FS MATHS : BINOMIAL SQUARE EXPANSION & EXPANSION OF TWO BRACKETS

FS MATHS is a simple method how to solve Binomial Square Expansion & Expansion Two Brackets 

STEP BY STEP...LET WATCH IN YOU TUBE

Indices & Logarithm

Let's make it easy to solve 'Indices Equation'...u just remember 3 types of indices equation...

First : By equating the base

a^x = a

Example:
Solve the equation of 2^(x+1) = 4^x

Solution2^(x+1) = 4^x
2^(x+1) = 2^(2x)
x+1 = 2x
Therefore, x = 1

Second : Using Logarithm
a^x = b

Example:

Solve the equation 5^x = 8

Solution:

5^x = 8
log 5^x = log 8
x log 5 = log 8
x = log8/log5

Therefore, x = 1.292


Third : Factorise

a^x +ba^x + c
Example :

Given 2^(2x) - 5(2^x) + 4 = 0. Find the value of x

Solution:

2^(2x) - 5(2^x) + 4 = 0
(2^x)^2 - 5(2^x) + 4 = 0

Replace y = 2^x

y^2 - 5y + 4 = 0
(y - 1)(y - 4) = 0
y = 1 or y = 4
2^x = 1 2^x = 4
2^x = 2^0 2^x = 2^2
x = 0 x = 2

Indices And Logarithms

If a is a real number and n is a positive integer, then

The number a is called base and n is called index.

LAWS OF INDICES

LAWS OF LOGARITHMS

Assume that x and y is a real number

ARGAND DIAGRAM

MODULUS AND ARGUMENT OF COMPLEX NUMBER




ADDITION AND SUBTRACTION OF COMPLEX NUMBERS ON
ARGAND DIAGRAM





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Examples Of Indices Equation

Complex Number

COMPLEX NUMBERS
- A complex number is written in the form of a + bi where a and b are real numbers.
- a is called real part and bi is called imaginary part
- i = and i^2 = -1
- Generally, ( -1 )^even no. = 1
( -1 )^odd no. = -1

For the quadratic equation, ax^2 + bx + c =0, we are use the formula below to solve the equation

Addition and Subtraction of Complex Numbers

If z = x + yi and w = u + vi,

thus,
z + w = x + yi + u + vi
= (x + u) + (y + v)i
z – w = x + yi - u + vi
= (x - u) + (y - v)i

Multiplication of Complex Numbers

i) If z = x + yi and w = u + vi,

thus,

zw = (x + yi)( u + vi)
= (x + yi)(u) + (x + yi)(vi)

ii) If z = x + yi and w = x – yi,

thus

zw = ( x + iy) ( x – iy )
= x ^2 - (yi)^2
= x^ 2 + y^2 ( real number)

So, w is known as complex conjugate of z


Division of Complex Numbers

i. If z = x + yi and w = u + vi

thus,


where u – vi is conjugate of w

ii. For the division process, the denominator must be a real number


Equality of Complex Numbers

i) Let say z = x + yi and w = u + vi where z = w

thus,

x + yi = u + vi
x – u = (v – y)i

ii) Therefore, x + yi = u + vi if and only if x = u and y = v

Pascal Triangle

We note that the coefficients (the numbers in front of each term) follow a pattern.

(a + b)^0 1
(a + b)^1 1 1
(a + b)^2 1 2 1
(a + b)^3 1 3 3 1
(a + b)^4 1 4 6 4 1
(a + b)^5 1 5 10 10 5 1
(a + b)^6 1 6 15 20 15 6 1


You can use this pattern to form the coefficients.

Notes :

- There are n + 1 terms.
- the exponent of a decrease by 1 from term to term while the exponent of b increases by 1
- __a^n +__a^(n-1)b+__a^(n-2)b^2+__a^(n-3)b^3+.............+ ___b^n


Examples :

Expand (x + 3)^4 by using the Pascal Triangle


Solution :
Step (1) : Draw a Pascal Triangle ( Refer above)

Step (2) : Create a formula of an expansion (there are n + 1 terms...so we have four terms)

(a + b)^4 = ___a^n + ___a^(n-1)b + ___a^(n-2)b^2 +___a^(n-3)b^3 + __b^4

Step (3) : Replace a = x, b = 3 and n = 4 into step 3. Also put the coefficient (refer Pascal

Triangle)on the underline in the formula

So,

(x + 3)^4 = 1x^4 + 4x^3(3) + 6x^2(3^2) + 4x(3^3) + 1(3^4)
= x^4 + 12x^3 + 54x^2 + 108x + 81

Rules of Differentiation for Algebraic Function

1 - Derivative of a constant function.

The derivative of f(x) = c where c is a constant.

f '(x) = 0

Example :

f(x) = 5 , then f '(x) = 0

2 - Derivative of a power function.

The derivative of f(x) = x^n where n is a constant real number.

f '(x) = n x ^(n- 1)

Example :

f(x) = x^7

then,

f '(x) = 7 x^(7-1)

= 7x^6

3 - Derivative of the sum of functions

The derivative of f(x) = g(x) + h(x) is given by

f '(x) = g '(x) + h '(x)

Example:

f(x) = 3x^4 + 2x

let g(x) = 3x^4 and h(x) = 2x

then,

f '(x) = g '(x) + h '(x)

= 12x^3 + 2

4 - Derivative of the difference of functions.

The derivative of f(x) = g(x) - h(x) is given by

f '(x) = g '(x) - h '(x)

Example:

f(x) = 5x - x^-2

let g(x) = 5x and h(x) = x^-2

then,

f '(x) = g '(x) - h '(x)

= 5 -(-2x^-3)
= 5 + 2^-3

5 - Derivatives of a composite functions

Example :

f(x) = (2x^3 + 5)^4

let a = 2, k = 4 and n = 3

thus,

f'(x) = kanx^(n-1)(ax^n + b)^(k-1)

= 4(2)(3)x^2(2x^3 + 5)^3

= 24x^2 (2x^3 + 5)^3

6 - Derivative of the product of two functions

The derivative of f(x) = g(x) h(x) is given by

f '(x) = g(x) h '(x) + h(x) g '(x)

Example:

7 - Derivative of the quotient of two functions

Example :

First Principles

Consider that y = f(x) and point P (x , y ) on a curve as at the figure 2.1.

If x increase to and y increase to , thus the new coordinates is becomes
When Q approaches the point P, will approaches to zero. And it’s written as

Therefore, from the limit idea, derivatives represent the slope of curve at a point.

So,

And the First Principles Formulae is

Example :

Differentiate the function below.

Solution :