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Q. If $f\left(x\right)=\log_{e} \left(\frac{1-x}{1+x}\right), \left|x\right|<1 , $ then $ f\left(\frac{2x}{1+x^{2}}\right) $ is equal to :

JEE MainJEE Main 2019Relations and Functions

Solution:

$f\left(x\right)=\log_{e} \left(\frac{1-x}{1+x}\right), \left|x\right|<1 $
$ f\left(\frac{2x}{1+x^{2}}\right) =\ell n \left(\frac{1- \frac{2x}{1+2x^{2}}}{1+ \frac{2x}{1+x^{2}}}\right) $
$ =\ell n \left(\frac{\left(x-1\right)^{2}}{\left(x+1\right)^{2}}\right)=2\ell n \left|\frac{1-x}{1+x}\right| =2f\left(x\right)$