QJEEM19Ma46f8: JEE Main 2019 Mathematics - Class 11 JEEM19Ma46f8 Solution: If f ( x ) = log e ( 1 − x 1 + x ) ,

Question

Verified Solution • 2025–26 Session moderate

If

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

,

| x | < 1

then

f (\frac{2 x}{1 + x^{2}})

is equal to

Accepted Answer

Given,

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

f (\frac{2 x}{1 + x^{2}})

=

\ln (\frac{1 - \frac{2 x}{1 + x^{2}}}{1 + \frac{2 x}{1 + x^{2}}})

=

\ln (\frac{x^{2} - 2 x + 1}{x^{2} + 2 x + 1})

=

\ln {(\frac{1 - x}{1 + x})}^{2}

=

2 \ln (\frac{1 - x}{1 + x})

= 2f(x)

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