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Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercise Exercis e Exercis e Exercis e Exercis e Exercis e

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1

PAPER 1 No

(3472 / 1)

Items

Comments

1

No. of questions

25 questions

2

Total Marks

……80 marks………….

3

Time

…….2 hours

4

Level of difficulty

Low : 6 :

PAPER 2 No

(Answer ALL)

Moderate 3

: :

High 1

(3472 / 2)

Items

Comments

1

Three sections

Section A

Section B

Section C

2

No of questions

6

5

4

Answer

Choose

Choose

…ALL…..

…four….

…two…..

40 marks

20 marks

(need to answer 12 questions ) 3

Total marks (100)

40 marks

4

Time

………2 ½ hours……………..

5

Level of difficulty

Low : 4

2

Moderate : 3

High 3

FORMAT&COMPONENT

1 mark



1.5

minutes

Check answers * Extra Time = ………………………………….

ALGEBRA

1. 2. 3. 4. 5. 6. 7.

Functions Quadratic Equations Quadratic Functions Indices & Logarithms Simultaneous Equations Progressions Linear Law

Geometry

1. Coordinate Geometry 2. Vector

Calculus

1. Differentiation 2. Integration

Triginometry

STATISTICS

Social Science

Science & Technology

1. Circular Measures 2. Trigonometric Functions 1. 2. 3. 4.

Statistics Probability Permutation & Combination Distribution Probability

1. Index Numbers 2. Linear Programming 1. Solutions to Triangle 2. Motion in a Straight Line.

3

FORM 4 TOPICS

P(X)

Learn with your heart and you’ll see the wonders …

4

FORM 4 TOPICS

TOPIC 1 : FUNCTIONS

a

b image = ………………..



f(a) = b 



Given f (x) and gf(x). Find g(x) .  Thus, g(x) = f ( x) 



object = ……………………..

3x  5 , x 1

gf f

-1

X1

TOPIC 2: QUADRATIC EQUATIONS

[ ax2 + bx + c = 0 ]

Types of roots

>0 b2 – 4ac

- two distinct real roots - intersects at two points

=0

- two equal roots - touches / tangent

<0

- no root - does not intersect - f(x) is always positive

0

Real roots

Sum and Product of Roots [ ax2 + bx + c = 0 ] Sum of roots, ( + ) =

b a

Product of roots ( ) =

c a

x2 –

Sx + P =0

TOPIC 3: QUADRATIC FUNCTIONS

General form

1.

ax2 + bx + c

f (x) =

Same value of a

Similarity

c = y-intercept.

Difference

Specialty

-

Able to find: shape y intercept 5

CTS form =

a( x + p)2 + q Same value of a q = max/min value of f(x) Able to find - turning point ( - p , q)

FORM 4 TOPICS

2.

Sketch Graphs:

y = ax2 + bx + c

(a) From the graph i

value of a

:

ii

value of b2-4ac:

iii type of roots : iii y-intercept iv 3.

Positive <0 No roots

C

:

Equation of axis of symmetry :

X= 

b 2a

Inequalities : Solving 2 inequalities  use Graph

(x – a) (x – b) > 0

(x – c) (x – d) < 0

1. Let the right hand side = 0 - factorise 2. Find the roots of the equation 3. Sketch the graph 3. Determine the region 

c

positive or negative

a

x
d

b

, x rel="nofollow"> b,

……………………….

c<x
…………….………

TOPIC 4: SIMULTANEOUS EQUATIONS

Substitution



Use ……………………………….. method



To find the ………………………………… points between a straight line and a curve.

intersection

6

FORM 4 TOPICS

TOPIC 5: INDICES & LOGARITHM

Use Index rule :

(i)

x+1

5

x

. 125

=

1 25

Change the base to ………………………. the same number

5x +1. 5 3x = 5–2 x + 1 + 3x = – 2 x = – INDEX Use log

:

Use substitution : or can be factorise

Insert log on both sides

(ii)

