Symmetry and Monotonicity of Inverse Tangent Function

To determine the properties of the function $g(u) = 2\tan^{-1}(e^u) - \frac{\pi}{2}$, we analyze its symmetry (parity) and monotonicity.

1. Parity of $g(u)$:

Substitute $-u$ in place of $u$:
$g(-u) = 2\tan^{-1}(e^{-u}) - \frac{\pi}{2}$

Using the inverse trigonometric identity $\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$ for $x > 0$, and noting that $e^u > 0$ for all $u \in (-\infty, \infty)$, we have:
$\tan^{-1}(e^{-u}) = \tan^{-1}\left(\frac{1}{e^u}\right) = \frac{\pi}{2} - \tan^{-1}(e^u)$

Substitute this back into the expression for $g(-u)$:
$g(-u) = 2\left(\frac{\pi}{2} - \tan^{-1}(e^u)\right) - \frac{\pi}{2}$
$g(-u) = \pi - 2\tan^{-1}(e^u) - \frac{\pi}{2}$
$g(-u) = \frac{\pi}{2} - 2\tan^{-1}(e^u)$
$g(-u) = -\left(2\tan^{-1}(e^u) - \frac{\pi}{2}\right) = -g(u)$

Since $g(-u) = -g(u)$ for all $u \in (-\infty, \infty)$, the function $g(u)$ is an odd function.

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2. Monotonicity of $g(u)$:

Differentiate $g(u)$ with respect to $u$:
$g'(u) = \frac{d}{du}\left(2\tan^{-1}(e^u) - \frac{\pi}{2}\right)$
$g'(u) = 2 \cdot \frac{1}{1 + (e^u)^2} \cdot \frac{d}{du}(e^u)$
$g'(u) = \frac{2e^u}{1 + e^{2u}}$

Since $e^u > 0$ and $1 + e^{2u} > 0$ for all real numbers $u$:
$g'(u) > 0 \quad \forall u \in (-\infty, \infty)$

Since the derivative is strictly positive everywhere on its domain, $g(u)$ is strictly increasing in $(-\infty, \infty)$.

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Conclusion:

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

Correct Option: C