Intervensi Add Math Bpk

7/30/2019 Intervensi Add Math Bpk

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BAHAN PROGRAM INTERVENSI

PPSMI UNTUK MURID

TINGKATAN LIMA

ADDITIONAL MATHEMATICS

2012


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MINIMUM SYLLABUS REQUIREMENT

1. FUNCTIONS

Determine domain , codomain , object, image and range of relation.

1. Diagram 1 shows the relation between setPand set Q.

a. State the following:

i. Domain =

ii. Codomain =

iii. Objects =

iv. Images =

v. Range =...

vi. Object of 9 = ....

vii. Image of 2 =

b. Represent the above relation using

i. a set of ordered Pairs

ii. a Cartesian graph

3

2

-2

5

4

3

9

SetP Set QDiagram 1

-3 1

1


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Classifying the types of relations

State the type of the following relations

a)

..

b )

.

c)

..

d)

..

e) { ( 3 , 1 ) , (9 , 1 ) , (12 , 2 ) , (15 , 2) }.

..

f )

4

16

36

x

Type of number

3

24

9-3

x x

x

2

2

x

4

6

x

3

24

-29

-3

2

9

Prime

-3Even

Odd

1 2 3 40

1

2

3

4

5

SetP

Set Q


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Inverse function

1. Givenf(x) = 2x + 5 , findf-

(x)2. Giveng(x) = 2

3

x, findg

-1(x)

3. Given that ,4

4)( xxf findf-1 (2) 4. Giveng(x) = x43 , find i. g

-1(x)

ii. g-1

(4)

5. Givenf(x) = 3 - 2x , findf-

(x)6. Given that ,

52)(

xxg findg(3)


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7. Given thatf(x) = 2 - 3x , findf-

(1)8. Given that ,

2

3)(

xxf findf(1)

9. Given that ,3

12)(1

xxg findg(x)

10. Given thatf-

(x) = 2 - 3x , findf(1)


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2. QUADRATIC EQUATIONS

To express a given quadratic equation in general form ax2

+ bx + c = 0 and stating thevalue ofa , b and c.

Example 1

x2

= 5x9

x25x + 9 = 0

Compare with the general formax

2+ bxc = 0

Thus, a = 1, b = -5 and c = 9

Example 2

4x =x

xx 22

4x(x) =x22x

4x2

-x22x = 0

3x22x = 0

Compare with the general formThus, a = 3, b = - 2 and c = 0

Exercisesi. Express the following equation in general form and state the values ofa, b and c.

ii. Use the values ofa, b and c in the formulaa

acbbx

2

42

1. 3x =x2

52. (2x + 5) =

x

7

3. x(x + 4 ) = 3 4. (x1)(x + 2) = 3


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5.x

4=

x

x

5

3

6. x +px = 2x - 6

7. px (2x) =x4m 8. (2x1)(x + 4) = k(x1) + 3

9. (72x + 3x2) =

3

1x

10.7x1 = x

xx 22


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Forming quadratic equations from given roots

Example 1

3 , 2

x = 3 , x = 2

x - 3 = 0 ,x-2 = 0

(x3 )(x2 ) = 0

x

2

5x + 6 = 0Example 21, - 3

x = 1 , x = -3x1 = 0 , x + 3 = 0

(x1 ) (x + 3 ) = 0

x2

+ 2x3 = 0

a) 4 , - 7

b) - 6 , - 2

c) 2 ,3

1

d)

5

1,

3

2

e)3

1,

2

1

f) 4 , 0


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Finding Sum of Roots ( SOR ) and Product of Roots ( POR )

If and are roots of a quadratic equation ax2

+ bx + c = 0 ,

then sum of roots = + =

a

b

and product of roots = =a

c

No Quadratic EquationSum of Roots

( + )

Product of Roots

( )

