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KEPERLUAN PENGETAHUAN/KEMAHIRAN
TOPIKAL 1. Recognising a Quadratic Function

## 2. Shapes of graph of Quadratic Function

1. FUNCTIONS [ f(x) = ax 2 + bx + c ]
If a > 0
1. Representing a relation
- representing a relation using arrow diagrams,
ordered pairs, graphs * min
- Using the conept of domain, codomain, If a < 0 mak
object, image and range *

2. Classifying a relation
- one to one
one to many 3. Relation between the position of Quadratic
many to one Function Graphs and its roots:
many to many i. graph intersects the x-axis at two points
b2-4ac > 0
3 Expressing functions ii graph does not intersects the x-axis
b2-4ac < 0
4. Determining composite functions iii graph touches the x-axis
- fg, gf, f 2, f 3, ........ b2-4ac = 0
- determining g from the composite function
fg or gf 4. Finding the min @ max value of QF using the
completing the square method:
5 Determining the inverse function [ f(x) = ax 2 + bx + c General form
- Let f -1 (x) = y, then x = f (y) = a(x + p )2 + q after completing the
- the condition of an inverse function square]
If a>0 min vakue = q
axis of symmetry : x = -p
min point = (-p, q)
2. QUADRATIC EQUATIONS If a<0 nilai max value = q
axis 0f symmetry : x = -p
Basic Knowledge : mak point. = (-p, q) ]
a. Expansion
b. Factorisation (common factor, difference of two 5. Sketching Quadratic Function Graph
squares, table / cross multiply) 6 steps :
i Determine the shape (identify a)
1. General form of a quadratic equation ii Determine the position (evaluate b2-4ac)
iii. Completing the square (find max@ min
2. Determining roots of a quadratic equation point and axis of symmtery)
ax 2 + bx + c = 0 using iv. Solve f(x) = 0 - find point of intersection
i. factorisation with x-axis)
ii completing the square v. Find f(0) - find point of intersection
− b ± b 2 − 4ac with y-axis)
iii. formula x = vi. Plot the points and connect them with a
2a smooth curve

## 3. Forming QE from given roots by expansion 4. Quadratic Inequalities

Ranges of Quadratic Inequalities
4. Relation between two roots: - Using line number
Given ax 2 + bx + c = 0 , then i. Factorise
−b c ii State two values of x
SOR = , POR = ii Use suitable method to find the correct
a a
position / area
5. Forming QE from given roots using formula : iii State the range
x 2 - (SOR)x + (POR) = 0

## 6. Types of roots for QE a b

i. roots are distint : b2-4ac > 0 x<a x>b
ii. roots are equal : b2-4ac = 0 a<x<b
iii. roots are not real : b2-4ac < 0
4. SIMULTANEOUS EQUATIONS

## 3. QUADRATIC FUNCTIONS Basic Knowledge :

(a). Simultaneous linear equation
Basic knowledge: - elimination
- substitution
(b). Solving Quadratic equation
- factorization
- formula 6 COORDINATE GEOMETRY

## 1. Simultaneous equations in two unknowns : Basic Knowledge :

one linear equation and one non-linear equations. - coordinates
3 basic types of simultaneous equations: - straight line
i. 2x + y = 0
xy + y2 - 8 = 0 1. Distance between two points
ii. 3x + 2y = 1 AB = ( x1 − x 2 ) 2 + ( y1 − y 2 ) 2
4 y 3x
− = −2
x y 2. Mid points of two given points
iii. x2 + xy + y2 = 2x + 3y = 7
 x1 + x2 y1 + y2 
Mid point AB =  , 
2. Simultaneous equations in Daily Problems:  2 2 
Finding perimeter, area and the coordinates of
points of intersections between a curve and a 3. Coordinate of a point dividing a line segment
straight line. according to a given ratio m:n
Main steps:
i. Understand the problem and choose the  mx1 + nx2 my1 + ny 2 
P=  , 
correct symbol to represent the unknown.  m+n m+n 
ii. Decide on a plan
- form linear and non linear equation 4. Area of a polygon
- solve the equations by substitutions
iii. Carry out the plan 1
x1 x 2 . . . x n x1
L= 2
y1 y 2 . . . y n y1

