Question

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

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

Answer 1

Answer: (B)

We have, $f(x)=\log \left(x+\sqrt{x^{2}+1}\right)$
$ \therefore f(-x)=\log \left(-x+\sqrt{x^{2}+1}\right) \times \frac{\left(x+\sqrt{x^{2}+1}\right)}{\left(x+\sqrt{x^{2}+1}\right)}$
$=\log \left(\frac{-x^{2}+x^{2}-1}{x+\sqrt{x^{2}+1}}\right)=-\log \left(x+\sqrt{x^{2}+1}\right) $
$\Rightarrow$ $f(-x)=-f(x)$
$ \therefore f(x)$ is an odd function.
(i) If $f(-x)=f(x)$, is an even function.
(ii) If $f(x+T)=f(x)$, is a periodic function with period $T$