Explanation

(a)
Given, $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$
$\therefore\ \ \ \ \ \ f( - x) = \frac{e^{- 2x} - 1}{e^{- 2x} + 1} = - f(x)$
$\therefore\ \ \ \ \ f(x)$ is an odd function
Now, $f^{'}(x) = \frac{4e^{2x}}{\left( 1 + e^{2x} \right)^{2}} > 0,\ \forall x \in R$
$\Rightarrow \ \ \ f(x)$ is an increasing function