Question

Determine whether the following is even or odd

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

• A. $f\left(x\right)$ is even function
• B. $f\left(x\right)$ is odd function
• C. $f\left(x\right)$ is neither even nor odd
• D. $f\left(x\right)$ is periodic function

Answer 1

Answer: (B)

$f\left(x\right)$ is odd function

$f\left(x\right)=\log (x+\sqrt{1+x^2})$
$f(-x)=\log (-x+\sqrt{1+x^2})$
$f\left(x\right)+f(-x)=\log \left(\right(x+\sqrt{1+x^2}\left)\right(-x+\sqrt{1+x^2}\left)\right)$
$=\log (1+x^2-x^2)$
$=\log \left(1\right)$
$=0$
Hence, $f\left(x\right)$ is odd.