QJEEM05Mad0f8: JEE Main 2005 Mathematics - Class 11 JEEM05Mad0f8 Solution: Let , be a function defined by , then is both one-one and onto…

Question

Verified Solution • 2025–26 Session moderate

Let

f:( - 1,1) \to B

, be a function defined by

f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}

,
then

f

is both one-one and onto when B is the interval

Accepted Answer

Given

\,\,f\left( x \right) = {\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) = 2{\tan ^{ - 1}}x

for

x \in \left( { - 1,1} \right)

If

\,\,x \in \left( { - 1,1} \right) \Rightarrow {\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 4},{\pi \over 4}} \right)

\Rightarrow 2{\tan ^{ - 1}}x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)

Clearly, range of

f\left( x \right) = \left( { - {\pi \over 2},{\pi \over 2}} \right)

For

f

to be onto, co-domain

=

range

\therefore

Co-domain of function

= B = \left( { - {\pi \over 2},{\pi \over 2}} \right).

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