The Equation $11^x + 13^x + 17^x =19^x $
Has
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.
Define, $f(x) = \big(\frac{11}{19}\big)^x + \big(\frac{13}{19}\big)^x + \big(\frac{17}{19}\big)^x - 1$. Clearly $f$ is decreasing continuous function on whole $\mathbb{R}$. Therefore, $f$ can have at most one zero. See that $f(1) = \big(\frac{11}{19}\big) + \big(\frac{13}{19}\big) + \big(\frac{17}{19}\big) - 1 > 0$ and for any sufficiently large $a\in \mathbb{R}$, $f(a) = \big(\frac{11}{19}\big)^a + \big(\frac{13}{19}\big)^a + \big(\frac{17}{19}\big)^a - 1 < 0$. Hence, by intermediate value theorem, there exists at least one zero in the interval $(1, a)$. Thus, $f$ has exactly one zero.
Let $f(x)=11^x+13^x+17^x-19^x$. Then $f(3)=1582>0$, $f(4)=-3589<0.$ By Intermediate-value theorem $f(x)=0$ has at least one root in $(3,4)$. Let $x_0$ be that root.
Suppose that $x_0+\epsilon$ is another root. Let $\epsilon >0.$ Then $$19^{x_0+\epsilon}=11^{x_0+\epsilon}+13^{x_0+\epsilon}+17^{x_0+\epsilon}< (11^{x_0}+13^{x_0}+17^{x_0})19^{\epsilon}=19^{x_0+\epsilon}$$ with contradiction. Similarly, for $\epsilon<0$ case we can reach a contradiction. So there is one 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)<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)<0 for all x. Hence a=b. So f(x) has only one real root.
