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Add Maths Formulae List: Form 4 (Update 18/9/08)
01 Functions
Absolute Value Function
Inverse Function
If
f ( x ), if f ( x ) ≥ 0
f ( x)
y = f ( x ) , then f −1 ( y ) = x
Remember:
Object = the value of x
Image = the value of y or f(x)
f(x) map onto itself means f(x) = x
− f ( x), if f ( x ) < 0
General Form
ax 2 + bx + c = 0
−b ± b 2 − 4ac
x=
2a
where a, b, and c are constants and a ≠ 0.
*Note that the highest power of an unknown of a
When the equation can not be factorized.
Nature of Roots
Forming Quadratic Equation From its Roots:
If α and β are the roots of a quadratic equation
α +β =−
b
a
αβ =
c
a
b 2 − 4ac
b 2 − 4ac
b 2 − 4ac
b 2 − 4ac
x 2 − (α + β ) x + αβ = 0
or
x − ( SoR ) x + ( PoR ) = 0
SoR = Sum of Roots
PoR = Product of Roots
2
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1
>0
=0
<0
≥0
⇔ two real and different roots
⇔ two real and equal roots
⇔ no real roots
⇔ the roots are real
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Completing the square:
General Form
f ( x) = ax 2 + bx + c
f ( x) = a ( x + p)2 + q
where a, b, and c are constants and a ≠ 0.
(i)
(ii)
(iii)
(iv)
*Note that the highest power of an unknown of a
the value of x, x = − p
min./max. value = q
min./max. point = (− p, q)
equation of axis of symmetry, x = − p
Alternative method:
a > 0 ⇒ minimum ⇒ ∪ (smiling face)
f ( x) = ax 2 + bx + c
a < 0 ⇒ maximum ⇒ ∩ (sad face)
a > 0 and f ( x) > 0
the value of x, x = −
(ii)
min./max. value = f (−
(iii)
equation of axis of symmetry, x = −
b
x < a or x > b
b
)
2a
b
2a
Nature of Roots
a > 0 and f ( x) < 0
⇔ intersects two different points
at x-axis
2
b − 4ac = 0 ⇔ touch one point at x-axis
b 2 − 4ac < 0 ⇔ does not meet x-axis
b 2 − 4ac > 0
a
b
2a
(i)
a
b
a< x<b
04 Simultaneous Equations
To find the intersection point ⇒ solves simultaneous equation.
Remember: substitute linear equation into non- linear equation.
http://www.one-school.net/notes.html
2
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05 Indices and Logarithm
Laws of Indices
Fundamental if Indices
Zero Index,
a0 = 1
a m × a n = a m+n
Negative Index,
a −1 =
1
a
a m ÷ a n = a m−n
( a m ) n = a m× n
a
b
( ) −1 =
b
a
Fractional Index
1
an
= a
m
an
= a
( ab) n = a n b n
n
n
a n an
( ) = n
b
b
m
Fundamental of Logarithm
Law of Logarithm
log a y = x ⇔ a x = y
log a mn = log a m + log a n
log a a = 1
log a
log a a x = x
log a mn = n log a m
log a 1 = 0
http://www.one-school.net/notes.html
m
= log a m − log a n
n
Changing the Base
3
log a b =
log c b
log c a
log a b =
1
logb a
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06 Coordinate Geometry
Distance Between Point A and C =
(x1 − x2 )2 + (x1 − x2 )2
Gradient of line AC, m =
y2 − y1
x2 − x1
Or
⎛ y − int ercept ⎞
Gradient of a line, m = − ⎜
⎟
⎝ x − int ercept ⎠
Parallel Lines
Perpendicular Lines
When 2 lines are parallel,
When 2 lines are perpendicular to each other,
m1 = m2 .
