JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: A

2f(x)

Option A (

2f(x)

) is correct because substituting

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

into the function and simplifying yields twice the original function.

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