8  3x = 7

(iii)

3 + 3 = 12 n 3 . 3 + 3n = 12 a(3) + a = 12 4a = 12 a = 3 n 3 =3  n=1

(iv)



n+1

2n

............................ IND

n

Steps of solutions 1. separate the index 2. substitute

n

3 + 5. 3 = 6 a2 + 5a – 6 = 0 (a – 1) (a + 6) = 0 a = 1 , a = –6 3n = 1 3n  –6 n=0

LOGARITHM : Use Rules of logarithms to simplify or to solve logarithmic equations log2 (x + 9) = 3 + 2 log2 x log2 (x + 9) – log2 x log2 (x + 9) x2

2

=3

x + 9 = 23 x2 x + 9 = 8x 2 8x2 – x – 9 = 0 (8x -9) (x + 1) = 0 x –1, 7

x=



=3

FORM 4 TOPICS

TOPIC 6: COORDINATE GEOMETRY 

Distance





Ratio theorem



Mid point





Equation of straight line y = mx + c



Area (positive)



Gradient

Arrange anticlockwise

m1 = m2

- parallel

Equation of locus : Rhombus :

: general form

ax + by + c = 0

: gradient form

y = mx + c y x  1 a b

: intercept form

m1 m2 = –1

- perpendicular



…use distance formula………………….

……its diagonal are perpendicular to each other ……….

Parallelogram, square, rectangle, rhombus. its diagonal share the same mid point

TOPIC 7: STATISTCS EFFECT ON CHANGES TO DATA The change in values when each data is

Mean

Mode

Median

Range

Interquartile range



Variance

Added with k

+k

+k

+k

unchange

unchange

unchange

unchange

Multiplied by m

m

m

m

m

m

m

TOPIC 8: CIRCULAR MEASURES  

 radian = ……… 180  …… For s = r and A = ½ r2 ,



the value of  is in ……radian…….

Area of segment = ½ r2 (  - sin )

 

Shaded angle =



rad

8

 –

m

2

FORM 4 TOPICS

TOPIC 9:

DIFFERENTIATION

gradient of normal mN mT = –1 gradient of tangent

mT =

Tangent

equation of normal

dy dx

equation of tangent

y – y1 = m(x – x1)

Rate of change

dy = dy  dx dt dx dt

Small Changes

 y = dy   x dx

Applications

minimum

d2y

maximum

d2y

dx 2

approximate value y ORIGINAL + y

 0

Turning points

dy dx = 0

TOPIC 10: 

dx 2

 0

SOLUTION OF TRIANGLES

Sine Rule - Ambiguous Case  two possible angles  acute and obtuse angle



Cosine Rule



Area

=  ab sin C

TOPIC 11:

INDEX NUMBERS





I A,



Given that the price index of an item is 120. If the price index increases at the same rate in the next year, what will be the new price index of the item?

B

I B, C = I

A, C

………………………………120 

120 …………………………………………… 100

………………………………………………………………………………………. 9

:

10

FORM 5 TOPICS

TOPIC 1: PROGRESSIONS

Janjang Aritmetik

Janjang Geometri 4, 3, 2.25, ……., ….

Examples

:

20, 15, 10, …..., ….

Uniqueness

:

d = T2 – T1 Sn 

Others : Given Sn find Tn

Example:

T r= 2 T1

n( a  l ) 2

S =

Given Sn = n( 3 + 2n), find T8.

