Question

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

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

Answer 1

The correct option is B

an odd function

Explanation for the correct option:

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

A function

is odd if $f\left(-x\right)=-f\left(x\right)\Rightarrow f\left(x\right)+f\left(-x\right)=0$

is even if $f\left(-x\right)=f\left(x\right)\Rightarrow f\left(x\right)+f\left(-x\right)=2f\left(x\right)$

We have,
$\begin{array}{rcl}f\left(-x\right)& =& \log \left[\left(-x\right)+\sqrt{{\left(-x\right)}^2+1}\right]\\ & =& \log \left[-x+\sqrt{x^2+1}\right]\end{array}$

Therefore,
$\begin{array}{c}f\left(x\right)+f\left(-x\right)=\log \left[x+\sqrt{x^2+1}\right]+\log \left[-x+\sqrt{x^2+1}\right]\\ =\log \left[\left(x+\sqrt{x^2+1}\right)\left(-x+\sqrt{x^2+1}\right)\right]\\ =\log \left({\left(\sqrt{x^2+1}\right)}^2-x^2\right)\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ =\log \left(x^2+1-x^2\right)\\ =\log \left(1\right)\\ =0\end{array}$

Thus, $f\left(x\right)$ is odd.

Hence, option(B) i.e. an odd function is correct.