## 5. Gradient of a straight line

5. INDICES AND LOGARITHMS
y 2 − y1
m=
Basic Knowledge : x 2 − x1
Index - positive, negative zero and fractions
6. Equation of a straight line
1 Value of numbers in the form of integer and (i) y = mx + c
fractional indices (ii) y - k = m ( x - h )
2. Law of indices x y
(iii) + =1
am x an = a m+n a b
am ÷ an = a m - n
(a m ) n = a mn 7. Point of intersetion of two straight lines
[ use simulaneous equations ]
3 Logarithms
- Definition 8. Parallel lines : m1=m2
- Finding logarithm of a number
9. Perpendicular lines : m 1 . m 2 = -1
4. Laws of logarithm
log a xy = log a x + log a y 10. Peperpendicular Bisector
x m 1 . m 2 = -1
log a y = log a x - log a y use midpoints
log a x m = m log a x 11 Equation of locus involving distance
i. Locus of a moving point that is constant
distance from a fixed point.
5 Changing the base of logarithm ii. The ratio of the distance of a moving point
log c b from the two points is constant.
log a b = iii Masalah yang berkaitan dengan lokus
log c a

## 6 Solving equation involving indicecs 7. STATISTICS

i. comparison of indices
ii. comparison of bases Basic Knowledge : Min, mode, median
iii. using logarithm
1 Measures of Central Tendency
7. Solving Logarithmic Equations - mean , mode, median
i. using definiion of logarithm
ii. changing both sides of the equation to single 2 Measures of Dispersion
logarithm with the same base.
- range, interquartile range, variance, standard [x new value, x original mean]
deviation If x = x , both mean in the original and
new
3. Ungrouped Data set of data are still the same
Mean : x=
∑x If x > x , mean in new set of data is more
N than original data
Mode : the value which occurs the most If x < x , mean in new set of data is less
number of times than
Median : the centre value (Q2 )when the set original data
of data is arranged in ascending or
descending order 4 Most suitable measure of central tendency
Range : largest value - smallest value (a) Under normal conditions, mean is chosen
Interquartile range : Q3 - Q1 because every value is taken into account -
a good overall picture of data
∑x  ∑x 
2 2

Variance: σ 2
= −  (b) Exteme value exists, median is the better
N  N  choice
 
(c) If a certain value occurs frequently, mode is
Standard deviation : σ = var ians
the better choice
4. Group Data 5. Effect on range, interquartile range, variance,
Mean : x=
∑ fx standard deviation
∑f (a) Data is changed uniformly
New range : c x (original range)
Modal class: the class that has the highest New interquartile range
frequency : c x original interquartile
Mode : estimate mode from histogram New standard deviation
(graph of frequency against upper : c x original standard deviation
limit) New variance : c2 x original variance
Median : i. Formula (b) Extreme values in data
 N2 − F  range will incresed - unsuitable
Med (Q2) = L +   c interquartile range - unchanged, suitable
Variance will increased - unsuitable
 f  standard deviation - most suitable
ii. OGIVE (graph of cumulative (c) Every data is added or subtrated with a
frequency aganist upper limit) constant value
Range : mid point of highest class - mid - measures of dispersion unchanged
point of lowest class
Interquartile range
: i. Formula = Q3 - Q1 8. CIRCULAR MEASURES
 N +F
Q1 = L +  4 c Basic knowledge :
 f  - Perimeter of a circle [ P = 2πr ]
  - Area of a circle [ L = πr 2 ]

