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8/12/2019 SPM Add Maths Formula List Form 4
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Add Maths Formulae List: Form 4 (Update 18/9/08)
01 Functions
Absolute Value Function Inverse Function
f (x ), iff (x)0 Ify =f (x), then f
1(y )=x
f (x) Remember:f(x),iff(x) 0 two real and different rootsa ab 4ac = 0 two real and equal roots
The Quadratic Equation b 4ac
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03 Quadratic Functions
General Form Completing the square:
f (x )=ax 2+bx +c (x)=a(x +p )2+q
where a, b, and care constants and a0. (i) the value ofx, x = p
*Note that the highest power of an unknown of a(ii) min./max. value = q(iii) min./max. point = (p, q)
quadratic function is 2. (iv) equation of axis of symmetry, x =p
Alternative method:
a >0minimum(smiling face)
f (x)=ax
2
+bx +ca 0and f (x)>0 a >0and f (x)0 intersects two different points
atx-axis
b24ac= 0 touch one point atx-axis
a b a bb
24ac
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05 Indices and Logarithm
Fundamental if Indices Laws of Indices
Zero Index, a0
=1 am
an
=am
+n
a
1 1
aman=amn
Negative Index, =a
a 1
=
b ( am
)n=a
mn
( )b a
Fractional Index
1
= na
( ab)n=a
nb
n
an
m (a)n= an
a = a
bn
b
Fundamental of Logarithm Law of Logarithm
log amn=log am+loganlogay=xax
=y
log aa=1 m
logaax
=x
log an=log amlogan
log amn= nlog am
log a1 =0Changing the Base
logab= logcb
logca
logab=1
logba
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06 Coordinate Geometry
Distance and GradientDistance Between Point A and C =
( x1x2)2
+( x1x2)2
Gradient of line AC, m =y2y1xx
Or2 1
Gradient of a line, m = y intercept
x intercept
Parallel Lines Perpendicular Lines
When 2 lines are parallel, When 2 lines are perpendicular to each other,
m =m2. m1m2= 11
m1= gradient of line 1
m2= gradient of line 2
Midpoint A point dividing a segment of a line
x +x y +y A point dividing a segment of a line
Midpoint, M =1 2
,1 2
nx +mx ny +my2 2 2 2P = 1 , 1
m +n m +n
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Area of triangle:
Area of Triangle
1
= 2
A = ( x1y2+x2y3 +x3y1)(x2y1+x3y2+x1y3)2
Form of Equation of Straight LineGeneral form Gradient form I ntercept form
ax +by +c =0 y =mx +c x+y =1a
m = gradientb
bc =y-intercepta =x-intercept m = a
b =y-intercept
Equation of Straight LineGradient (m) and 1 point (x1, y1) 2 points, (x1, y1) and (x2, y2) given x-intercept and y-intercept given
given
y y1=m(x x1) y y1 =y2y1 x+y =1x x a bx x
1 2 1
Equation of perpendicular bisectorgets midpoint and gradient of perpendicular line.
Information in a rhombus:
A B (i) same lengthAB=BC=CD=AD
(ii) parallel linesmAB=mCD or mAD=mBC
(iii) diagonals (perpendicular)mACmBD= 1
(iv) share same midpointmidpointAC= midpoint
DC
BD
(v) any pointsolve the simultaneous equations
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Remember:
y-interceptx=0
cuty-axisx=0
x-intercepty =0
cutx-axisy=0
**point lies on the linesatisfy the equationsubstitute the value ofxand ofyof the point into the
equation.
