UPSC NDA I 2015 Mathematics Paper

Concept:The required area is bounded by the x‑axis ($y=0$), the vertical line $x=\sqrt{3}$, and the circle $x^2+y^2=4$ in the first quadrant.The region lies under the circle from $x=\sqrt{3}$ to $x=2$.Explanation:The circle $x^2+y^2=4$ has radius $2$.Substitute $x=\sqrt{3}$ into the circle to find the intersection point in the first quadrant:$(\sqrt{3})^2 + y^2 = 4 \Rightarrow 3+y^2=4 \Rightarrow y^2=1 \Rightarrow y=1$.So the point $A(\sqrt{3},1)$ lies on both the line and the circle.The area is the integral of $y$ from $x=\sqrt{3}$ to $x=2$ (where the circle meets the x‑axis at $(2,0)$).From the circle, $y=\sqrt{4-x^2}$ for the upper half.Area $= \int\limits_{\sqrt{3}}^{2} \sqrt{4-x^2}\,dx$.Using the formula $\int\sqrt{a^2-x^2}\,dx = \frac{1}{2}x\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C$ with $a=2$, we evaluate:$= \left[ \frac{1}{2}x\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{\sqrt{3}}^{2}$.At $x=2$: $\frac{1}{2}(2)\sqrt{0} + 2\sin^{-1}(1) = 0 + 2\cdot\frac{\pi}{2} = \pi$.At $x=\sqrt{3}$: $\frac{1}{2}(\sqrt{3})\sqrt{4-3} + 2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} + 2\cdot\frac{\pi}{3} = \frac{\sqrt{3}}{2} + \frac{2\pi}{3}$.Subtract: $\pi - \left(\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right) = \pi - \frac{2\pi}{3} - \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.Answer:$\frac{\pi}{3}-\frac{\sqrt{3}}{2}$ (Option A).