Question

The even function of the following is:

• A. $f\left(x\right)=\frac{a^x+a^{-x}}{a^x-a^{-x}}$
• B. $f\left(x\right)=\frac{a^x+1}{a^x-1}$
• C. $f\left(x\right)=x.\frac{a^x-1}{a^x+1}$
• D. $f\left(x\right)={\log }_2(x+\sqrt{x^2+1})$

Answer 1

The correct option is C $f\left(x\right)=x.\frac{a^x-1}{a^x+1}$

Consider an even function $f\left(x\right)$.
Then, $f\left(x\right)$ follows the following relation.
$f\left(x\right)=f(-x)$.

Consider option A
$f\left(x\right)=\frac{a^x+a^{-x}}{a^x-a^{-x}}$
$f(-x)=\frac{a^{-x}+a^x}{a^{-x}-a^x}$
$=-\left[\frac{a^x+a^{-x}}{a^x-a^{-x}}\right]$ $=-f\left(x\right)$

This is an odd function and not an even function.
Consider option B
$f\left(x\right)=\frac{a^x+1}{a^x-1}$
$f(-x)=\frac{a^{-x}+1}{a^{-x}-1}$
$=\frac{1+a^x}{1-a^x}$ $=-\left[\frac{a^x+1}{a^x-1}\right]$ $=-f\left(x\right)$

This is also an odd function and not an even function.
Consider option C
$f\left(x\right)=x\left(\frac{a^x-1}{a^x+1}\right)$
$f(-x)=-x\left(\frac{a^{-x}-1}{a^{-x}+1}\right)$ $=-x\left(\frac{1-a^x}{1+a^x}\right)$
$=x\left(\frac{a^x-1}{a^x+1}\right)$ $=f\left(x\right)$

This is an even function.

Consider option D
$f\left(x\right)={\log }_2(x+\sqrt{x^2+1})$
$f(-x)={\log }_2(-x+\sqrt{x^2+1})$
Now
$f\left(x\right)+f(-x)={\log }_2(x+\sqrt{x^2+1})+{\log }_2(-x+\sqrt{x^2+1})$
$={\log }_2\left[\right(x+\sqrt{x^2+1}\left)\right(-x+\sqrt{x^2+1}\left)\right]$
$={\log }_2(-x^2+(x^2+1\left)\right)$ $={\log }_2\left(1\right)$ $=0$

Hence
$f\left(x\right)+f(-x)=0$
Or
$f\left(x\right)=-f\left(x\right)$.
Hence it is an odd function.