Modul Add Math Perling

7/29/2019 Modul Add Math Perling

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ADDITIONAL MATHEMATICS MODULE, SMK SERI PERLING

1

SMK SERI PERLINGMODUL KECEMERLANGAN

PREPARED BYEN AZRI AZMI BIN ABD AZIZ

PANEL OF ADD MATHS


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QUADRATIC FUNCTIONSInequalitiesExample

Find the range of values ofxfor which 0152

2

xx SolutionMethod 1

01522 xx

Let 1522 xxxf

= 53 xx

When 0

xf

053 xx

3x or 5

For 01522 xx

5x or 3x

Method 2

Using a number line

Check sign ( +ve or ve ) of any region

The signs will be alternate

3 0xf 5

-3 5

+ve -ve

+ve


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Look at the question :

If > : look at the +ve region

If < : look at the ve region

For 0152

2

xx

5x or 3x

SIMULTANEOUS EQUATIONSGuidance Example

1 Arrange the linear equation such that one of

the two unknowns becomes the subject of

the equation.

(avoid fraction if possible)

x + 2y= 1

x=

2 Substitute the new equation from step 1 into

the non-linear equation .

Simplify and express in the form

ax2+ bx + c = 0.

( )2+ 4y

2= 13

= 0

3 Solve the quadratic equation by factorisation,

completing the square or by using the

formula (2y 3)( ) = 0,

y=2

3or

4 Substitute the values of the unknown

obtained in step 3 into the linear equation. Wheny= 2

3,

x= 1 2( ) =

Wheny= ,

x=


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INDICES AND LOGARITHMINDICES

Solve each of the following equations

Examples Exercises

1. 33x = 81

33x

= 34

3x = 4

x =3

4

1. 9x

= 271-x

2. 2x . 4x+1 = 64

2x. 2

2 (x+1)= 2

6

x+ 2x+ 2 = 6

3x = 4

x =3

4

2. 4x. 8

x-1= 4

3. 0168 1 xx

022 143 xx

143 22 xx

3. 5x- 25

x+1= 0


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7

44322

xx

3x= 4x+ 4

x= - 4

4.32

116 x

5

4

2

12

x

54 22 x

4x= -5

4

5x

4.x

x

32

18 1

COORDINATE GEOMETRYDetermine whether two lines are parallel / perpendicular

Examples Solution

1. Determine whether the straight lines

2yx= 5 and x 2y= 3 are parallel.

2yx= 5,

y= 52

1x ,

2

11 m

x 2y= 3

y= 32

1x ,

2

12 m

Since 21 mm , therefore the straight lines 2yx= 5 and x


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2y= 3 are parallel.

2. Determine whether the straight lines

3yx 2 = 0 andy+ 3x+ 4 = 0 are

perpendicular.

3yx 2 = 0

y=32

31 x , 3

11 m

y+ 3x+ 4 = 0

y= 3x 4, 32 m

)3(3

121 mm = -1.

Hence, both straight lines are perpendicular.


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9

Equation of a straight line

The equation of a straight line may be expressed in the following forms:

i) The general form : ax+ by+ c= 0

ii) The gradient form : y = mx + c ; m = gradient , c = y-intercept

iii) The intercept form :a

x+

b

y= 1 , a = x-intercept , b = y-intercept

a) If given the gradient and one point:

1yy = )( 1xxm

Eg. Find the equation of a straight line that

passes through the point (2,-3) and has a

gradient of4

1.

1yy = )( 1xxm

)2(4

1)3( xy

144 xy

E1. Find the equation of a straight line that

passes through the point (5,2) and has a

gradient of -2.

y = -2x + 12

E2. Find the equation of a straight line that

passes through the point (-8,3) and has a

gradient of4

3.

4y = 3x + 36

Gradient = m

P(x1, y1)


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b) If two points are given :Note : You may find the gradient first, then use

either (a) y = mx + c

or (b) yy1 = m( xx1)

or

(c)1

1

xx

yy

=

12

12

xx

yy

Eg. Find the equation of a straight line that

passes through the points (-3, -4) and (-5,6)

)3(

)4(

x

y=

)3(5

)4(6

2

10

3

4

x

y

y+ 4 = -5 ( x+ 3 )

y= -5x- 19

c) The x-intercept and the y-intercept are given:

m= -

errceptx

ercepty

int

int

Equation of straight line is :

a

x+

b

y= 1

Note : Sketch a diagram to help you !

