Question

Prove that $g\left(x\right)=\log (x+\sqrt{1+x^2})$ is an odd function.

Answer 1

Any function, $f\left(x\right)$ is an odd function if and only if $f(-x)=-f\left(x\right)$

Given,

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

Now, $g(-x)=\log (-x+\sqrt{1+(-x{)}^2})$

$\Rightarrow g(-x)=\log (\sqrt{1+(x{)}^2}-x)$

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

$\Rightarrow g(-x)=-\log (\frac{1}{\sqrt{1+x^2}-x}\times \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x})$

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

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

$\Rightarrow g(-x)=-g\left(x\right),\forall x$

This gives $g\left(x\right)$ is an odd function.