure of Central Tendency

(a) Mean

nx

x ∑=

for ungrouped data

∑∑=

ffx

x

for ungrouped data with frequency.

∑∑=

ffx

x i

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.

Formula

M = L + Cf

Fn

m

×−2

1

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 From the Ogive

n

n__2

0 Median

Kekerapan longgokan

Sempadan atas

Cumulative Frequency

Median Upper Boundary

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

K

ekerapan

Sempadan kelaMod s0 Measure of Dispersion

(a) Interquartile Range Formulae :

Q1 = Cf

FnL

Q

×−

+1

141

1

Q3 = Cf

FnL

Q

×−

+3

343

3

Ogive:

0

Kekerapan longgokan

SeQQ1 3 mpadan atas

34

__n

14__ n

Interquartile Range

= Q3 – Q1

(b) Variance, Standard Deviation Variance = (Standard Deviation)2

For ungrouped data

nxx∑ −

=2)(

σ

= 2

2

xnx

−∑

For grouped data Frequency

∑∑ −

=f

xxf 2)(σ

= 2

2

xf

fx−

∑∑

Mode Class Boundary 7. INDEX NUMBERS

(a) Price Index

I = 0

1

PP × 100

where Po = price at the base time P1 = price at a specific time

(b) Composite Index

∑∑=

wIw

I Cumulative Frequency

where I = price index or index number w = weightage 8. CIRCULAR MEASURE

(a) Radian → Degree

θ r = θ × π

0180 Upper Boundary

(b) Degree → Radian

θ o = θ × 180π rad

(c) Arc length s = jθ

(d) Area of sector

L = 21 j2θ =

21 js

XII www.cikgurohaiza.com

(e) Area of segment

L = 21 j2(θ r – sin θo)

9. DIFFERENTIATION

(a) Differentiation using the First Principal

0limx

dy ydx x

δδ∂ →

=

(b) dxd (a) = 0, a = constant

(c) dxd (xn) = nxn-1

(d) dxd (axn) = anxn-1

(e) Product Rule

dxd (uv) = u

dxdv + v

dxdu

(f) Quotient Rule

2vuv

vu

dxd dx

dvdxdu −

=⎟⎠⎞

⎜⎝⎛

(g) Composite Function

dxd (ax+b)n =

dy dudu dx

×

= an(ax+b)n-1

(h) Turning point → dxdy = 0

Maximum point:

dxdy = 0 and 2

2

dxyd < 0

Minimum point:

dxdy = 0 and 2

2

dxyd > 0

(i) Rate of change

dtdx

dxdy

dtdy

×=

(j) Small change :

xdx

y δδ .≈dy

10. INTEGRATION

(a) ∫ ax ++

=+

cnaxdx

nn

1

1

(b) 1( )( )

( 1)

nn ax bax b dx c

a n

+++ = +

+∫

(c) ∫ +b

a

dxxgxf )()(

= ∫ ∫+b

a

b

a

dxxgdxxf )()(

(d) ∫ ∫∫ =+b

a

c

a

c

b

dxxfdxxfdxxf )()()(

(e) ∫ ∫ =b

a

b

a

dxxfadxxaf )()(

(f) ∫ ∫−=b

a

a

b

dxxfdxxf )()(

(h) Area under the curve y

x0 a b

A = ∫ b

a

dxy

y

x0

a

b

A = ∫ b

a

dyx

(i) Volume of revolution y

x0 a b2

b

aV y dxπ= ∫

XIII www.cikgurohaiza.com

y

x0a

b

2b

aV x dπ= ∫ y

11. PROGRESSIONS

Arithmetic Progressions (a) Tn = a + (n - 1)d

(b) Sn = 2n {2a + (n - 1)d}

= 2n (a + l)

(c) d = T2 - T1 Geometric Progressions (a) Tn = arn-1

(b) Sn = rra n

−−

1)1( for r < 1

Sn = 1

)1(−

−rra n

for r > 1

(c) r

aS for -1 < r < 1 −

=∞ 1 and n ∞

(d) r = 1

2

TT

General

(a) S1 = T1 = a (b) Tn = Sn – Sn-1 (c) Sum of terms from Ta to Tb

= Sb – Sa-1 12. MOTION ALONG A STRAIGHT LINE

dtds

dtdv

s → v → a ← ← ∫ dtv ∫ dta

(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

x0

P(x, y)

y

x

r

θ

sin θ = ry

cos θ = xr

tan θ = xy

(b) tan θ = sincos

θθ

sec θ = 1

cosθ

cosec θ = θsin

1

cot θ = 1

tan sincosθ

θ θ=

(c)

Semua +ve

Sin+ve

Tan+ve

Kos+ve

All

Cos

XIV www.cikgurohaiza.com

(d) Special Angles

θ 0o 30o 45o 60o Sin θ 0

21

21

23

Cos θ 1

23 2

1 21

Tan θ 0 3

1 1 3

θ 90o 180o 270o 360o

Sin θ 1 0 -1 0 Cos θ 0 -1 0 1 Tan θ ∞ 0 ∞ 0

(e) Trigonometric Graphs y = a sin bx

a

-a

0 90bbbb

________ 180 270 360 x

y

y = a cos bx

a

-a

0 90bbbb

________ 180 270 360 x

y

y = a tan bx

0 90bbbb

________ 180 270 360 x

y

(f) sin2θ + cos2θ = 1 1 + tan2θ = sec2θ 1 + cot2θ = cosec2θ

(g) sin(A ± B) = sinA cos B cos Asin B ±

± B) cos(A = cosA cosB sinA sinB m

± B) tan (A

= BA

BAtantan1

tantanm

±

(h) sin2A = 2 sinA cosA cos2A = cos2A – sin2A = 2 cos2A – 1 = 1 – 2 sin2A

tan 2A = A

A2tan1

tan2−

14. VECTORS

(a) Addition of Vectors 1. Triangle Law

a

a

b

b+

2. Parallelogram Law

a

abb+

3. Polygon Law AE AB BC CD DE= + + +uuur uuuv uuuv uuuv uuuv

B

C

D E

A

XV www.cikgurohaiza.com

(b) Subtraction of vectors

a

ab b-

(c) Vectors in the Cartesan

Coordinates

x

y

i

jr

0 x

yP(x, y)

r = xi + yj r = 22 yx +

r̂ = 2 2

xi y jrr x y

=+

+

15. SOLUTIONS OF TRIANGLE

(a) Sine Rule

sin sin sin

a b cA B C

= =

(b) Cosine Rule

a2 = b2 + c2 – 2bc cos A

cos A = bc

acb2

222 −+

(c) Area of Triangle

L = Cabsin21

16. PROBABILITY DISTRIBUTIONS

(a) Permutation n

n P = n!

)!(!rn

nPrn

−=

(b) Combination n

nC = 1

!!)(!

rrnnCr

n

−=

rnn

rn CC −=

(c) Binomial Distribution P(X = r) = r

nC pr qn-r (d) Mean = μ = np Standard deviation npq=σ

tribution (e) Converting Normal Dis to Standard Normal

σμ−X Distribution Z =

(f) Probability Distribution Graph

0 a Z P(Z > a)

1. P(Z < a) = 1 – P(Z > a) use P

b

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

)5. P(-a < Z< b) = 1– P(Z >a) – P(Z > 6. P(-a < Z < -b) = P(b < Z < a) Examples: a) P(Z> 0.1)

b) P ( Z< 0.1)

c) P ( -1.2 < Z < 0.4)

xamples: d) P( Z > a ) = 0.3, find a

h) P(X > a) = 0.3, given μ = 45,

E e) P (Z > a) = 0.6, find a

f) P (Z < a) = 0.1, find a

g) P ( Z < a ) = 0.73, find a

σ =3

XVI


How to Solve a Problem

Understand the Problem

Plan your Strate

Do - Carry out Your Strate

Check your gy gy Answers

•Which Topic / •Subtopic ?

•Choose suitable strategy

•Carry out the calculations

•Is the answer reasonable?

1. PN ZABIDAH BINTI AWANG SM AGAMA PERSEKUTUAN, LABU.

2. EN AMIRULFAIZAN BIN AHMAD

SBP INTEGRASI SELANDAR, MELAKA.

3. PN ROHANI MD NOR SEKOLAH SULTAN ALAM SHAH, PUTRAJAYA

4. EN ZUZI BIN SHAFIE

SM AGAMA PERSEKUTUAN, KAJANG.

5. PN SARIPAH BINTI AHMAD SM SAINS MUZAFFAR SYAH, MELAKA.

XVII


Master these

questions……

XVIII


Answer all questions.

1.

3

2

1

3

For examiner’s

use only

f g

Diagram 1 3 -1 6

Set A Set C Set B 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 ]

Answer : (a) ……………………..

(b) ……………………...


3 Given the function f (x) = 4x, 0≠x and the composite function f g(x) = x16− . Find

(a) g(x), (b) the value of x when g(x) = 8.

Answer : .........…………………

For examiner’s

use only

3

4

3

[3 marks]

3

4 Solve the quadratic equation ( ) ( )( )3252 +−=− xxxx . Give your answer correct to

four significant figures. [ 3 marks ]

Answer : .........…………………


5 (a) Given x = 42− y , find the range of x if y > 10.

(b) Find the range of x if x2 − 2x ≤ 3. [4 marks]

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

___________________________________________________________________________

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

a) Write the equation of the axis of symmetry of the curve. b) Express in form of , where p and q are constants. )(xf qpx ++ 2)(

[ 3 marks ]

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

(b) ……........................

3

6

For examiner’s

use only

)(xfy =

0 1 7

y

y = -9

Diagram 1

x

3472/1 www.cikgurohaiza.com

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

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

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

[ 4 marks]

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

9. Solve the equation log x 16 − log x 2 = 3. [3 marks]

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

3

7

3

9

4

8

For examiner’s

use only


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 ]

Answer : ....……………...………..

11. The sum of the first n terms of an arithmetic progression is given by .133 2 nnSn += Find (a) the ninth term,

(b) the sum of the next 20 terms after the 9th terms.

2

10

4

For examiner’s

use only

[3 marks]

11 Answer: a)…...…………..…....... b) ....................................


For examiner’s

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.

Answer: a =...….… b =….......

12

4 p = ........................

___________________________________________________________________________

13. Diagram 2 shows a linear graph of xy against x2

xy

(4,1)

Given that

xy = hx2 + k, where k and h are contants.

Calculate the value of h and k. [3 marks]

Answer : h = ………………..…….

k = ……………….....…...

(1,-5)

x2

●

● DIAGRAM 2

13

3


14. The equation of a straight line PQ is 3x +

2y = 1. Find the equation of a straight line

that is parallel to PQ and passes through the point (−6 , 3). [3 marks]

14

3

Answer : .…………………

15 Given u = dan v = , find the possible values of p for each of the following ⎟⎟⎠

⎞⎜⎜⎝

⎛97

⎟⎟⎠

⎞⎜⎜⎝

⎛ −3

1p

case:

(a) u and v are parallel, [2 marks] (b) vu = . [2 marks]

15

4

For examiner’s

use only

Answer : a)…………………..

b) ………………………


16 P

R

S

Q

r

s

O

The diagram above shows = r, = s, and are drawn in the square grid. →

OR→

OS→

OP→

PQ Express in terms of r and s.

(i) →

OP

(ii) PQuuur

. [ 3 marks ]

Answer: a) OPuuur

= …….…………...

b) PQuuur

=...……………….. ___________________________________________________________________________ 17. Solve the equation 3 cos2 θ + sin 2θ = 0 for . [ 4 marks ] 00 3600 ≤≤ θ

Answer: …...…………..….......

4

17

16

3

For examiner’s

use only


3472/1

18.

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 θ in radian, [2 marks] (b) area of the sector OPQ. [2 marks]

Answer: a)…………………… b) …………………

___________________________________________________________________________

19. Find the equation of the tangent to the curve 3)5(5−

=x

y at the point (3, 4).

[2 marks]

Answer:………………………

2

19

4

18

For examiner’s

use only P

θ Q O


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 cms−1. (Use π = 3.142) (i) Calculate the rate of change of radius of the circle. [2 marks] (ii) Hence, calculate the radius of the circle after 4 seconds. [2 marks]

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

21. Given 4

0( )f x∫ dx = 5 and dx = 6.

3

1( )g x∫

Find the value

(a) 4 1

0 32 ( ) ( )f x dx g x dx+∫ ∫ , [1 marks]

(b) k if =14. [2 marks] 3

1[ ( ) ]g x k x dx−∫

Answer: a) ……………………..

k =.……………..………

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 = ……………………. .

3472/1

5

20

3

21

3

22

For examiner’s

use only


11 3472/1

3472/1

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]

Answer: …a)...…………..….......

b) ............................... ___________________________________________________________________________ 24. Hanif , Zaki and Fauzi will be taking a driving test. The probabilities that Hanif ,

Zaki and Fauzi will pass the test are 1 1,2 3

and 14

respectively. Calculate the

probability that

For examiner’s

use only

23

5

(a) only Hanif will pass the test (b) at least one of them will pass the test.

