Question

If $f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)$ for $x_1,x_2ϵ$ (-1, 1), then f(x) is

• A. $log\frac{1-x}{1+x}$
• B. $log\frac{2+x}{1-x}$
• C. $ta{n}^{-1}\frac{1-x}{1+x}$
• D. $ta{n}^{-1}\frac{1+x}{1-x}$

Answer 1

The correct option is A $log\frac{1-x}{1+x}$

We have, $f\left(x\right)=log\frac{1-x}{1+x}$
$f\left(x_1\right)-f\left(x_2\right)=log\frac{1-x_1}{1+x_1}-log\frac{1-x_2}{1+x_2}$
$=log\frac{(1-x_1)(1+x_2)}{1+x_1-x_2-x_1{x}_2}$
Also, $f\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)=log\frac{1-\frac{x_1-x_2}{1-x_1{x}_2}}{1+\frac{x_1-x_2}{1-x_1{x}_2}}$
$=log\frac{1-x_1{x}_2-x_1+x_2}{1-x_1{x}_2+x_1-x_2}$
$\therefore$ (a) is true