Question

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

• A. True
• B. False

Answer 1

The correct option is A True

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

Now

$f(-x)$

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

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

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

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

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