m1 × m2 = −1
m1 = gradient of line 1
m2 = gradient of line 2
Midpoint
A point dividing a segment of a line
A point dividing a segment of a line
⎛ x1 + x2 y1 + y2 ⎞
,
⎟
2 ⎠
⎝ 2
Midpoint, M = ⎜
http://www.one-school.net/notes.html
⎛ nx + mx2 ny1 + my2 ⎞
P =⎜ 1
,
⎟
m+n ⎠
⎝ m+n
4
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Area of triangle:
Area of Triangle
=
1
2
A=
1
( x1 y2 + x2 y3 + x3 y1 ) − ( x2 y1 + x3 y2 + x1 y3 )
2
Form of Equation of Straight Line
General form
ax + by + c = 0
Intercept form
y = mx + c
x y
+ =1
a b
c = y-intercept
Equation of Straight Line
Gradient (m) and 1 point (x1, y1) 2 points, (x1, y1) and (x2, y2) given
given
y − y1 y2 − y1
y − y1 = m( x − x1 )
=
x − x1 x2 − x1
a = x-intercept
b = y-intercept
m=−
b
a
x-intercept and y-intercept given
x y
+ =1
a b
Equation of perpendicular bisector ⇒ gets midpoint and gradient of perpendicular line.
Information in a rhombus:
A
D
B
C
http://www.one-school.net/notes.html
same length ⇒ AB = BC = CD = AD
parallel lines ⇒ mAB = mCD or mAD = mBC
diagonals (perpendicular) ⇒ mAC × mBD = −1
share same midpoint ⇒ midpoint AC = midpoint
BD
any point ⇒ solve the simultaneous equations
(i)
(ii)
(iii)
(iv)
(v)
5
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Remember:
y-intercept ⇒ x = 0
cut y-axis ⇒ x = 0
x-intercept ⇒ y = 0
cut x-axis ⇒ y = 0
**point lies on the line ⇒ satisfy the equation ⇒ substitute the value of x and of y of the point into the
equation.
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
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 , y 2 ) with
a ratio m : n is
PA = r
PA m
=
PB n
( x − x1 ) 2 + ( y − y1 ) 2 = r 2
PA = PB
( x − x1 ) + ( y − y1 ) 2 = ( x − x2 ) 2 + ( y − y2 ) 2
2
( x − x1 ) 2 + ( y − y1 ) 2 m 2
=
( x − x2 ) + ( y − y 2 ) 2 n 2
More Formulae and Equation List:
At
http://www.one-school.net/notes.html
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.
6
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07 Statistics
Measure of Central Tendency
Ungrouped Data
Mean
x=
Without Class Interval
Σx
N
x=
x = mean
Σx = sum of x
x = value of the data
N = total number of the
data
Median
Σ fx
Σf
2
2
=
2
TN + TN
2
When N is an odd number.
TN + T N
+1
m=
2
When N is an even
number.
2
2
Σ fx
Σf
x = mean
f = frequency
x = class mark
m = TN +1
When N is an odd number.
m=
x=
x = mean
Σx = sum of x
f = frequency
x = value of the data
m = TN +1
Grouped Data
With Class Interval
+1
2
When N is an even number.
(lower limit+upper limit)
2
⎛ 1N −F⎞
⎟⎟ C
m = L + ⎜⎜ 2
⎝ fm ⎠
m = median
L = Lower boundary of median class
N = Number of data
F = Total frequency before median class
fm = Total frequency in median class
c = Size class
= (Upper boundary – lower boundary)
Measure of Dispersion
Ungrouped Data
variance
x2
∑
σ =
2
N
−x
2
σ = variance
Standard
Deviation
Σ(x − x )
σ=
N
σ=
2
Σx
− x2
N
2
http://www.one-school.net/notes.html
Grouped Data
Without Class Interval
With Class Interval
fx 2
∑
σ =
∑f
2
−x
σ = variance
σ=
σ=
Σ(x − x )
fx 2
∑
σ =
∑f
2
Σx 2
− x2
N
−x
σ = variance
2
N
7
2
σ=
σ=
Σ f (x − x)
2
Σf
Σ fx 2
− x2
Σf
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
+k
−k
×k
÷k
Measures of
Mean, median, mode
Central Tendency
Range , Interquartile Range
Measures of
Standard Deviation
dispersion
Variance
×k
×k
× k2
No changes
No changes
No changes
08 Circular Measures
Terminology
×
xo = ( x ×
180
π
π
180
180
x radians = ( x ×
) degrees
D
π
degrees
×
π
180D
Remember:
???
