add maths module form 4 & 5
DESCRIPTION
TRANSCRIPT
FORMAT&COMPONENT
1
TO EXCELL in
You need to…
set TARGET
familiar with FORMAT of PAPER
do EXERCISESExercise Exercise Exercise Exercise Exercise
Exercise Exercise Exercise Exercise ExerciseExercise Exercise Exercise Exercise Exercise
Exercise Exercise Exercise Exercise ExerciseExercise Exercise Exercise Exercise Exercise
Exercise Exercise Exercise Exercise ExerciseExercise Exercise Exercise Exercise Exercise
Exercise Exercise Exercise Exercise ExerciseExercise Exercise Exercise Exercise Exercise
Exercise Exercise Exercise Exercise ExerciseExercise Exercise Exercise Exercise Exercise
MMyy TTAARRGGEETT
2
PAPER 1 (3472 / 1)
No Items Comments
1 No. of questions 25 questions (Answer ALL)
2 Total Marks……80 marks………….
3 Time…….2 hours
4 Level of difficulty Low : Moderate : High
6 : 3 : 1
PAPER 2 (3472 / 2)
No Items Comments
1 Three sections Section A Section B Section C
2 No of questions 6 5 4
(need to answer 12questions )
Answer
…ALL…..
Choose
…four….
Choose
…two…..
3 Total marks (100) 40 marks 40 marks 20 marks
4 Time………2 ½ hours……………..
5 Level of difficultyLow :4
Moderate :3
High3
FORMAT&COMPONENT
3
* Extra Time = ………………………………….
ALGEBRA
1. Functions2. Quadratic Equations3. Quadratic Functions4. Indices & Logarithms5. Simultaneous Equations6. Progressions7. Linear Law
Geometry1. Coordinate Geometry2. Vector
Calculus1. Differentiation2. Integration
Triginometry1. Circular Measures2. Trigonometric Functions
STATISTICS
1. Statistics2. Probability3. Permutation & Combination4. Distribution Probability
Social Science1. Index Numbers2. Linear Programming
Science & Technology1. Solutions to Triangle2. Motion in a Straight Line.
11 mmaarrkk mmiinnuutteess1.5
Check answers
FORM 4 TOPICS
4
Learn with your heart
and you’ll see the wonders …
P(X)
FORM 4 TOPICS
5
a b
X 1
TOPIC 1 : FUNCTIONS
f(a) = b object = …………………….. image = ………………..
Given f (x) and gf(x). Find g(x) . Thus, g(x) =
1
53)(
x
xxf ,
TOPIC 2: QUADRATIC EQUATIONS [ ax2 + bx + c = 0 ]
Types of roots
Sum and Product of Roots [ ax2 + bx + c = 0 ]
Sum of roots, ( + ) =a
bProduct of roots ( ) =
a
cx2 – S x + P = 0
TOPIC 3: QUADRATIC FUNCTIONS
General form CTS form
f (x) = ax2 + bx + c = a( x + p)2 + q
Similarity Same value of a Same value of a
Differencec = y-intercept. q = max/min value
of f(x)
Specialty
Able to find:
- shape
- y intercept
Able to find
- turning point
( - p , q)
- two distinct real roots- intersects at two points
- two equal roots- touches / tangent
- no root- does not intersect
- f(x) is always positive
1.
gf f -1
b2 – 4ac
> 0
=0
< 0
0 Real roots
FORM 4 TOPICS
6
2. Sketch Graphs: y = ax2 + bx + c
(a) From the graph
3. Inequalities : Solving 2 inequalities
1. Let the right hand side = 0- factorise
2. Find the roots of the equation3. Sketch the graph
3. Determine the region
positive or negative
…
TOPIC 4: SIMULTANEOUS EQUATIONS
Use ……………………………….. method
To find the …………………………………
x
Substitution
intersection
i value of a : Positive
ii value of b2-4ac: < 0
iii type of roots : No roots
iii y-intercept : C
use Graph
(x – a) (x – b) > 0 (x – c) (x – d) < 0
a b
……………………. …………….………
points between a straight line and a curve.
ivEquation of axis ofsymmetry :
X =a2
b
c d
< a , x > b, c < x < d
FORM 4 TOPICS
TOPIC 5: INDICES & LOGARITHM
Use Index rule : (i) 5x+1 . 125x =25
1……………………….