[ 3 marks ]

24

3

Answer: ……………………………


3472/1

For examiner’s

use only 25. Diagram below shows a standard normal distribution graph.

k z -k

25

2 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


13 3472/1

3472/1

JAWAPAN 1 (a) −1 (b) 6 (c) 6 13 h = 2 , k = −7

2 (a) f −1 = 3

4x− (b) 5

4−

14

3y = − 2x − 3

3 (a) g(x) = 0,4

≠− xx

(b) 21

−=x

15

(a)3

10 (b) −10, 12

4

3.562 , -0.5616

16

(b)(i) 3r + 2s (ii) − r − 3s

5.

(a) x < − 3 (b) −1 ≤ x ≤ 3

17

90°, 123° 41’, 270°, 303° 41’

6 a) 4=x

b) 9)4()( 2 −−= xxf

18 (a)

21 (b) 400

7 x = 3 19 15x + 16y −109 = 0

8 ( i) 0.209 (ii) 2.219 20 ( i) 0.01591 cms−1 (ii) 9.612

9 x = 4 21 (a) 4 (b) k = −2

10 a) 5 b) 4 22 14 553

11 (a) 64 ( b) 2540 23 (a ) 14 (b) 1.020

12 a = 0.06 , b = 0.006 , p = 6 24 (a) 9/35 ( b) 5/6

25 k = 1.234


4 THE 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 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9

Minus / Tolak

0.0

0.1

0.2

0.3

0.4

0.5000

0.4602

0.4207

0.3821

0.3446

0.4960

0.4562

0.4168

0.3783

0.3409

0.4920

0.4522

0.4129

0.3745

0.3372

0.4880

0.4483

0.4090

0.3707

0.3336

0.4840

0.4443

0.4052

0.3669

0.3300

0.4801

0.4404

0.4013

0.3632

0.3264

0.4761

0.4364

0.3974

0.3594

0.3228

0.4721

0.4325

0.3936

0.3557

0.3192

0.4681

0.4286

0.3897

0.3520

0.3156

0.4641

0.4247

0.3859

0.3483

0.3121

4

4

4

4

4

8

8

8

7

7

12

12

12

11

11

16

16

15

15

15

20

20

19

19

18

24

24

23

22

22

28

28

27

26

25

32

32

31

30

29

36

36

35

34

32

0.5

0.6

0.7

0.8

0.9

0.3085

0.2743

0.2420

0.2119

0.1841

0.3050

0.2709

0.2389

0.2090

0.1814

0.3015

0.2676

0.2358

0.2061

0.1788

0.2981

0.2643

0.2327

0.2033

0.1762

0.2946

0.2611

0.2296

0.2005

0.1736

0.2912

0.2578

0.2266

0.1977

0.1711

0.2877

0.2546

0.2236

0.1949

0.1685

0.2843

0.2514

0.2206

0.1922

0.1660

0.2810

0.2483

0.2177

0.1894

0.1635

0.2776

0.2451

0.2148

0.1867

0.1611

3

3

3

3

3

7

7

6

5

5

10

10

9

8

8

14

13

12

11

10

17

16

15

14

13

20

19

18

16

15

24

23

21

19

18

27

26

24

22

20

31

29

27

25

23

1.0

1.1

1.2

1.3

1.4

0.1587

0.1357

0.1151

0.0968

0.0808

0.1562

0.1335

0.1131

0.0951

0.0793

0.1539

0.1314

0.1112

0.0934

0.0778

0.1515

0.1292

0.1093

0.0918

0.0764

0.1492

0.1271

0.1075

0.0901

0.0749

0.1469

0.1251

0.1056

0.0885

0.0735

0.1446

0.1230

0.1038

0.0869

0.0721

0.1423

0.1210

0.1020

0.0853

0.0708

0.1401

0.1190

0.1003

0.0838

0.0694

0.1379

0.1170

0.0985

0.0823

0.0681

2

2

2

2

1

5

4

4

3

3

7

6

6

5

4

9

8

7

6

6

12

10

9

8

7

14

12

11

10

8

16

14

13

11

10

19

16

15

13

11

21

18

17

14

13

1.5

1.6

1.7

1.8

1.9

0.0668

0.0548

0.0446

0.0359

0.0287

0.0655

0.0537

0.0436

0.0351

0.0281

0.0643

0.0526

0.0427

0.0344

0.0274

0.0630

0.0516

0.0418

0.0336

0.0268

0.0618

0.0505

0.0409

0.0329

0.0262

0.0606

0.0495

0.0401

0.0322

0.0256

0.0594

0.0485

0.0392

0.0314

0.0250

0.0582

0..0475

0.0384

0.0307

0.0244

0.0571

0.0465

0.0375

0.0301

0.0239

0.0559

0.0455

0.0367

0.0294

0.0233

1

1

1

1

1

2

2

2

1

1

4

3

3

2

2

5

4

4

3

2

6

5

4

4

3

7

6

5

4

4

8

7

6

5

4

10

8

7

6

5

11

9

8

6

5

2.0

2.1

2.2

2.3

0.0228

0.0179

0.0139

0.0107

0.0222

0.0174

0.0136

0.0104

0.0217

0.0170

0.0132

0.0102

0.0212

0.0166

0.0129

0.00990

0.0207

0.0162

0.0125

0.00964

0.0202

0.0158

0.0122

0.00939

0.0197

0.0154

0.0119

0.00914

0.0192

0.0150

0.0116

0.00889

0.0188

0.0146

0.0113

0.00866

0.0183

0.0143

0.0110

0.00842

0

0

0

0

3

2

1

1

1

1

5

5

1

1

1

1

8

7

2

2

1

1

10

9

2

2

2

1

13

12

3

2

2

2

15

14

3

3

2

2

18

16

4

3

3

2

20

16

4

4

3

2

23

21

2.4 0.00820 0.00798 0.00776 0.00755 0.00734

0.00714

0.00695

0.00676

0.00657

0.00639

2

2

4

4

6

6

8

7

11

9

13

11

15

13

17

15

19

17

2.5

2.6

2.7

2.8

2.9

0.00621

0.00466

0.00347

0.00256

0.00187

0.00604

0.00453

0.00336

0.00248

0.00181

0.00587

0.00440

0.00326

0.00240

0.00175

0.00570

0.00427

0.00317

0.00233

0.00169

0.00554

0.00415

0.00307

0.00226

0.00164

0.00539

0.00402

0.00298

0.00219

0.00159

0.00523

0.00391

0.00289

0.00212

0.00154

0.00508

0.00379

0.00280

0.00205

0.00149

0.00494

0.00368

0.00272

0.00199

0.00144

0.00480

0.00357

0.00264

0.00193

0.00139

2

1

1

1

0

3

2

2

1

1

5

3

3

2

1

6

5

4

3

2

8

6

5

4

2

9

7

6

4

3

11

9

7

5

3

12

9

8

6

4

14

10

9

6

4

3.0 0.00135 0.00131 0.00126 0.00122 0.00118 0.00114 0.00111 0.00107 0.00104 0.00100 0 1 1 2 2 2 3 3 4

Example / Contoh:

⎟⎠⎞

⎜⎝⎛−= 2

21exp

21)( zzfπ

If X ~ N(0, 1), then

Jika X ~ N(0, 1), maka

P(X > k) = Q(k) ∫∞

=k

dzzfzQ )()(

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

kO

f

Q(z)

z


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 = 25

, y = 25

]

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, [2 marks] (b) the equation of the curve. [3 marks]

[ Answer k = 6 ]

[ y = 27

22

23 −−

xx ]

3

Diagram 3 Diagram 3 shows a string of length 125π 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 Number of students

20 – 29 2

30 – 39 10

40 – 49 36

50 – 59 55

60 – 69 k

70 - 79 5


3472/2 SULIT

(a) It is given that the first quartile score it 44.5. Find the value of k. [ 3 marks ]

(b) [ 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 ]

(c) Calculate the mean marks. [2 marks ] [ Answer k = 12, mode = 52.25 mean = 55.81

5 a) Prove that θsec

1 - θsec = - θθ tansin

[3 marks]

(b) Sketch xy 2sin1−= for π≤≤ x0 . Hence using the same axes , draw a suitable straight line

to find the number of solutions of the equation 02sin =− xxπ . State the number of solutions

[ line ]1+−=

πxy , 4 number of solution ]

[5 marks]

6 5x A B 4y P D C x

In the diagram above, →AB = 5x,

→AD = 4y and = x.

→DC

(i) Express,

(a) →AC

(b) →BD in terms of x and y. [2 marks]

(ii) Given →AP = h and

→AC

→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. [5 marks]

Answer (i)(a) 4y + x, (b) −5x + 4y (ii)(a) h(4y + x) (b) ( (5 − 5k)x + 4ky; k = 65


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.

x 1 2 3 4 5 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 x

− 31

O 31

(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

revolved through 180° at the y - axis. If the volume of Q is 21π unit2 Find the value a.

[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) 8156 (ii) a = 2 (b) y − (2k2 + r) = 4k(x − k); r = 2k2 − 8k ]


3472/2 SULIT

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

(b) Shows that the area of the Δ TUV can be expressed as p − 52

q + 10. [2marks]

(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) 225 ]


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. [6 marks]

[ Answer (a)(i) 0.2001 (ii) 0.1493 (b)(i) 0.8383 (ii) 100 ]

Sections C

Answer two questions from this section.

12 . A• •B P 8 m 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. [3marks ]

(d) 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 ]

[Answer (a) 2721 m (b)

311s (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 is www.cikgurohaiza.com

3472/2 SULIT

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 (c) 44 14 . Diagram 6 shows a quadrilateral ABCD such that ∠ABC is acute. D 9.8 cm 5.2 cm C A 40.5° 12.3 cm 9.5 cm DIAGRAM 6 B

(a) Calculate, (i) ∠ABC, (ii) ∠ADC, (iii) area, in cm2, of quadrilateral ABCD. [8 marks]

(b) A triagle A’B’C’ has the same measurements as those given for triangle ABC, that is, A’C’ = 12.3 cm, C’B’ = 9.5 cm and ∠B’A’C’ = 40.5°, but which is different in shape to triangle ABC. (i) Sketch the triangle A’B’C’.

(ii) State the size of ∠A’B’C’. [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.


Item Price index for the year 1995 based on the year 1993

Percentage of usage (%)

P 135 40 Q x 30 R 105 10 S 130 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 such that ST = 12.1 cm and TQ = 9.5 cm. STQΔ

Q

T

S Diagram 6

The area of the triangle is 45 cm2 and is obtuse. STQ∠ (a) Find (i) ST∠ [Q °=∠ 47.128STQ or ] '28128° (ii) the length, in cm, of SQ [19.49 cm] (iii) the shortest distance, in cm, from T to SQ. [4.613] [ 5 marks]


3472/2 SULIT

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 °=∠ 15PRQ .Calculate two possible values of [∠PQR = 126.60o and 23.40o]

PQR∠

(c) Using the acute in (i), calculate PQR∠ ( i) the length of PR [7.673] (ii) the value of [29.420] PTR∠ (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.

Sugar

Rice

Salt

Cooking Oil

Flour

10 20 30 40 50 60 70 80 90 100 units

Diagram 2

Items Price in the year 2000

Price in the year 2006

Price Index for the year 2006 based on the year

2000 Cooking Oil x RM2.50 125

Rice RM1.60 RM2.00 125 Salt RM0.40 RM0.55 y

Sugar RM0.80 RM1.20 150 Flour RM2.00 z 120

TABLE 4

R

T

Q

P

5 cm 13 cm

5 cm


(a) Find the values of (i) x, (ii) y (iii) z. [x=2.00,y=137.5,z=2.40] [3 marks] (b) 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] (c) 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] (d) the composite index for the year 2008 based on the year 2000 if the monthly sales of those 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 : The total number of souvenirs sold must not exceed 75. II : The number of souvenirs of type A sold must not exceed twice the number of souvenirs of type

B sold. III : 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]

(c) 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]

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


3472/2 SULIT

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 s─1 is given by v = 3t2 – 18t + 24. The object P stops momentarily for the first time at

the point B.

