Question

Which among the following is/are odd functions?

• A. $g\left(x\right)=\frac{a^x-1}{a^x+1}$
• B. $f\left(x\right)=x\left(\frac{e^x-1}{e^x+1}\right)$
• C. $h\left(x\right)=\log (x^6+\sqrt{x^2+1})$
• D. $k\left(x\right)=x(x^2-|x|)$

Answer 1

Answer: (A), (D)

$k\left(x\right)=x(x^2-|x|)$

Clearly $h\left(x\right)=\log (x^6+\sqrt{x^2+1})$ is an even function because it has even powers of $x$

For $f\left(x\right)=x\left(\frac{e^x-1}{e^x+1}\right)$
$f(-x)=-x\left(\frac{e^{-x}-1}{e^{-x}+1}\right)=-x\left(\frac{1-e^x}{e^x+1}\right)=f\left(x\right)$
$\therefore$ $f\left(x\right)$ is even function.

For $g\left(x\right)=\frac{a^x-1}{a^x+1}$
$g(-x)=\frac{a^{-x}-1}{a^{-x}+1}\Rightarrow g(-x)=\frac{1-a^x}{1+a^x}\therefore g(-x)=-g\left(x\right)$
$\therefore g\left(x\right)$ is odd function.

For $k\left(x\right)=x(x^2-|x|)$
$k(-x)=-x(x^2-|x|)\Rightarrow k(-x)=-k\left(x\right)$
$\therefore k\left(x\right)$ is odd function.