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Add Maths Formulae List: Form 4 (Update 18/9/08)

01 Functions

Absolute Value Function Inverse Function

f ( x), if f ( x) ≥ 0f ( x) − f ( x), if f ( x) < 0

If y = f ( x) , then f −1 ( y) = x

Remember:Object = the value of xImage = the value of y or f(x)f(x) map onto itself means f(x) = x

02 Quadratic Equations

General Form

ax 2 + bx + c = 0

where a, b, and c are constants and a ≠ 0.

Quadratic Formula

2

x = −b ± b − 4 ac

2a

When the equation can not be factorized.

Forming Quadratic Equation From its Roots:

If α and β are the roots of a quadratic equationα + β = − b αβ = ca a

The Quadratic Equation

x2 − (α + β ) x + αβ = 0or

x2 − (SoR) x + ( PoR) = 0

Nature of Roots

b 2 − 4 ac > 0 ⇔ two real and different rootsb 2 − 4 ac = 0 ⇔ two real and equal rootsb 2 − 4 ac < 0 ⇔ no real rootsb 2 − 4 ac ≥ 0 ⇔ the roots are real

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ONE-SCHOOL.NET03 Quadratic Functions

General Form

f ( x) = ax2 + bx + c

where a, b, and c are constants and a ≠ 0.

*Note that the highest power of an unknown of aquadratic function is 2.

a > 0 ⇒ minimum ⇒ ∪ (smiling face)

a < 0 ⇒ maximum ⇒ ∩ (sad face)

Completing the square:

f ( x) = a( x + p)2 + q

(i) the value of x, x = − p(ii) min./max. value = q(iii) min./max. point = (− p, q)(iv) equation of axis of symmetry, x = − p

Alternative method:

f ( x) = ax2 + bx + c

Quadratic Inequalities

a > 0 and f ( x) > 0 a > 0 and f ( x) < 0a b a b

x < a or x > b a < x < b

Nature of Roots

b2 − 4ac > 0 ⇔ intersects two different points at x-axis

b2 − 4ac = 0 ⇔ touch one point at x-axisb2 − 4ac < 0 ⇔ does not meet x-axis

04 Simultaneous Equations

To find the intersection point ⇒ solves simultaneous equation.

Remember: substitute linear equation into non- linear equation.

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05 Indices and Logarithm

Fundamental if Indices

Zero Index, a0 = 1Negative Index, a−1 = 1

a

( a

)−1 = b b a

1Fractional Index a n = n a

ma n = n am

Laws of Indices

am × an = am + n

am ÷ a n = a m − n

(am )n = am×n

(ab)n = anbn

a an

( )n =b bn

Fundamental of Logarithm

loga y = x ⇔ a x = yloga a = 1loga a x = xloga 1 = 0

Law of Logarithm

loga mn = loga m + loga

n

log m = log m − log na n a a

log a mn = n log a m

Changing the Base

log b = logc ba log ac

log b = 1

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Midpoint A point dividing a segment of a line

Midpoint, M = ⎛ x1 + x2 , y1 + y2 ⎞⎝ ⎠ A point dividing a segment of a line

P = ⎛ nx1 + mx2 ,

ny1 + my2 ⎞⎜ ⎟⎝ m + n m + n ⎠

12

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06 Coordinate Geometry

Distance and GradientDistance Between Point A and C =(x1 − x2 )2 + (x − x2 )Gradient of line AC,

Or

m = y2 − y

1

x2

− x1

⎛ y − int ercept ⎞Gradient of a line, m = − ⎜ ⎟⎝ x − int ercept ⎠

Parallel Lines Perpendicular Lines

When 2 lines are parallel,

m1 = m2 .

When 2 lines are perpendicular to each other,

m1 × m2

= −1

⎜2 2

⎟

Area of triangle:

Area of Triangle

1=

2

A = 12

( x1 y2 + x2 y3 + x3 y1 ) − ( x2 y1 + x3 y2 + x1 y3 )Form of Equation of Straight Line

General form Gradient form Intercept form

ax + by + c = 0 y = m x + c

m = gradient

x y+ = 1a b

a = x-interceptb = y-intercept

m = − b a

Equation of Straight LineGradient (m) and 1 point (x1, y1)given

y − y1 = m( x − x1

)

2 points, (x1, y1) and (x2, y2) given

y − y1 = y2 − y1

x-intercept and y-intercept given

x + y = 1a b

Equation of perpendicular bisector ⇒ gets midpoint and gradient of perpendicular line.

Information in a rhombus:

A B(i) same length ⇒ AB = BC = CD = AD(ii) parallel lines ⇒ mAB = mCD or mAD = mBC

(iii) diagonals (perpendicular) ⇒ mAC × mBD

= −1

(iv) share same midpoint ⇒ midpoint AC = midpoint

D BDC (v) any point ⇒ solve the simultaneous equations

Remember:

y-intercept ⇒cut y-axis ⇒x-intercept ⇒cut x-axis ⇒

x = 0x = 0y = 0

y = 0**point lies on the line ⇒ satisfy the equation ⇒ substitute the value of x and of y of the point into theequation.

