Let the function g : (-infty, infty) to left( | vedclass

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(C) Given $g(u) = 2 	an^{-1}(e^u) - \frac{\pi}{2}$.To check for odd/even,we evaluate $g(-u)$:$g(-u) = 2 	an^{-1}(e^{-u}) - \frac{\pi}{2}$$= 2 	an^{

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odd and is strictly increasing in $(-\infty, \infty)$

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(C) Given $g(u) = 2 	an^{-1}(e^u) - \frac{\pi}{2}$.To check for odd/even,we evaluate $g(-u)$:$g(-u) = 2 	an^{-1}(e^{-u}) - \frac{\pi}{2}$$= 2 	an^{-1}\left(\frac{1}{e^u}ight) - \frac{\pi}{2}$Using the identity $	an^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we have $	an^{-1}\left(\frac{1}{e^u}ight) = \cot^{-1}(e^u) = \frac{\pi}{2} - 	an^{-1}(e^u)$.Substituting this back:$g(-u) = 2 \left[ \frac{\pi}{2} - 	an^{-1}(e^u) ight] - \frac{\pi}{2}$$= \pi - 2 	an^{-1}(e^u) - \frac{\pi}{2}$$= \frac{\pi}{2} - 2 	an^{-1}(e^u)$$= - \left( 2 	an^{-1}(e^u) - \frac{\pi}{2} ight) = -g(u)$.Since $g(-u) = -g(u)$,the function is odd.Now,check for monotonicity by finding the derivative:$g'(u) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot e^u = \frac{2e^u}{1 + e^{2u}}$.Since $e^u > 0$ for all $u \in (-\infty, \infty)$,$g'(u) > 0$.Thus,$g(u)$ is strictly increasing in $(-\infty, \infty)$.

Let the function $g : (-\infty, \infty) \to \left( - \frac{\pi}{2}, \frac{\pi}{2} \right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is -

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