Question

Consider the following functions are odd function in their default domains
(i) $\frac{2^x-1}{2^x+1}$
(ii) $\frac{x^2+1}{x\sin x}$
(iii) $\ln \left(\frac{1+x}{1-x}\right)$
(iv) $x{e}^{\left|x\right|+\cos x}$
Which of these is/are odd

• A. (i) and (iii)
• B. (i) and (iv)
• C. all four
• D. (i), (iii) and (iv)

Answer 1

The correct option is D (i), (iii) and (iv)

$\left(i\right)$

$f\left(x\right)=\frac{2^x-1}{2^x+1}$

$f(-x)=\frac{2^{-x}-1}{2^{-x}+1}=\frac{1-2^x}{1+2^x}$

$⟹f(-x)=-\frac{2^x-1}{2^x+1}=-f\left(x\right)$

Hence is an odd function.

$\left(ii\right)$

$f\left(x\right)=\frac{x^2+1}{x\sin x}$

$f(-x)=\frac{(-x{)}^2+1}{(-x)\sin (-x)}$

$⟹f(-x)=\frac{x^2+1}{x\sin x}=f\left(x\right)$

Hence is an even function.

$\left(iii\right)$

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

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

$⟹f(-x)=-\ln \left(\frac{1+x}{1-x}\right)=-f\left(x\right)$

Hence is an odd function.

$\left(iv\right)$

$f\left(x\right)=x{e}^{|x|+\cos x}$

$f(-x)=(-x)e^{|-x|+\cos (-x)}$

$f(-x)=-x{e}^{|x|+\cos x}=-f\left(x\right)$

Hence is an odd function.

Hence (i), (iii) and (iv) are odd functions.

Answer-(D)