Question

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

• A. even
• B. odd
• C. neither even nor odd
• D. even for only some values of $x$

Answer 1

Answer: (B)

odd

$f\left(x\right)=\log (x+\sqrt{1+x^2)}$
$f(-x)=\log (-x+\sqrt{1+x^2})$
$f(-x)=\log \left[\right(\sqrt{1+x^2}-x\left)\left(\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}\right)\right]$
$f(-x)=\log \left[\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}\right]$
$f(-x)=\log \left[\frac{1}{\sqrt{1+x^2}+x}\right]$
$f(-x)=-\log (\sqrt{1+x^2}+x)$
$\Rightarrow f(-x)=-f\left(x\right)$
Hence, $f$ is an odd function.