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Q.
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 :
Bihar CECEBihar CECE 2003
Solution:
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.