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

Answer: (A)

Increasing

$f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$
$f\left(x\right)=\frac{[e^{2x}+1]\left[2e^{2x}\right]-[e^{2x}-1]\left[2e^{2x}\right]}{{[e^{2x}+1]}^2}$
$f^1\left(x\right)=\frac{2e^{2x}[e^{2x}+1-e^{2x}+1]}{{[e^{2x}+1]}^2}$
$f^1\left(x\right)=\frac{4e^{2x}}{{[e^{2x}+1]}^2}$
Now, ${[e^{2x}+1]}^2>0$ and $e^{2x}>0$
$\therefore f^1\left(x\right)>0$
$\therefore f\left(x\right)$ is increasing function.