Question

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

• 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 nature of the given function:

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

On differentiating it with respect to $x$, we get
$f\text{'}\left(x\right)=\frac{\left(e^{2x}+1\right)\left(2e^{2x}\right)-\left(e^{2x}-1\right)\left(2e^{2x}\right)}{{\left(e^{2x}+1\right)}^2}$ ; $\left[\mathbf{\because }\frac{\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}\frac{\mathbf{u}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{v}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}{\mathbf{v}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{2}}}\right]$
$\Rightarrow$$f\text{'}\left(x\right)=\frac{2e^{2x}\left(e^{2x}+1-e^{2x}+1\right)}{{\left(e^{2x}+1\right)}^2}$
$\Rightarrow$$f\text{'}\left(x\right)=\frac{4e^{2x}}{{\left(e^{2x}+1\right)}^2}$
Now, As we know

${\left(e^{2x}+1\right)}^2>0$ and $e^{2x}>0$ ; $[\because$exponential functions are always $>0$$]$

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

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

Hence, Option ‘A’ is Correct.