Eg. The x-intercept and the y-intercept of thestraight line PQare 4 and -8 respectively. Find

the gradient and the equation ofPQ.

m PQ =

errceptx

ercepty

int

int

=

4

8

= 2

Equation :4x +

8y = 1

82 xy At the x-axis, y = 0

x

y

O

-

4


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STATISTICS

Finding median using formula

The number of vehicles that pass by a toll plaza from 1 p.m to 2 p.m. for 60

consecutive days is shown in the table below.

Number of vehicles Number of days5950 4

6960 10

7970 24

8980 16

9990 6

Calculate the median of the number of cars using formula.Solution :

Number ofvehicles

Number of days(f)

Cumulativefrequency

5950 4 4

6960 10 (14)7970 (24) 38

8980 16 ( )

9990 6 ( )

Step 1 : Median class is given by = 302

60

2

TTTn

Median lies in this

interval


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Convert measurements in radians to degrees and vice versa.

Arc length of a circle

rs where s= arc of a circle

r= radius of a circle

= angle subtended at the center ( in radian )

Area of a sector

Complete the table below, given the areas and the radii of the sectors and angles subtended.

2

2

1rA , is in radians

DIFFERENTIATION:

o180

Radian Degrees

1 rad =

o180= __________

1o

= rad180

= _________

o180


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9. Given xxy 43 2 , find the value ofdx

dy

when x=2.

10. Given 21)( xxxf , find the value of).1('and)0(' ff

INDEX NUMBER

Index number or price index ,I

1000

1 Q

QI where Q0 = quantity or price at base time

Q1 = quantity or price at specific time

Composite index ,

i

ii

W

WI

I where Ii = index number

Wi = weightage


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QUESTION

Solution

x 2 3 4 5 6

y 2 9 20 35 54

The above table shows the experimental values of two variables, xandy. It

is know thatxandyare related by the equation

y= px2+ qx

a) Draw the line of best fit forx

yagainst x

a) From your graph, find,

i) p

ii) q

STEP 1

y= px2+ qx

x

y=

x

px2+

x

qx

x

y= px + q

Y= mX + c

Table

Non- linear

Reduce the non-linear

The equation is divided throughout by x

To create a constant that is free from x

Linear form


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Note : For teachers reference

STEP 2

x 2 3 4 5 6

y 2 9 20 35 54

x

y

1 3 5 7 9

STEP 3

construct table

Using graph paper,

- Choose a suitable scale so that the graph

drawn is as big as possible.

- Label both axis

- Plot the graph of Y against X and draw


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STEP 4

Gradient , p =26

19

= 2

y- intercept = q= -3

From the graph,

find p and q

Construct a right-angled triangle,

So that two vertices are on the line of

Determine the y-intercept, q


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INTEGRATION

Integration of xn :

1.

3 13

3 1

xx dx c

= 44

xc

Integration of axn :

Note : m dx mx c , m a constant

3 136 6.

3 1

xx dx c

=4

6.4

xc

=43

2

xc

10dx 10x + c

1

, 11

nn axax dx c n

n

1

, 11

nn xx dx c n

n


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1 1

8 8.1 1

xx dx c

= 28.2

xc

= 24x c

3

3

22dx x dx

x

=

3 1

2.3 1

xc

=

2

2.2

xc

=2

1c

x

To Determine Integrals of Algebraic Expressions.

Note : Integrate term by term. Expand & simplify the given expression where necessary.

Example :2(3 4 5)x x dx =

3 23 45

3 2

x xx c

= x32x

2+ 5x + c


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2. Find the magnitude for each of the vectors

Example

3 ~ ~2i j

2 23 2

13 unit

3. Find the magnitude and unit vector for each of the following

Example

~ ~ ~

3 4r i j

Solution :

2 2

~

~ ~ ~

Magnitude, 3 4

= 5

1unit vector, r, (4 3 )

5

r

i j


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TRIGONOMETRIC FUNCTIONS

To sketch the graph of sine or cosine function , students are encouraged to follow the steps below.