INDEX
Use log : (ii)
Use substitution : (iii)or can be factorise
(iv)
LOGARITHM : Use Rules of logarithms to
Change the base to
the same number
5x +1. 5 3x = 5–2
x + 1 + 3x = – 2
7
8 3x = 7 ............................ IND
3n+1
+ 3n
= 123n . 3 + 3n = 12
a(3) + a = 124a = 12a = 3
3n = 3 n = 1
32n
+ 5. 3n
= 6a2 + 5a – 6 = 0(a – 1) (a + 6) = 0
a = 1 , a = –63n = 1 3n –6n = 0
simplify or to solve logarithmic equations
log2 (x + 9) = 3 + 2 log2 x
log2
(x + 9) – log2
x 2 = 3
log2
(x + 9) = 3
x 2
x + 9 = 23
x 2
x + 9 = 8x 2
8x2 – x – 9 = 0
(8x -9) (x + 1) = 0
x –1, x =
Steps of solutions
1. separate the index
2. substitute
Insert log on both
sides
x = –
FORM 4 TOPICS
8
TOPIC 6: COORDINATE GEOMETRY
Distance Ratio theorem
Mid point Equation of straight line y = mx + c
Area (positive)Arrange
anticlockwise: general form ax + by + c = 0
Gradient : gradient form y = mx + c
- parallel m1
= m2
: intercept form 1b
y
a
x
- perpendicular m1
m2= –1
Equation of locus : …use distance formula………………….
Rhombus : ……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 invalues wheneach data is
Mean Mode Median Range Interquartilerange Variance
Added with k + k +k +k unchange unchange unchange unchange
Multiplied by m m m m m m m m 2
TOPIC 8: CIRCULAR MEASURES
…… radian = ………
For s = r and A = ½ r2 , the value of is in ……radian…….
Shaded angle = –
Area of segment = ½ r2 ( - sin )
rad
180
FORM 4 TOPICS
TOPIC 9: DIFFERENTIATION
gradient of normal
gradient of tangent
Tangent
equa
Rate of change
Applications
Small Changes
Turning points
TOPIC 10: SOLUTION OF TRIANGLE
Sine Rule
- Ambiguous Case two possible a
Cosine Rule
Area = ab sin C
TOPIC 11: INDEX NUMBERS
I A, B I B, C = I A, C
Given that the price index of an itemrate in the next year, what will be th
………………………………120
………………………………………
mN mT = –1
dx
dy= 0
9
equation of normal
tion of tangent
minimum
maximum
S
ngles acute
is 120. If the prie new price index
100
120 …………
…………………
mT =dx
dy
y
dy = dydt dx
y = dydx
2
dx
d
2
dx
d
approximate valuey ORIGINAL + y
– y1 = m(x – x1)
dxdt
x
02
y
and obtuse angle
ce index increases at the sameof the item?
…………………………………
…………………………….
02
y
10
:
FORM 5 TOPICS
T
TOPIC 1: PROGRESSIONS
Janjang Aritmetik Janjang Geometri
Examples : 20, 15, 10, …..., …. 4, 3, 2.25, ……., ….
Uniqueness : d = T2
– T1
r =1
2
T
T
Others :2
)( lanSn
S =
r
a
1
Given Sn find Tn Example: Given Sn = n( 3 + 2n), find T8.
Thus, T8 = S8
– S7
Find the sum from the3rd term until the 7th
term.
S 3 to 7 = S7
– S2
TOPIC 2: LINEAR LAW
Convert to linear form
Y =
ay = x +x
bxy =
xhx
py xy =
T + 1 = a2 + k
1T=
y = axb log y =
y = k px log y =
11
1 T2 T3 T4 T5 T6 T7
m X + c
a
1x2 +
a
b
h x + p
a + k
b log x + log a
log p x + log k
+ + + + + +
TOPIC 3: INTEGRATIONS
To find THE EQUATION OF A CURVE given dy/dx
dx
dy= ……gradient function ……………………
Equation of CURVE, dxfunctiongradienty }{ the integrated function must have c
AREA under a curve: Show how you would find the area of the shaded region.
x = y2 – y
1. Find the intersectionpoint.when x = 2, y = 4
2. Find the areaArea under curve + area
= 2
0
2x dx + (1)(4)
intersection, x = 1thus, y = 4(1) – 12 = 3
Shaded Area:Area under curve – area
= 4
1
2)4( xx dx – (3)(3)
1. Expand yy = x3 – 3x2 + 2x
Shaded Area:
Area above = 1
0dxy
Area below = 1
2dxy
Total area
1. Formula
1
0dyxA
VOLUME : Show the strategy to find the generated volume.
revolved about x-axis revolved about x-axis revolved about y-axis revolved about y-axis
V
wh
y = x2
str. line: y = –4x + 12
y = 4x - x2
str. line: y = –x + 4 y = x2– 1 x = y2 – 1
13–
y = x2
2 3 1 40 1 2
1
0
str. line: y = –4x + 12
y = 4x - x2
str.line: y = –x + 4 y = x (x–1) (x–2)
12
dxy2 dxyV 2 dyxV 2
dyxV 2
ere y2 = (x2)2 where y2 : (4x – x2)2 where x2 = y + 1 where x2 = (y2 – 1)2
2
0
22 dx)x(
+ Volume cone
4
1
22 )4( dxxx
– volume cone
I= 2
1
dy)1y( 1
1
22 dy)1y(
–1412 3
4 –
21
FORM 5 TOPICS
13
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, OT = ……AB………………………………………...