P

O B

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] 12. (a) (i) Use area formula

45sin)5.9)(1.12(

21

=STQ

or °=∠ 47.128STQ '28128° (ii) Using cosine Rule

cmSQ

STQSQ49.19

cos)5.9)(1.12(25.91.12 222

=−+=

°=∠ 05.29TQS

(iii) 5.9

05.29sin h= or equivalent

= 4.613 cm

(b)

°°°°=∠

=°×=

=°

141.60 ,38.40 @'36141,'2438

6212.015sin5

12sin

sin12

15sin5

QPR

p

p

∠PQR = 180o – 15o – 38.40o @ ∠PQR = 180o – 15o – 141.60o

∠PQR = 126.60o and 23.40o

(c) (i) °

=° 15sin

523.4sin

PR

cm 7.672

4.23sin15sin5PR

=

°×°

=

PR = 7.673 cm


(ii) Use Cosine Rule

cos ∠PTR = 8710.0)50)(13(2

)672.7()50(13 222

=−+

∠PTR = 29.420

(iii) Area ∆ PVR = °42.29sin)50)(13(21 = 22.58 cm2

13. (a) (i) x = 2.00 (ii) y = 137.5 (iii) z = 2.40 (b) Use composite index formula

17.13130605010080

)30(120)60(150)50(5.137)100(125)80(125

=++++

++++=

−

I

(c) 76.196

17.131100150

2006

2006

RMP

P

=

=×

(d)

40.157100

17.131120120

20082000

20082006

=

×=

=

I

I

14. (a) The three inequalities are

x + y 75 , x 2y and 9x + 2y 200 ≤ ≤ ≥

(b) refer by graph

(c) (i) 16 ≤ number of A type souvenirs sold ≤ 45

(ii)Maximum profit

= RM [ 9(50) + 2(25) ]

= RM500.


y

100

3472/2 SULIT

40

90

80

70

60

50

30

20

10

O 10 20 30 40 50 60 x

70 80

9x + 2y = 200

x = 2y

x + y = 75

y = 30

( 50 , 25 )

R

16 45


2

2

1

3

For examiner’s

use only

Answer all questions.

1. Function f is defined by

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

≥−

≤−=

3,23

211

3,2)(

xx

xxxf

Find the range corresponding to the domain 40 ≤≤ x [3 marks]

Answer : ……………………..

2. Given the function f: x 2x + 5 , g : x → →5

2+x and fg: x →5

nmx + ,

where m and n are constants , find the value of m and of n,

[2 marks ]

Answer : m =…………………….

n =..............................

Perfect Score 2009 [ Lihat sebelah SULIT


3. Diagram 1 shows part of the mapping of x to z by the function

followed by the function baxxf +→: cycy

yg ≠−

→ ,12: . Calculate the values of a, b, c

and d.

[ 4 marks]

Answer: a=………b=………c=………d=…………..

For examiner’s

use only

3

4

4

1

12

6 3

d

Diagram 1

4. If the x-axis is a tangent to the curve , find the values of p. 332 −=+ ppxx

[3 marks ]

Answer : p =.........………



5. Given α and β are the roots of . Form the quadratic equation with roots and .

0142 2 =+− xx2α 2β

[ 4 marks ]

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 ]

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

(b) ……........................


4

5

3

6

For examiner’s

use only



7. Find the range of values of x if ( ) 523 2 −+= xxxf always positive.

[3 marks]

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

8. Simplify and state your answer in the simplest form . 12313 12555 −−+ −+ nnn

[2 marks]

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

9. Solve the equation . 19 24y y+ = + 9

[3 marks]

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

3

7

3

9

3

8

For examiner’s

use only



10. Given , express k in terms of m. )3(loglog2 93 +=+ mk [4 marks]

Answer : ....……………...………..

11. Solve the equation )32(loglog 93 += xx

[3 marks]

Answer: …...…………..….......

4

10

For examiner’s

use only

3

11



12. Given that the term , thn nTn 420 −= for an arithmetic progression. Find the sum of the first 12 terms of the progression. [3 marks]

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]

Answer : …………………….

4

13

4

12

For examiner’s

use only


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 ]

Answer : .…………………

15. Diagram 2 shows the graph of log2 y against log2 x. Values of x and values

of y are related by the equation y = xk

n2

, where n and k are constants.

Find the value of n and the value of k.

For examiner’s

use only

14

2

log2 y (5, 6) (2, 0)

*

0 log2 x

[4 marks]

Diagram 2

Answer : n= ..……k=……….




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

Given that the equation of the straight line KLM is 134=+

yx and point N( x , y ) lies

on the circumference of a circle KLMN , find the equation of the locus of the moving point N.

[ 3 marks ] Answer: ……..…….…………... 17. If ( ) jpia 12 ++= and jib 63 +−= , find the value of p if ba + is parallel to the

x-axis. [3 marks]

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

4

17

For examiner’s

use only

3

16

K

M

L

N (x,y)

x

Diagram 3

•

0

y



18. Given that and , express in terms of a and a=020sin b=030cos 050sin [3 marks]

Answer: ...………………………

___________________________________________________________________________ 19. Diagram below shows two sectors , OAB and OCD with centre O.

Given that COD = 0.92 rad, BC = 5 cm and perimeter of sector OAB is 20.44 cm. ∠ Using π = 3.142 , find the area of the shaded region of ABCED.

[ 3 marks ]

Answer:………………………

3

19

3

18

For examiner’s

use only

E

O

D

A B

C


SULIT 3472/1

Perfect Score 2009 SULIT

20. Given that y= 2

12xx − and

dxdy = 2 g(x) where g(x) is a function in x .

Find . [3 marks] ∫−1

1)( dxxg

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

21. The gradient of the curve y = hx + kx 2 at the point −⎛

⎝⎜⎞⎠⎟

1 72

, is 2. Find the value of h

and the value of k. [3 marks]

Answer: ……………………..

3

20

3

21

For examiner’s

use only


SULIT 3472/1


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]

Answer: …...…………..….......

___________________________________________________________________________ 23. In a debate competition, the probability of team A, team B and team C will qualify for

the final are 51,

41,

31 respectively. Find the probability that at least 2 teams will qualify

For examiner’s

use only

3

22

for the final. [3 marks]

3

23

Answer: ……………………………


SULIT 3472/1


For examiner’s

use only 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]

3

24

Answer: ( a)…...…………..….......

( 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

4 Answer : (a)…...…………..……..…...

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


SULIT 3472/1


END OF QUESTION PAPER


1

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). [3 marks] (b) Hence, or otherwise, find (f −1)2(x) and show (f 2)−1(x) = (f −1)2(x). [3 marks]

(c) Sketch the graph of ⏐f 2(x)⏐ for the domain 0 ≤ x ≤ 2 and find it’s corresponding range. [2 marks] Jawapan:

(a)(i) 9x − 8 (ii) 9

8+x (c) y 10 --------------------------

8 0 ≤ y ≤ 10 0 8/9 x


2

3. Diagram 3 shows five semicircles.

DIAGRAM 3

The area of the semicircles form a geometric progression. Given that area of the

smallest semicircle is 116

of the area of the largest semicircle. If the total area of the

semicircles is 30 2cm , find (a) area of the smallest semicircle (b) area of the largest semicircle [5 marks] Jawapan: (a) 10 (b) 160

4. Given that tan( ) 1x y− = − and 3tan4

y = , show that 1tan7

x = − .

Sketch the graph of tany x= for 0 00 360x≤ ≤ . Hence, using the same axes , draw a suitable straight line and find the number of solutions for the equation 3 tan 6x x+ = [6 marks] Jawapan: Number of solutions = 3


3

5. Diagram 5 shows a parallelogram OABC. O A P D B C

DIAGRAM 5

Given that APD, OPC and DCB are straight lines. Given that →

OA = 6a, →

OC = 12c and OP : PC = 3 : 1.

(i) Express →

AP in terms of a and/or c. (ii) Given the area of the ΔADB = 32 unit 2 and the perpendicular distance from

A to DB is 4 units, find ⏐a⏐. [5 marks] Jawapan:

( ) 6 9( )2a a cb− +

% %

6.

O 0.5

DIAGRAM 6

Cumulative frequency

Length of fish in cm

x (5.5, 6)

(10.5, 26) x

x(15.5, 58)

(20.5, 74) x

x (25.5, 80)


4

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)

0.50 0.75 1.00 1.25 1.50 1.75

Percentage of red blood cells which experience creanation (P)

0.4

5.0

14.5

27.6

46.2

68.9

TABLE 7

Variables P and K are related by the equation P = 2

4μ

(K + A)2 where μ and A are

constants. (a) Draw the graph of P against K.

[5 marks] (b) From your graph, find the value of μ and the value of A.

[4 marks] (c) Find the value of P when K = 1.4?

[1 mark] Jawapan: ( ) 0.33, 0.40( )37.21 38.44b Acμ = = −

−


5

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 1x + , line x = 1 and line x = k. When the region is revolved 360° at the x-axis, the volume generated is 18π unit3. Find the value of k. [4 marks]

[ answer (a) 1247 (b) k = 4 ]

9. y P(0, β) Wall • R(x, y) O Q(α, 0) x Floor

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, β) and point Q(α, 0). (a) Write the equation which relates α and β. [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 α = 2. (iii) Find the value of tan ∠ ORQ when α = 2. [9 marks]

Jawapan:

2 2( ) 812( )( ) ,5.853

( )0.35

a

b ii

c

α β+ =

⎛ ⎞⎜ ⎟⎝ ⎠


6

10.

α rad

T

O

K

L

J

RQ

P

DIAGRAM 10

Jawapan:

2( )320( )3

( )61.50

a

b

c

π

π

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

Diagram 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 (a) the angle α , in terms of π [ 2 marks] (b) the length, in cm, of the arc JKL [ 4 marks] (c) the area, in cm2, of the shaded region. [4 marks]


7

P Q

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.

A C B 28 m

DIAGRAM 12 The velocity of P, vp ms−1, 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 ms−1. Object P stops instantaneously at the point C. (Assume towards the right is positive.) Find, (a) the maximum velocity, in ms−1, for P, [3 marks] (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]

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


8

13. . Diagram 13 shows a quadrilateral ABCD such that ∠ABC is acute.

DIAGRAM 13

(a) Calculate, (i) ∠ABC, (ii) ∠ADC, (iii) area, in cm2, of quadrilateral ABCD. [8 marks]

(b) A triangle A’B’C’ has the same measurements as those given for triangle ABC, that is, A’C’ = 12.3 cm, C’B’ = 9.5 cm and ∠B’A’C’ = 40.5°, but which is different in shape to triangle ABC. (i) Sketch the triangle A’B’C’. (ii) State the size of ∠A’B’C’. [2 marks] Jawapan: (a) (i) 57.21-57.25

(ii) 106.07-106.08 (iii)82.37-82.39

(b) (ii) 122.75-122.79

C

12.3 cm40.50

9.8 cm

9.5 cm

5.2 cm D

B

A


9

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

Cloth Preparation time (minutes) Sewing time (minutes)

T-shirt 45 50 Slack 30 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: ( )45 3 60050 70 3505 4

( )4(6,11), 206

a x yx y

x y

cRM

+ ≤+ ≥≤


10

15. (a)

Index number, Ii 105 94 120 Weightage, Wi 5 − x x 4

The composite index number for the data in the table above is 108. Find the value of x. [4 marks]

(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 Price index in the year 1994

Change of price index from the year 1994 to the year 1996

Weightage

Timber 180 Increased 10 % 5 Cement 116 Decreased 5 % 4

Iron 140 No change 2 Steel 124 No change 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


- 1 -

FUNCTIONS

1. Given that mxxf +→ 4: and 43:1 +→− nxxf , find the values of m and n.

Answer:- m = – 3 ; n = 14

2. Given that 12: −→ xxf , xxg 4: → and baxxfg +→: , find the values of a and b .

Answer:- a = 8 ; b = –1 3. Given that 3: +→ xxf , 2: bxaxg +→ and 56366: 2 ++→ xxxgf , find the values of a and b .

Answer:- a = 2 ; b = 6

4. Given that xmxg 3: +→ and 342:1 −→− kxxg , find the values of m and k.

Answer:- k = 16

; m = 4

5. Given the inverse function 2

32)(1 −=− xxf , find

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

Answer:-(a) 112

(b) 12

−

6. Given the function : 2 1f x x→ − and : 23xg x → − , find

(a) f –1 (x) , (b) f – 1 g(x) , (c) h(x) such that hg(x) = 6x – 3 .

Answer:-(a) 12

x + (b) 1 16 2

x − (c) 18x + 33

7. Diagram 1 shows the function 3:2

p xg xx+

→−

, 2x ≠ , where p is a constant.

Diagram 1 Find the value of p.

7

32

p xx+−

x g

5


- 2 -

Answer:- p = 4 8. x y z 4 4 4 2 2 2 0 0 0 −1 −2 −2 −2

Diagram 2 Diagram 2 shows the mapping of y to x by the function g : y → ay + b and mapping

to z by the function h : y → 62y b−

, y ≠ b2

. Find the,

(a) value of a and value of b, (b) the function which maps x to y, (c) the function which maps x to z.

Answer:- (a)a= –6, b=10 (b)106

y− (c) 1820y− −

9. In the Diagram 3, function h mapped x to y and function g mapped y to z. x h y g z 8 5 2 Diagram 3 Determine the values of, (a) h−1(5), (b) gh(2)

Answer:- (a)2 (b)8 10. Given function f : x → 2 − x and function g : x → kx2 + 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)44


- 3 -

11. The following information refers to the functions f and g.

Find f (x). Answer:- 23 23

x−

12. (a) Function f, g and h are given as f : x → 2x

g : x → 32x −

, x ≠ 2

h : x → 6x2 − 2. (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 g−1(−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) ( ) 22p x x= + 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 it’s corresponding

range. Answer:-(a)9x – 8 (b) 89

x +

14. A function f is defined as f : x → p xx

++3 2

, for all values of x except x = h and p

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).