1. Determine the angle to be labeled on the x-axis.

eg : Function angle

y= sin x x = 90o

y= cos 2x2x= 90o

x= 45o

y= sin x2

3 x

2

3= 90

o

x= 60o

2. Calculate the values ofyfor each value ofxby using calculator

eg : y= 1 2 cos 2x

x 0 45 90 135 180 225 270 315 360

y -1 1 3 1 -1 1 3 1 -1

. Plot the coordinates and sketch the graph

45 90 135 180 225 270 315 360x

y

3

2

1


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PERMUTATIONS AND COMBINATIONS

1. The number of ways of arranging all the

alphabets in the given word.

Solution:

6! = 6.5.4.3.2.1

= 720

2. The number of ways of arranging four of the

alphabets in the given word so that last alphabet

is S

Solution:

The way to arrange alphabet S = 1

The way to arrange another 3 alphabets=5P

3

The number of arrangement = 1 x5P

3= 60

3. How many ways to choose 5 books from 20

different books

Solution:

The number of ways=20

C5

= 15504

4. In how many ways can committee of 3 men

and 3 women be chosen from a group of 7 men

and 6 women ?

Solution:

The numbers of ways =7

C3

x6C

3

= 700


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Alternative method

Black

Yellow

Black

Yellow

Black

Yellow

10

6

10

4

10

4

10

4

6

10

6

PROBABILITY

Question Answer

1.

The above figure shows six numbered

cards. A card is chosen at random.

Calculate the probability that the

number on the chosen card

(a) is a multiple of 3 and a factor of

12

(b) is a multiple of 3 or a factor of

12.

Let

A represent the event that the number on the chosen card

is a multiple of 3, and

B represent the event that the number on the chosen card

is a factor of 12.

A = {3, 6, 9}, n(A)= 3

B = {2, 3, 4, 6}, n(B) = 4

A B = {3, 6}

A B = {2, 3, 4, 6, 9}

(a) P(A B) =3

1

6

2 .

(b) P(A B) =6

5

P(A B) = P(A) + P(B) P(A B)

=6

2

6

4

6

3

=6

5.

2. A box contains 5 red balls, 3 yellow

balls and 4 green balls. A ball is

chosen at random from the box.

Calculate the probability that the balls

drawn neither a yellow nor a green.

P (yellow) =3

12.

P(green) = 412

P(yellow or green) =3

12+

4

12=

7

12.

3. Box C contains 4 black marbles

and 6 yellow marbles. A marbles is

chosen at random from box C, its

2 3 4 6 8 9

10

4


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(c) P(Z - 0.23)

Solution

(c) P(Z - 0.23) = 1 P(Z < - 0.23)= 1 P(Z > 0.23)

= 1 0.40905

= 0.59095

(d) P(Z > - 1.512)

Solution

(d) P(Z < - 1.512) = P(Z > 1.512)

= 0.06527

-1.512 1.512

-0.230.23


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(e) P(0.4 < Z < 1.2)

Solution(e) P(0.4 < Z < 1.2) = P(Z > 0.4) P(Z > 1.2)

= 0.3446 0.1151

= 0.2295

(f) P(- 0.828 < Z - 0. 555)

Solution

(f) P(- 0.828 < Z - 0. 555) = P(Z > 0.555) P(Z > 0.828)

= 0.28945 0.20384

= 0.08561

(g) P(- 0.255 Z < 0.13)

Solution

(g) P(- 0.255 Z < 0.13) = 1 P(Z < - 0.255) P(Z > 0.13)

= 1 P(Z > 0.255) P(Z > 0.13)

= 1 0.39936 0.44828

= 0.15236

0.4 1.2

0.4

1.2

-0.828 -0.555 0.555 0.828


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SummaryNormal Distribution

Type 1P( Z > positive no)P ( Z > 1.2 ) = 0.1151

.....................................................Type 2

P(Z < negative no)

P ( Z < - 0.8 ) = P (Z > 0.8)= 0.2119

.....................................................Type 3

P ( Z < positive no)

P ( Z < 1.3 )= 1 P ( Z>1.3)= 1 0.0968= 0.9032

Type 6

P (Negative no < Z < Negative no )

P ( -1.5 < Z < - 0.8 )

= P ( 0.8 < Z < 1.5 )

= P ( Z > 0.8 ) P ( Z > 1.5 )

= 0.2119 0.0668 = 0.1451

......................................................

