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

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

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

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

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

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