1. x 3x + 2 = 03 2

2. x + 5x + 3 = 0- 5 3

3. x + 2x -6 = 0

4. x 7x - 8 = 0

5. x + 2x - 5 = 0

6. 2x 6x + 7 = 0

7. 3x 7x - 9 = 0

8. 2x + 4 - 3x = 0

9. 3x + 3x + 2 = 0

10. 4x 3x + 2 = 0

11. 2x 3x - 2 = 0

12. 3x + 9x + 2 = 0

13. 4x 3x - 2 = 0

14. 3x + 9x + 2 = 0

15 2x + 5x - 2 = 0

16. 3x - 12x - 2 = 0

17. 2x - 9x - 2 = 0


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3. QUADRATIC FUNCTIONS

Inequalities

Example

Find the range of values ofx for which 01522 xx

SolutionMethod 1

01522 xx

Let 1522 xxxf = 53 xx

When 0xf 053 xx

3x or 5

For 01522 xx 5x or 3x

Method 2

Using a number line

Check sign ( +ve orve ) of any region

The signs will be alternate

Look at the question :If > : look at the +ve region

If < : look at theve region

For 01522 xx 5x or 3x

3 0xf 5

-3 5

+ve -ve +ve


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Exercises

A. Factorisation

B. Solve the following inequalities

a) 3x2

x < 0 b) x2

7x + 10 > 0

c) 2x2 + 5x7 0 d) x22x 8

e) 2x2 +x > 3 f) x (x1 ) > 12

g) 2 (x22 ) < 7x h) 3x25x + 4 > 3x2


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4. SIMULTANEOUS EQUATIONS

Guidance Example

1 Arrange the linear equation such thatone 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

+ 4y2

= 13

= 0

3 Solve the quadratic equation byfactorisation, completing the square or

by using the formula

(2y3)( ) = 0,

y =23 or

4 Substitute the values of the unknown

obtained in step 3 into the linearequation.

Wheny =2

3,

x = 12( ) =

Wheny = ,

x =

Exercises

1. Expand the following expression

1 (x + 3 )2 3 3x x

2 6 9x x

2 (x + 6 )

2 12 36x x

3 (x 5 )2

2 10 25x x

4 (x 7 )2

2 14 49x x

5 ( 3x + 4 )

29 24 16x x

6 ( 6 x )2

236 12x x


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7 ( 4x 5 )2

216 40 25x x

8 ( 3 5x )2

29 30 25x x

9 ( 2x + 5 )

24 20 25x x

10 2 ( 2x 5)2

28 40 50x x

11 3 ( 5 4x )2

275 120 48x x

12 5 ( 2 3x )2

220 60 45x x

13 5 ( 3 4x )2

245 120 80x x

14 22 3

42

x

24 12 9x x

15 2 ( 3x 5 )2

218 60 50x x

16 21 2

33

x

21 4 4

3

x x


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17 3 ( 8 x )2

2192 48 3x x

18 21 3

22

x

21 6 9

2

x x

2. Factorise the following:

1 x + 3x = 2 2x + 10x =

3 2 12 36x x = 42 10 25x x =

5 x + 14x + 49 = 6 x - 16x + 64

7 x + 8x + 15 = 8 x + 7x + 12=

9 x - 9x + 20 = 10 x - 11x + 28 =


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11 x + 2x - 15 = 12 x + 5x -14 =

13 x - 6x - 16 = 14 x - 5x - 24 =

15 2x + 17x + 21 = 16 3x - 14x + 8 =

17 2x + 9x -35 = 18 2x - 7x - 30 =

19. 6x - 19x + 10 = 20 5x + 13x + 6 =


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3. Solve the following simultaneous equations

1. x +y = 6 andx +y = 20

Ans:x=2,y=4x=4,y=2

2. 4x +y = -8 andx +xy = 2

Ans :x=-2,y=0

x=-3,y=4

3. 2p + q = 3 and 4p + 3q = 13

Ans:

2

1,

4

7

2,2

1

qp

qp

4. 2x - 3y = 4 and x xy + y

= 16

Ans:

4,8

7

12,

7

4

yx

yx


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5. 1032and532

yxyx

Ans:

3,2

1

3

2,4

yx

yx

6. 2y +x =y2 +x25 = 5

Ans :x=3,y=1x=-1,y=3


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5. INDICES AND LOGARITHM

INDICES

Solve each of the following equations

Examples Exercises1. 33x = 81

33x

= 34

3x = 4

x =3

4

1. 9x

= 27-x

2. 2x . 4x+1 = 64

2x. 22 (x+1) = 26x + 2x + 2 = 6

3x = 4

x =3

4

2. 4x

. 8x -

= 4

3. 0168 1 xx

022 143 xx 143 22 xx

44322 xx

3x = 4x + 4

x = - 4

3. 5x

- 25x

= 0

4.32

116 x

54

2

1

2

x

54 22 x

4x = -5

4

5x

4.x

x

32

18 1


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6. COORDINATE GEOMETRY

Determine whether two lines are parallel / perpendicular

Examples Solution

1. Determine whether the straight lines2yx = 5 andx2y = 3 are parallel.

2yx = 5,

y = 52

1x ,

2

11 m

x2y = 3

y = 32

1x ,

2

12 m

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

andx2y = 3 are parallel.

2. Determine whether the straight lines3yx2 = 0 andy + 3x + 4 = 0 are

perpendicular.

3yx2 = 0

y =3

2

3

1x ,

3

11 m

y + 3x + 4 = 0

y=3x4, 32 m

)3(3

121 mm = -1.

Hence, both straight lines are perpendicular.

Exercises Solution

3. Determine whether the straight lines

y3x + 5 = 0 and 2y - 5x + 4= 0 are

parallel.

4. Determine whether the straight lines

3yx + 8 = 0 andy + 3x - 2 = 0 areperpendicular.


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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 thatpasses 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 agradient of -2.

y = -2x + 12


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

gradient of4

3.

4y = 3x + 36

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

Gradient = m

P(x1, y1)


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passes through the points (2, -1) and (3,0)

y = x - 3


passes through the points (-4,3) and (2,-5)

4x + 3y +7 = 0

c) The x-intercept and the y-intercept aregiven:

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

the straight linePQ are 4 and -8respectively. Find the gradient and the

equation ofPQ.

m PQ =

errceptx

ercepty

int

int

=

4

8

= 2

Equation :4

x+

8

y= 1

82 xy

E2. The x-intercept and the y-intercept ofthe straight line PQ are -6 and 3

respectively. Find the gradient and theequation ofPQ.

2y = x+6

E3. Thex-intercept of a straight lineAB is -5 and its gradient is -3. Find they-intercept

of the straight lineAB and the equation of

AB.

3x + 5y +15 = 0

At the x-axis, y =0At the y-axis, x =0

x

y

O

-

4


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7. 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 ( )

Step1 : Median class is given by =30

260

2

TTTn

Therefore, the median class is 7970

Step 2 : Median = cf

Fn

Lm

2

= (___)

24

142

60

( __ )

= 76.17

L = lower boundary of the medianclass = 69.5

n = 60f F = cumulative frequency before the

median class =14fm = frequency of the median class

=24c = size of the median class

= upper boundary lower

boundary

Median lies in this

interval


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To estimate the mode using a histogram

Modal class = 7970

(c)

Class boundary Number of days(frequency)

49.5 59.5 4

59.569.5 10

69.579.5 24

79.589.5 16

89.599.5 6

(c) The histogram is shown below


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49.5 59.5 69.5 79.5 89.5 99.5 Number of vehicles

Estimated mode = 76

Frequency

5

10

15

20

25


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** what happen to measures of central tendency / measures of dispersion when + , - ,

x

Exercises:

1. The table below shows the heights of 40 students in cm.

Height( cm ) 150154 155159 160164 165169 170174 175 - 179

Frequency 4 9 12 8 5 2

a. Find the medianb. Construct a histogram and hence, find the estimated mode

2. The table below shows the marks obtained by 50 students in a test.

Mark 10-19 20-29 30-39 40-49 50-59 60-69 70-79

Number of students 3 6 13 10 7 7 4

a. Find the medianb. Construct a histogram , and hence find the estimated mode


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8. CIRCULAR MEASURE

Convert measurements in radians to degrees and vice versa.

Convert the following angles in radians to degrees and minutes.

a. 1.5 rad b. 0.63 rad

c. rad2

d. rad

2

3

Convert the following angles to radians.

a. 50 b. 124.3

c. 72 35 d. 285 21

o180

Radian Degrees

1 rad =

o180 = __________

1o= rad

180

= _________

o180


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Arc length of a circle

rs where s = arc of a circler= radius of a circle

= angle subtended at the center ( in radian )

Find the length of arc.