 3N + F  1. Radian (rad)
Q3 = L +  4 c π rad. = 180o
 f 
  2
ii. OGIVE A
∑ fx  ∑ fx 
2
2 r
Variance: σ2 = −  θ P
∑f ∑f 
  O
r segment
Standard deviation : σ = var ians
B
3 Effects on Mode , mean, median r = radius
(a) Data being change uniformly AB = chord
New mean = c + original mean or APB = Arc length
c x original mean OAPB = sector
New mode = c + original mode or
c x original mode 3. Arc length of a circle.
New median = c + original median or s = r θ (θ rad)
c x original median
(b) Extreme values exist 4 Perimeter of segment of a circle
- mean will be affected
- mode and median unchanged θ
Perimeter of segment AB = 2r sin + rθ
(c) Certain data is added or removed 2
5 Area of sector [maximum @ minimum point]
1 r2θ (θ rad ) lmain steps:
L= 2
2 dy
1. Find
dx
, ddx 2y
6 Area of segment of a circle dy
1 r2θ - 1 r2 sin θ 2. Let = 0, find the values x & y
= dx
2 2
d2y
3. Substitute the value of x in
dx 2
9. DIFFERENTIATION d2y
If > 0, the point is minimum
dx 2
1. Limit
d2y
dy δy If < 0, the point is maximum
= lim dx 2
dx δx →∞ δx b. Daily problems
1. Form suitable equations
2. First derivative for polynomial using first priciple 2. repeat steps in (a).

## 3. First derivative for polynomial functions iv. Rate of change

i. Using formula [involve time]
dy
y = axn ,
dx
= anxn-1 @
dt
= ×
dy dy dx
dx dt
f(x) = axn , f ' (x) = anxn-1
v. Small changes & approximation
ii. y = u + v δy
dy
=
du + dv δx
=
dy
dx
dx dx dx Approximate value of y = y + δy

iii y = u - v
dy
=
du - dv 10. SOLUTIONS OF TRIANGLE
dx dx dx
1 Finding unknown sides or angles of triangle
iv. Product of two polynomial A
dy
y = uv , = u dv + v du c b
dx

## v. Quotient of two polynomial B a C

u ,
y=
v
dy
dx
= vdu− udv
v2 (a) Sine Rule
a b c
= =
vi Composite function sin A sin B sin C
dy
y = [f(x)]n , = n [f(x)] n-1 f ' (x) (i) Use to solve a triangle if given either
dx
i. two angles and one side
ii. two sides and the non included angle
4. Find values of first derivatives
(ii) Ambigous Cases of a Triangle
5. Second derivatives of function
d2y b
: , f '' (x)
dx 2 a
θ
5. Application / use of differentiation
dy - may occur when two sides (where
i. = gradient of a curve / tangent a<b ) and a non included angle θ are
dx given
dy
ii. Equation of tangent : y - k = (x - h)
dx (b) Cosine Rule
a 2 = b 2 + c 2 – 2ab sin A
−1
Equation of normal : y - k = dy (x - h)
dx Use to solve a triangle if given either
iii. Maximum/Minimum i three sides
a. Turning point ii two sides and the included angle
(c). Area of a triangle iv. Expressing the recurring decimal in the form
1 of a geometric series
= ab sin C
2
3 To determine the n th term of any progression :
1 Tn = Sn - Sn - 1
= ac sin B
2
1
= bc sin A
2
13. LINEAR LAW
4. Problems in 3 dimensions involve Basic Knowledge :
- Pythagoras theorams - Straight line
- Trigonometrical ratios - gradient
- Equation of a straight line y = mx + c,
- plotting coordinates
11. INDEX NUMBER 1. Lines of best fit
a All plotted points should lie as close as
1. Index number I possible
Q to the line
I = 1 ×100 b The line should pass through as many points as
Q0
possible
I index number c The number of points above and below the line
Q1 quantity at specific time should be about the same and at about the
Q2 quntity at base time same distance from the line.