Type 7

P ( negative no < Z < postive no )

P ( -1.2 < Z < 0.8 )

Type 1P ( Z > K ) = less than 0.5

P ( Z > K ) = 0.2743

K = 0.6

......................................................Type 2

P ( Z < K ) = less than 0.5

P( Z < K ) = 0.3446

P ( Z > - K ) = 0.3446- K = 0.4

K = - 0.4

-0.255 0.13

0.13-0.255


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......................................................Type 4.

P( Z > negative no)

P ( Z > - 1.4 )= 1 P ( Z < -1.4 )= 1 0.0808= 0.9192

....................................................Type 5

P( positive no < Z < positive no)

P ( 1 < Z < 2 )= P ( Z > 1 ) P ( Z > 2 )= 0.1587 0. 0228= 0.1359

= 1 P ( Z > 0.8) P ( z < -1.2 )

= 1 P ( Z > 0.8 ) P ( Z > 1.2 )

= 1 0.2119 0.1151

=0.673

.......................................................Type 3

P( Z < K ) = more than 0.5P ( Z < K ) = 0.8849P ( Z > K ) = 1 0.8849

= 0.1151K = 1.2

......................................................Type 4

P ( Z > K ) = more than 0.5

P ( Z > K ) = 0.7580P( Z < K ) = 1 0.7580 = 0.2420P ( Z > -k ) = 0.2420

- K= 0.7K = - 0.7


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LINEAR PROGRAMMING

Problem interpretation and the formation of the relevant equations or inequalities

The table below shows the mathematical expressions for the different inequalities used.

Mathematical Expressions Inequality

a y greater than x xy

b y less than x xy

c y not more than x xy

d y not less than x xy

e The sum ofx and y is not more than k kyx

f y is at least k times the value ofx kxy

g yexceeds xat least k y x k

Example:

A company delivers 900 parcels usingx lorries and y vans. Each lorry carries 150 parcels while each van

carries 60 parcels. The cost of transporting the parcels using a lorry is RM 60 while that of a van is RM 40 .

The total cost spent on transportation is not more than RM 480.

(a) Write down two inequalities other than x0 andy0 , that satisfy all of the above conditions.

Solution

(a)3025

90060150

yx

yx..

2423

4804060

yx

yx.

The two inequalities that satisfy the given conditions are :

3025 yx and 2423 yx

1

2


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SLOT 1


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CHAPTER 1: FUNCTIONSECTION A

1. The function fis defined as 4,4

3)(

x

xxf . Find

i) )2(f

ii) 1)( xf

iii) xxf )(

2. Given that the function 47: xxg . Find the value ofpif 25)2( pg .

3. The functiongis defined as kxxxg ,12

3)( . Find the value of

i) k.

ii)

2

3g

4. Figure below shows an arrowed diagram that represent qpxxf 2)( .

x f(x)

1 1

-2 -5

Find the value ofpand ofq.


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SECTION B

1. Given that f : x12

3

x, x - and g: x 3x+1 . Find fg(x).

[3 marks]

2. Given that the function f(x) =2 xand the function fg(x) =13x2. Find the function g.[3 marks]

3. The functions fandgare defined as f: x 3x+ 4 and g: x x2+ 6 respectively. Find

the values ofxfor which fg=gf [4 marks]

4. Given thath: x 4/x, x 0 , and hg : x 8x, find

(a) g(x)

(b) the value ofxfor whichgh(x) = 6 [3 marks ]

SECTION CInverse Functions.

1. Function fdefined as2

1,

12

3:

x

x

xxf , find ).(

1 xf

2. Given that 3,3

2:

x

x

xxf , find the value of ).2(

1f

3. Function fandgare defined by 93: xxf and 32: xxg , find .)(1gf

4. Given baxxf : and2

1:1

xxf find the value ofaand b.

