Question

Check whether the function $f\left(x\right)=\log (x+\sqrt{1+x^2})$ is odd or even.

Answer 1

Given the function,

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

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

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

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

$=-f\left(x\right)$.

So the function $f\left(x\right)$ is an odd function.