S8 – S7

Thus, T8 =

Find the sum from the 3rd term until the 7th term.

a 1 r

S7 – S2

S 3 to 7 =

+ T1

+ T2

+ T3

+ T4

+

+

T5

T6

TOPIC 2: LINEAR LAW Convert to linear form

b x

ay = x + y

p h x x

T + 1 = a 2 + k

b

y = ax

y=kp

x

Y

=

m

X

+

c

xy

=

1 a

x2

+

b a

y x

=

h

x

+

p

T 1 

=

a



+

k

log y

=

b

log x

+

log a

log y

=

log p

x

+

log k

11

T7

TOPIC 3: INTEGRATIONS  To find THE EQUATION OF A CURVE given dy/dx

dy = ……gradient function …………………… dx Equation of CURVE,

y

 {gradient function} dx

 the integrated function must have

c

AREA under a curve: Show how you would find the area of the shaded region.



str. line: y = –4x + 12 y = x2

2

1

0

4

1

2 0

intersection, x = 1 thus, y = 4(1) – 12 = 3

1. Expand y y = x3 – 3x2 + 2x

1. Formula

A Shaded Area:

Shaded Area: 2. Find the area Area under curve + area  Area under curve – area  4 2 2 2 = x dx +  (1)(4) = (4xx ) dx –  (3)(3) 1 0





x = y2 – y

1

y = 4x - x2

3

1. Find the intersection point. when x = 2,  y = 4

y = x (x–1) (x–2)

str.line: y = –x + 4

1

0

x dy

1

Area above =

0 y dx

Area below =

 2 y dx

1

Total area

 VOLUME : Show the strategy to find the generated volume.

revolved about x-axis

y=x

revolved about x-axis

y = 4x - x

2

x = y2 – 1

y = x2– 1

str. line: y = –x + 4

str. line: y = –4x + 12

revolved about y-axis

revolved about y-axis

2

1

3–

4– 2 V 

y

2 dx

where y2 = (x2)2





2

0

1

3

(x 2 ) 2 dx

+ Volume cone

V 

y

1 2

4 2 dx

V 

where y2 : (4x – x2)2



4

1

x

2 dy

where x2 = y + 1

(4 x  x 2 ) 2 dx

I= 

– volume cone 12



2

1

( y  1) dy

–1 V 

x

2 dy

where x2 = (y2 – 1)2



1

( y 2  1) 2 dy

1

FORM 5 TOPICS

TOPIC 4: VECTORS  If vectors a and b are parallel, thus, ………a = k b …………………….………..  If OA = a and OB = b , thus, AB = … OB - OA……= b – a ………………..….  If T is the mid point of AB thus,  Given m = 2i + 3j i)

m+n

OT = ……AB………………………………………...

and n = i – 4j find,

= ……2i + 3j + ( i – 4j ) = 3i – j ……………. 32  ( 1)2 

ii) m + n = ……

10 …………………………………………..

iii) unit vector in the direction of m + n = …… 1 

1 (3i  j ) ……………………… 10

  2

  2  1 

  3

  AB =        If A(1, 3) and B(–2, 5) find AB : …OA =   , OB =    5  – 3  =  2  3   5       

can also be written in the form of i and j.

 Given CD = 2h x + 5y and CD = 8x – 2hky , find the value of h and of k. 2h x + 5y

= 8x – 2hky ………………………

compare coef. of x ,

TOPIC 5 :

(comparison method)

compare coef of y.

TRIGONOMETRIC FUNCTIONS

 Solving equation : SIMPLE Solve: 1. Separate coefficient of trigo

2 cos 2x =

cos 2x =

2. Determine the quadrant 3. Find the reference angle

3

for 0  x  360

3 2

2x = 30 0  x  720

4. Find new range (if necessary). 5. Find all the angles

positive values  1st and 4th quadrant

2x = 30, 330, 390, 690 x = 15, 165, 195, 345

13



Solving Equation : Using Identity WHEN? sin 2x cos x = sin x cos 2x cos x = 0

………Angles are not the same………

sin2 x + 3cos x = 3 2 sec2 x + tan2 x = 5

………have different functions ……….

 Proofing: Use Identity Remember : tanA 

sinA cosA

, tan 

1 . cot

 Use of Trigo Ratios:

, sec 2x =

1 1 , cosec A = sin A cos 2x

Examples: If sin A =  , A is not acute,

From the question given, 1.

Determine the quadrant involved.

……second……………….

2.

