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JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

If

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

, then

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

will be equal to

View answer and explanation

Correct option: D

2f\left( x \right)

Option D (

2f\left( x \right)

) is correct because substituting

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

into

f(x)

and simplifying results in

2f(x)

.

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f:(-\infty, 2] \longrightarrow(-\infty, 4]

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f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

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S=\left\{x: f(x)=f^{-1}(x)\right\}