## 2. Equation of line of best fit

2. Composite Index 3. Apply linear law to non-linear equation
__ ∑ Ii W i a Reduce non linear relations to a linear form
I = W Y = mX + c
∑ i b Determine values of constants of non linear
Ii index number for items i relations given
Wi corresponding weightage for items i i lines of best fit
ii data
12. PROGRESSIONS c Obtains information from
i lines of best fit
ii equations of linea of best fit
1. Arithmatic Progression
i. Terms in AP
Tn = a + (n - 1)d
14. INTEGRATION
ii. Sum of the first n terms in an AP
n (2a + (n - 1)d ) or 1 Integration is the opposite of differentiation
Sn =
2 dy
If = f(x), then y = ∫ f ( x ) dx
n (a + l ) dx
Sn =
2
2. Indefinite Integration
2. Geometric Progression dy
i If =k
i. Terms in GP dx
Tn = ar n - 1 then y = kx + c
ii. Sum of the first n terms in a GP dy
a (1 − r n ) ii. If = ax n
Sn = r<1 or dx
1−r ax n +1
∫ ax dx = + c , n ≠ -1
n
then y =
a (r n − 1) n +1
Sn = r>1
r −1
( ax + b) n +1
∫ (ax + b) dx = +c , n ≠
n
iii. Sum to Infinity iii.
a (n +1)
a
S∞ = -1
1−r
iv iv. Volume of Revolutions
a when the area bounded is rotated 360o
∫ f ( x ) + g( x ) dx = ∫ f ( x ) dx + ∫ g( x ) dx about the x-axis
b
V = π∫ y 2 dx
a
2 Definite Integration
b
i. ∫a
f ( x )dx = [g( x ) ]ba b. when the area bounded is rotated 360o
= g(b) − g(a)
V= π∫ab x 2 dy
b b
ii ∫a k f ( x ) dx = k ∫a f ( x ) dx

b a 15 VECTOR
iii ∫a f ( x ) dx = −∫b f ( x ) dx

iv 1 Vector
c - A quantity which has
∫a f ( x ) dx = ∫ab f ( x ) dx + ∫bc f ( x ) dx i arah
ii magnitud (Teoram Pithagoras)

## 3. Application of Integration 2 Vector Notation : AB , a , AB, a

i. Equation of curves from function of gradient
dy dy 3 Neagative vector
y = ∫( ) dx where =gradient - AB = BA
dx dx
of curve
4 Zero vector
0 has zero magnitude

## ii. Area under a curve 5. Equal vector

a bounded by the x-axis from x = a to x = - same magnitude and direction
b
y 6 Parallel Vectors
AB dan CD are parallel if and only if
AB =k CD , k is a constant
A
7 Non Parallel Vectors
x If the vectors a and b are not parallel, then
a b
b
ha = kb can only be true if h = k = 0
A= ∫
a
ydx
8 Collinear Vectors
A, B and C are collinear if and pnly if
uuur uuur uuur
AB = m BC = n AC

9 Vector Operations
a. Multiplying with scalars
- ka is a vector with magnitude k times of
b. bounded by y-axis from y=a to y = b  a
y If k>0, then ka and a have the same
a direction
A If k<0, then ka and a are in the opposite
b direction

x b Addition of vectors
b - Triangle Law
A= ∫a
xdy - Parallelogram Law
- Polygon Law
[ The value of A will negative when the
shaded area is below the x-axis or on the c Subtraction of vector
left-hand side of the y-axis ] a - b = a + (-b)

c. Area between the curve and a straight 10 Vectors in the Cartesan plane
line
b a. If i and j are unit vectors in the directions of
A= ∫a [ f ( x ) ± g( x ) ] dx the x- and y-axes, then
x 1
PQ =   = xi + yj secant (sec) =
y cos
 PQ  = x 2 +y 2 1
cosecant (cosec) =
sin
b Unit Vectors in given direction. 1 cos
cotangent (cot) = or
tan sin
If r = xi + yj, the unit vector in the
direction of r is 5 Reflex angle (angle > 180o )

∧ r 6 Complementary angle
=
r r ∠p and ∠q are complementary if ∠p + ∠q = 90o
1
= ( xi + y j ) sin θ = cos (90 - θ)
x + y2
2 cos θ = sin (90 - θ)
1 x  tan θ = cot (90 - θ)
=   cot θ = tan (90 - θ)
2 2 y 
x +y sec θ = cosec (90 - θ)
cosec θ = sec (90 - θ)
c Operations with vectors in terms of i and j
(Adding, subtracting, multiply by scalars)
- apply operations to components
separately or use the concept of
matrices. 7 Values of six trig functions

