COMEDK 2025 Afternoon Solved Paper

Concept: The domain is found by ensuring the argument of $\sin^{-1}$ lies in $[-1,1]$ and the square‑root radicand is non‑negative.Explanation:First, $\sin^{-1}(2x)$ is defined only when $-1 \le 2x \le 1$.This gives $-\frac12 \le x \le \frac12$.Second, the square root requires $\sin^{-1}(2x) + \frac{\pi}{6} \ge 0$.So $\sin^{-1}(2x) \ge -\frac{\pi}{6}$.Because $\sin^{-1}$ is an increasing function, applying $\sin$ to both sides preserves the inequality: $2x \ge \sin\left(-\frac{\pi}{6}\right) = -\frac12$.Thus $x \ge -\frac14$.Combine both conditions: $x$ must satisfy $x \ge -\frac14$ and $x \in \left[-\frac12, \frac12\right]$.The intersection is $\left[-\frac14, \frac12\right]$.Answer: Option A: $\left[-\frac14, \frac12\right]$.