Question

The function $\frac{e^{2x}-1}{e^{2x}+1}$ is

• A. Increasing
• B. Decreasing
• C. Even
• D. Strictly increasing

Answer 1

The correct option is D

Strictly increasing

Explanation for the correct option

The given function, $f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$.

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left(\frac{e^{2x}-1}{e^{2x}+1}\right) \\ \Rightarrow f\text{'}\left(x\right)=\frac{\left(e^{2x}+1\right){\displaystyle \frac{d}{dx}}\left(e^{2x}-1\right)-\left(e^{2x}-1\right){\displaystyle \frac{d}{dx}}\left(e^{2x}+1\right)}{{\left(e^{2x}+1\right)}^2}\left[\because \frac{d}{d\mathrm{x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v}{\displaystyle \frac{d\mathrm{u}}{d\mathrm{x}}}-\mathrm{u}{\displaystyle \frac{d\mathrm{v}}{d\mathrm{x}}}}{{\mathrm{v}}^2}\right] \\ =\frac{\left(e^{2x}+1\right)\times \left(2e^{2x}\right)-\left(e^{2x}-1\right)\times \left(2e^{2x}\right)}{{\left(e^{2x}+1\right)}^2} \\ =\frac{\left(2e^{4x}+2e^{2x}\right)-\left(2e^{4x}-2e^{2x}\right)}{{\left(e^{2x}+1\right)}^2} \\ =\frac{4e^{2x}}{{\left(e^{2x}+1\right)}^2} \\ \Rightarrow f\text{'}\left(x\right)={\left[\frac{2e^x}{\left(e^{2x}+1\right)}\right]}^2 \end{gathered}$$

There exists no real value of $x$, such that $\frac{2e^x}{\left(e^{2x}+1\right)}=0$.

Thus, $f\text{'}\left(x\right)\ne 0$.

As the square of an expression is always positive. thus ${\left[\frac{2e^x}{\left(e^{2x}+1\right)}\right]}^2>0$.

Hence, $f\text{'}\left(x\right)>0$ for all real values of $x$

So, the given function is strictly increasing.

Therefore, the function $\frac{e^{2x}-1}{e^{2x}+1}$ is strictly increasing.

Hence the correct option is option (D) i.e. strictly increasing