D
http://www.one-school.net/notes.html
O
???
8
÷k
÷k
÷ k2
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Length and Area
A = area
s = arc length
θ = angle
l = length of chord
Arc Length:
s = rθ
Length of chord:
l = 2r sin
Area of Sector:
θ
A=
2
Area of Triangle:
1 2
rθ
2
A=
1 2
r sin θ
2
Area of Segment:
A=
09 Differentiation
Differentiation of a Function I
Gradient of a tangent of a line (curve or
straight)
y = xn
dy
= nx n−1
dx
dy
δy
= lim ( )
dx δ x →0 δ x
Example
y = x3
Differentiation of Algebraic Function
Differentiation of a Constant
y=a
dy
=0
dx
dy
= 3x 2
dx
a is a constant
Differentiation of a Function II
y = ax
dy
= ax1−1 = ax 0 = a
dx
Example
y=2
dy
=0
dx
http://www.one-school.net/notes.html
Example
y = 3x
dy
=3
dx
9
1 2
r (θ − sin θ )
2
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Differentiation of a Function III
Chain Rule
y = ax n
y = un
dy
= anx n−1
dx
dy dy du
=
×
dx du dx
Example
y = 2 x3
Example
y = (2 x 2 + 3)5
dy
= 2(3) x 2 = 6 x 2
dx
u = 2 x 2 + 3,
y = u5 ,
Differentiation of a Fractional Function
u and v are functions in x
therefore
therefore
du
= 4x
dx
dy
= 5u 4
du
dy dy du
=
×
dx du dx
= 5u 4 × 4 x
1
xn
Rewrite
y=
= 5(2 x 2 + 3) 4 × 4 x = 20 x(2 x 2 + 3) 4
y = x−n
Or differentiate directly
y = (ax + b) n
dy
−n
= − nx − n−1 = n+1
dx
x
dy
= n.a.(ax + b) n −1
dx
Example
1
y=
x
y = x −1
−1
dy
= −1x −2 = 2
x
dx
y = (2 x 2 + 3)5
dy
= 5(2 x 2 + 3) 4 × 4 x = 20 x(2 x 2 + 3) 4
dx
Law of Differentiation
Sum and Difference Rule
y =u±v
u and v are functions in x
dy du dv
=
±
dx dx dx
Example
y = 2 x3 + 5 x 2
dy
= 2(3) x 2 + 5(2) x = 6 x 2 + 10 x
dx
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10
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Product Rule
Quotient Rule
y = uv
u and v are functions in x
dy
du
dv
= v +u
dx
dx
dx
y=
u
v
dy
=
dx
Example
y = (2 x + 3)(3 x 3 − 2 x 2 − x)
u and v are functions in x
v
du
dv
−u
dx
dx
v2
Example
x2
y=
2x +1
u = x2
v = 2x +1
du
dv
= 2x
=2
dx
dx
du
dv
−u
v
dy
= dx 2 dx
dx
v
dy (2 x + 1)(2 x) − x 2 (2)
=
(2 x + 1) 2
dx
u = 2x + 3
v = 3x3 − 2 x 2 − x
du
dv
=2
= 9 x2 − 4x − 1
dx
dx
dy
du
dv
=v
+u
dx
dx
dx
3
2
=(3 x − 2 x − x)(2) + (2 x + 3)(9 x 2 − 4 x − 1)
Or differentiate directly
y = (2 x + 3)(3x3 − 2 x 2 − x)
dy
= (3x3 − 2 x 2 − x)(2) + (2 x + 3)(9 x 2 − 4 x − 1)
dx
4 x2 + 2 x − 2 x2 2 x2 + 2 x
=
=
(2 x + 1) 2
(2 x + 1) 2
Or differentiate directly
x2
y=
2x +1
dy (2 x + 1)(2 x) − x 2 (2)
=
dx
(2 x + 1) 2
4 x2 + 2 x − 2 x2 2 x2 + 2 x
=
=
(2 x + 1) 2
(2 x + 1) 2
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11
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Gradients of tangents, Equation of tangent and Normal
Gradient of tangent at A(x1, y1):
dy
dx
Equation of tangent: y − y1 = m( x − x1 )
Gradient of normal at A(x1, y1):
mnormal = −
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
dy
tangent of the line (for a curve) is the value of
dx
when x = x1.