## 16. TRIGONOMETRIC FUNCTIONS i. Without calculator or mathematical tables

a. Draw the quadrant diagram
Basic Knowledge b Draw the right angle triangle in the
- Pythagoras Theoram correct quadrant
c Use Pythagoras Teoram to calculate the
1 Positive angle - rotating ray is turned anti length
clockwise about the origin from the fixed d State the ratio
positive x-axis
Negative angle - rotating ray is turned clockwise
about the origin from the fixed positive x-axis ii. Special angles (without calculator)
a [0o, 90o, 180o, 360o.......]
2 1 revolution = 360o = 2π radian sine 1
0 0
3 Every angle θ has a corresponding reference
angle α -1
y cosine 0
rotating ray -1 1
0
α θ
x tangent ∞ 1
0 0
5 Signs of three trigonometric ratios
-∞
Sine +ve All +ve
b 45o, 30o, 60o
Q2 Q1 Langkah-langkah:
Q3 Q4 - Lukis rajah sukuan
- Lukis segitiga bersudut tegak dalam
Cosine +ve Tangent +ve sukuan dan lengkapkan sisi guna
Teoram Pithagoras
4 Definition of six trigonometric functions - Nyatakan tanda
opposite - Cari nilai (jawapan dalam bentuk
sine (sin) = "SOH"
hypotenuse pecahan terendah)
a d ja c e n t
cosine (cos) = "CAH" 2. Penyelesaian persamaan trigo.
h y p o te n u s e Langkah-langkah
opposite - tentukan sukuan - lukis rajah
tangent (tan) = "TOA" - cari sudut tirus
- dapatkan nilai sudut dalam sukuan
=
sin berkenaan
cos - Tentukan julat dalam mana sudut berada.
Teknik-teknik peny. persamaan: Hasilan yang belum dikenalpasti
- Takrif / Asas
- Pemfaktoran 2 Taburan Binomial
- Identiti Asas i Pemboleh ubah rawak diskret
- rumus sudut majmuk, berganda, separuh Pemboleh ubah rawak yg mpunyai nilai yg
terbilang dan terhingga
3. Pembuktian Identiti Trigo menggunakan ii Cubaan Bernoulli (cubaan Binomial)
i. Takrif - cubaan yg hanya mempunyai dua
ii. Identiti Asas kesudahan yg mungkin iaitu kejayaan atau
iii. rumus sudut majmuk / berganda kegagalan
iii Pemboleh ubah rawak diskret Binomial
4. Graf fungsi trigo - bilangan kejayaan yg dicerap dlm satu
i. Melakar dan melukis graf bagi sebarang cubaan binomial
fungsi trigo iv Kebarangkalian utk mendapat r kali kejayaan
ii. Menentukan bil. peny. persamaan trigo P(X=r) = n C r p r q n-r
dengan melakar graf p - kebarangkalian kejayaan
iii Mencari peny. hampir bagi fungsi trigo q - kebarangkalian kegagalan
dengan melukis graf. p+q=1, 0<p<1
r = 0, 1, 2, 3 ........ n
v Taburan Binomial
17 PILIHATUR DAN GABUNGAN Taburan kebarangkalian bagi semua
pemboleh ubah rawak diskret binomial
1 Prinsip pendaraban vi Graf Taburan Binomial
Graf kebarangkalian lawan pembolehubah
2 Pilihatur rawak diskret
i. Bilangan pilihatur bagi n benda = n ! vii Bagi satu taburan binomial
ii Bilangan pilihatur bagi r benda drp n benda min = np
= n P r @ guna kaedah ' rumah ' varians = npq
iii Pilihatur bersyarat - tetapkan syarat terlebih sisihan piawai = npq
dahulu 3 Taburan Normal
i Pemboleh uabah rawak selanjar
3 Gabungan Pemboleh ubah rawak yg mpunyai bilangan
i Bilangan gabungan r benda drp n benda nilai tak terhingga dalam satu lingkungan
=nCr ii Graf Taburan Normal
ii Gabungan bersyarat - tetapkan syarat f(x)
terlebih dahulu

