JEE MATHEMATICS MCQ: FUNCTIONS

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Application +4 | -1

f(x)=\log _e\left(\frac{1-x}{1+x}\right),|x|<1

, then

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

is equal to

View answer and explanation

Correct option: A

2 f(x)

Option A (

2 f(x)

) is correct because substituting

\frac{{2x}}{{1+x^2}}

into the function and simplifying using logarithm properties results in

2f(x)

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