Question

The minimum value of $\sqrt{e^{x^2}-1}$ is

• A. $1$
• B. $0$
• C. $2$
• D. None of these.

Answer 1

The correct option is B

$0$

Find the minimum value of the given function

Given : $f\left(x\right)\hspace{0.17em}=\sqrt{e^{x^2}-1}$

Differentiate the function with respect to $x$,

$f\text{'}\left(x\right)\hspace{0.17em}=\frac{\left(x\right)e^{x^2}}{\sqrt{e^{x^2}-1}}\left[\frac{d}{dx}{x}^n=n.x^{n-1},\frac{d}{dx}{e}^y=e^y\frac{dy}{dx}\right]$

We can see that,

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

$f\text{'}\left(x\right)<0$ if $x<0$.

Therefore at $x=0$ function have minimum value.

Now, evaluate the minimum value of the function

$$\begin{gathered} f\left(x\right){\hspace{0.17em}}_{min}=\sqrt{e^{0^2}-1} \\ \Rightarrow f\left(x\right){\hspace{0.17em}}_{min}=\sqrt{1-1} \\ \Rightarrow f\left(x\right){\hspace{0.17em}}_{min}=0 \end{gathered}$$

Therefore, the minimum value of the function is $0$.

Hence option $\left(B\right)$ is the correct answer.