Question

Determine whether $g\left(x\right)=\sin \log [x+\sqrt{x^2+1}]$ is an even function or odd or neither .

Answer 1

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

$=\log \left[\frac{1}{\sqrt{1+x^2}+x}\right]$

$\Rightarrow f(-x)=-f\left(x\right)$
Thus $f\left(x\right)$ is an odd function.

Now $g\left(x\right)=\sin \left\{f\right(x\left)\right\}$

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

$=-\sin \left\{f\right(x\left)\right\}$ (Since, sine function is an odd function)

$\Rightarrow g(-x)=-g\left(x\right)$
Hence $g\left(x\right)$ is an odd function.