SECTION D


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1. Given the f : x 3x 1 and g : x x + 7 . Find

(a) the image of 4 under f

(b)the value of f-1 g (3)(c) f

-1g

-1(x)

2. The function f is defined by f : x 3x + 4 and function g is defined g : x x2 + 6.Find the value of

(a) fg(3)

(b)gf-1(1)

3. Functions f and g are defined by f : x 2x + 3 and g : x 8

13

x

x, x 8. Find the

value of gf-1

(5)

4. A function f is defined by f : x 1

23

x

x, x 1. Find the value of

(a) ff-1(2)

(b)f-1

(4)


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k =2

1

ii)1)

2

3(2

3)(

xg = 4

3

4 1 qp , q =1 p (1)

4p + q = 5 (2)

substitute (1) into (2)

4p +(1 p)= 5

3p= 6

p = 2 , q= 3

SECTION B:1. f(x) =

12

3

xg(x) = 3x + 1

fg(x) = f (3x + 1)

= 1)13(2

3

x

=36

3

x


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= 2/1,12

1

x

x

2. f(x) = 2 - x

fg(x) = 1 3x2

2 g(x) = 1 3x2

1 + 3x2= g(x)

3. fg = gf

3( x2+ 6) + 4 = (3x + 4)

2+ 6

3x2 +18 + 4 = 9x2 + 24x + 16 + 6

6x2+ 24x = 0

6x( x + 4) = 0

x = 0, -4

4. h(x) = 4 hg(x) = 8x

X

(a) h [g(x) ] = 8x

4 = 8x

g(x)

g(x) = 4

8x

SECTION C1. Let yxf ).(1

So xyf ).(

(b) gh(x) = 6

6

)4

(8

4

x

x = 48


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2

1,

21

3)(,

21

3

3)21(32

23

)12(3

12

3

1

xx

xxfTherefore

x

xy

xxyxxyy

xxyy

yxy

xy

y

2. Let )2(1 fy

Then 2).( yf

8,

8

262

622

23

2

1

fThus

y

yy

yy

y

y

3. )93()( xgxgf

156

3)93(2

x

x

Let 156 xy

6

15

yx

6

15)()(, 1

xxgfThus

4. baxxf :

yxLetf :1


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a

bxy

bayx

yfx

)(

Hencea

bxxf )(1

a

bxx

2

1

Thus a = 2 and b = 1

SECTION D:1. (a) 11

(b)3

11

(c)3

6x

2. (a) 49

(b) 7

3. -

3

14

4. (a) -12

(b)7

6


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SLOT 2


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47

Find,

(a) the value of p

(b) the equation of the axis of symmetry

(c) the value of n and k

18. Given that qpxaxx 22 )(563 , find the values of

a) a b) p c) q

19. The quadratic function 12)5(2)(2 xxf has a minimum value at point A.

Find the coordinates of point A.

20. Diagram 1 shows the graph of the quadratic function2128)( xxxf .

Find the values of a and b.

Answer:

1. x2 x -6 = 0

2.. x = -7 ,23

3.. x =2

3 , 3

4.. p = 3 , q = -15

5. -5

6. )3

2,

2

1(

7.

8

25k

8. 5m = 2n

9. 3 , - 5

10. 0.553 , 1.447

11. -1 , 4

)(xf

x


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12. k < -3 , k > 5

13. 4 ,7

4

14. p > 4

15

15. min value = 7 , x =3

2

16. h = - 4 , k = 3

17. a) 11 b) x = -3 c) n = 3 , k = 2

18. a = 2 ; p = 1 ; q = 2

19. (-5,13)

20. a = 2 ; b = 6


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2 3( 2)x y 1.4

y

x

2 ( 2) 19 5 kyx k y yx

1, 3 2A y B x

x

4 .C y

CHAPTER 4: SIMULTANEOUS EQUATIONS

1. Solve the simultaneous equations 2x+y= 6 and x2+ xy= 8.

2. Solve the simultaneous equations x2+y= 2x+y= 7

3. Solve the simultaneous equations 5x 3y = x2 2xy = 21

4. Solve the simultaneous equations

5. Solve the simultaneous equations

6.

.

Give your answers correct to 4 significant figures.

7. Given that (h, 2) is a solution for the simultaneous equation

wherebyhand kare constants. Determine the values ofhand k.

8. Given that a curve and a straight line

Without drawing their graphs, calculate their points of intersection.

9. Given that and

Find the values ofxandysuch that 2A = 3B = 2C.

.

1y

xx

2 6x y and

22x y and ( ) 7.y y x

3 5 1x y 3

2 2yx

and


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(2 ,3 )p q

22 8 2 4xy y x y

10. Given that is a solution for the simultaneous equation

wherebypand qare constants. Determine the values ofpand q.


