Find Number Of Roots of Equation $11^x + 13^x + 17^x =19^x $ The Equation
$11^x + 13^x + 17^x =19^x  $

Has 


No Real Roots
Only One Real Roots
Exactly Two Real Roots 
More than Two Real Roots 


What I have done is 

The function $f(x)=11^x + 13^x + 17^x -19^x  $ is strictly increasing and being always positive it have no roots
I'm afraid your answer is wrong. 

One way to see that is consider the function

$$f(x)=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$$

is strictly decreasing and $f(-\infty)=\infty$, $f(\infty)=-1$.

So there is only ONE real root.
Let y= f(x)= (11÷19)^x+(13÷19)^x+(17÷19)^x-1.
When we put x=0 then  y=2 and when x->infinity then  y->-1 so graph of f(x) cuts the x axis. Hence it has atleast one real root. So option 1 is wrong.
Also f'(x)&lt;0 for all x
Let us suppose that f(x) has two real roots a,b such that   f(a)=0 and f(b)=0. So f:[a,b]->R is continuous in [a,b] and differentiable in (a,b) then by mean value theorem, there exist c in (a,b) such that 
(f(b)-f(a))=(b-a)f'(c) => (b-a)f'(x)=0 but f'(x)&lt;0 for all x. Hence a=b. So f(x) has only one real root.