Given m = 2i + 3j and n = i – 4j find,
i) m + n = ……2i + 3j + ( i – 4j ) = 3i – j …………….
ii) m + n = …… 10)1(3 22 …………………………………………..
iii) unit vector in the direction of m + n = …… )ji3(10
1 ………………………
If A(1, 3) and B(–2, 5) find AB : …OA =
3
1, OB =
5
2 AB =
5
2–
3
1=
2
3
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 ……………………… (comparison method)
compare coef. of x , compare coef of y.
TOPIC 5 : TRIGONOMETRIC FUNCTIONS
Solving equation : SIMPLE
Solve: 2 cos 2x = 3 for 0 x 360
1. Separate coefficient of trigocos 2x =
2
3positive values 1st and 4th
quadrant2. Determine the quadrant
3. Find the reference angle 2x = 30
4. Find new range (if necessary). 0 x 720
5. Find all the angles2x = 30, 330, 390, 690
x = 15, 165, 195, 345
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 = 32 sec2 x + tan2 x = 5 ………have different functions ……….
Proofing: Use Identity
Remember :
cot
.1tan,
Acos
AsinAtan , sec 2x =
x2cos
1, cosec A =
Use of Trigo Ratios: Examples:
From the question given, If sin A = , A is not acute,
1. Determine the quadrant involved. ……second……………….
2. Determine the values of the other trig. fxn cos and tan = negative
in the quadrant. cos A =53 , tan A = –
find sin 2A
3. Do you need to use identity? sin 2A = 2 sin A cos A
4. Substitute values = 2 ( ) (53 )
=25
24
Sketch Graphs
y = a sin b x + ca =a cos b x + ca tan b x + c b =
c =
Basic Graphs
2
1-
–1-
Asin
1
y = sin x
max / minimum point
number of basic shape between 0 and 2
14
increase / decrease translation of the
2
1-
–1-
y = cos x y = tan
2
x
FORM 5 TOPICS
15
TOPIC 6: PERMUTATIONS & COMBINATIONS
Permutations = …order of arrangement is important Combinations =…order not important….
Three committee members of a society are to bechosen from 6 students for the position ofpresident, vice president and secretary. Findthe number of ways the committee can be
chosen.
Permutations: 6P3
Three committee members of a society are to bechosen from 6 students. Find the number ofways the committee can be chosen.
Combination: 6C3
with condition:Find the number of different ways the lettersH O N E S T can be arranged if it mustbegin with a vowel.
2 5 4 3 2 1
conditionvowels = 2 choices
Find the number of ways 11 main players of afootball team can be chosen from 15 localplayers and 3 imported players on the
condition that not more than 2 importedplayers are allowed.
condition 2 Import.
Case : (2 Import, 9 local) or 3C2.15C9
(1 import 10 locals) or + 3C1.15C10
( 0 import 11 locals) + 3C0.15C11
TOPIC 7: PROBABILITY
P(A) =)S(n
)A(n
- 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 = 1 – (lose and lose)
TOPIC 8: PROBABILITY DISTRIBUTION
BINOMIAL DISTRIBUTION
- The table shows the binomial probability distribution ofan event with n = 4 .
-
-
NO
- Fo
- T
- T
TO
M
re
st
m
______
0.30.25
0.2
0.15
0.1
0.05
P(X)
X = r r = 0 r = 1 r = 2 r = 3 r = 4
16
total = 1
formula: P(X = r) = n C r p r q n – r
mean, = np standard deviation = npq
RMAL DISTRIBUTION
rmula :
XZ
ype 1 : Given value of X find the value of Z find th[use formula] [use
ype 2 : Given the probability Find the value of Z Find i[use log book] [use
PIC 9: MOTION IN STRAIGHT LINES
Displacement, s Velocit
s = dtvd
dv
aximum velocity -dt
dv
turn to O s = 0 -
ops momentarily v = 0
ax. acceleration
P(X) 0.2 0.15 0.3 0.25 0.1
0 1 2 3 4 X=rGraph of Binomial Prob Distribution
variance = npq
e probabilitycalculator]
ts value of X .formula]
y, v Acceleration, a
t
s
dt
dva =
2
2
dt
sd
0 a = 0
-
0dt
da
FORM 5 TOPICS
17
TOPIC 10: LINEAR PROGRAMMING
Given:
(i) y > x – 2 (ii) x + y 5 (iii) 4x y
(a) Draw and shade the region, R, that satisfy the three inequalities on the graph paper providedusing 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.
–3 –2 –1 0 1 2 3 4 5 6
1
2
3
4
5
6
–1
–2
–3
–4
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
y = x – 2
y + x = 5
4x = y
R