Question

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

• A. an increasing function
• B. a decreasing function
• C. an even function
• D. None of these

Answer 1

Answer: (A)

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