UPSC NDA I 2015 Mathematics Paper

Concept:Area bounded by the x‑axis, a line, and a circle in the first quadrant, computed using integration.Explanation:The circle is $x^{2}+y^{2}=4$ (radius $2$).The line is $x=\sqrt{3}\,y$.Find their intersection in the first quadrant: substitute $x$ into the circle: $(\sqrt{3}y)^{2}+y^{2}=4 \Rightarrow 3y^{2}+y^{2}=4 \Rightarrow y^{2}=1 \Rightarrow y=1$. Then $x=\sqrt{3}$.The required region consists of two parts:1. Triangle $OAB$ with vertices $O(0,0)$, $A(\sqrt{3},1)$, and $B(\sqrt{3},0)$. Its area is $\frac{1}{2}\times\sqrt{3}\times1=\frac{\sqrt{3}}{2}$.2. Region under the circle from $x=\sqrt{3}$ to $x=2$ (where the circle meets the x‑axis). The circle gives $y=\sqrt{4-x^{2}}$.Area $=\int_{\sqrt{3}}^{2}\sqrt{4-x^{2}}\,dx = \left[\frac{1}{2}x\sqrt{4-x^{2}}+2\sin^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2}$.At $x=2$: $0+2\sin^{-1}(1)=2\cdot\frac{\pi}{2}=\pi$.At $x=\sqrt{3}$: $\frac{1}{2}\sqrt{3}\sqrt{4-3}+2\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}+2\cdot\frac{\pi}{3}=\frac{\sqrt{3}}{2}+\frac{2\pi}{3}$.So the area under the circle is $\pi - \left(\frac{\sqrt{3}}{2}+\frac{2\pi}{3}\right) = \frac{\pi}{3}-\frac{\sqrt{3}}{2}$.Total area = triangle area + area under circle = $\frac{\sqrt{3}}{2} + \left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$.Answer:$\frac{\pi}{3}$ (Option A).