Equation of Locus( use the formula of distance)The equation of the locus of a moving point P( x, y) which

is always at a constant distance (r) from a fixed point A ( x1 , y1 ) is

PA = r2 2 2( x − x1 ) + ( y − y1 ) = r

The equation of the locus of a moving point P( x, y) which is

always at a constant distance from two fixed points A ( x1 , y1 ) and B ( x2 , y2 ) with a ratio m : n is

PA = mPB n

( x − x ) 2 + ( y − y )2 m2 1 1 =( x − x ) + ( y − y )2 n 22 2

The equation of the locus of a moving point P( x, y) which is always

equidistant from two fixed points A and B is the perpendicular bisector of the straight line AB.

PA = PB

( x − x )2 + ( y − y )2 = ( x − x )2 + ( y − y )21 1 2 2

More Formulae and Equation List:

SPM F o rm 4 Physics - Formulae List SPM Form 5 Physics - Formulae List

SPM Form 4 Chemistry - List of Chemical ReactionsSPM Form 5 Chemistry - List of Chemical Reactions

All at 0OF 4DIPPM OFU

Measure of Central Tendency07 Statistics

Ungrouped DataGrouped Data

Without Class Interval With Class IntervalMean

x = Σx

N

x = meanΣx = sum of xx = value of the dataN = total number of thedata

x = Σ fxΣ f

x = meanΣx = sum of xf = frequencyx = value of the data

x = Σ fxΣ f

x = meanf = frequencyx = class mark

= (lower limit+upper limit)

2Median

m = TN +12

When N is an odd number.

TN

+ TN

+1m = 2 2

2When N is an even

m = TN +12

When N is an odd number.

TN

+ TN

+1m = 2 2

2When N is an even number.

⎛ 1 N − F ⎞m = L + ⎜ 2 ⎟ C⎝ f m ⎠

m = medianL = Lower boundary of median classN = Number of dataF = Total frequency before median classfm = Total frequency in median classc = Size class

= (Upper boundary – lower boundary)

Measure of Dispersion

Ungrouped DataGrouped Data

Without Class Interval With Class Interval

variance2σ 2 = ∑ x − x

2

2σ 2 = ∑ fx − x 2

2σ 2 = ∑ fx − x 2

StandardDeviation

σ = variance

Σ ( x − x )2σ =N2σ = Σx − x 2

N

σ = variance

2Σ ( x − x )σ =N2σ = Σx − x 2

N

σ = variance

Σ f ( x − x )2σ = Σ fΣ fx2σ = − x 2

The variance is a measure of the mean for the square of the deviations from the mean.

The standard deviation refers to the square root for the variance.

Effects of data changes on Measures of Central Tendency and Measures of dispersion

Data are changed uniformly with+ k − k × k ÷ k

Measures ofCentral Tendency

Mean, median, mode + k − k × k ÷ kMeasures of dispersion

Range , Interquartile Range No changes × k ÷ kStandard Deviation No changes × k ÷ kVariance No changes × k2 ÷ k2

Terminology08 Circular Measures

Convert degree to radian: Convert radian to degree:

× 180xo = ( x × π

180)radiansπ

radians degrees

x radians = ( x × 180 ) degreesπ

× π180D

Remember:

180D = π

rad

???0.7 rad

O

???

1.2 rad

D

360D = 2π rad

r = radiusA = areas = arc lengthθ = anglel = length of chord

Arc Length:

s = rθ Length of chord:

θl = 2r sin

2

Area of Sector:

1 2A = r θ2

Area of Triangle:

1 2A = r sin θ2

Area of Segment:

A = 1 r 2 (θ − sin θ )

2

d

Length and Area

09 Differentiation

Gradient of a tangent of a line (curve or straight)

Differentiation of a Function I

y = xn

dy = m (δ y

)

dy = nxn−1

lidx δ x→0 δ

x

Differentiation of Algebraic FunctionDifferentiation of a Constant

y = a a is a constant

dy = 0dx

Exampley = 2dy = 0dx

Exampley = x3

dy = 3x2

dx

Differentiation of a Function II

y = ax

dy = ax1−1 = ax0 = adx

Exampley = 3x

dy = 3dx

ONE-SCHOOL.NETDifferentiation of a Function III Chain Rule

y = axn y = u n u and v are functions in x

dy = anxn−1

dx

dy = dy × du

dx du dx

Exampley = 2x3

dy = 2(3) x2 = 6x2

dx

Exampley = (2x2 + 3)5

u = 2x2 + 3, therefore

du = 4x dx

Differentiation of a Fractional Function

y = 1xn

y = u5 , therefore

dy = dy × du

dx du dx= 5u 4 × 4x

dy = 5u 4

du

Rewrite

y = x− ndy = −nxdx

Example

y = 1x

y = x−1

− n−1 = −n

xn+1

= 5(2x2 + 3)4 × 4x = 20x(2 x2 + 3)4

Or differentiate directlyy = (ax + b)n

dy = n.a.(ax + b)n −1

dx

y = (2x2 + 3)5

dy = 5(2x2 + 3)4 × 4x = 20 x(2x2 + 3)4

dy = −1x−2 = −1 dxdx x2

Law of Differentiation

Sum and Difference Rule

y = u ± v u and v are functions in x

dy = du ± dv

dx dx dx

Exampley = 2x3 + 5x2

dy = 2(3) x2 + 5(2) x = 6x2 + 10xdx

ONE-SCHOOL.NETProduct Rule Quotient Rule

y = uv u and v are functions in x

dy = v du + u

dv

dx dx dx

y = u v u and v are functions in x

v du − u

dv

Exampley = (2x + 3)(3x3 − 2x2 − x)