18 KEBARANGKALIAN MUDAH

1 Mengetahui konsep a b µ x
i Ujikaji - satu proses @ tindakan memerhati * Lengkung bersimetri pada paksi x = µ
sesuatu * Jumlah luas rantau di bawah lengkung =
ii Ruang sampel (S) - set semua kesudahan yg 1
mungkin bagi satu ujikaji * P(a < X < b) = luas di bawah lengkung
iii Peristiwa - suatu subset bagi ruang sampel antara x = a dan x = b
yg memuaskan syarat tertentu
iii Graf Taburan Normal Piawai
n (A) f(z)
2. P (A) =
n (S)

## 4 Peristiwa saling eksklusif

[peristiwa A dan B tidak boleh berlaku serentak 0 a b z
A∩B = φ ] * Lengkung bersimetri pada paksi-f(z)
P(A∩B) = 0 * Jumlah luas rantau di bwh lengkungan =
P(A∪B) = P(A) + P(B) 1
5. Peristiwa tak bersandar * P(a < Z < b) = luas rantau dibawah
[Kebarangkalian peristiwa B berlaku tidak lengkung antara z = a dan z = b
bergantung kepada kesudahan peristiwa A}
P(A∩B) = P(A) x P(B) 4 Pemboleh ubah taburan normal, X boleh
ditukarkan kepada pemboleh ubah taburan
X −µ
normal piawai Z dengan rumus Z =
19 TABURAN KEBARANGKALIAN σ

## 1. Pemboleh ubah rawak

5 Nilai P(Z > z) boleh ditentukan dgn sifir > rantau atas / kanan garis
kebarangkalian hujung atas taburan normal < rantau bawah / kiri garis
piawai >, < garis putus-putus
≥ ,≤ garis penuh
20. GERAKAN GARIS LURUS 2. Menentukan ketaksamaan yang menakrifkan
sesuatu rantau
1. Rajah GGL 1 3. Menafsirkan masalah dan membentuk
sesaran : s = f(t) ketaksamaan yang berkenaan
4. Menentukan nilai mak/min sesuatu masalah
ds Langkah-langkah :
halaju : v= i. Tafsir masalah dengan menentukan
dt
pembolehubah yang terlibat
ii. Bentuk sistem ketaksamaan @ persamaan
iii. Bina rantau R yang memuaskan ketaksamaan
pecutan : a=
dv @ d 2 s itu
dt dt 2 iv. Tentukan nilai mak/min k = ax + by di
bawah kekangan tertentu.
2. Rajah GGL 11
pecutan : a = f(t)

halaju : v= ∫adt

sesaran : s= ∫vdt
3. Sifat-sifat GGL
i. s = f(t) sesaran selepas t saat
s = 0 zarah berada/kembali di 0
s > 0 zarah berada di kanan 0
s < 0 zarah berada di kiri 0
Jumlah jarak - kaedah garis nombor
Jarak dalam saat ke n = S n - S n - 1
Sesaran mak/min :
- s = f(t)
ds = 0
-
dt
- cari t, ganti dalam s
- ujian mak/min

## ii. v = f(t) halaju selepas t saat

Halaju seragam - zarah bergerak dengan laju
malar
v = 0 zarah berhenti seketika
v > 0 zarah bergerak ke kanan
v < 0 zarah bergerak ke kiri
halaju mak/min :
- v = f(t)
dv = 0
-
dt
- cari t, ganti dalam v
- ujian mak/min

## iii. a = f(t) pecutan selepas t saat

Pecutan seragam - halaju berubah dengan kadar
yang sama
a = 0 halaju seragam
a > 0 halaju menokok terhadap masa
a < 0 halaju menyusut terhadap masa

ketaksamaan
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