Preliminary Concepts (CALCULUS)

THESE ARE THE FUNDAMENTAL CONCEPTS

Fraction expression,

if, 

We already know we cant divide any number by " zero "
i am trying to divide but could not find any solution and reach the conclusion hahaha ... infinity (∞)(not defined)

so first thing we do is to find the values for which denominator becomes zero.. these values certainly will not be in the domain of solution.

take a example

1.

what we do, we cancel the term  from both the sides and reach the conclusion that x = 2.... at least I do the same .....:)

but is it really possible for above equation that x = 2 is the solution??

if we put x = 2 on above equation,

what will we get ?

denominator of 2nd term and fourth term will become zero... and become infinity (not defined)...
and you cant subtract infinity from infinity...

so we reach the conclusion that x ≠ 2

for any other value other than x = 2, we can cancel  from both the sides... and get x = 2 which is not possible for above equation.
Therefore the Equation has no solution

but, if the equation is like,



=> x ≠ 2
and for x is not equal to 2, you can cancel  from both sides and you will get x = 3, which will be the solution of this equation.



2. Euclid Theorem: "Equal Things are equal to each other"

Examples: Find the no. of solutions of  ?

What is called solution????

Solution are the values of variable 'x' for which the given equation is TRUE...

   1.  

     first see the Left Hand Side i.e. L.H.S.

       x²   , square of any no. is positive and of zero is zero

     therefore, x²  ≥ 0

     |x| is modulus function which is always positive or equal to zero if x = 0

    Therefore, |x| ≥ 0

     => x²  + 4|x| + 5  ≥ 5

    Therefore, LHS never be Zero

     For no value of "x" the above expression is zero

     NO SOLUTION....

2. 2^x + 6^x  = 7^x

LHS = Always Even Number
RHS = Always Odd Number

Means LHS never equal to RHS

No solution


3. |x|^{(x^2 + 2x + 3)}  = - 5

LHS = always positive because |x| is positive and any power of positive number is positive

No solution

4. Value of √x is always positive

√(x²  + 2x + 1) = -5

Not possible because LHS will always be positive.
No Solution

<sup>but 

if</sup> x² = 4
x has two values

i.e. x = + 2

graph of √x

Y-axis - VERTICAL AXIS... gives the value of y
X-axis - HORIZONTAL AXIS... gives the value of x


  we can see from the graph that for every value of y there is only one positive x ... this means y = √x is always positive. We will later study it in the functions.

it is clear from graph y = x²  that for every value of y there are two values of x
 if we draw a line parallel to x -axis it will cut the graph on two points....

5. √-a x √-b = - √ab

√ai x √bi = √ab i²  = - √ab



6. Always try to avoid extra calculation

x = y

=> logx = logy

or sin^-1x  =  sin^-1y

first equation is true for all x and y ..

but second equation is true only for x > 0 and y >0 and third equation is true for -1 ≤ x ≤ 1 and -1 ≤ y ≤ 1 

7. For two positive numbers

A.M ≥ G.M.

e.g. 2^x + 2^{-x =} sinx

take LHS = 2^x + 2^-x 

 2^x ≥ 0 and 2^-x ≥ 0  i.e. both are positive

therefore, for these two positive numbers 
A.M ≥ G.M.

=> 

or, 2^x + 2^-x ≥  2 => LHS ≥  2
             
and RHS = sinx 
and we know -1 ≤sinx ≤ 1 => -1 ≤RHS ≤ 1

therefore LHS will never be equal to RHS
NO SOLUTION

8. 5^x + 5^-x  = sine^x
clearly LHS will always be greater than 2 and RHS will never be greater than 1
NO SOLUTION