1. 2.

Complete the table below by finding the values of, r or s.

r s

1. 1.5 rad 9 cm

2. 14 cm 30 cm

3. 2.333 rad 35 cm

0.5 rad

8 c m

Q

P

O

152

6.4 cmO

BA


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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

Area of sector Radius Angle subtended

1. 38.12 cm 50

2. 90 cm 9.15 cm

3. 72 cm =1.64 rad

4. 18 cm2 6.5 cm

5. 200 cm 1.778 rad

6. 145 cm 8 cm


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9. DIFFERENTIATION:

1. y = 10

dx

dy=

2. y =5

x

dx

dy=

3. f(x) = -2 3x

f (x)=4. y =

x

7

dx

dy=

5.33

1)(

xxf

f (x)=

6. xxy 24

dx

dy=

7.

x

xx

dx

d5

12

2

2

dx

dy

xxy )23(

9. Given xxy 43 2 , find the value of

dx

dywhenx =2.

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

Always change

a fractional

function to the

negative index

before finding

differentiation

8.


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11. INDEX NUMBER

Index number or price index ,I

100

0

1

Q

QI where Q0 = quantity or price at base time

Q1 = quantity or price at specific time

Composite index ,

i

ii

W

WII where Ii = index number

Wi = weightage

The table shows the price of 3 types of goods: A, B and C in the year 2005 and 2006.

Types of good

Price Price index in 2006

(Base year = 2005)2005 2006

A RM 1.20 RM 1.60 z

B x RM 2.30 110

C RM 0.60 y 102

Find the value ofx,y and z

Calculate the composite index for each of the following data

Index number, I 120 110 105

Weightage, W 3 4 3


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1. PROGRESSIONS

Finding the nth term of an AP and a GP

Arithmetic Progression Geometric Progression

Tn = a + (n1 ) d Tn = ar

n - 1

1. Find the 9th term of the arithmetic

progression.2, 5 , 8 , ..

Solution:

a = 2d = 5-2=3

9 2 (9 1)3T

= _______

2. Find the 11th term of the arithmeticprogression.

53, ,2, ........

2

3. For the arithmetic progression

0.7, 2.1 , 3.5, .. ,find the 5th term .

4. Find the thn term of the arithmetic

progression

14,6 ,9, .....

2

5. Find the 7 th term of the geometric

progression.

- 8, 4 , -2 , ..

Solution:

a = - 8 r =8

4

=

2

1

T7 = (-8)(2

1

)7-1

=8

1


progression.

16, -8, 4,


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7. For the geometric progression

9

4,

3

2, 1 , .. ,find the 9 th term .


progression

50, 40, 32.

Find the sum to infinity of geometric progressions

Find the sum to infinity of a given

geometric progression below:

Example:

2 26, 2, , ,.......

3 9

a = 6

2 1

6 3r

1

6=

11- -3

9=

2

aS

r

1. 24, 3.6, 0.54, .

2. 81, -27,9, ..

3.1 1 1

, , ,.......2 4 8

..

1

aS

r

sum to infinity

a = first term

r = common ratio

S


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Express the following recurring decimal as a fraction in its simplest form

1.

.

3.0

Use.

3.0

= 0.3333..

= 0.3 + 0.03 + 0.003 + .

where a = 0.3 and r = 0.1

2. 7.0

3. 25.0

4. 96.0

r

aS

1

1.01

3.0

9.0

3.0

3

1


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2. LINEAR LAW

Steps to plot a straight line

Using a graph paper.

QUESTION

x 2 3 4 5 6y 2 9 20 35 54

The above table shows the experimental values of two variables,xandy. It is know thatx andy are related by the equation

y = px2

+ qx

a) Draw the line of best fit forx

yagainst x

a) From your graph, find,

i) pii) q

Table

Identify Y and X from part (a)

Construct a table

Follow the scale given.Label both axes

Line of best fit

Determine : gradientm

Y-intercept c

Non- linear


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Solution

STEP 1

y = px2

+ qx

x

y=

x

px 2+

x

qx

x

y= px + q

Y= mX + c

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

Reduce the non-linear

To the linear form

The equation is divided throughout by xTo create a constant that is free from x

On the right-hand side i.e, q

Linear form

Y = mX + c

construct table

Using graph paper,

- Choose a suitable scale so that the graphdrawn is as big as possible.

