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,consider $g(-u) = 2 	an^{-1}(e^{-u}) - \frac{\pi}{2}$.Since $e^{-u} = \frac

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Odd and strictly increasing on $(-\infty, \infty)$

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(C) Given $g(u) = 2 	an^{-1}(e^u) - \frac{\pi}{2}$.To check for odd/even,consider $g(-u) = 2 	an^{-1}(e^{-u}) - \frac{\pi}{2}$.Since $e^{-u} = \frac{1}{e^u}$,we have $g(-u) = 2 	an^{-1}\left(\frac{1}{e^u}ight) - \frac{\pi}{2}$.Using the identity $	an^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$,we know $	an^{-1}\left(\frac{1}{x}ight) = \cot^{-1}(x)$ for $x > 0$.Thus,$g(-u) = 2 \cot^{-1}(e^u) - \frac{\pi}{2}$.Now,$g(u) + g(-u) = 2(	an^{-1}(e^u) + \cot^{-1}(e^u)) - \pi = 2(\frac{\pi}{2}) - \pi = 0$.Since $g(-u) = -g(u)$,the function is odd.To check for monotonicity,find 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 $u \in \mathbb{R}$,$g'(u) > 0$ for all $u \in \mathbb{R}$.Therefore,$g(u)$ is strictly increasing on $(-\infty, \infty)$.

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

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