JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 12)
Let $$\Delta$$ be the area of the region $$\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$$. Then $${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$$ is equal to
$$2\sqrt 3 - {1 \over 3}$$
$$2\sqrt 3 - {2 \over 3}$$
$$\sqrt 3 - {4 \over 3}$$
$$\sqrt 3 - {2 \over 3}$$
Explanation
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Required area $=2 \int\limits_{1}^{3} 2 \sqrt{x} d x+\int_{3}^{\sqrt{21}} \sqrt{\left(21-x^{2}\right)} d x$
$=2\left(\left[\left.2\left(\frac{x^{3 / 2}}{3 / 2}\right)\right|_{1} ^{3}\right]+\left[\frac{x}{2} \sqrt{21-x^{2}}+\frac{21}{2} \sin ^{-1}\left(\frac{x}{\sqrt{21}}\right)\right]_{3}^{\sqrt{21}}\right)$
$=2 \sqrt{3}+\frac{21 \pi}{2}-\frac{8}{3}-21 \sin ^{-1} \sqrt{\frac{3}{7}}=\Delta$
$\therefore \frac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)=\sqrt{3}-\frac{4}{3}$
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