- Label both axis

- Plot the graph of Y against X and drawthe line of best fit


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x

y

12

10

x

8

x

6

x

4

x2

x

2 3 4 5 6

- 2

- 4

1x

0


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Exercise

1. Table 1 shows the values of two variables,x andy , obtained from an experiment. The

variablesx andy are related by the equationpx

rpxy , wherep and r are constants.

x 1.0 2.0 3.0 4.0 5.0 5.5

y 5.5 4.7 5.0 6.5 7.7 8.4

Table 1

a. Plotxy against x2, by using a scale of 2 cm to 5 units on both axes.

Hence , draw the line of best fit.

b. Use the graph from (a) to find the value ofi. pii. r

2. Table 2 shows the values of two variables,x andy , obtained from an experiment. Variablesx

andy are related by the equationy = pkx+ 1

, wherep and k are constants.

x 1 2 3 4 5 6

y 4.0 5.7 8.7 13.2 20.0 28.8

Table 2

a. Plot log y against ( x + 1 ) , using a scale 2 cm to 1 unit on the ( x + 1 ) axis and 2 cm to0.2 unit on the log y axis.Hence, draw line of best fit

b. Use the graph from (a) to find the value of

i. pii. k

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 best fit, calculate the gradient, p

Determine the y-intercept, q

from the straight line graph


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3. INTEGRATION

Integration of xn

:

1.

3 13

3 1

xx dx c

=

4

4

xc

2. 5x dx 3.9x dx

4. 3x dx 5.

2x dx 6. x dx

Integration of axn

:

Note : m dx mx c , m a constant

1.

3 136 6.

3 1

xx dx c

=

4

6.4

xc

=43

2

xc

2. 410x dx 3.34x dx

4.. 10dx 10x +c 5. 12

dx 6.. 3dx

7.

1 1

8 8.1 1

xx dx c

=

2

8.

2

xc

=24x c

8. 6x dx 9. 3x dx

10. 312x dx 11.28x dx 12.

510x dx

1

, 11

nn axax dx c n

n

1

, 11

nn xx dx c n

n


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13.3

3

22dx x dx

x

=

3 1

2.3 1

xc

=2

2.2

xc

=2

1c

x

14.5

5

88dx x dx

x

=

15.4

12dx

x

16.3

2

5dx

x 17.

2

3x dx

18. 20.9x dx

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

= x32x2 + 5x + c

1. (6 4)x dx

=

2. 2(12 8 1)x x dx

=

3. 3( 3 2)x x dx

=

4. (3 2)x x dx

=

5. (2 1)(2 1)x x dx

=

6. ( 2)( 3)x x dx

=


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7. 2(3 2)x dx

=

8.2

(2 1)(2 1)x xdx

x

=

9.

2

2

6 4xdx

x

=

10.

2

2

(3 4)xdx

x

=

11. 2(2 1)x x dx

=

12. 2(2 )x dx

=

Definite Integral

1. Given that2

1( ) 3f x dx and

2

3( ) 7f x dx . Find

(a) 2

1the value of k if ( ) 8kx f x dx

(b) 3

15 ( ) 1f x dx

Answer : (a) k =22

3

(b) 48

2. Given that4

0( ) 3f x dx and

4

0( ) 5g x dx . Find

(a)4 0

0 4( ) ( )f x dx g x dx (b)

4

03 ( ) ( )f x g x dx

Answer: (a)15(b) 4


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Finding equation of a curve given gradient function of a curve and one point

1. Find the equation of the curve that passes through ( 2,-6) and has the gradient

function )3(2 xx

dx

dy

2. A curve with 6 axdx

dy, passes through (2,1). At this point , the gradient is 4. Find

a. the value ofa

b. equation of the curve


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3. Find the equation of the curve that has the gradient of 2x + 1 and passes through