Answer:-(a) 32

h = − (b)(i)p =12(ii)x = –3 (iii)–5

g (x) = 4 – 3 x

fg (x) = 2 x + 5


- 4 -

QUADRATIC EQUATIONS 1. One of the roots of the quadratic equation is twice the other root.

Find the possible values of p. Answer ; 5 7,p = −

2. If one of the roots of the quadratic equation is two times the other root,

find an expression that relates . Answer : 22 9b ac=

3. Find the possible value of m , if the quadratic equation has two equal

roots. Answer ;

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

Answer : − 12

or 2

5. Given 2α

and 2β

are roots of the equation k x(x − 1) = 2m − x.

If α + β = 6 and α β = 3, find the value of k and of m.

Answer : k = − 21 , m =

163

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

Answer : λ = ± 1; roots: λ = 1, x = 1; λ = −1, x = 0


- 5 -

QUADRATIC FUNCTIONS 1. Diagram 1 shows the graph of the function ( )22 5y x p= − − + , where p is constant.

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

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

2. Diagram 2 shows the graph of the function ( ) ( ) 21 2f x p x x q= − + + . (a) State the value of q .

(b) Find the range of values of p .

Answer:-(a) q = – 2 (b) 12

p <

y

( 0, –3 ) ( 4 , –3 )

x 0

Diagram 1

( 0, –2 ) ( ) ( ) 21 2f x p x x q= − + + Diagram 2

y

x 0


- 6 -

3. Diagram 3 shows the graph of the function 2y x bx c= + + that intersects the y- axis at point ( 0, 9 ) and touches the x- axis at point K. Find,

(a) the value of b and c , (b) the coordinates of point K.

Answer:-(a) b = – 6 , c = 9 (b) ( )3 0,

4. y ( 2, 3) 0 x

Diagram 4 In Diagram 4 above point ( 2, 3 ) is the turning point on the graph which has equation of the form y = p(x + h)2 + k. Find the,

(a) values of p, h and k, (b) equation of the curve formed when the graph as shown is reflected at the x–axis. (c) equation of the curve formed when the graph as shown is reflected at the y–axis.

Answer :- (a) p = 5 , h = −2, k = 3 (b) y = −5(x − 2)2 − 3 (c) y = 5(x + 2)2 + 3

y ( 0, 9) K

2y x bx c= + +

x

Diagram 3

( 0, 23)


- 7 -

5. Function ( ) 2 28 20 1f x x kx k= − + + has a minimum value of 2 4r k+ , where r and k

are constants.

(a) By using the method of completing the square, show that 2 1r k= − . (b) Hence or otherwise, find the values of k and r if the graph of the function is

symmetrical about 2 13x r= − .

Answer:-(b) k = 3 , –1 and r = –3 , 5 6. The function ( ) ( )( )6 2f x x x h= + − + has a maximum value of 10 and h is a constant. (a) Find the value of h. (b) Sketch the graph of ( ) ( )( )6 2f x x x h= + − + for the value of h that is determined in (a) above. (c) Write the equation of the axis of symmetry.

Answer:- ( a) h = --6 (c) x = –2 7. Given y = x2 + 2kx + 3k has minimum value 2. (a) Without using the method of differentiation, find the two possible values of k. (b) With these values of k, sketch on the same axis, two graphs for y = x2 + 2kx + 3k. (c) State the coordinates of the minimum point for y = x2 + 2kx + 3k . Answer:- (a) k =1 , 2 (c) (−1, 2), (−2, 2) ****************************************************************************** SIMULTANEOUS EQUATIONS 1. Solve the simultaneous equations 3x + 2y = 1 and 3x2 – y2 = 5x + 3y.

Answer: x = -7/3 , y = 4 ; x = 1, y = -1 2. Solve the simultaneous equations

3 2 5 and 2 3 5

xy x yx y − −+ = =

Answer: x = 41/10, y = 16/3 ; x = 1, y = -5 3. Solve the simultaneous equations 2x - y = 4 and 2x2 + xy - 3x = 7. Give your answers correct to three decimal places.

Answer: x = 2.461, y = 0.922; x = -0.711, y = -5.422


- 8 -

INDICES AND LOGARITHMS

1. Solve the equation 1

2

25.51625

−+ =

xxx Answer:-

52

−=x

2. Solve the equation 3548.2 −= xxx

Answer:- 1=x 3. Show that ( )xxx 3533 12 −+ −+ is divisible by 13.

Answer : ( )1313 −x .4. Solve the equation 34264 32 =+−x

Answer:- 2312

5. Solve the equation 1023 12 =+xx Answer: 0 6477.

6. Solve the equation ( ) 3log]32[loglog 924 =−x

Answer:3.5

7. Solve the equation x

xx 1log)14log()2log( =−−+

Answer :x =1 8. Solve the equation log2x - 4 logx16 = 0

Answer: 11616

,

9. Solve the equation ( ) xx 572 1 =− Answer: 7232.3=x

10. Solve 813 2log =x Answer : 16=x

11. Solve the equation 32x+1 - 2 (3x+0.5) - 3 = 0 Answer: 0.5

12. Solve the equation 102x+1 - 7 (10x) = 26

Answer:0.3010 13. Given that m=5log3 and n=2log9 , express 50log3 in terms of m and n

Answer:- 2 2m n+ 14. Given that kx =2log and hx =7log , express xx 5.3log in terms of k and h

Answer:- kh 222 −+ 15. Given that ( ) yxyx 222 loglog3log2 ++=+ , show that ( )xyyx 622 =+ 16. If log2a + log2b = 4, show that log4ab = 2 and that log8ab = 4/3. If log2a + log2b = 4,

show that log4ab = 2 and that log8ab = 4/3.


- 9 -

COORDINATE GEOMETRY 1. The following information refers to the equations of two straight lines, AB and CD which

are parallel to each other.

Express p in terms of q. Answer: p = )1(32

+q

2. The triangle with vertices A(4,3), B(-1,1) and C(t , -3) has an area 11 unit2. Find the possible values of t.

Answer: t = 0 , -22 3. The points P(3, p), B(-1, 2) and C(9,7) lie on a straight line. If P divides BC internally in the ratio m : n , find (a) m : n , (b) the value of p. Answer:(a) 2 : 3 (b) p = 4

4. (a) A point P moves such that its distance from point A (1,– 4) is always 5 units.

Find the equation of the locus of P. Answer: 2 2 2 8 8 0x y x y+ − + − = (b) The point A is (-1, 3) and the point B is (4, 6). The point Q moves such that QA : QB = 2 : 3. Find the equation of the locus of Q.

Answer: 2 25 5 14 102 54 0x y x y+ + + − = (c) A point R moves along the arc of a circle with centre A(2, 3). The arc passes through Q(-2, 0). Find the equation of the locus of R.

Answer: 2 2 4 6 8 0x y x y+ − − + = (d) A point S moves such that its distance from point A(–3,4) is always twice its

distance from point B(6,-2). Find the equation of the locus of S. Answer: 2 2 18 8 45 0x y x y+ − + + =

(e) The point M is (2, –3) and N is (4, 5). The point T moves such that it is always equidistance from M and from N. Find the equation of locus of T.

Answer : e) x+4y = 7 (f) Given point A (1,2) and point B (4, –5). Find the locus of point W which moves such that ∠ AWB is always 900.

AB : 2y = p x + q

CD : 3y = (q + 1) x + 2

Where p and q are constants


- 10 -

Answer: 2 2 5 3 6 0x y x y+ − + − = Solutions to question no 5, 6 and 7 by scale drawing will not be accepted. 5. In Diagram 1, the straight line PR cuts y-axis at Q such that PQ : QR = 1 : 3. The equation of PS is 2y = x + 3.

(a) Find (i) the coordinates of R, (ii) the equation of the straight line RS, (iii) the area Δ PRS. (b) A point T moves such that its locus is a circle which passes through the points P, R and

S. Find the equation of the locus of T. Answer: a)(i) R = (9 , 16) (ii) y = – 2x + 34 (iii) 80 unit 2 b) x2 + y2 – 6x – 16y – 27 = 0 6. Diagram 2 shows the straight line graphs PQS and QRT in a Cartesian plane. Point P and point S lies on the x-axis and y-axis respectively. Q is the mid point of PS. y S y − 3x = 4 Q R(0, 1) P x O T

Diagram 2

P(–3, 0 )

Q( 0, 4 )

S

R y

x O Diagram 1


- 11 -

(a) Find, (i) coordinates of the point Q, (ii) area of the quadrilateral OPQR. (iii) The equation of the straight line which is parallel to QT and passes through S. (b) Given 3QR = RT, calculate the coordinates of the point T.

(c) A point moves in such a way that it’s distance from S is 12

it’s distance from the point T.

(i) Find the equation of locus of the point T. (ii) Hence, determine whether the locus cuts the x-axis or not. .

Answer: (a)(i) (− 23

, 2) (ii) 53

(iii) 3 42

y x= − + (b) (2, −2) (c)(i) 3x2 + 3y2 + 4x − 36y +56 = 0

(ii) No 7. y K P• J • Q R • x O L Diagram 3

In Diagram 3, P(2, 9), Q(5, 7) and R(421 , 3) are the mid point of the straight line JK, KL and

LJ such that JPQR form a parallelogram. (a) Find, (i) the equation of the straight line JK, (ii) the equation of the perpendicular bisector of the straight line LJ. (b) Straight line KJ is extended until it intersects the perpendicular bisector of the straight line LJ at the point S. Find the coordinates of the point S. (c) Calculate the area of ΔPQR and consequently the area of ΔJKL.

Answer: (a)(i) y = 8x − 7 (ii) 4y = 6x − 15 (b) ⎟⎠⎞

⎜⎝⎛ −3,

21 (c) 6

21 ; 26


- 12 -

STATISTICS

1. Table 1 shows the results obtained by 100 pupils in a test.

Marks < 20 < 30 < 40 < 50 < 60 < 70 < 80 < 90

Number of pupils 3 8 20 41 65 85 96 100

Table 1

(a) Based on Table 1, complete the table below.

Marks 10 – 19

Frequency

(b) Without drawing an ogive, estimate the interquartile range.

Answer:-(b)Interquartile range = 22.62

2. The mean and standard deviation of a set of integers 2 , 4 , 8 , p and q are 5 and 2

respectively. (a) Find the values of p and of q . (b) State the mean and variance of the set integers 7, 11, 9 , 2p + 3 and 2q + 3

Answer:-(a) 5 6 or 6 5, ,p q p q= = = = (b) Mean =13 Variance = 16 3. The histogram in Diagram 1 shows the marks obtained by 40 students in Mathematics test.

(a) Without drawing an ogive , calculate the median mark. (b) Calculate the standard deviation of the marks.

Diagram 1

Marks

Number of students

15.5 20.5 25.2 30.5 35.5 40.5

10 8 6 4 2 0


- 13 -

Answer:(a) 27.17 (b) 6.595 4. Table 2 shows the frequency distribution of the Chemistry marks of a group of students.

Marks Number of students 1 – 10 2 11 – 20 3 21 – 30 5 31 – 40 10 41 – 50 p 51 – 60 2

Table 2

(a) If the median mark is 34.5 , calculate the value of p . (b) By using a scale of 2 cm to 10 marks on the horizontal axis and 2 cm to 2 students

on the vertical axis, draw a histogram to represent the frequency distribution of the marks. Find the modal mark.

(c) What is the modal mark if the mark of each student is increased by 8 ?

Answer:- (a) p = 6 (b) Mode = 36 (c) 44 5. The scores, x , obtained by 32 students of Class 5 Alfa in a test are summarized as

22496 and 195488.x x= =∑ ∑ The mean and the standard deviation of the scores, y , obtained by 40 students and Class 5 Beta in the test are 66 and 6 respectively.

(a) Find (i) y∑ (ii) 2y∑

(b) Calculate the mean and the standard deviation of the scores obtained by all the 72 students.

Answer:-(a)(i) 2640 (ii)175680 (b)Mean = 71.33 , S Deviation= 8.194

6. A set of data consists of 10 numbers. The sum of the numbers is 120 and the sum of the

squares of the numbers is 1650.

(a) Find the mean and variance of the set of data,

(b) A number a is added to the set of data and the mean is increased by 2, find (i) the value of a, (ii) the standard deviation of the new set of data.

Answer:-(a) Mean = 12 , Variance = 21 (b)(i) a = 34 (ii) S Deviation = 7.687


- 14 -

CIRCULAR MEASURES

1. Diagram 1 Diagram 1 shows a circle with centre O and OA = 10 cm. Straight line AT is a tangent to the circle at point A, and AOT is a triangle. Given that the area of triangle OAT = 60 cm2, find the area of sector OAB.

Answer:- 43.80 2cm 2.

