Verified Solution • 2025–26 Session moderate QJEEM25Maf840: JEE Main 2025 Mathematics - Class 12 JEEM25Maf840 Solution: The area of the region, inside the circle and outside the… The area of the region, inside the circle (x−23)2+y2=12(x-2 \sqrt{3})^2+y^2=12(x−23)2+y2=12 and outside the parabola y2=23xy^2=2 \sqrt{3} xy2=23x is : Multiple Choice Options A 3π−83 \pi-83π−8 B 6π−86 \pi-86π−8 C 3π+83 \pi+83π+8 D 6π−166 \pi-166π−16 Step-by-Step Verified Solution amp; Required area =2∫023(43x−x2−23x)dxamp;=2∫023(12−(x−23)2−23x)dx\begin{aligned} & \text { Required area }=2 \int_0^{2 \sqrt{3}}\left(\sqrt{4 \sqrt{3} x-x^2}-\sqrt{2 \sqrt{3} x}\right) d x \\\\ & =2 \int_0^{2 \sqrt{3}}\left(\sqrt{12-(x-2 \sqrt{3})^2}-\sqrt{2 \sqrt{3} x}\right) d x\end{aligned}amp; Required area =2∫023(43x−x2−23x)dxamp;=2∫023(12−(x−23)2−23x)dx =2[x−23212−(x−23)2+122sin−1(x−2323)−23x3232]023\begin{array}{r}=2\left[\frac{x-2 \sqrt{3}}{2} \sqrt{12-(x-2 \sqrt{3})^2}+\frac{12}{2} \sin ^{-1}\left(\frac{x-2 \sqrt{3}}{2 \sqrt{3}}\right)\right. \left.-\frac{\sqrt{2 \sqrt{3}} x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^{2 \sqrt{3}}\end{array}=2[2x−2312−(x−23)2+212sin−1(23x−23)−2323x23]023 amp;=2{3π−8}amp;=6π−16 sq. units. \begin{aligned} & =2\{3 \pi-8\} \\\\ & =6 \pi-16 \text { sq. units. }\end{aligned}amp;=2{3π−8}amp;=6π−16 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
