If f ( x ) = log e ( 1 - x 1 + x ) , | x | 1 , then f ( 2 x 1 + x 2 ) is equal to

Question

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

Smart Achievers · Accepted Answer

(4)

Here, $f (x) = \log_{e} (\frac{1 - x}{1 + x})$

Now, $f (\frac{2 x}{1 + x^{2}}) = \log_{e} (\frac{1 - \frac{2 x}{1 + x^{2}}}{1 + \frac{2 x}{1 + x^{2}}})$

$= \log_{e} (\frac{1 + x^{2} - 2 x}{1 + x^{2} + 2 x}) = \log_{e} {(\frac{1 - x}{1 + x})}^{2}$

$= 2 \log_{e} (\frac{1 - x}{1 + x}) = 2 f (x)$

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