Verified Solution • 2025–26 Session moderate QJEEM22Mabd0a: JEE Main 2022 Mathematics - Class 12 JEEM22Mabd0a Solution: The area of the region is equal to : The area of the region {(x,y):∣x−1∣≤y≤5−x2}\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}{(x,y):∣x−1∣≤y≤5−x2} is equal to : Multiple Choice Options A 52sin−1(35)−12\frac{5}{2} \sin ^{-1}\left(\frac{3}{5}\right)-\frac{1}{2}25sin−1(53)−21 B 5π4−32\frac{5 \pi}{4}-\frac{3}{2}45π−23 C 3π4+32\frac{3 \pi}{4}+\frac{3}{2}43π+23 D 5π4−12\frac{5 \pi}{4}-\frac{1}{2}45π−21 Step-by-Step Verified Solution A=∫−11(5−x2−(1−x))dx+∫12(5−x2−(x−1))dxA = \int\limits_{ - 1}^1 {\left( {\sqrt {5 - {x^2}} - (1 - x)} \right)dx + \int\limits_1^2 {\left( {\sqrt {5 - {x^2}} - (x - 1)} \right)dx} }A=−1∫1(5−x2−(1−x))dx+1∫2(5−x2−(x−1))dx A=2(x25−x2+52sin−1x5)−2x∣01+x25−x2+52sin−1x5−x22+x∣12\left. {A = 2\left( {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right) - 2x} \right|_0^1 + \left. {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }} - {{{x^2}} \over 2} + x} \right|_1^2A=2(2x5−x2+25sin−15x)−2x01+2x5−x2+25sin−15x−2x2+x12 =(5π4−12)= \left( {{{5\pi } \over 4} - {1 \over 2}} \right)=(45π−21) sq. units Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
