The range of $f(x) = \tan x + \frac{1}{2}\sin^{-1} x$ is:
$(-\pi, \pi)$
$\left[-\frac{3\pi}{4}, \frac{3\pi}{4}\right]$
$\left(-\frac{3\pi}{4}, \frac{3\pi}{4}\right)$
$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Step-by-Step Solution
Key Concept: The domain of f(x) = tan x + (1/2)sin⁻¹ x is restricted to [-1, 1] (from sin⁻¹ x), not all of ℝ. Since f is continuous and strictly increasing on [-1, 1], the range equals [f(-1), f(1)] = [tan(-1) + (1/2)sin⁻¹(-1), tan(1) + (1/2)sin⁻¹(1)] = [-π/4 - π/4, π/4 + π/4] = [-π/2, π/2].
Given domain of $f(x)$ is $[-1,1]$ and $f(x)$ is continuous and increasing, we find that $\tan^{-1}x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$ and $\frac{1}{2}\sin^{-1}x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$, both matching the required range.
Correct Answer: 4