JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 12)
Let the function $$g:\left( { - \infty ,\infty } \right) \to \left( { - {\pi \over 2},{\pi \over 2}} \right)$$ be given by
$$g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}.$$ Then, $$g$$ is
$$g\left( u \right) = 2{\tan ^{ - 1}}\left( {{e^u}} \right) - {\pi \over 2}.$$ Then, $$g$$ is
even and is strictly increasing in $$\left( {0,\infty } \right)$$
odd and is strictly decreasing in $$\left( { - \infty ,\infty } \right)$$
odd and is strictly increasing in $$\left( { - \infty ,\infty } \right)$$
neither even nor odd, but is strictly increasing in $$\left( { - \infty ,\infty } \right)$$
Explanation
Given that,
$$g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$$ for $$u \in ( - \infty ,\infty )$$
$$g( - u) = 2{\tan ^{ - 1}}{e^{ - u}} - {\pi \over 2} = 2{\cot ^{ - 1}}({e^u}) - {\pi \over 2}$$
$$ = 2\left( {{\pi \over 2} - {{\tan }^{ - 1}}({e^u}) - {\pi \over 2}} \right)$$
$$ = - 2{\tan ^{ - 1}}({e^u}) + {\pi \over 2}$$
$$ = - g(u)$$
$$\therefore$$ $$g(u) = - g(u)$$
$$ \Rightarrow g(u)$$ is an odd function
We have, $$g(u) = 2{\tan ^{ - 1}}({e^u}) - {\pi \over 2}$$
$$g'(u) = {{2{e^u}} \over {1 + 2{e^{2u}}}}$$
$$g'(u) > 0,\forall u \in R$$ [$$\because$$ $${e^u} > 0$$]
So, $$g'(u)$$ is increasing function.
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