Equation of Locus
( use the formula of The equation of the locus of a The equation of the locus of a moving
distance) moving pointP(x,y) which is point P(x,y) which is always
The equation of the locus of a always at a constant distance equidistant from two fixed pointsAandB
moving pointP(x,y)which from two fixed points is the perpendicular bisector of theis always at a constant A (x,y)andB (x
2,y2) with straight lineAB.
distance (r) from a fixed point1 1
a ratio m:nisA (x1,y1)is
(xx)2
PA =PB+ (yy 2
PA m+ (yy) = (xx)
2PA =r =
1 1 2PB n
(xx1)2+(yy1)
2=r
2 (xx1)2+ (yy1)2 =
m 2
(xx2) +(yy2)2 n
2
More Formulae and Equation List:
SPM Form 4 Physics - Formulae List
SPM Form 5 Physics - Formulae List
SPM Form 4 Chemistry - List of Chemical Reactions
SPM Form 5 Chemistry - List of Chemical Reactions
All atOne-School.net
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Measure of Central Tendency
07 Statistics
Ungrouped DataGrouped Data
Without Class Interval With Class IntervalMean
x =x
x=fx
x =fx
N f f
x =mean x =mean x =mean
x= sum ofx x= sum ofx f =frequency
x =value of the data f =frequency x =class mark
N = total number of the x =value of the data(lower limit+upper limit)data =
2
Medianm =TN+1 m =TN+1 1 N F
22 2 m =L +
fmC
WhenNis an odd number. WhenNis an odd number.
m = medianT N +TN T N +TN L = Lower boundary of median class
m = +1 m = +1 N = Number of data2 2 2 22 2 F = Total frequency before median class
WhenNis an even WhenN is an even number. fm= Total frequency in median class
number. c = Size class
= (Upper boundary
lower boundary)
Measure of Dispersion
Ungrouped DataGrouped Data
Without Class Interval With Class Interval
variance
2
=
x2 2
2
=
fx2x
2
2
=
fx 2 2
N x f f x
= variance = variance = variance
Standardxx
2 (xx)2
f(xx)2
= =DeviationN
=N f
x 2 2 x2 2 2 = N x = N x = fx x
f
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The varianceis a measure of the meanfor the square of the deviationsfrom the mean.
The standard deviati onrefers to the square rootfor the variance.
Effects of data changes on Measures of Central Tendency and Measures of dispersion
Data are changed uniformly with
+ k k k kMeasures of Mean, median, mode + k k k k
Central Tendency
Measures ofRange , Interquartile Range No changes k k
Standard Deviation No changes k kdispersion
Variance No changes k k
08 Circular MeasuresTerminology
Convert degree to radian:
Convert radian to degree:
x
o=(x )radians
180D 180
x radians=(x
180 )degrees
radians degrees
180
D
Remember:
180 =rad 1.2 radO
???0.7 rad ???360 =2rad
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Length and Area
r = radius
A = area
s = arc length = angle
l = length of chord
Arc Length: Length of chord: Area of Sector: Area of Triangle: Area of Segment:
s =r 1 2 1 2 1 2
l =2r sin2 A = 2r A = 2 sin A =2r (sin )
09 Differentiation
Gradient of a tangent of a line (curve or
straight)
dy= lim (ydx x0x
Differentiation of Algebraic Function
Differentiation of a Constant
y =a a is a constant
dy
dx
= 0
Example
y =2
dydx= 0
Differentiation of a Function I
y =xn
dydx= nx
n1
Example
y =x3
dydx= 3x
2
Differentiation of a Function II
y =ax
dydx= ax
11= ax
0= a
Example
y =3x
dy
dx= 3
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Differentiation of a Function III
y =axn
dydx= anx
n1
Example
y =2x3
dydx= 2(3)x
2= 6x
2
Differentiation of a Fractional Function
y =x
1n
Rewrite
y =xn
dydx= nx
n1=x
n
n+1
Example
y =1x
y =x1
dy
dx= 1x2=
x
12