dy = dx dxdx v2

u = 2 x + 3

v = 3x3 − 2x2 − x

Examplex2

du = 2 dv = 9 x2 − 4x − 1

y =2 x + 1

dx dxdy = v

du + u dv

dx dx dx

u = x2

du = 2 x

v = 2x + 1dv = 2

=(3x3 − 2x2 − x)(2) + (2x + 3)(9x2 − 4 x

− 1)

Or differentiate directly

dx dx

v du − u

dv dy = dx dx dx v2

y = (2x + 3)(3x3 − 2x2 − x)

dy 3 2 2

dy =dx

(2 x + 1)(2 x) − x2

(2)

(2 x + 1)2= (3x

dx− 2x − x)(2) + (2x + 3)

(9x

− 4x − 1)

4 x2 + 2 x − 2

x2

=(2x +

1)2

2x2 + 2 x=(2 x + 1)2

Or differentiate directlyx2

y =2 x +

1

dy (2 x + 1)(2 x) − x2 (2)=dx (2 x + 1)2

4 x2 + 2 x − 2 x2

= (2x + 1)22x2 + 2 x= (2 x + 1)2

ONE-SCHOOL.NETGradients of tangents, Equation of tangent and Normal

Gradient of tangent at A(x1, y1):

dy = gradient of tangentdx

Equation of tangent: y − y1 = m( x − x1 )

Gradient of normal at A(x1, y1):

1m

normal = −

If A(x1, y1) is a point on a line y = f(x), the gradient of the line (for a straight line) or the gradient of the

tangent of the line (for a curve) is the value of dy

1− dydx

mtangent= gradient of normal

when x = x1.dx Equation of normal : y − y1 = m( x − x1 )

Maximum and Minimum Point

Turning point ⇒ dy = 0dx

At maximum point,dy = 0dx

d 2 y < 0dx2

At minimum point ,dy = 0dx

d 2 y > 0dx2

dA dA= ×dr

dr

dt dt

Rates of Change Small Changes and Approximation

Chain rule

If x changes at the rate of 5 cms -1 ⇒ dx = 5dt

Decreases/leaks/reduces ⇒ NEGATIVES values!!!

Small Change:

δ y ≈ dy ⇒ δ y ≈ dy × δ xδ x dx dx

Approximation:ynew

= yoriginal + δ

y

= y + dy × δ xoriginal

dx

δ x = small changes in x

ONE-SCHOOL.NET10 Solution of Triangle

Sine Rule:

a = b = c

sin A sin B sin C

Use, when given�2 sides and 1 non included

angle�2 angles and 1 side

a a B

AA

b 180 – (A+B)

Cosine Rule:

a2 = b2 + c2 – 2bc cosA2 2 2b = a + c – 2ac cosB

c2 = a2 + b2 – 2ab cosC

b2 + c2 − a2

cos A =2bc

Use, when given�2 sides and 1 included angle�3 sides

a a c

Ab b

Area of triangle:a

Cb

1A = a b sin C

2

C is the included angle of sides aand b.

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Case of AMBIGUITY If ∠C, the length AC and length AB remain unchanged,the point B can also be at point B′ where ∠ABC = acute

Aand ∠A B′ C = obtuse.If ∠ABC = θ, thus ∠AB′C = 180 – θ .

180 - θ θ Remember : sinθ = sin (180° – θ)C B′ B

Case 1: When a < b sin ACB is too short to reach the side opposite to C.

Outcome:No solution

Case 2: When a = b sin ACB just touch the side opposite to C

Outcome:1 solution

Case 3: When a > b sin A but a < b.CB cuts the side opposite to C at 2 points

Outcome:2 solution

Case 4: When a > b sin A and a > b.CB cuts the side opposite to C at 1 points

Outcome:1 solution

Useful information:In a right angled triangle, you may use the following to solve the

c problems.b (i) Phythagoras Theorem: c = a2 + b2θ

a Trigonometry ratio:(ii)

sin θ = b , cosθ = a , tan θ = bc c a

(iii) Area = ½ (base)(height)

ONE-SCHOOL.NET11 Index Number

Price Index Composite index

I = P1 × 100P0

I = Price index / Index number

I = ΣWi I iΣWi

I = Composite IndexW = Weightage

I = Price index

I A,B × I B,C = I A,C ×100