Explanation

(a)
Given, $2\cos{\left( e^{x} \right) = 5^{x} + 5^{- x}}$
Since, $\cos{e^{x} \leq 1\ \ \Rightarrow \ 2\cos{e^{x} \leq 2}}$ …(i)
And $\frac{5^{x} + 5^{- x}}{2} \geq \sqrt{5^{x}.5^{- x}}$
$\Rightarrow \ \ 5^{x} + 5^{- x} \geq 2$
$\therefore$ LHS $\leq 2$, RHS$\geq 2$
Now, $5^{x} + \frac{1}{5^{x}} = 2$ at $x = 0$
But, at $x = 0$
$2\cos{e^{x} \neq 2}$
Hence, no solution will exist