Determine the values of the other trig. fxn in the quadrant.

cos and tan = negative cos A =  3 , tan A = –  5

find sin 2A 3.

Do you need to use identity?

4.

Substitute values

sin 2A = 2 sin A cos A = 2 ( ) (  3 ) 5

= 

24  25

Sketch Graphs y = a sin b x + c a = max / minimum point a cos b x + c a tan b x + c b = number of basic shape between 0 and 2 c = increase / decrease translation of the Basic Graphs y = sin x 1-

1-

2

2

–1-

y = tan x

y = cos x

–1-

14

2

FORM 5 TOPICS

TOPIC 6: PERMUTATIONS & COMBINATIONS Permutations = …order of arrangement is important Three committee members of a society are to be chosen from 6 students for the position of president, vice president and secretary. Find the number of ways the committee can be chosen. Permutations: 6 3

Combinations =…order not important….

Three committee members of a society are to be chosen from 6 students. Find the number of ways the committee can be chosen.

P

C

Combination: 6 3

with condition: Find the number of different ways the letters H O N E S T can be arranged if it must begin with a vowel.

2

5

4

3

2

Find the number of ways 11 main players of a football team can be chosen from 15 local players and 3 imported players on the condition that not more than 2 imported players are allowed.

1

condition  2 Import. Case : (2 Import, 9 local) or 3C2. 15C9 3 15 (1 import 10 locals) or + C1. C10 3 15 ( 0 import 11 locals) + C0. C11

condition vowels = 2 choices

TOPIC 7: PROBABILITY 

P(A) =

n ( A) n ( S)

- Probability event A or B occurs = P(A) + P(B) - Probability event C and D occurs =

P(A) . P(B)

 Considering several cases: Probability getting the same colors = Example: (Red and Red) or (Blue and blue) Probability of at least one win in two matches = (win and lose) or (lose and win) or (win and win) Or using compliment event

=

15

1 – (lose and lose)

TOPIC 8:

PROBABILITY DISTRIBUTION P(X)

BINOMIAL DISTRIBUTION

_

0.3 0.25 _ 0.2 _ 0.15 _ 0.1 _ 0.05 _

- The table shows the binomial probability distribution of an event with n = 4 . X=r

r=0

r=1

r=2

r=3

r=4

P(X)

0.2

0.15

0.3

0.25

0.1

0 1

2

3

4

X=r

Graph of Binomial Prob Distribution

total = 1 - formula: P(X = r) = n C r p r q n – r - mean,  = np

standard deviation =

variance = npq

npq

NORMAL DISTRIBUTION Z 

- Formula :

X  

Given value of X  find the value of Z  find the probability [use formula] [use calculator]

- Type 1

:

- Type 2

: Given the probability Find the value of Z  Find its value of X . [use log book] [use formula]

TOPIC 9: MOTION IN STRAIGHT LINES Displacement, s s= Maximum velocity return to O



Velocity, v v

v dt

ds dt

Acceleration, a a

d 2s dv = dt dt 2

-

dv 0 dt

a=0

s=0

-

-

v=0

stops momentarily

da 0 dt

max. acceleration

16

FORM 5 TOPICS

TOPIC 10: LINEAR PROGRAMMING Given:

x+y  5

(ii)

(iii) 4x  y

(i)

y >x–2

(a)

Draw and shade the region, R, that satisfy the three inequalities on the graph paper provided using 2 cm to 2 units on both axes.

(b)

Hence, find, in the region R, the maximum value of 2x + y where x and y are integers.

2 possible maximum points (x, y intergers) (1, 4) and ( 3, 2) . Point (3, 1) cannot be taken because it is not in R (it’s on dotted line)

2x + y = 2(1) + 4 = 6 = 2(3) + 2 = 8  the max value

6 4x = y

5 4 3

y=x–2

2 R

1

–3

–2

–1

0

1

2

–1

3

4

5

6 y+x=5

–2 –3 –4

17