Question

If $f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_2{x}_2}\right)$ for $x_1,x_2\in [-1,1]$, then $f(x)$ is equal to :

• A. ${\tan }^{-1}\frac{(1+x)}{(1-x)}$
• B. ${\tan }^{-1}\frac{(1-x)}{(1+x)}$
• C. $\log \frac{(1+x)}{(1-x)}$
• D. none of these

Answer 1

Answer: (B)

We have, $f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)$
Since, $x_1,x_2\in [-1,1]$
Let $x_1=-1$ and $x_2=1$, then
$f(-1)-f(1)=f\left[\frac{-1-1}{1+1}\right]$
$=f(-1)$
$\Rightarrow f(1)=0$
Now take, $f(x)={\tan }^{-1}\left(\frac{1-x}{1+x}\right)$
$f(1)={\tan }^{-1}\left(\frac{1-1}{1+1}\right)={\tan }^{-1}0$
$=0$
Which is satisfied.