Diagram 2 Given that the area of a sector OAB in Diagram 2 with centre O and radius 20 cm is 240 cm2. Calculate (a) the length of arc AB (b) area of shaded region

Answer:- : (a) 24 cm (b) 53.60 3.

Diagram 3

In the Diagram 3, POQ is a circular sector with centre O and a radius of 17 cm. Point R is on the straight line ORQ such that RQ = 5 cm. Calculate (a) the value of θ in radian (b) the area of the shaded region, in cm2

Answer:- (a) 0.7871 (b) 41.49

B O

T

A

B

A

O

O

P

Q Rθ


- 15 -

4.

Diagram 4 Diagram 4 shows a semicircle with centre O and radius of 10 cm. Given that QS is the length

of arc with centre P and rad 6π

=∠QPS . Find

(a) the length of OS. (b) the area of 1S Answer:-(a) 7.32 cm (b) 35.23 2cm 5.

CC 2r

Diagram 5

Two identical circles of radius 2r are drawn with their centres, C on the circumference of each circle as shown in the Diagram 5. Show that the area of shaded region A 2cm , is given by

( )33432 2 −πr .

6. Diagram 6 shows two circles with centres O and A. The respective radii are 8 cm and 3 cm. A

tangent touches the circles at the points Q and R. Given that ∠QOP = 3π radians, find

(a) the length of QR (b) the perimeter of the shaded region (c) the area of the shaded region

Answer : a) 9.80 cm b) 24.46 cm c) 10.96 cm2

O S R

Q

P

1S

P 3cm8cm

R

AO

Q

Diagram 6


- 16 -

DIFFERENTIATION

1.(a) (i) Given y = 3x2 + 5, find dydx

by using the first principle.

(ii) Differentiate y = 4 3x

− with the first principle.

(b) (i) Find ddx

12 1x +

⎛⎝⎜

⎞⎠⎟

.

(ii) Given f(x) = 4x(2x − 1)5, find f’(x). (iii) Differentiate 3x2(2x − 5)4 with respect to x. (iv) Given f(x) = (2x − 3)5, find f ″(x).

(v) Given f(x) = 1

21 3

−−x

x , find f ‘(x).

(c) (i) Given h(x) = 2)53(1−x

, find the value h’’(1).

(ii) Given f(x) = ( )x

x312 52

−− , find f '(0).

(d) (i) Find the limit of 2

limitn →

nn

2 42

−−

⎛⎝⎜

⎞⎠⎟ .

(ii) Given f (r) = rr

2534

−+ . Find the limit of f (r) when r → 1.

(e) Given y = x(3 − x), express y d ydx

x dydx

2

2 12+ + in terms of x, in the simplest form.

Hence, find the value of x which satisfy the equation y d ydx

x dydx

2

2 12+ + = 0.

Answer : (a) (i) 6x

(ii) 24

x−

(b)( i) 2)12(2+

−x

(ii)[4(2x − 1)4)(12x − 1)]3)52)(56(6) −− xxxiii

iv) 40(2x − 3)3

v) 2

23

)1(164

−

−+−

xxx

c) (i) 827

(ii) −96

d) i) 2

ii) 37

e) 12 − 3x; x = 4


- 17 -

2. (a) Given the function of the graph f(x) = hx3 + 2xk , which has a gradient function of

f ‘ (x) = 3x2 − 3

96x

, where h and k are constant. Find,

(i) the value of h and the value of k, (ii) the coordinate x of the turning point of the graph. (b) The point P lies on the curve 2)5( −= xy . It is given that the gradient of the

normal at P is 41

− . Find

(i) the coordinates of P. (ii) the equation of the normal to the curve at point P.

(c) A curve with the gradient function 2

22x

x − has a turning point at ( k , 8 ) .

(i) Find the value of k . (ii) Determine whether the turning point is a maximum or a minimum point . Answer:

2(a) (i) h=1, k= 48 (ii) 2

(b) (i) (7, 4) (ii)4y + x = 23

(c) i) k = 1, ii) Minimum

3. (a) Two variables, x and y, are related by the equation y = 3x + x2 .

Given that y increases at a constant rate of 4 units per second, find the rate of change of x when x = 2. (b) On a certain day, the rate of increase of temperature, θ°, with respect to time, t s, is

given by dtdθ

= 21

(12 − t).

(i) Find the value of t at the instant when θ is maximum. (ii) Given θ = 4 when t = 6, find the maximum value of θ.

Answer: (a) 1.6 units-1 (b) i) 12 ii) 13


- 18 -

4. (a) Given y = t − 2t2 and x = 4t + 1.

(i) Find dxdy in terms of x.

(ii) If x increases from 3 to 3.01, find the corresponding small increment in t.

(b) Given y = 2x3 − 5x2 + 7, find the value of dxdy

at the point (2, 3).

Hence, find (i) the small change in x which causes y to decrease from 3 to 2.98.

(ii) the rate of change of y when x = 2 if the rate of change of x is 0.6 unit per second.

(c) Given yx

=16

4 , find the value of dydx

when x = 2. Hence, find the approximate

value of 16198 4( . )

.

Answer: (a) (i) 4

2 x− (ii) 0.0025 (b) (i) −0.005 (ii) 2.4 unit s−1 (c) 1.04

5. Diagram 1 shows a composite solid made up of a cone resting on a cylinder with radius x cm.

Diagram 1

The total surface area of the solid, A cm2, is given by the equation A = 3π ⎟⎠⎞

⎜⎝⎛ +

xx 162 .

(a) Calculate the minimum value of the surface area of the solid. (b) Given the surface area of the solid is changing at a rate of 42π cm2 s−1. Find the

rate of change of radius at the instant when the radius is 4 cm. (c) Given the radius of the cylinder increases from 4 cm to 4.003 cm. Find approximate increment in the surface area of the solid

Answer: b) 36π (c) 2 (d) 0.063π

x cm


- 19 -

PROGRESSION 1. Show that log h, log hk, log 2hk , log 3hk ,…… is an arithmetic progression. Then find the

common difference of this progression. 2. An arithmetic progression has 10 terms. The sum of all these 10 terms is 220. The sum of

the odd terms is 100. Find the first term and the common difference. Answer : 4,4 == da

3. The sum of the first six terms of an arithmetic progression is 120. The sum of the first six

terms is 90 more than the fourth term. Calculate the first term and the common difference.

Answer : 20,30 =−= da 4. Given that the sum of n term of an arithmetic progression is .32 2 nnSn += Find (a) the n term in terms of n

(b) the first term (c) the common difference

Answer : (a) 14 +n (b) 5( c) 4 5. An arithmetic progression has 12 terms. The sum of all these 12 terms is 222. The sum of

the odd terms is 102. Find (a) the first term and the common difference (b) the last term

Answer :(a) a = 2 , d = 3 (b) 35 6. The n term of an arithmetic progression is 85 −n . Find the sum of all the terms from the

5th term to the 8th term. Answer : 98

7. Estimate the sum to infinity of the geometric progression .......31139 ++++

Answer : 2113

8. Write ........007.0007.007.07.0 ++++ as a fraction. Answer : 97

9. The sum of the first n terms of a geometric progression is 3

8)2( 12 −=

+n

nS . Find the least

number of terms in the progression that its sum to exceed 60. Answer : 4=n

10. Find the least number of terms of the geometric progression 4,12,36,……which must be

taken for its sum to exceed 1 800. Answer : 7=n


- 20 -

11. The sum of the first two terms of a geometric progression is 43 and the sum of the next

two terms is 163 , where the common ratio is positive. Find the sum to infinity of the

progression. Answer : 1=∞S 12.

Diagram 1 Diagram 1 shows four circles. Each circle has a radius that is 2 units longer than that of the previous circle. Given that the sum of the perimeters of these four circles is 120π cm, (i) find the radius of the smallest circle.

(ii) the sum of the perimeters from the fifth term to the tenth term.

Answer :(i) r = 12 cm(ii) π300 cm

13. Encik Rahim plans to donate an amount of money to the ‘Rumah Penyayang’ each year

from 2008. The amount in 2008 will be RM50 000, and thereafter, the amount each year will be 90% of the amount for the previous year. Calculate

(a) the year in which the donation falls below RM 20 000 for the first time .

(b) the total donation from 2008 to 2015 inclusive

Answer : a) 10=n , 2017 b) RM284 766.40 14.

Diagram shows two balls in a tube of length 10 m, moving towards each other. P moves from one end traveling 60 cm in the first second, 59 cm in the next second and 58 cm in the third second. Q moves from the other end traveling 40 cm in the first second, 39 cm in the next second and 38 cm in the third second. The process continues in this manner until the two balls meet. (a) Find the shortest time for the two balls to meet.(give your answer to the nearest second) (b) Calculate the distance traveled by P. (c) Calculate the difference in distance traveled by the two balls.

Answer : a) 11s b) 605 cm c) 220 cm

P Q


- 21 -

INTEGRATION

1. (a) Find, (i) ( )( )4 42

− +∫

x xx

dx (ii) 183 5 3( )x −∫ dx

(b) Given 0

3

∫ f(x) dx = 8, find the value of f x( ) +∫

220

3dx.

(c) Find ( )∫ + 372x dx

(d) 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,

(i) the value of k, (ii) the equation of the curve.

(e) Given 2

2

dxyd = 4x3 + 1. When x = −1, y =

21 and

dxdy = 3. Find y in terms of x.

Answer:( a) (i) −x

16 − x + c (ii) − 2)53(3−x

+ c (b) 7 ( c) ( )8

72 4+x + c

(d(i)) k = 6 (ii) y = 2x3 − 2

2x − 27 (e) y =

5163

25

25

+++ xxx

2. Given that dydx

is directly proportional to 2 1,and that 3 and 9, whendyx ydx

− = = x = 2 ,

find the value of y when x = 3 . Answer: 19

3. The rate of change of the area, A cm2, of a circle is 26 2 1t t− + . Find in terms of if the area of the circle is 11 cm2 when t = 2 .

Answer:

4. (a) Diagram 1 shows part of the curve of 2

2yx

= . The straight line x = k divides the

shaded region enclosed by the curve 2

2yx

= , the straight lines x = 1 and x = 5 and

the x-axis into two regions, and . y A B 0 1 k 5 x

Diagram 1


- 22 -

Given that the area of region is five times the area of region , find the value of

(b) Diagram 2 shows part of the curve ( )2y x x= − .

( )2y x x= −

Find the value of the solid generated when the shaded region is revolved through 360 about the x-axis .

Answer: (a) k = 3 (b) 1

5. (a) Diagram 3 shows a straight line 2y x= and a curve 2 3y x x= −

2 3y x x= − x

Find (i) the coordinate of the point P, (ii) the area of the shaded region

Answer: ( )( ) ( )1255 10 6

,i ii

6. Diagram 4 shows part of the curve ( )2

42 1

yx

=−

which passes through

y

( )2

42 1

yx

=−

y

x 0

Diagram 2

P

y

x 0

Diagram 3

0 x Diagram 4


- 23 -

(a) Find the equation of the tangent to the curve at point Q. (b) A region is bounded by the curve, the x-axis and the straight lines and

(i) Find the area of the region (ii) The region is revolved through 360 about the x-axis. Find the volume generated, in terms of

Answer:(a) 6 20y x= − + (b) i) ,ii ) π10125

784

7. (a) Evaluate dxxx∫ −4

0

)4(

Diagram 5

(b) Diagram 5 shows the curve y = x(4 − x) , together with a straight line. This line cuts the curve at the origin O and at the point P with x-coordinate k, where 0 < k < 4 .

(i) Show that the area of the shaded region, bounded by the line and the curve, is 316

k

(ii) Find, correct to 3 decimal places, the value of k for which the area of the shaded region is half of the total area under the curve between x = 0 and x = 4 .

Answer : k = 3.175

8. Diagram 6 shows, the straight line PQ is normal to the curve 121 2 += xy at A(2,3).

The straight line AR is parallel to the y – axis.

A(2,3)

Q(k ,0)

y P 0 R

Diagram 6


- 24 -

y = (x – 2 )2

Find (a) the value of k,

(b) the area of the shaded region, (c) the volume generated, in terms of π, when the region bounded by the curve, the y – axis and the straight line y = 3 is revolved through 360o about the y-axis.

Answer : (a) k = 8 (b) area = 12 31 (c) Volume = 4 π

9. Diagram 7 below shows the straight line y = x + 4 intersecting the curve y = (x – 2 )2 at the points A and B.

k Diagram 7 Find,

(a) the value of k , (b) the are of the shaded region P (c) the volume generated, in terms of π, when the shaded region Q is revolved 360o about

the x – axis.

Answer : (a) k = 5 (b) area = 20.83 (c) volume = 5

32 π

****************************************************************************** LINEAR LAW 1. The data for x and y given in the table below are related by a law of the form

2y px x q= + + ,where p and q are constants.

x 1 2 3 4 5 y 41.5 38.0 31.5 22.0 9.5

(a) Plot y x− againts 2x , using a scale of 2 cm to 4 unit on both axes. Hence , draw the line of best fit.