)3,2

1(

4. The gradient function of a curve which passes throughA ( 1 , -12 ) is 3x26x.

Find the equation of the curve


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4. VECTOR

Vector in the cartesian coordinates

1. State the following vector in terms in ~i and ~j and also in Cartesian coordinates

Example Solution

~

22

0OA i

~

03

3OB j

~ ~

3 4

3

4

OP p i j

Exercise Solutions

(a) OP

= (b) OQ

(c) OR

(d) OS

(e) OT

(f)OW

~

j 5

4

3

2

1

543210

~

p

~i

B

P

A

1

4

3

2

1

2

SR

P

Q

-1-3 -2 -1

T

W

31 4

~i

~

j

-2

O


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2. Find the magnitude for each of the vectors

Example

3~ ~

2i j

2 23 2

13 unit

(a)~~

52 ji

(b)~~

125 ji (c)~~ji

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

(a)~ ~ ~

2 6r i j

(b)~

6

3a

(c)

~

1

2h


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45

* Given 2 parallel vectors and find the unknown in one of the vector ( vector AB ,

vector CD )

* 3 points which are collinear , finding the unknown using vector ( vector AB ,

vector BC )

SPM 2003/no. 12 / paper 1.

1. Diagram 2 shows two vectors, and QOOP .

Express

(a) OP in the formx

y

,

(b) OQ in the formxi +yj. [ 2 marks]

P(5, 3)

y

Q(-8, 4)

xO


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46

5. TRIGONOMETRIC FUNCTIONS

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

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

eg : Function angle

y = sinx x = 90o

y = cos 2x 2x = 90o

x = 45o

y = sin x2

3 x

2

3= 90

o

x = 60o

2. Calculate the values ofy for each value ofx by using calculator

eg : Sketch the grapha. y = sin 2xb. y = cos 2xc. y = 1 + sin 2xd. y = 12 sin x

y = 12 cos 2x

x 0 45 90 135 180 225 270 315 360

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

3. Plot the coordinates and sketch the graph

45 90 135 180 225 270 315 360x

y3

2

1


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marks are awarded for

shape

max and minimum

periodic ( the last angle and the middle angle )

Exercises

1. Sketch the graphs ofy = 1 + sin 2x for 0 x 180o

2. Sketch the graph ofy = 2 cos x2

3for0 x 2.


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6. 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 SSolution:

The way to arrange alphabet S = 1

The way to arrange another 3 alphabets= 5

P3

The number of arrangement = 1 x 5 P3= 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

x 6 C3

= 700

5. Four out of the letters from the wordBESTARI are arranged in a row. Find the

possible different arrangements.

6. An excursion group consisting of 4 malesand 4 females is to be chosen from 8 males

and 7 females. Find the number of ways theexcursion group can be formed.

7. Find the number of ways to arrange 7

students in a row.

8. The Mathematics teacher would like tochoose three students out of ten candidatesto form school quiz team. Find the number

of ways the teacher can do it.


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Alternative method

Black

Yellow

Black

Yellow

Black

Yellow

10

6

10

4

10

4

10

4

10

6

10

6

7. PROBABILITY

Question Answer

1.

The above figure shows six

numbered cards. A card is chosen

at random. Calculate theprobability that the number on the

chosen card

(a) is a multiple of 3 and afactor of 12

(b) is a multiple of 3 or a factorof 12.

Let

A represent the event that the number on the chosencard is a multiple of 3, andB represent the event that the number on the chosen

card is a factor of 12.

A = {3, 6, 9}, n(A)= 3B = {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, 3yellow balls and 4 green balls. A

ball is chosen at random from the

box. Calculate the probability thatthe balls drawn neither a yellow

nor a green.

P (yellow) =3

12

.

P(green) =4

12

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 colour is noted and the marbles

is noted and the marbles is

returned to the box. Then a

second marbles is chosen.Determine the probability that

(a) both the marbles are black.(b) the two balls are of different

colours.

(c) at least one of the ballschosen is yellow.

(a) P(black black)=10

4

10

4 =

25

4

(b) P(same colours)= P(black black) + P(yellow yellow)

2 3 4 6 8 9

10

4


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50

=

25

4+

10

6

10

6=

25

13.

