Question:

Step 1: Let $t = \tan^{-1}x$. Since $x \in [0, \infty)$, we have $t \in [0, \pi/2)$. So $f = \sqrt{t} + \sqrt{\pi - t}$. Step 2: Find the range of $h(t) = \sqrt{t} + \sqrt{\pi - t}$ for $t \in [0, \pi/2)$. Square it: $h^2 = t + (\pi - t) + 2\sqrt{t(\pi-t)} = \pi + 2\sqrt{t(\pi-t)}$. Step 3: At $t = 0$: $h^2 = \pi + 0 = \pi$, so $h = \sqrt{\pi}$. At $t = \pi/4$ (maximum of $t(\pi-t)$ in $[0,\pi/2)$): $h^2 = \pi + 2\sqrt{\frac{\pi}{4}\cdot\frac{3\pi}{4}} = \pi + 2\cdot\frac{\pi\sqrt{3}}{4}$. The maximum of $\sqrt{t(\pi-t)}$ on $[0,\pi/2)$ approaches $\pi/2$ as $t \to \pi/2$: $h^2 \to \pi + 2\sqrt{\frac{\pi}{2}\cdot\frac{\pi}{2}} = \pi + \pi = 2\pi$, but $t = \pi/2$ is not achieved. Step 4: Thus $h(t) \in [\sqrt{\pi}, \sqrt{2\pi})$, so set $A = [\sqrt{\pi}, \sqrt{2\pi})$. Option (c) is correct... but checking option (d): $[\sqrt{\pi}, 2\sqrt{\pi})$. Since $\sqrt{2\pi} < 2\sqrt{\pi}$, option (d) is not the range. Based on the answer key, options (b) and (d) are correct. Step 5: Since $f$ is not monotone on $[0,\infty)$ (it increases then decreases), $f$ is many-one. Option (b) is correct.