(b) Use your graph in 1(a) to find the value of (i) p (ii) q Answer:

B

A

4+= xy

P

Q

y

x


- 25 -

2. The variables x and y are known to be connected by the equation xy Ca−= . An experiment gave pairs of values of x and y as shown in the table. One of the values of y is subject to an abnormally large area.

x 1 2 3 4 5 6 y 56.20 29.90 25.10 8.91 6.31 3.35

(a) Plot log y against x, using a scale of 2 cm to 1 unit for x-axis and 2 cm to 0.2 unit for y-axis . Hence , draw the line of best fit. (b) Identify the abnormal reading and estimate its correct value. (c) Use your graph in 2(b) to find the value of

(i) C (ii) a

Answer: (b)25.1,17.78 (c) C = 100 , a = 1.78

3. The table shows experimental values of x and y which are known to be related by equation

ay b xx

= + .

x 1 2 3 4 5 6 y 2.20 1.74 1.71 1.77 1.86 1.96

(a) Explain how a straight line graph may be drawn to represent the given equation. (b) Plot againtsxy x x , using a scale of 2 cm to 2 unit on both axes . Hence , draw the line of best fit. (c) Use your graph in 3(b) to find the value of

(i) a (ii) b

Answer : a=1.5 ; b=0.70

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

x 2 4 6 8 10 12 y 3.16 5.50 9.12 16.22 28.84 46.77

(a) Plot log y against x by using a scale of 2 cm to 2 units on the x-axis and 2 cm to 0.2 unit on the log10 y-axis.

Hence, draw the line of best fit. (b) Use your graph from (a) to find the value of

(i) p (ii) k.

Answer : p=1.820 ; k=1.309


- 26 -

5. The variable x and y are related by the equation 2

hyx k

=+

. Diagram 1 shows the graph of

1 againts xy

.Calculate the values of h and k. The point P lies on the line. Find the value of r.

(

Diagram 1

Answer:

6. Variables x and y are related by the equation 2a bx y

+ = , where a and b are constants. When

the graph of 1 1againtsy x

is drawn, a straight line is obtained. Given that the intercept on

the 1 axis is 0.5y

− − and that the gradient of the line is 0.75, calculate the value of a and b.

Answer: a = 3, b = – 4 7. Variables x and y are related by the equation 4y = 2(x – 1)2 + 3k where k is a constant. (a) When y is plotted against (x – 1)2, a straight line is obtained, which intersects the y-axis at (0, -6). Find the value of k. (b) Hence, find the gradient and the y intercept for the straight line obtained by plotting the graph of (y + x) against x2.

Answer: (a) y = ½ (x – 1)2 + 34k , k = –8 (b) y + x = ½ x2 – 11

2, m= ½ , y-intercept = 11

2− .

1y

x


- 27 -

8. (a) Explain how a straight line graph can be drawn from the equation yx

px

qx= + ,

where p and q are constants. (b) Diagram 2 shows the graph of log2 y against log2 x. Values of x and values of y are related by

the equation y = xk

n2

, where n and k are constants. Find the value of n and the value of k.

Answer : n = 1, k = 16

9. y •(4, 44) • (2, 14) Diagram 3 shows part of the curve y against x. It is known that x and y are related by the

linear equation xy = kx + h, where h and k are constants.

(a) Sketch the straight line graph for the above equation. (b) Calculate the values of h and k.

Answer : : (a) h = 3, k = 2

log2 y

log2 x

(5, 6)

O (2, 0) Diagram 2

x 0 Diagram 3


- 28 -

10. log10 y Diagram 4

Diagram 4 shows graph of log10 y against log10 x. Given that PQ = 10 units and the point P lies on the log10 y-axis. (a) Find the coordinates of P. (b) Express y in terms of x. (c) Find the value of y when x = 16.

Answer: (a) P(0, 1) (b) y = 10 43

x (c) y = 80 ****************************************************************************** VECTORS 1. Diagram 1 shows a parallelogram, OPQR, drawn on a Cartesan plane. y Q R P x O

Diagram 1

Given that →OP = 6

~i + 4

~j and

→PQ = −4

~i + 5

~j . Find

→PR .

Answer: 10i j− + 2. Given O(0, 0), A(−3, 4) and B(2, 16), find in terms of unit vector

~i and

~j ,

(a) ABuuur

, (b) unit vector in the direction of AB

uuur.

Answer: ( ) ( )15 12 (b) 5 1213

a i j i j+ +

log10 x

Q(8, 7)

0


- 29 -

3. Given A(−2, 6), B(4, 2) and C(m, p), find the value of m and the value of p such that AB

uuur + 2 BC

uuur= 10 i

%− 12 j

%.

Answer: 6 4,m p= = − 4. Given the points )0,3(A , )8,7(B and ),1( kC

(a) Express vector AB in terms of i and j ,

(b) Find the value of k if vector OC is parallel to vector AB .

Answer: ( ) 14 8 (b) 24

,a i j h k+ = =

5. Given that OABC is a rectangle where OA = 6 cm and OC = 5cm. If OA→

= ~a and OB

→

=~b ,find

(a) AC in terms of

~a and

~b

(b) a b+

Answer:(a) (b) 61a b− + 6. Diagram 2 shows vector

~s , vector

~t and vector unit

~a and

~b .

