Question

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

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

Answer 1

The correct option is A

Increasing

Explanation for the correct option:

Find the extrema of the given function:

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

$f\text{'}\left(x\right)=\frac{[e^{2x}+1]\left[2e^{2x}\right]-[e^{2x}-1]\left[2e^{2x}\right]}{{[e^{2x}+1]}^2}$ $\left[\mathbf{\because }\frac{\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}}\right]$

$f\text{'}\left(x\right)=\frac{2e^{2x}[e^{2x}+1-e^{2x}+1]}{{[e^{2x}+1]}^2}$

$f\text{'}\left(x\right)=\frac{4e^{2x}}{{[e^{2x}+1]}^2}$

Now, ${[e^{2x}+1]}^2>0$ and $e^{2x}>0$

$\Rightarrow$$f\text{'}\left(x\right)>0$

$\therefore f\left(x\right)$ is increasing function.

Hence, Option ‘A’ is Correct.