(c) 1P(both blacks) = 125

4=

25

21

4. A box contains 3 red balls , 5yellow balls and 2 blue balls. Aball is drawn at random from the

box. Find the probability that the

ball is not blue in colour.

5. The probability that Alia qualifies

for the final of a track event is5

2

while the probability that Aisha

qualifies is3

1. Find the probability

that

a. both of them qualifies for thefinal

b. only one of them qualifies for thefinal.

6. A box contains 10 yellow marbles

and y blue marbles. If a marble ispicked randomly from the box, the

probability of getting a blue marble

is

7

2. Find the value ofy.


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51

8. PROBABILITY DISTRIBUTIONS

Example 1 :

Find the value of each of the following probabilities by reading the standardised normal

distribution table.

(a) P(Z > 0.934)

(b) P(Z 1.25)

Solution

(b) P(Z 1.25) = 1P(Z > 1.25)= 10.1057

= 0.8944

1.251.25


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(c) P(Z - 0.23)

Solution

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

= 1P(Z > 0.23)= 10.40905= 0.59095

(d) P(Z > - 1.512)

Solution

(d) P(Z < - 1.512) = P(Z > 1.512)= 0.06527

(e) P(0.4 < Z < 1.2)

Solution

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

= 0.34460.1151

= 0.2295

-1.512 1.512

-0.230.23

0.4 1.2

0.4 1.2


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53

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

Solution

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

= 0.289450.20384= 0.08561

(g) P(- 0.255 Z < 0.13)

Solution

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

= 10.399360.44828

= 0.15236

-0.828 -0.555 0.555 0.828

-0.255 0.13 0.13-0.255


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54

Example 2 :

Find the value of each of the following :

(a) P(Z z) = 0.2546

(b)P(Z < z) = 0.0329(c) P(Z < z) = 0.6623(d)P(z < Z < z 0.548) = 0.4723

Solution

(a)P(Z z) = 0.2546Score-z = 0.66

(b)P(Z < z) = 0.0329Score-z = -1.84

(c) P(Z < z) = 0.66231 - P(Z > z) = 0.6623

P(Z > z) = 10.6623

= 0.3377

Score-z = 0.419

(d) P(z < Z < z 0.548) = 0.47231P(Z < z)P(Z > 0.548) = 0.4723

1P(Z < z)0.2919 = 0.4723P(Z < z) = 10.29190.4723

= 0.2358

Score-z = -0.72

z

0.2546


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Summary

Normal DistributionType 1

P( 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 )

= 1P ( Z>1.3)

= 10.0968

= 0.9032

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

.

Type 4.

P( Z > negative no)

P ( Z > - 1.4 )

= 1P ( Z < -1.4 )

= 10.0808

= 0.9192

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

Type 5

P( positive no < Z < positive

no)

P ( 1 < Z < 2 )

= P ( Z > 1 ) P ( Z > 2 )= 0.15870. 0228

= 0.1359

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 )

= 0.21190.0668 = 0.1451

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

.

Type 7

P ( negative no < Z < postive no )

P ( -1.2 < Z < 0.8 )

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

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

1.2 )

= 1

0.2119

0.1151

=0.673

Type 1

P ( 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

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

Type 3

P( Z < K ) = more than 0.5

P ( Z < K ) = 0.8849

P ( Z > K ) = 1 0.8849

= 0.1151K = 1.2

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

Type 4

P ( Z > K ) = more than 0.5

P ( Z > K ) = 0.7580

P( Z < K ) = 1 0.7580 = 0.2420

P ( Z > -k ) = 0.2420

- K= 0.7

K = - 0.7


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Exercises

A. If z is standard normal variable, find the value of each of the following.

1. P(Z > 1.25 ) = 2. P(Z < 1.136 ) =

3. P(Z > -2.18 ) = 4. P ( -0.93 < Z < 1.02 ) =

5. P ( - 2.04 < Z < - 1.63 ) = 6. P ( 0 < Z < 1.228 ) =

B. Find the z-score of each of the following

1. P(Zz) = 0.75

3. P(Z z)=0.6044 4. P(Z z)= 0.8032

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