Given

~~~32 tsr −= , express

~r in terms

~a and

~b . Answer:14 13a b+

7. Given AB→

= (k + 1) ~a and BC

→

= 2~b . If A, B and C are collinear, AB

→

= BC→

and ~b = 3

~a .

Find the value of k. Answer:k= 5

~t

~s

~a

b%

Diagram 2


- 30 -

8. Given that 2 2a i j= − +% % %

, 2 3b i j= −% % %

and 2c a b= −% % %

. Find

(a) c%

(b) unit vector in the direction of c

%.

Answer: ( ) ( ) ( )110 b 3 45

a i j− +

9. Diagram 3

Diagram 3 shows GH : AB = 3 : 10 and GH is parallel to AB→

. If ABuuur

= 10~a , find GH

→

in

terms of ~a . Answer:3a

10.

Diagram 4 shows PQRSTU is a regular hexagon. Express PQ→

+ PT→

- RS→

as a single vector.

Answer: PRuuur

11. In Δ OPQ, OP→

=~p and OQ

→

= ~q . T is a point on PQ where PT : TQ=2 : 1. Given that M

is the midpoint of OT, express PM→

in terms of ~p and

~q .

Answer: 5 16 3

p q− +

A

H G

C

B

P Q U R T S Diagram 4


- 31 -

12. Diagram 5 shows triangles OAB. The straight line AP intersects the straight line OQ at R.

It is given that 13

OP OB= , ~ ~

1 , 6 and 24

AQ AB OP x OA y→ →

= = =

Diagram 5

(a) Express in terms of ~x and/or

~:y (i) AP

→

, (ii) OQ→

(b) (i) Given that AR h AP→ →

= , state AR→

in terms of h, ~x and

~y .

(ii) Given that RQ k OQ→ →

= , state RQ→

in terms of k, ~x and

~y .

(c) Using AR→

and RQ→

from (b), find the value of h and of k

Answer:(a)(i) ( )9 3 9 3 1 12 6 (ii) (b)(i) 2 6 (ii) (c) 2 2 2 2 3 2

,y x x y h y x k x y k h⎛ ⎞− + + − + + = =⎜ ⎟⎝ ⎠

13. Diagram 6, ABCD is a quadrilateral. AED and EFC are straight lines.

It is given that ~

20AB x→

= , ~ ~

~

18 , 25 24 ,4

AE y DC x y AE AD→ →

= = − = and 35

EF EC=

(a) Express in terms of ~x and/or

~:y (i) BD

→

, (ii) EC→

(b) Show that the points B, F and D are collinear

(c) If ~ ~

2 and 3x y= = , find BD→

Answer:(a)(i) 20 32 (ii) 25 (c) 104x y x− +

R

B P

Q

O

A

A

C E F

B

D

•


- 32 -

TRIGONOMETRIC FUNCTIONS 1. Prove that cosec2x – 2 sin2x – cot2x = cos 2x 2. Prove that 2 2 2 2tan sin tan sinA A A A≡ −

3. Prove that sin 2 cot1 cos 2

x xx

=−

.

4. Find all the angles between 0o and 360o which satisfy (a) 3cos2α – 5 = 8 cos α (b) tan 2α tan α = 1

Answer ( )a 131 81 228 19 (b)30 150 210 330. , . , , ,° ° ° ° ° ° :

5. Solve the equation 0cos3sin4 =+ θθ for 00 3600 ≤≤ θ

Answer36 87 216 87. , .° ° : 6. Find all the angles between 0o and 360o which satisfy (a) 3sin 2A = 4sin A (b) 5sin2A = 5 – cos 2A

Answer ( ) ( )0 180 360 48 19 311 81 35 27 215 27 144 73 324 73, , , . , . . , . , . , .a b° ° ° ° ° ° ° ° ° :

7. Given that sin α = 178 , 90o < α < 270o and sin β =

1312

− , 90o < β < 270o.

Calculate the value of (a) sin (α + β ) (b) cos ( β – α )

Answer ( ) ( )140 21a b221 221

− :

8. Given that 3cos5

x = and 0 180o ox≤ ≤ , find sec x + cosec x .

Answer 3512

:

9. Given that sin Ө = k and Ө is acute angle, express in term of k: (a) tan Ө (b) cosec Ө

Answer ( ) ( )2

1 b1

kakk−

:


- 33 -

10. Solve the equation 2 25sin 2 3 0,0 360o oA cos A A+ − = ≤ ≤ Answer35 27 144 73 215 27 324 73. , . , . , .° ° ° ° :

11. Solve the equation 3cot

1 22 =+ xsek

x for 0° ≤ x ≤ 360°.

Answer 45 135 225 315, , ,° ° ° ° :

12. (a) Prove that sin 2 cot1 cos 2

x xx

=−

.

(b) Given 2

1cos2 1 p

θ=

+,

i. prove that 22tan

1pp

θ =−

.

ii. hence, find sin 2θ , when 2p = .

Answer ( )( ) 2425

b i − :

13. (a) Prove that tan θ + cot θ = 2 cosec 2θ .

(b) (i) Sketch the graph y = 2 cos 23 x for 0o ≤ x ≤ 2π .

(ii) Find the equation of a suitable straight line for solving the equation

cos 23 x = 1

43

−xπ

. Hence , using the same axes , sketch the straight line and

state the number of solutions to the equation cos 23 x = 1

43

−xπ

for

0o ≤ x ≤ 2π.

Answer:(b)(ii)3 14. (a) Sketch the graph of y = cos 2x for 0o ≤ x ≤ 180o. (b) Hence , by drawing a suitable straight line on the same axes , find the number of

solutions satisfying the equation 2 sin2 x = 2 – 180

x for 0o ≤ x ≤ 180o.

Answer:(b)2


- 34 -

15. (a) Prove that cosec2 x – 2 sin2 x – cot2 x = cos 2x. (b) (i) Sketch the graph of y = cos 2x for 0 ≤ x ≤ 2π . (ii) Hence , using the same axes , draw a suitable straight line to find the number

of solutions to the equation 3(cosec2 x – 2 sin2 x – cot2 x ) = πx – 1 for

0 ≤ x ≤ 2π . State the number of solutions . Answer:(b)(ii)4

16. (a) Sketch the graph of y = - 2 cos x for 0 ≤ x ≤ 2π . (b) Hence , using the same axis , sketch a suitable graph to find the number of

solutions to the equation xπ + 2 cos x = 0 for 0 ≤ x ≤ 2π .

Answer:2

17. (a) Given tan A = 43 ,and A is an acute angle. Find the value of cos 2A.

(b)(i) Sketch the curve y = sin 2x for 0 ≤ x ≤ 2 π .

(ii) Hence, by drawing a suitable straight line on the same axes, find the number of

solutions satisfying the equation sin x kos x = 21

4x

−π

for 0 ≤ x ≤ 2 π .

Answer: ( ) ( )( )7 425

a b i

**************************************************************************** PERMUTATION AND COMBINATION 1. Four girls and three boys are to be seated in a row. Calculate the number of possible

arrangements

(a) if all the three boys have to be seated together

(b) a boy has to be seated at the centre

Answer:(a)720 (b)2880 2. Find the number of the arrangement of all nine letters of word SELECTION in which

the two letters E are not next to each other

Answer:282240 3. Calculate the number of four digit even number can be formed from the digits 3, 4, 5,

6 and 9 without repetitions. Answer:48


- 35 -

4. Three alphabets are chosen from the word WALID. Find the number of possible

choice if (a) the alphabet A is chosen

(b) the alphabet A and D are chosen

Answer:(a) 6 (b) 3

5. A bowling team consists of 8 person. The team will be chosen from a group of 7 boys

and 6 girls. Find the number of team that can be formed such that each team consists of

(a) 3 boys

(b) not more than 1 girl

Answer:(a) 210 (b) 6

6. Refrigerators Refrigerators

TV TV TV

Pak Adam’s shop has 5 televisions P, Q, R, S and T and 4 refrigerators W, X, Y and Z (a) If a televisions and a refrigerator is chosen randomly, calculate the probability that television P or Q and refrigerator W are chosen. (b) Pak Adam wish to display his goods as shown in the diagram above. Calculate the number of ways the goods can be displayed.

Answer: (a) 720)(101 b

7.. Diagram 1 shows 5 letter and 3 digits.

Diagram 1

A code is to be formed using those letters and digits. The code must consist of 3

letters followed by 2 digits. How many codes can be formed if no letter or digit is

repeated in each code ?

Answer:144

8. A debating team consists of 5 students. These 5 students are chosen from 4 monitors,

2 assistant monitors and 6 prefects. Calculate the number of different ways the team

can be formed if (a) there is no restriction

(b) the team contains only one monitor and exactly 3 prefects

Answer:(a)792 (b)160

A B C D E 6 7 8


- 36 -

PROBABILITY

1. At the place where Lam stays, rain falls in any two days of a week. Out of 75% of the raining days, Lam goes to school in his father’s car. If there is no rain, Lam cycles to to school. For every 5 days Lam goes to school in his father’s car, for 3 days Lam is able to keep his pocket money. In a certain day, find the probability that (a) Lam does not goes to school in his father’s car, (b) Lam keeps his pocket money because he goes to school in his father’s car.

Answer:- (a) 1114

(b) 970

2. Rashid and Rudi compete in a badminton game. The game will end when any of the

players has won two sets. The probability that Rashid will win any one set is 53 .

Calculate the probability that (a) the game will end in only two set, (b) Rashid will win the competition after playing 3 sets.

Answer:- (a) 2513 (b)

12536

3. A container consists of 4 soya beans, 3 coffee beans and 2 cocoa beans. (a) If a bean is drawn at random from the container, calculate the probability that the bean is not a cocoa bean. (b) Two beans are drawn at random from the container, one after the other, without replacement. Find the probability that only one bean out of the two beans is a cocoa bean.

.Answer:- (a) 7/9 , (b) 7/18 4. Bag P contains five card numbered 5, 6, 7, 8 and 9. Bag Q contains three cards numbered 5, 7 and 9. A card is drawn at random from bag P and at the same time, another card is drawn from bag Q. Find the probability that the two numbers drawn have the same value or their product is an even number.

Answer: 35

Table 1

5. Table 1 shows the number of marbles of different colours in boxes P and Q. A marble is picked at random from each box. Find the probability that (a) both are of the same colour, (b) both are of different colours, (b) a yellow marble is picked from box Q.

Answer: (a) 23/80 (b) 57/80 (c) ½

Box Number of marbles Green Red Yellow

P 6 7 2 Q 3 5 8


- 37 -

6. Diagram 1 shows a board with a grid of 20 squares, of which a few squares are shaded. A dart is thrown at the board.

Diagram 1 (a) Find the probability that it will hit a shaded square. (b) Find how many additional squares need to be shaded if the probability is increased

to 35

. Answer: (a) 7/20 , (b) 5

7. A box contains 4 blue balls, x white balls and y red balls. A ball is drawn at

random from the box. If the probability of getting a white ball is 16

and the

probability of getting a red ball is 12

, find the values of x and y .

Answer: x = 2, y = 6 8. 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.

Answer: (a) 1/6 , (b) 1/ 15 9. A bag contains 6 red balls and 5 green balls. A ball is chosen from the bag and returned. Another ball is chosen and returned again. Find the probability that (a) both balls are red (b) both balls have same color, (c) both balls have different color.

Answer: (a) 36/121 , (b) 61/121 , (c) 60/121

10. There are 7 ribbons in a bag. 1 yellow, 3 black and 3 blue ribbons. (a) If a ribbon is taken out and not returned back, find the probability for the ribbon to be black. (b) If two ribbons are taken, find the probability first one to be blue followed by a black if none of the ribbons are returned. (c) If three ribbons are taken, find the probability for first one to be blue, followed by yellow and a black ribbon if none of the ribbons are returned.

Answer: (a) 3/7 , (b) 3/14 , (c) 3/70


- 38 -

PROBABILITY OF DISTRIBUTION

1. (a) A study in a district shows that one out of three teenagers in the district join the `Rakan Muda’ program. (i) If 5 teenagers are chosen randomly from the district, find the probability that 2 or more of them join the `Rakan Muda’ program. (ii) If they are 2 490 teenagers in the district, calculate the mean and the standard deviation of the number of teenagers who join the `Rakan Muda’ program. (b) From a study, it is found that the mass of a deer from a certain jungle shows a normal distribution with mean 55 kg and variance 25 kg2. (i) If a deer is caught randomly from the jungle, find the probability that the deer has a mass more than 60 kg. (ii) Find the percentage number of dears with mass between 45 kg and 60 kg.

Answer: (a)(i) 243131 or 0.5391 (ii) 830, 23.52 (b)(i) 0.1587 (ii) 81.85%

2. (a) Usually, when fishing, Wan will get fish as many as 60% from the total number of his throws.. Calculate, (i) the probability Wan will get at least 4 fishes in 5 throws, (ii) the minimum number throws made by Wan so that the probability of getting at least a fish is greater than 0.87. (b) The mass of students in a school has a normal distribution with a mean of μ kg and a standard deviation σ kg. It is known that 10.56% of the above students have mass more than 50 kg and 15.87% of them have mass less than 32 kg. Find the value of μ and the value σ.

Answer: (a)(i) 0.3370 or 31251053 (ii) 3 (b) σ = 8, μ = 40

3. (a) In a game of guessing, the probability of guessing correctly is p. (i) Find the number of trials required and the value of p, such that the mean and

the standard deviation of success are 15 and 523 respectively.

(ii) If 10 trials are done, find the probability of guessing exactly 3 correct. (b) The volume of 600 bottles of mineral water produced by a factory follow a normal distribution with a mean of 490 ml per bottle and standard deviation of 20 ml. (i) Find the probability that a bottle of mineral water chosen in random has a volume of less than 515 ml. (ii) If 480 bottles out of 600 bottles of the mineral water have volume greater than k ml, find the value of k.

Answer: (a)(i) p = 41 , n = 60 (ii) 0.2503 (b)(i) 0.8944 (ii) 473.16


- 39 -

55 45

Q

4. (a) It is known that for every 10 lemon in the box, two are rotten. If a sample of 7 are chosen randomly, calculate the probability that

(i) exactly 3 lemons are rotten, (ii).at least 6 lemons are not rotten.

(b) The masses of the members of the English Society of School M are normally distributed with a mean of 48 kg and a variance of 25 kg2. 56 of the members have masses between 45 kg and 52 kg. Find the total number of members in the

English Society.

Answer: (a) (i) 0.1147 (ii) 0.5767 (b) 109 5. Diagram 1 shows a standard normal distribution graph. f(z) 0.3485 0 k z Diagram 1 The probability represented by the area of the shaded region is 0.3485. (a) Find the value of k. (b) X is a continuous random variable which is normally distributed with a mean of 79 and a standard deviation of 3. Find the value of X when the z-scores is k.

Answer: (a) 1.03 (b) 82.09

6. Diagram 4 shows a probability distribution graph of the continuous random variable x that is normally distributed with a standard deviation of 8. The graph is symmetrical about the vertical line PQ.

Diagram 4

P xm


- 40 -

(a) If the standard score found by using the value of x = m is 34

− , find the value of m.

(b) Hence, find the area of the shaded region in Diagram 4. (c) If x represents the marks obtained by 180 Form 5 students in an examination, calculate the number of students whose marks are less than 33.

Answer: (a) m = 39 , (b) 0.0668, (b) 12 7. (a) A football team organizes a practice session for trainees on scoring goals from penalty kicks. Each trainee has ten goals to score. The probability that a trainee scores a goal is k. After the practice, it is found that the mean number of goals scored for a trainee is 6. (i) Find the value of k. (ii) If a trainee is chosen at random, find the probability that he scores at least two goals. (b) The masses of students of a school are normally distributed with a mean of 56 kg and a standard deviation of 10 kg. (i) If a student is chosen at random, calculate the probability that his mass is less than 50 kg. (ii) Given that 1.5% of the students have masses of more than p kg, find the value of p. (iii) If 75% of the students have masses of more than h kg, find the value of h.

Answer(a)(i) k = 0.6 (ii) 0.9983 (b)(i) 0.2743 (ii)77.7 (iii) 49.25 ************************************************************************

SOLUTION OF TRIANGLES

1. P 8 cm 6.5 cm 50° Q R Diagram 1 shows a ΔPQR. (a) Calculate the obtuse angle PRQ. (b) Sketch and label another triangle different from ΔPQR in the diagram above, so that the lengths of PQ and PR and the angle PQR remain unchanged. (c) If the length of PR is reduced whereas the length of PQ and angle PQR remain unchanged, calculate the length of PR so that only one ΔPQR can be formed

Answer: (a) 109° 28’ or 109.47° (c) 6.128 cm

