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)

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