UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The equation $|y| = 1 - x^{2}$ represents two symmetric curves about the $x$-axis. The total area is found by integrating one region and using symmetry.Explanation:For $y \ge 0$, $|y| = y$, so $y = 1 - x^{2}$ (an inverted parabola with vertex at $(0,1)$).For $y < 0$, $|y| = -y$, so $-y = 1 - x^{2}$ $\Rightarrow$ $y = x^{2} - 1$ (an upward parabola with vertex at $(0,-1)$).The curves are symmetric about the $x$-axis and also symmetric about the $y$-axis.Thus the area bounded by $|y| = 1 - x^{2}$ is four times the area under $y = 1 - x^{2}$ in the first quadrant (region OABO).Area in first quadrant $= \int_{0}^{1} (1 - x^{2}) \, dx = \left[ x - \frac{x^{3}}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$ square units.Total area $= 4 \times \frac{2}{3} = \frac{8}{3}$ square units.Answer:$\frac{8}{3}$ square units (Option B).