Download - Spm Add Maths Formula List Form4
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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 ⎠
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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
⎟
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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
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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
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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
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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
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D
360D = 2π rad
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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
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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
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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
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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
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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
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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)
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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