JEE Main 202329 Jan, Morning shiftMathematicsHardIntegral Calculus369× since 2002Let A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2}A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2} and B={(x,y)∈R×R:0≤y≤min{2x,4−(x−1)2}}. B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }B={(x,y)∈R×R:0≤y≤min{2x,4−(x−1)2}}. . Then the ratio of the area of A to the area of B isAOption A. ππ+1\frac{\pi}{\pi+1}π+1πBOption B. π−1π+1\frac{\pi-1}{\pi+1}π+1π−1COption C. ππ−1\frac{\pi}{\pi-1}π−1πDOption D. π+1π−1\frac{\pi+1}{\pi-1}π−1π+1CheckSkip · show solution