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2

2

2

2

(2)5 5 2

3

2

( 2) 19 5

( ) ( 2)(2) 19 5

3( ) ( 2)(2) 19 5

2

3 28 0

( 7)( 4) 0

7 4

3( 7) 3(4)

2 221

2

ky ky

x h

hk

x k y

h k

hh

h h

h h

h or h

k k

6

2 2

3 2 3(3 2 ) 2(4 )

2 1

6

12 2 2( ) 2(4 )

12 2(4 )

2 1

6

4 6 2 0 2 3 1 0

(2 1)( 1) 0

11

2

12( ) 1

2(1) 12 6 6

B C x y

yx

A C y yx

y yy

y y y y

y y

y or y

x x

1 1

3 2

2

2

2

1 44

3( 2)

3 ( 4) 2

3 18 0

( 6)( 3) 0

6 3

6 4 3 4

10 1

(6,10) ( 3, 1)

yy x

x

x y

x x

x x

x x

x or x

y y

and

2

2

2

2 4 2 2(3 ) 4

3 2

2 8 4

2(2 )(3 ) 8(3 ) 4

12 (3 2) 72 4

29 6 1 0

(3 1)(3 1) 0

1

3

13( ) 2 3

3

x y p q

p q

xy y

p q q

q q q

q q

q q

q

p

7. 9.

8. 10.


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INDICES AND LOGARITHMS

1. Given that (3x) (4

x) = 144, find the value of x.

2. If 2x= y, express 128(4

x 3) in terms of y.

3. Solve the equation 92x + 1

27x = 0.

4. Solve the equation 9x + 1

= 10(3x).

5. Solve 4x + 3

= 1.

6. Evaluate each of the following.

(a) log3 0.125 (b) log43.8 (c) log2 4.187

7. Given that log7 2 = 0.3562, log7 3 = 0.5646 and log7 5 = 0.8271, find the value of each of

the following.

(a) log2 42 (b) log14 30

8. Given that log5 p = q, express each of the following in terms of q.(a) logp 625 (b) logp 5

9. Given log3 p = m, express the following in terms of m.

(a) log3 9p (b) logp 81

10. Given logm 3 = x and logm 2 = y, express the following in terms of x and or y.

(a) log6 m2

(b) log8 m

11. Given 3 log3

x = 2log3

y. Express x in terms of y.

12. Given logm 2 = x and logm 5 = y. Express logm 12.5m in terms of x and y.

13. Solve 4x5log = 64.


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60

COORDINATE GEOMETRY1. The distance between two pointsA(2, 7) and B(p, -1) is 10 units

2. Find the posiibles

values ofp.

2. A pointTdivides the lines segment joining points P(5, -2) and Q(-5, 13) internally inthe ratio 2: 3. Find the coordinate of pointT.

3. The coordinates of the triangleABCareA(m, 0), B(-3, 5) and C(3, 2). Given that the

area of the triangleABCis 24 units2. Find the possible values ofm.

4. Diagram 1 shows a straight lineJKL. PointsJand Llie on they-axis and x-axis

respectively. A coordnate ofKis (5, a).

Find the value of a.

5. Diagram 2 shows a quadrilateral PQRS.

Given the coordinates ofP, Qand Rare (-3, -1), (1, 4) and (7, 3) respectively. If the

midpoint of the lines PRand QSmeet atM. Find

(a) the coordinatesofMand S,

y

K(5, a)

L

8J

x

DIAGRAM 1

S

R

Q

P

y

x

DIAGRAM 2


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13. Find the equation of the line passing thriugh the x-intercept3

10and parallel to the

line 146

yx

.

14.

Find the equation of the line parallel to 1312

yx

and its y-intercept is -2. If (p, 4 p)lies on this line, find the value ofp.

15. The liney= ax+ bis perpendicular to the line 3x+ 2y= 18 and passing through the

point (15, 4). Find the value ofaand ofb.

16. Find the equation of the locus of point P such that the ratio of its distances from points

A(-3, -4) and B(7, 4) is 2 : 3.


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6. 11, 13, 13, 14, 15, 16, kis a set of 7 positive integers arranged in the ascending order.

Given that the difference between the mean and the first quartile is 1.

