Question

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

Answer 1

The correct option is C

$f\left(x\right)=x\frac{\left(a^x-1\right)}{a^x+1}$

Explanation for Correct Option:

For an even function, $f\left(-x\right)=f\left(x\right)$.

Option $C$,

$f\left(x\right)=x\frac{\left(a^x-1\right)}{a^x+1}$

$\begin{array}{rcl}f\left(-x\right)& =& -x\frac{\left(a^{-x}-1\right)}{a^{-x}+1}\\ & =& -x\frac{\left(1-a^x\right)}{1+a^x}\\ & =& x\frac{\left(a^x-1\right)}{a^x+1}\end{array}$

So, $f\left(-x\right)=f\left(x\right)$.

Explanation for Incorrect Option:

Option $A$

$f\left(x\right)=\frac{a^x+a^{-x}}{a^x-a^{-x}}$

$\begin{array}{rcl}f\left(-x\right)& =& \frac{a^{-x}+a^x}{a^{-x}-a^x}\\ & =& \frac{a^{2x}+1}{1-a^{2x}}\end{array}$

So, $f\left(-x\right)\ne f\left(x\right)$.

Option $B$,

$f\left(x\right)=\frac{a^x+1}{a^x-1}$

$\begin{array}{rcl}f\left(-x\right)& =& \frac{a^{-x}+1}{a^{-x}-1}\\ & =& \frac{a^x+1}{1-a^x}\end{array}$

So, $f\left(-x\right)\ne f\left(x\right)$.

Option $D$,

$f\left(x\right)={\log }_2\left(x+\sqrt{x^2+1}\right)$

$\begin{array}{rcl}f\left(-x\right)& =& {\log }_2\left(-x+\sqrt{{\left(-x\right)}^2+1}\right)\\ & =& {\log }_2\left(-x+\sqrt{x^2+1}\right)\end{array}$

So, $f\left(-x\right)\ne f\left(x\right)$.

Hence, option $C$ is the correct answer.