Question

The function $\mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\right)$, is

• A. neither an even nor an odd function
• B. an even function
• C. an odd function
• D. a periodic function

Answer 1

The correct option is C

an odd function

Step 1. Explanation for correct option

For option C

Determine the function is odd, even, or periodic form.

It is given that the function $\mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\right)$.

Now,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left(-\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)$

For odd function: A function is odd if $\mathrm{f}\left(-\mathrm{x}\right)=-\mathrm{f}\left(\mathrm{x}\right)$ for all $\mathrm{x}\in \mathrm{R}$.

We have to check out the function,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x}\right)\left(\frac{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

Use the property $\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2$, we get,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\frac{1+{\mathrm{x}}^2-{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=-\log \left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right)$

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=-\mathrm{f}\left(\mathrm{x}\right)$

The $\mathrm{f}\left(\mathrm{x}\right)$ is an odd function.

Step 2. Explanation for Incorrect options

For option A .

Since the function is odd, option A is incorrect

For option B

It is given that the function $\mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\right)$.

Now,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left(-\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}\right)$

For even function: A function is even if $\mathrm{f}\left(-\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)$ for all $\mathrm{x}\in \mathrm{R}$.

We have to check out the function,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\sqrt{1+{\mathrm{x}}^2}-\mathrm{x}\right)\left(\frac{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

Use the property $\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)={\mathrm{a}}^2-{\mathrm{b}}^2$, we get,

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\frac{1+{\mathrm{x}}^2-{\mathrm{x}}^2}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=\log \left[\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}}\right)\right]$

$\Rightarrow$ $\mathrm{f}\left(-\mathrm{x}\right)=-\log \left(\sqrt{1+{\mathrm{x}}^2}+\mathrm{x}\right)$

$\Rightarrow$ $$\begin{gathered} \mathrm{f}\left(-\mathrm{x}\right)=-\mathrm{f}\left(\mathrm{x}\right) \\ \ne \mathrm{f}\left(\mathrm{x}\right) \end{gathered}$$

The $\mathrm{f}\left(\mathrm{x}\right)$ is an not even function.

So option B is incorrect

For option D

For $\mathrm{f}\left(\mathrm{x}\right)=\log \left(\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\right)$

$\mathrm{f}\left(\mathrm{x}+\mathrm{c}\right)\ne \mathrm{f}\left(\mathrm{x}\right)$

Option $\mathrm{D}$ is incorrect answer, because, a function is periodic if, for some given constant $\mathrm{c}$, $\mathrm{f}\left(\mathrm{x}+\mathrm{c}\right)=\mathrm{f}\left(\mathrm{x}\right)\forall \mathrm{x}$, the smallest such positive value of $\mathrm{c}$ is called the period of the function.

Hence, option $\left(\mathrm{C}\right)$ is correct answer.