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QJEEM23Maa6d8: JEE Main 2023 Mathematics - Class 12 JEEM23Maa6d8 Solution: Let be the area of the region . Then is equal to

Let Δ\Delta be the area of the region {(x,y)∈R2:x2+y2≤21,y2≤4x,x≥1}\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\} . Then 12(Δ−21sin⁡−127){1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right) is equal to

Multiple Choice Options

  • A 23−132\sqrt 3 - {1 \over 3}
  • B 23−232\sqrt 3 - {2 \over 3}
  • C 3−43\sqrt 3 - {4 \over 3}
  • D 3−23\sqrt 3 - {2 \over 3}

Step-by-Step Verified Solution

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 48 English Explanation
Required area =2∫132xdx+∫321(21−x2)dx=2 \int\limits_{1}^{3} 2 \sqrt{x} d x+\int_{3}^{\sqrt{21}} \sqrt{\left(21-x^{2}\right)} d x

=2([2(x3/23/2)∣13]+[x221−x2+212sin⁡−1(x21)]321)=2\left(\left[\left.2\left(\frac{x^{3 / 2}}{3 / 2}\right)\right|_{1} ^{3}\right]+\left[\frac{x}{2} \sqrt{21-x^{2}}+\frac{21}{2} \sin ^{-1}\left(\frac{x}{\sqrt{21}}\right)\right]_{3}^{\sqrt{21}}\right)

=23+21π2−83−21sin⁡−137=Δ=2 \sqrt{3}+\frac{21 \pi}{2}-\frac{8}{3}-21 \sin ^{-1} \sqrt{\frac{3}{7}}=\Delta

∴12(Δ−21sin⁡−1(27))=3−43\therefore \frac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)=\sqrt{3}-\frac{4}{3}

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