Question

Which of the following function is an odd function?

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

Answer 1

Answer: (A)

(a) $f(x)=\sqrt{1+x+x^2}-\sqrt{1-x+x^2}$
$f(-x)=\sqrt{1-x+x^2}-\sqrt{1+x+x^2}=-f(x)$
$\therefore f(x)$ is an odd function
(b) $f(x)=x\left(\frac{a^x+1}{a^x-1}\right)$
$\Rightarrow f(-x)=(-x)\left(\frac{a^{-x}+1}{a^{-x}-1}\right)$
$=(-x)\left(\frac{1+a^x}{1-a^x}\right)$
$=x\left(\frac{a^x+1}{a^x-1}\right)=f(x)$
$\therefore$ It is an even function
(c) $f(x)=\log \left(\frac{1-x^2}{1+x^2}\right)$
$\Rightarrow f(-x)=\log \left(\frac{1-x^2}{1+x^2}\right)=f(x)$
$\therefore$ It is an even function
(d) $f(x)=k\Rightarrow f(-x)=k=f(x)$
$\therefore$ It is an even function