1
− dy
1
mtangent
dx
Equation of normal : y − y1 = m( x − x1 )
Maximum and Minimum Point
Turning point ⇒
At maximum point,
dy
=0
dx
2
d y
<0
dx 2
http://www.one-school.net/notes.html
dy
=0
dx
At minimum point ,
dy
=0
dx
12
d2y
>0
dx 2
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Rates of Change
Chain rule
Small Changes and Approximation
Small Change:
dA dA dr
=
×
dt dr dt
dx
=5
dt
Decreases/leaks/reduces ⇒ NEGATIVES values!!!
If x changes at the rate of 5 cms -1 ⇒
δ y dy
dy
≈
⇒ δ y ≈ ×δ x
δ x dx
dx
Approximation:
ynew = yoriginal + δ y
= yoriginal +
dy
×δ x
dx
δ x = small changes in x
δ y = small changes in y
If x becomes smaller ⇒ δ x = NEGATIVE
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13
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10 Solution of Triangle
Sine Rule:
Cosine Rule:
Use, when given
 2 sides and 1 non included
angle
 2 angles and 1 side
a
A
b
B
a
a2 = b2 + c2 – 2bc cosA
b2 = a2 + c2 – 2ac cosB
c2 = a2 + b2 – 2ab cosC
a
b
c
=
=
sin A sin B sin C
a
Area of triangle:
cos A =
C
b
b2 + c2 − a 2
2bc
A=
Use, when given
 2 sides and 1 included angle
 3 sides
A
180 – (A+B)
a
a
http://www.one-school.net/notes.html
b
14
C is the included angle of sides a
and b.
c
A
b
1
a b sin C
2
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If ∠C, the length AC and length AB remain unchanged,
the point B can also be at point B′ where ∠ABC = acute
and ∠A B′ C = obtuse.
If ∠ABC = θ, thus ∠AB′C = 180 – θ .
Case of AMBIGUITY
A
180 - θ
Remember : sinθ = sin (180° – θ)
θ
C
B
B′
Case 1: When a < b sin A
CB is too short to reach the side opposite to C.
Case 2: When a = b sin A
CB just touch the side opposite to C
Outcome:
No solution
Case 3: When a > b sin A but a < b.
CB cuts the side opposite to C at 2 points
Outcome:
1 solution
Case 4: When a > b sin A and a > b.
CB cuts the side opposite to C at 1 points
Outcome:
2 solution
Outcome:
1 solution
Useful information:
c
b
a
θ
In a right angled triangle, you may use the following to solve the
problems.
(i) Phythagoras Theorem: c = a 2 + b 2
(ii)
Trigonometry ratio:
sin θ = bc , cos θ = ac , tan θ =
(iii) Area = ½ (base)(height)
http://www.one-school.net/notes.html
15
b
a
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11 Index Number
Price Index
Composite index
I =
P1
× 100
P0
I=
I = Price index / Index number
I = Composite Index
W = Weightage
I = Price index
P0 = Price at the base time
P1 = Price at a specific time
I A, B × I B ,C = I A,C ×100
http://www.one-school.net/notes.html
Σ Wi I i
Σ Wi
16
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