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JEE Main 202329 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Area Under Curves

JEE Main 2023 · 29 January · Morning Shift · Mathematics · Area Under Curves

Let Δ be the area of the region \ (x, y) ∈ ℝ ^ 2 : x ^ 2 + y ^ 2 ≤...

Let Δ\Delta be the area of the region \left\{\left(x, y\right) \in {\mathbb{R}}^{2}: {x}^{2} + {y}^{2} \le 21, {y}^{2} \le 4x, x \ge 1\right\}.Then. Then\dfrac{1}{2}\left(\Delta - 21\sin^{- 1}\dfrac{2}{\sqrt{7}}\right) is equal toLet Δ be the area of the region \ (x, y) ∈ ℝ ^ 2 : x ^ 2 + y ^ 2 ≤ 21, y ^ 2 ≤ 4x, x ≥ 1\ . Then (1)/(2)(Δ - 21 ^ - 1 2

Options

  1. A23−132\sqrt{3} - \dfrac{1}{3}
  2. B3−23\sqrt{3} - \dfrac{2}{3}
  3. C23−232\sqrt{3} - \dfrac{2}{3}
  4. D3−43\sqrt{3} - \dfrac{4}{3}

Correct answer

D. 3−43\sqrt{3} - \dfrac{4}{3}

Step-by-step solution

Plotting the diagram of x2+y2≤21 and y2≤4x{x}^{2} + {y}^{2} \le 21 \text{ and } {y}^{2} \le 4x for x≥1x \ge 1 we get,

Let Δ be the area of the region \ (x, y) ∈ ℝ ^ 2 : x ^ 2 + y ^ 2 ≤ 21, y ^ 2 ≤ 4x, x ≥ 1\ . Then (1)/(2)(Δ - 21 ^ - 1 2

Now from above diagram required Area Δ=\Delta =2∫132xdx+2∫32121−x2dx2{∫}_{1}^{3}2\sqrt{x}dx + 2{∫}_{3}^{\sqrt{21}}\sqrt{21 - {x}^{2}}dx

⇒Δ=83(33−1)+21sin⁡−1(27)−63⇒ \Delta = \dfrac{8}{3}\left(3\sqrt{3} - 1\right) + 21\sin^{- 1}\left(\dfrac{2}{\sqrt{7}}\right) - 6\sqrt{3}

So, 12(Δ−21sin⁡−1(27))\dfrac{1}{2}\left(\Delta - 21\sin^{- 1}\left(\dfrac{2}{\sqrt{7}}\right)\right)

=12(83(33−1)+21sin⁡−1(27)−63−21sin⁡−1(27))= \dfrac{1}{2}\left(\dfrac{8}{3}\left(3\sqrt{3} - 1\right) + 21\sin^{- 1}\left(\dfrac{2}{\sqrt{7}}\right) - 6\sqrt{3} - 21\sin^{- 1}\left(\dfrac{2}{\sqrt{7}}\right)\right)

=12(83(33−1)−63)= \dfrac{1}{2}\left(\dfrac{8}{3}\left(3\sqrt{3} - 1\right) - 6\sqrt{3}\right)

=23−832= \dfrac{2\sqrt{3} - \dfrac{8}{3}}{2}

=3−43= \sqrt{3} - \dfrac{4}{3}

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