JEE MAIN - Mathematics (2005 - No. 1)
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
$$f\left( x \right) = {\tan ^{ - 1}}{{2x} \over {1 - {x^2}}}$$,
then $$f$$ is both one-one and onto when B is the interval
$$\left( {0,{\pi \over 2}} \right)$$
$$\left[ {0,{\pi \over 2}} \right)$$
$$\left[ { - {\pi \over 2},{\pi \over 2}} \right]$$
$$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$
Explanation
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 $$ = B = \left( { - {\pi \over 2},{\pi \over 2}} \right).$$
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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