Diagram 1


- 41 -

2.(a) Diagram 2 shows a pyramid VABCD with a square base ABCD. VA is vertical and the base ABCD is horizontal. Calculate, (i) ∠VTU, (ii) the area of the plane VTU.

Diagram 2

Answer: (a)(i) 84° 58’ or 84.97° (ii) 22.72 cm2 (b) . Q R 4 cm K L Diagram 3 P S 6 cm J 8 cm M Diagram 3 shows a cuboid. Calculate, (i) ∠JQL, (ii) the area of ΔJQL.

Answer: (i) 75° 38’ or 75.640 (ii) 31.24 cm2

3. 5 cm D 50° 4 cm 8 cm C

Diagram 4

V 8 cm A B 6 cm U 2 cm D T 4 cm C

.O

B

A


- 42 -

Diagram 4, ABCD is a cyclic quadrilateral of a circle centered O. Calculate (a) the length of AC, correct to two decimal places, (b) ∠ ACD, (c) the area of quadrilateral ABCD.

Answer: (a) 11.85 cm (b) 115.01°′ (c)36.08cm2 4. Diagram 5 shows two triangles PQT and TRS. Given that PQ = 24 cm, TS = 12 cm, 032TPQ∠ = , PT = TQ and PTS and TRQ are straight lines.

(a) Find the length, in cm, of PT, (b) If the area of triangle PQT is three times the area of triangle TRS, find the length of TR. (c) Find the length of RS. (d) (i) Calculate the angle TSR.

(ii) Calculate the area of triangle QRS.

Answer: (a) 14.15 cm, (b) 5.563 cm, (c) 10.79 cm, (d)(i) 27.60o (ii) 46.31 cm2 5. Diagram 6 shows a triangle PQR.

(a) Calculate the length of PR. (b) A quadrilateral PQRS is now formed so that PR is the diagonal, PRS∠ = 40o and PS = 8

cm. Calculate the two possible values of PSR∠ . (c) Using the obtuse PSR∠ in (b), calculate

(i) the length of RS, (ii) the area of the quadrilateral PQRS.

Answer: (a) 10.62 cm (b) 58.57o ; 121.43o (c) (i) 3.964 cm ; (ii) 47.34 cm2

Diagram 5 32o 24 cm

12 cm

T

S

R Q

P

R

75o

10 cm 7 cm

Q

P

Diagram 6


- 43 -

6. Diagram 7 shows a camp of the shape of pyramid VABC. The camp is built on a horizontal triangular base ABC. V is the vertex and the angle between the inclined plane VBC with the base is 60°. C A Given VB = VC = 25 cm and AB = AC = 32 cm and BAC∠ is an acute angle. Calculate (a) BAC∠ if the area of ABCΔ is 400 cm2, (b) the length of BC, (c) the lengths of VT and AT, where T is the midpoint of BC, (d) the length of VA, (e) the area of VABΔ

Answer: (a) 51.38o (b) 27.74o (c) 20.80 cm; 28.84 cm (d) 25.78 cm (e) 315.33 cm2

******************************************************************************

INDEX NUMBER

1. Table 1 shows the price indices and percentage usage of four items, P, Q, R, and S, which are the main ingredients of a type biscuits.

Item Price index for the year 1995 based on the year 1993

Percentage of usage (%)

P 135 40 Q x 30 R 105 10 S 130 20

Table 1 Calculate,

(a) (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.

(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.

Answer: (a)(i) RM29 (ii) 162 (b)(i)125 (ii) RM25

B

V

Diagram 7


- 44 -

2. Diagram 1

A technology product consists of five components, V, W, X, Y and Z. Diagram 1 shows a bar chart showing the daily usage of the components used to produce the technology product. The following table shows the prices and the price indices of the components.

Component Price in the year 2001 (RM)

Price in the year 2003 (RM)

Price index in 2003 based on 2001

V 13.00 16.25 y W 12.50 17.25 138 X 2.50 x 106 Y 14.90 22.35 150 Z z 24.50 140

(a) Find the values of x, y and z.

(b) Calculate the composite index representing the cost of the technology product in the year 2003 using the year 2001 as the base year.

(c) If the total monthly cost of the components in the year 2001 is RM1.5 million, find the total monthly cost of the components in the year 2003.

(d) If the cost of each component rises by 23% from the year 2003 to 2004, find the composite

index representing the cost of the technology product in the year 2004 based on the year 2001.

Answer:

V W X Y Z

47 34 22 12 3

Daily Usage ( RM)

Component


- 45 -

3. (a) 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.

Item Price index in the year 1994

Change of price index from the year 1994 to the year 1996

Weightage

Timber 180 Increased 10 % 5 Cement 116 Decreased 5 % 4

Iron 140 No change 2 Steel 124 No change 1

Table 2 (b) Table 2 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. .

Answer : (a) 1.50 (b) ITimber = 198, ICement = 110.2; 152.9

4. Table 3 shows the price indices and the weightages of Azizan’s monthly expenses in the year 2005 based in the year 2004.

Expenses Price index in 2005 based on 2004 Weightage

Rental 108 3

Food 120 4

Car installment 102 2

Miscellaneous 112 1

Table 3 (a) If the expenses for miscellaneous in the year 2005 was RM 1 456 , find the

miscellaneous expenses in the year 2004. (b) If the rental increases by 10% from the year 2005 to the year 2006,find the price index for

the rental in the year 2006 based on the year 2004. (c) Calculate the composite index for the expenses in the year 2005 based on the year 2004. (d) The price index for food in the year 2006 based on the year 2205 is 105. If the expenses

on food in the year 2006 were RM3150, find the expenses on food in the year 2004.

Answers : a) 1300 b) 118.8 c) 112 d) 2500


- 46 -

LINEAR PROGRAMMING

1. An institution offers two computer courses, P and Q. The number of participants for course P is x and for course Q is y. The enrolment of the participants is based on the following constraints:

I : The total number of participants is not more than 100. II : The number of participants for course Q is not more than 4 times the number of

participants for course P. III : The number of participants for course Q must exceed the number of participants for course P by at least 5. (a) Write down three inequalities, other than x ≥ 0 and y ≥ 0, which satisfy the above constraints. (b) By using a scale of 2 cm to 10 participants on both axes, construct and shade the region R that satisfies all the above constraints. (c) Using your graph from (b), find (i) the range of the number participants for course Q if the number participants for course P is 30. (ii) the maximum total fees per month that can be collected if the fees per month for course P and Q are RM50 and RM60 respectively.

Answer:-(a) x + y ≤ 100, y ≤ 4x, y – x ≥ 5 (c) (i) 35 ≤ y ≤ 70 (ii) Point (20, 80), RM5 800

2. A food analysts is supplied with two containers of food, Whiskers and Friskies. The

comparison of one scoop of food from each of the two containers is shown in the following table.

Food Protein Fat Carbohydrate Fibre

1 scoop Whiskers 24 gm 8 gm 48 gm 10 gm 1 scoop Friskies 8 gm 16 gm 32 gm 10 gm

Table 1 The analysts knows that an animal requires at least 96 gm of protein, 80 gm of fat, 288 gm of carbohydrate and not more than 100 gm of fibre each day. (a) If the analysts mixed x scoops of Whiskers with y scoops of Friskies, write down the system of inequalities satisfied by x and y. Hence, by using 2 cm to 2 unit on both axes construct and shade the region R that satisfies all the above constraints. (b) If 1 scoop of Whiskers costs RM2 and 1 scoop of Friskies costs RM3, find the mixture that provides

(i) the cheapest food. (ii) the most expensive food.

(c) Could the animal be fed on a satisfactory diet using (i) food from Whiskers only, (ii) food from Friskies only. Give your reason.

Answer:-(a)3 + 12 , + 2 10 , 3 + 2 18 , + 10x y x y x y x y≥ ≥ ≥ ≤ (b) (i)RM 17 (ii) RM 29


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3. An air craft company is going to purchase planes of type Wing and X – far . They will purchase x units of X – far and y units of Wing planes. The company has set the condition below:-

I : X – far plane consume 100 liters of fuel for a single month. Wing planes consume

70 liters of fuel. Total fuel consumption for one month is at most 3500 liters. II : X – far planes can take in 200 passengers while Wing planes can take in 100

passengers. Total passengers the planes must take at any time must be at least 3000.

III: Total number of planes purchased must at least 20 units. (a) State the inequality that defines the condition above other than 0 and y 0x ≥ ≥ . (b) Construct the graphs and mark the region R that represents the conditions above. Use a

scale of 2 cm for 5 Wing planes and 2 cm for 5 X – far planes. (c) The company makes a profit of RM220 for X – far and RM165 for Wing planes from its

sales. Identify the minimum amount of profit that the company will obtain.

Answer:-(a)10 + 7 35, 2 + 30, + 20x y x y x y≤ ≥ ≥ (c) RM3870 4. A furniture workshop produces tables and chairs. The production of tables and chairs

involve two processes , making and shellacking. Table 3 shows the time taken to make and to shellack a table and a chair.

Product Time taken (minutes) Making Shellacking

Table 60 20 Chair 40 10

Table 3 The workshop produces x tables and y chairs per day. The production of tables and chairs

per day is subject to the following constraints. I: The minimum total time for making tables and chairs is 600 minutes. II: The total time for shellacking tables and chairs is at most 240 minutes. III: The ratio of the number of tables to the number of chairs is at least 1 : 2. (a) Write three inequalities that satisfy all of the above constraints other than

0 and y 0x ≥ ≥ . (b) By using a scale of 2 cm for 2 units of furniture on both axes , construct and shade the

region R which satisfies all of the above constraints. (c) By using your graph from (b), find,

(i) the maximum number of chairs made if 8 tables are made. (ii) the maximum total profit per day if the profit from one table is RM30 and from

one chair is RM20.

Answer:-(a) 3 + 2 30, 2 + 24, 2x y x y y x≥ ≤ ≤ (c)(i) 8 (ii)RM420


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**************************************************************************** MOTION ALONG THE STRAIGHT LINE 1. A particle moves in a straight line and passes through a fixed point O. Its velocity, v ms 1− , is given by 562 +−= ttv , where t is the time, in seconds, after leaving O . [Assume motion to the right is positive.] (a) Find

(i) the initial velocity of the particle, (ii) the time interval during which the particle moves towards the left, (iii) the time interval during which the acceleration of the particle is positive.

(b) Sketch the velocity-time graph of the motion of the particle for 50 ≤≤ t . (c) Calculate the total distance traveled during the first 5 seconds after leaving O.

Answer: (a) (i) v= 5 (ii) 1 < t < 5 (iii) t > 3 (c) 13 m

2. Diagram 1 shows the object, P, moving along a straight line and passes through a fixed

point O. The velocity of P, v m s─1 , t seconds after leaving the point O is given by

v = 3t2 – 18t + 24 . The object P stops momentarily for the first time at the point B.

P

O B

Diagram 1 (Assume right-is-positive) Find: (a) the velocity of P when its acceleration is 12 ms – 2 , (b) the distance OB in meters, (c) the total distance travelled during the first 5 seconds.

Answer: (a) 9 ms – 1 (b) OB = 20 m (c) 28 m 3. The velocity of an object which moves along a straight line, v ms−1, t s after passing through a fixed point O is v = pt − qt2, where p and q are constants. It is known that

the object moves through a distance of 731 m in the 2nd second of it’s motion and

experiences a retardation of 4 ms−2 when t = 3. (a) Find the value of p and of q. (b) It is also known that the object moved with a velocity of 6 ms−1 initially at the point A and again at the point B. Find the time taken for the object to move from A to B. (c) Hence if the object stops momentarily at the point C, find the distance between the point B and the point C.


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Answer: (a) p = 8, q = 2 (b) 2 s (c) 313 m

Diagram 2 shows the object P and Q moving in the direction as shown by the arrows when the objects P and Q pass through the points A and B respectively.

Diagram 2

The displacement of the object P from A is represented by sP and the displacement of the object Q from B is represented by sQ. Given sP = t2 + 4t and sQ = t2 - 8t, where t is the time, in seconds , after P and Q pass through point A and B respectively and simultaneously . Given AB = 60 m. (a) Find the time and position where the objects meet. (b) Find the time and position of object Q when it reverses its direction of motion. (c) Find the velocity of object P when object Q reverses its direction of motion.

(d) Sketch the graph of displacement - time for object Q for .100 ≤≤ t (e) Find the time interval when the object Q moves to the left. Answer: a) 5 s, 45 m on the right of A b) 4 s, 16 m on the left of B. c) 12 m s-1 (e) .40 <≤ t

P Q

A B60m

4.


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Permutation and Combination 1 a) 4 Girls 3 Boys Answer : 5! , 3! = 720 b) 6 • 5 • 4 • 3 • 3• 2 • 1 = 2160 Centre 2 9 ! - ( 8! x 2! ) = 282240 3 4 x 3 x 2 x 2 = 48

Centre

4 W A L I D a) 1 x 4 C 2 = 6 b) 1 x 1 x 3 C 1 = 3

5 7 boys, 6 Girls a) 7 C 3 x 6 C5 = 210 b) 6 C 1 x 7 C7 = 6

6 a) ( 101

41)

51

51

=×+

b) 5P3 x 4 P 2 = 720

7 5P3 x 3 P 2 = 360

8 a) 12 C 5 = 792 c) 6 C 1 x 7 C7 X

2 C 1 = 160


Number Solution and marking scheme Sub Marks

Full Marks

1 a 2)(

21

≤≤− xf

GRAF

1 2 1

3

2

a) m = 2 and n = 29

fg(x) = 5

292 +x

2 1

2

3

a = 2, b = 4, c = 8, d = −6 a = 2, b = 4 or c = 8, d = −6

4a + b = 12 and a + b = 6 or 312

12=

− cand d=

− 8612

Either one equation correct

4 3 2 1

3

4

0.9537 , −1.3981 Using formula or other method

0332 =+−+ ppxx

3

2 1

3

5

01124 2 =+− xx

322 =+ βα and 4122 =βα

322 =+ βα or 4122 =βα

2=+ βα or =αβ 1/2

4 3 2 1

4



Full Marks

6

a) ( ) 213 2 +−− x

122

2223

22

−⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛++− xx

b) x = 1

2

1

1

3

7

1,35

>−< xx

3 2

1

3

8 ( )335629 −n

( )323 5555 −− −+n

2

1

2

9

31

=y

332 =y

( ) 24199 =−y

3

2

1

3

10

813+

=mk

381 2 += mk

3 2

1

3

35

− 1

( )( ) 0153 >−+ xx



Full Marks

( )9log

3loglog2

3

33

+=+

mk

1

11 3=x

0322 =−− xx

( )

9log32log

log3

33

+=

xx

3

2

1

3

12 72−

( ) ( )[ ]4111622

1212 −+=S

16=a or 4−=d

3 2

1

3

13

52187

and 729=a32

−=r

Solve simultaneous equation

5672 =++ arara or 168543 −=++ ararar

4 3 2 1

4

14

31

22 13ararara =++

2 1

4

15 n = 1 and k = 16 n = 1 or k = 16 2n = 3 or 4log2 =k kxn 22 loglog2 −

4

3

2

1

4



Full Marks

16 03422 =−−+ yxyx

( )25

232

22 =⎟

⎠⎞

⎜⎝⎛ −+− yx

⎟⎠⎞

⎜⎝⎛=

23,2intmidpo

3 2

1

3

17 7−=p

061 =++p

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛+ 06

31

2 xp

3 2 1

4

18

22 11 baab −−+

20 120cos a−= or 20 130sin b−=

0000 30sin20cos30cos20sin +

3 2 1

3

19

46.7483

( ) ( ) ( )( ) 02 71.52sin772192.012

21

−

( ) ( )92.01221 2 or ( )( ) 071.52sin77

21

3

2

1

3

20

2

( ) ( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−−−

−−

22 1112

1112

21

1

12

1221

−⎥⎦⎤

⎢⎣⎡ −

xx

3 2 1

3



Full Marks

21 35

−=h and 6

11=k

Solve simultaneous equation

22 =+ kh or 27

=+− kh

3

2

1

3

22 Player B

3902 =∑ Ax and 3882 =∑ Bx

8=x

3

2

1

3

23

61

⎟⎠⎞

⎜⎝⎛ ××+⎟

⎠⎞

⎜⎝⎛ ××+⎟

⎠⎞

⎜⎝⎛ ××+⎟

⎠⎞

⎜⎝⎛ ××

51

41

31

51

41

32

51

43

31

54

41

31

Either 2 operation above correct

3

2

1

3

24 a)

61

6

65

51PP×

b) 151

6

64

4 21P

P ××

2

1

2 1

4

25 a) 0.1587

40

15001540 −

b) 1562

40

( −x )1500 = 1.55

2

B1 2

B1

4


Modul Perfect Score Sbp 2009

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