Q. The function $f(x) = \log\left( \frac{1 + x}{1 - x} \right)$ satisfies the equation
A) $f(x + 2) - 2f(x + 1) + f(x) = 0$
B) $f(x) + f(x + 1) = f\{ x(x + 1)\}$
C) $f(x) + f(y) = f\left( \frac{x + y}{1 + xy} \right)$
D) $f(x + y) = f(x)f(y)$
Explanation
(c)
$f\left( \frac{x + y}{1 + xy} \right) = \log\left( \frac{1 + \frac{x + y}{1 + xy}}{1 - \frac{x + y}{1 + xy}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \log\left( \frac{1 + xy + x + y}{1 + xy - x - y} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \log\left( \frac{(1 + x)(1 + y)}{(1 - x)(1 - y)} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \log{\left( \frac{1 + x}{1 - x} \right) + \log\left( \frac{1 + y}{1 - y} \right)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = f(x) + f(y)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.