UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The principal value of $\sin^{-1}x$ is the unique angle $\theta$ in the closed interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ such that $\sin \theta = x$.Explanation:Let $\sin^{-1}x = y$.Then $\sin y = x$.For the sine function to be invertible, its domain must be restricted to $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.Within this interval, $\sin y$ is one‑to‑one and covers all values from $-1$ to $1$.Thus the range of $\sin^{-1}x$ (principal value) is exactly $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.Both endpoints $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ are included because $\sin\left(-\frac{\pi}{2}\right) = -1$ and $\sin\left(\frac{\pi}{2}\right) = 1$.Answer:Option B: $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.