Question

The function $f\left(x\right)=e^{ax}+e^{-ax}$, $a>0$ is monotonically increasing for

• A. $–1<x<1$
• B. $x<–1$
• C. $x>–1$
• D. $x>0$

Answer 1

The correct option is D

$x>0$

Explanation for the correct option :

Given function $f\left(x\right)=e^{ax}+e^{-ax}$

On, differentiating with respect to $x$, we get,

We know that, for $f\left(x\right)$ to be increasing, $f\text{'}\left(x\right)>0$

$$\begin{gathered} f\text{'}\left(x\right)=e^{ax}+e^{-ax} \\ =a{e}^{ax}-a{e}^{-ax} \\ =a(e^{ax}-e^{-ax}) \end{gathered}$$

$f\text{'}\left(x\right)>0$ then the function is increasing.

$$\begin{gathered} \Rightarrow a(e^{ax}-e^{-ax})>0 \\ \Rightarrow e^{ax}-e^{-ax}>0 \\ \Rightarrow e^{ax}>e^{-ax} \\ \Rightarrow e^{2ax}>1 \end{gathered}$$

For $e^{2ax}$ to be greater than $1$, $x$ should be greater than $0$

Thus, $f\left(x\right)$ is strictly increasing for $x>0$.

Hence, the correct option is $\mathrm{Option}\left(\mathrm{D}\right)$.