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

Let the function g:(-∞, ∞) → (-π/2, π/2) be given by g(u) = 2tan⁻¹(eᵘ) - π/2. Then, g is

$g(u) = 2\tan^{-1}(e^{u}) - \dfrac{\pi}{2}$

$g:(-\infty,\infty)\to\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$\text{To find: parity (even/odd) and monotonicity of } g$

$u\to\infty:\ \tan^{-1}(e^u)\to\dfrac{\pi}{2} \;\Rightarrow\; g(u)\to\dfrac{\pi}{2}$

$u\to-\infty:\ \tan^{-1}(e^u)\to 0 \;\Rightarrow\; g(u)\to-\dfrac{\pi}{2}$

$\tan^{-1}a + \tan^{-1}\dfrac{1}{a} = \dfrac{\pi}{2} \quad (a>0)$

$g(-u) = 2\left(\dfrac{\pi}{2}-\tan^{-1}(e^u)\right) - \dfrac{\pi}{2} = -g(u)$

$g(0) = 2\cdot\dfrac{\pi}{4} - \dfrac{\pi}{2} = 0 \quad\checkmark$

$g'(u) = 2\cdot\dfrac{e^{u}}{1+e^{2u}} = \dfrac{2e^{u}}{1+e^{2u}}$

$g'(u) > 0 \ \forall u \;\Rightarrow\; g \text{ strictly increasing on } (-\infty,\infty)$

$\text{increasing with limits } \pm\dfrac{\pi}{2} \;\Rightarrow\; \text{range } \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$\boxed{g \text{ is odd and strictly increasing}}$

$\text{Answer: option (3)}$