Law of Differentiation
Sum and Difference Rule
y =u v u andv are functions inx
dydx=
dudx
dvdx
Example
y =2x3+5x
2
dydx= 2(3)x
2+ 5(2)x= 6x
2+10x
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Chain Rule
y =un u andv are functions inx
dy =dydudx du dx
Example
y =(2x2+3)
5
u =2x2+3, therefore =4x
dx
y =u5, therefore dy =5u4
dy =dydudu
dx du dx
= 5u44x
= 5(2x2+3)
44x=20x(2x
2+3)
4
Or differentiate directly
y =(ax +b)n
dydx= n.a.(ax+b)
n1
y =(2x2+3)
5
dydx= 5(2x
2+ 3)
4 4x= 20x(2x
2+3)
4
10
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Product Rule
y =uv u andv are functions inx
dy=vdu+udvdx dx dx
Example
y =(2x +3)(3x32x
2x)
u =2x +3 v =3x32x
2x
du= 2 = 9x2 4x1
dx dx
dy= v +u
dx dx dx
=(3x32x
2x)(2) +(2x+3)(9x
24x1)
Or differentiate directly
y =(2x +3)(3x32x
2x)
dy = (3x32x
2x)(2)+ (2x+ 3)(9x
2 4x1)
dx
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Quotient Rule
y = u andv are functions inx
vdu dv
dy= dx
dxdx v
Example
y =x
2
2x+1
u =x2 v =2x +1
du = 2x dv = 2dx dx
du dvdy
=v
dx udx
dx v2
dy=
(2x+1)(2x) x2(2)
dx (2x+1)2
=4x
2+2x2x
2=
2x2+ 2x
(2x+1)2 (2x+1)
2
Or differentiate directly
y = x22x+1
dy=
(2x+1)(2x) x2(2)
dx (2x+1)2
=4x
2+2x2x
2=
2x2+2x
(2x+1)2 (2x+1)
2
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Gradients of tangents, Equation of tangent and Normal
If A(x1,y1) is a point on a liney = 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 ofdy
dx
when x = x1.
Gradient of tangent at A( x1, y1):
dy
dx= gradient of tangent
Equation of tangent: yy1=m(xx1)
Gradient of normal at A( x1, y1):
mnormal
= m
1
tangent1 = gradient of normal
dy dxEquation of normal : yy1=m(xx1)
Maximum and Minimum Point
Turning point dy= 0dx
At maximum point, At minimum point ,
dy=
0
d y 0dx dx dx dx
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Rates of Change Small Changes and Approximation
Small Change:
Chain rule dA =dA dry dy dydt dr dt
x dxydxx
Ifxchanges at the rate of 5 cms-1
dx
= 5 Approximation:dt
Decreases/leaks/reduces NEGATIVES values!!! new original
=yoriginal
+ x
dx
x=small changes inx
y=small changes iny
Ifxbecomes smallerx=NEGATIVE
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10 Solution of Triangle
Sine Rule:
sina
A=
sinb
B=
sinc
C
Use, when given
2 sides and 1 non includedangle
2 angles and 1 side
a a B
AA
b 180 (A+B)
Cosine Rule:
a2= b
2+ c
22bc cosA
b2= a
2+ c
22ac cosB
c2
= a2
+ b2
2ab cosC
cosA=b2+c
2
a22bc
Use, when given
2 sides and 1 included angle 3 sides
a a c
bA
b
Area of triangle:
a
C
b
A =12a b sinC
C is the included angle of sidesa
and b.
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Case of AMBIGUITY
A
180 -
C B B
If C, the length ACand length AB remain unchanged,
the point Bcan also be at point Bwhere ABC= acute
andA BC= obtuse.IfABC= , thusABC= 180.
Remember : sin= sin (180)
Case 1: WhenabsinAbuta < b. Case 4: When a>bsinAanda > b.
CB cuts the side opposite to C at 2 points CB cuts the side opposite to C at 1 points
Outcome: Outcome:
2 solution 1 solution
Useful information:
In a right angled triangle, you may use the following to solve the
bc problems.
(i) Phythagoras Theorem: c= a2+b
2
a(ii)
Trigonometry ratio:
sin= b, cos= a, tan= bac c
(iii) Area = (base)(height)
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11 Index Number
Price Index Composite index
I = 1 100 WiIiI =P W
0 i
I =Price index/ Index number = Composite IndexIP0= Price at the base time W =Weightage
P1= Price at a specific time I =Price index
IA,BIB,C=IA,C100