Find the value ofk. [2 marks]

7. Table 2 shows the distribution of marks scored in the February monthly test.

Find the range of the marks. [2 marks]

TABLE 2

8. Table 3 shows the points obtained by survey done with a group of fresh

graduates. Find the range of value ofxif the median point is 4. [2 marks]

TABLE 3

9 . If the mean ofx- 2, x+ 4, 2x+ 1, 2x- 1, x+ 5 and x- 1 is 5. Find the value xand

hence determine the variance. [4 marks]

10. Table 4 shows a set of points scored during a contest. They are arranged in

the ascending order and ris a positive integer. Find for each value of the r, the

possible values of the mode. [2 marks]

TABLE 4


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5. Diagram 5 shows a quadrant of a circle OABC.

. Given the area of the sector OAB is 36 cm2, find,

a) the length of arc AB,

b) the angle in degrees.

6. Diagram 6 shows a sector with centre O and radius r.

Given that the length of the arc PQR is 46.5 cm, find the value of r.

7. Diagram 7 shows a circle with centre O.

O

B

A

C

12 cm

DIAGRAM 5

DIAGRAM 6


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Given that the radius is 7.2 cm and the length of the major arc AB is 35 cm, find the value of

.

8. Diagram shows a circular sector OPRQ with centre O.

Given

that the perimeter of the sector OPRQ is 50 cm, find,

a) the value of in radians,

b) the area of the shaded region.

DIAGRAM 7

20 cm

O

P

QR

DIAGRAM 8


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Given that OE = 6 cm, EF = 2 cm and EOH = 1.2 radian. Find the area of the

shaded region.

14. Diagram 14 shows OAC and OBD are two concentric sectors with centre O.

DIAGRAM 14

DIAGRAM 13


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Given that the lengths of the arc AC, the arc BD and AB are 11.2 cm, 18.2 cm and 5 cm

respectively, calculate

(a) the value of 0 in radians,

(b) the length of OA,

(c) the area of the shaded region.

15. Diagram 15 shows a sector OAC with centre O.

Given that the radius is 10 cm. Find,

(a) (i) the perimeter of the segment ABC,

(ii) the area of the shaded region.

(b) Express the area of the shaded region as a

percentage of the area of the sector OAC.

DIAGRAM 15


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9. Given thatf(x) = (x2 5x)(x+ 3). Find the possible values ofxiff(x) = 0.

10. Given thatf(x) = 2x3x2 6x. Find f(x).

x(x 4)

11. Giveny= (2t 3)2 and x= t2 + 2. Find dyin terms oft.

dx

12. Find the equation of the tangent to the curvey= 2x2x+ 3 that is parallel to the

straight liney= 3x 4.

13. Find the gradient of the normal to the curve y = (x+ 3)2atx= 2.

x

14. Given f(x) =px3+ 2x

2 5x+ 3 has turning point atx= -1. Find the value ofp.

15. Find the coordinates of the turning points for the curve y = 4x + 1/x.

16. Given that 3

3

4rV , find

dr

dV.

17. Differentiate74 )31( xx with respect to x.

18. Find

xdx

d

2

9.

19. Given that 1442 rh , find the value ofdh

drwhen 4h .

20. Differentiate12

5

x

xwith respect to x.

21. Given that kky 23 and 25 kx , finddx

dyin terms ofx.


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Answer:

1.) 7

2.) 0

3.) 24.) 25.) 56.) (a) 4x

5

(b) 1 + 10

x3

(c) 2x 2

7.) 3/2

8.) 6V 18

V3

9.) -5/3, 3

10.) -7/(x 2)311.) (4t 6)/t12.) y= 3x+ 1

13.) 4/5

14.) 3

15.) (1/2, 1), (-1/2, -7)

16.)2

4 r

17.) )334()31(63 xxx

18.)2)2(

9

x

19.)2

9

20.)2)12(

5

x

21.) )76(25

1x

22.)3

1

23.) 1/ 75


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3. In the Diagram 3, CDEFis a square and EFGis a straight line.

DIAGRAM 3

Calculate

a) FCH,

b) the length ofFG,

c) the area of quadrilateral CFGH.

4. In the Diagram 4, PQRis a straight line and QRSis an isosceles triangle.

DIAGRAM 4

Find

a) the length ofSQ,

b) SPQ,

c) the area of triangle SPR.

