If $\mathrm{f}(x)=\log _{\mathrm{c}}\left(\frac{1-x}{1+x}\right),|x| \lt 1$, then $\mathrm{f}\left(\frac{2 x}{1+x^2}\right)$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} f\left(\frac{2 x}{1+x^2}ight) & =\log _{\mathrm{e}}\left(\frac{1-\frac{2 x}{1+x^2}}{1+\frac{2 x}{1+x^2}}ight) \ & =\log _{\mathrm{e}}\left(\frac{(1-x)^2}{(1+x)^2}ight) \ & =2 \log _{\mathrm{e}}\left(\frac{1-x}{1+x}ight) \ & =2 \mathrm{f}(x)\end{aligned}$

If $\mathrm{f}(x)=\log _{\mathrm{c}}\left(\frac{1-x}{1+x}\right),|x| \lt 1$, then $\mathrm{f}\left(\frac{2…

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