Question

Given function $f\left(x\right)=\left(\frac{e^{2x}-1}{e^{2x}+1}\right)$ is

• A. Increasing
• B. Decreasing
• C. Even
• D. None of these

Answer 1

The correct option is A Increasing

$f\left(x\right)=\left(\frac{e^{2x}-1}{e^{2x}+1}\right)\Rightarrow \left(i\right)f(-x)=\left(\frac{e^{-2x}-1}{e^{-2x}+1}\right)=\frac{1-e^{2x}}{1+e^{2x}}\Rightarrow f\left(x\right)=-\frac{e^{2x}-1}{e^{2x}+1}=-f\left(x\right)$
f(x) is an odd function.
Again $f\left(x\right)=\left(\frac{e^{2x}-1}{e^{2x}+1}\right)\Rightarrow f^{\prime }\left(x\right)=\frac{4e^{2x}}{(1+e^{2x}{)}^2}>0\forall n\in R$
$\Rightarrow f\left(x\right)$ is an increasing function.