5. In the Diagram 5, LMNis a straight line.


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50Sin

12

50Sin

12

482.10

PS

2

1

885.6

35Sin

14

416.91Sin

416.91Sin

14

2

1

ANSWER1.

a) P= 180o 50o 42o = 88o

QR= Sin 88o= 15.655 cm

b) PQ= Sin 42o= 10.482 cm

Sin 50o=

PS= 10.482 x 0.7660 = 8.030 cm

c) The area ofPQR

= 10.482) (15.655) sin 50o

= 62.852 cm2

2.

a)

LM2= KL

2+ KM

2 2 (KL) (KM) cos MKL

= 122

+ 102

2 (12) (10) cos 35o

= 47.404LM= 6.885 cm

Sin KML = 12

KML= 88.584o

b) KMN= 180o 88.584o = 91.416o

Sin KNM= 10

KNM= 45.567

o

MKN= 180

o 45.567o 91.416o

= 43.017o

MN= Sin 43.017o

= 9.554 cm

c) The area of whole diagram

= (12) (14) sin (35o+ 43.017

o)

= 82.170 cm2

P

Q R

42o

50o

12 cm

S


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))((2

222

MNOM

ONMNOM

)16)(13(291613 222

10

46Sin

2

1

2

1

86Sin

12

))((2

222

PNNO

POPNNO

)374.11)(10(2

4374.1110 222

2

1

5.

a) Cos

OMN=

= = 0.8269

OMN= 34.216o

Sin LKM= 4

LKM= 16.722o

KML= 180o 46o 16.722o

= 117.278o

KMO= 180o

34.216o

117.278o

= 28.506

o

b) The area of the whole diagram

= Area ofKLM+ Area ofOMN

= (10) (4) sin 117.278o+

(13) (16) sin 34.216o

= 17.776 + 58.481

= 76.257 cm2

6.

a) MPN= 180o 71o 23o = 86o

NP= Sin 71o

= 11.374 cm

Cos PNO=

= = 0.938

PNO= 20.289o

b) MNO= 20.289o+ 23

o

= 43.289o

MO2= NO

2+ MN

2 2 (NO) (MN) cos MNO

= 102+ 12

2 2 (10) (12) cos 43.289o

= 69.303

MO= 8.325 cm

c) The area ofMNO

= (10) (12) sin 43.289

o

= 41.141 cm2


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SLOT 10


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b) If the price index from year 2002 to year 2004, for item Pincreases 10%, item Q

increases 20%, item Rdecreases 10% and item Ndecreases 5%, find the price of

each item in 2004.


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100

1200

p

I

13

1384

10

1032

I

3. a)

b) Q0 = Price of product in 2000

Q1 = Price of product in 2002

100 = 105

p= RM1 260

c)

Compound index number,

= 106 (to the nearest integer)

4. a)

x+ 2y= 117

2x+ 4y= 234 (1)

2x 3y= 88 (2)

(1) (2),

7y= 322

y= 46

2x 3 (46) = 88

x= 25

b)



75

100

Ii

92

138 3

2

1

4

wi

iw = 10

414

150

100

368

Iiwi

iiwI = 1 032

75

100

92

138

Price Index

2004

(2003 = 100)

K

L

M

Item

N

114

120

Ii

94

105 4

2

3

4

wi

iw = 13

420

228

360

376

Iiwi

ii

wI= 1 384

2

3

4

4

Weightage

R

S

T

U

Item


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101

x

x

9

1101090

50.2

1Q

14

1633

5.

a)

= 120

x= 1

b) Q0 = Price of product in 1996

Q1 = Price of product in 2001

For fruits,

100 = 120

Q1 = RM3.00 per 500 g

c)

6.a)

b) In year 2000,



In year 1998,

I

143

94

Ii

124

116 5

3

4

2

wi

iw = 14

580

429

376

248

Iiw

i

ii

wI= 1 633

143

94

124

116

Price Index

2000

(1996 = 100)

Transportation

Food

Clothes

Expenses

Rent

110

120

Ii

115

130 3

x

2

4

wi

iw = 9 +x

390

110x

240

460

Iiwi

iiwI = 1 090 + 110x


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SLOT 11

(EXAMINATION)PAPER 1

2 HOURS


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