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JEE Main 20249 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Area Under Curves

The parabola y^2=4 x divides the area of the circle x^2+y^2=5 in...

The parabola y2=4xy^2=4 x divides the area of the circle x2+y2=5x^2+y^2=5 in two parts. The area of the smaller part is equal to:

Options

  1. A13+5sin⁡−1(25)\frac{1}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)
  2. B13+5sin⁡−1(25)\frac{1}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)
  3. C23+5sin⁡−1(25)\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)
  4. D23+5sin⁡−1(25)\frac{2}{3}+\sqrt{5} \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)

Correct answer

C. 23+5sin⁡−1(25)\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)

Step-by-step solution

y2=4xx2+y2=5\begin{aligned} & y^2=4 x \\ & x^2+y^2=5\end{aligned}\thereforeAreaofshadedregionasshowninthefigurewillbeArea of shaded region as shown in the figure will beA1=∫014xdx+∫155−x2dx=43⋅[x32]01+[x25−x2+52sin⁡−1x5]15=13+5π4−52sin⁡−1(15)∴ Required Area =2A1=23+5π2−5sin⁡−1(15)=23+5(π2−sin⁡−115)=23+5cos⁡−115=23+5sin⁡−1(25)\begin{aligned} & \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} \\ & =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}\right]_1^{\sqrt{5}} \\ & =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & \therefore \text{ Required Area }=2 \mathrm{ A}_1 \\ & =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right) \\ & =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} \\ & =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{aligned}$
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