Q. If $f(x) = \log\left( \frac{1 + x}{1 - x} \right)$, then $f\left( \frac{2x}{1 + x^{2}} \right)$ will be equal to
A) $2f(x^{2})$
B) $f(x^{2})$
C) $2f(2x)$
D) $2f(x)$
Explanation
(d)
$f\left( \frac{2x}{1 + x^{2}} \right) = \log{\left\lbrack \frac{1 + \frac{2x}{1 + x^{2}}}{1 - \frac{2x}{1 + x^{2}}} \right\rbrack =}\log\left( \frac{1 + x}{1 - x} \right)^{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f\left( \frac{2x}{1 + x^{2}} \right) = 2f(x)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.