JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: C

2f\left( x \right)

Option C (2f(x)) is correct because by substituting (2x)/(1+x²) into f(x) and using logarithm properties, we can simplify the expression to 2f(x).

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