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JEE Main 202229 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
29 July
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2022 · 29 July · Morning Shift · Mathematics · Area Under Curves

The area of the region \ (x, y): |x - 1| ≤ y ≤ 5 - x ^ 2 \ is equal...

The area of the region {(x,y):∣x−1∣≤y≤5−x2}\left\{\left(x, y\right): \left|x - 1\right| \le y \le \sqrt{5 - {x}^{2}}\right\} is equal toThe area of the region \ (x, y): |x - 1| ≤ y ≤ 5 - x ^ 2 \ is equal to

Options

  1. A52sin⁡−1(35)−12\dfrac{5}{2}\sin^{- 1}\left(\dfrac{3}{5}\right) - \dfrac{1}{2}
  2. B5π4−32\dfrac{5\pi}{4} - \dfrac{3}{2}
  3. C3π4+32\dfrac{3\pi}{4} + \dfrac{3}{2}
  4. D5π4−12\dfrac{5\pi}{4} - \dfrac{1}{2}

Correct answer

D. 5π4−12\dfrac{5\pi}{4} - \dfrac{1}{2}

Step-by-step solution

Plotting region for ∣x−1∣≤y≤5−x2\left|x - 1\right| \le y \le \sqrt{5 - {x}^{2}}, we get,

The area of the region \ (x, y): |x - 1| ≤ y ≤ 5 - x ^ 2 \ is equal to

The intersection points of the curves x2+y2=5andy={x}^{2} + {y}^{2} = 5 and y =\left|x - 1\right| will be (−1,2) and  \left(- 1, 2\right)\text{\text{ and } }\left(2, 1\right)

Here chord ABAB subtends a right angle at the centre of the circle.
So, required area = area of ∆ABC+∆ ABC + area of segment of circle on chord ABAB

=\dfrac{1}{2}AC⋅BC+AC \cdot BC + [area of quarter circle −- area of ∆AOB∆ AOB]

=\dfrac{1}{2}\times \sqrt{2} \times 2\sqrt{2} + \left(\dfrac{90 °}{360 °} \times \pi{\left(\sqrt{5}\right)}^{2} -\dfrac{1}{2}×(5)\times {\left(\sqrt{5}\right)}^{2}\right)</p><p>=5π4−12</p></blockquote></p><p>=\dfrac{5\pi}{4} - \dfrac{1}{2}</p></blockquote>

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