Let A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x, y): |x| + |y| ≤ k\ . If 27 (Area A _ 1 ) = 5 (Area A _ 2 ), then k is equal to

Correct answer: 0 A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x, y): |x| + |y| ≤ k\ Area ( A _ 1 ) = 2[ ∫ _ 0 ^ 2 y ^ 2 dy + ∫ _ 2 ^ 4 (8 - 2y)dy] = 2[ ( y ^ 3 3 ) _ 0 ^ 2 + (8y - y ^ 2 ) _ 2 ^ 4 ] Area ( A _ 1 ) = 2 × (20)/(3) = (40)/(3) Area ( A _ 2 ) = 4 × (1)/(2) k ^ 2 Area ( A _ 2 ) = 2 k ^ 2 Given 27(Area A _ 1 ) = 5(Area A _ 2 ) ⇒ 27 × (40)/(3) = 5 × 2 k ^ 2 ⇒ k = 6

JEE Main 202227 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2022 · 27 June · Morning Shift · Mathematics · Area Under Curves

Let A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x,...

Let

{A}_{1} = \left\{\left(x, y\right): \left|x\right| \le {y}^{2}, \left|x\right| + 2y \le 8\right\}

and

{A}_{2} = \left\{\left(x, y\right): \left|x\right| + \left|y\right| \le k\right\} .

27

(Area

{A}_{1}

)

= 5

(Area

{A}_{2}

), then

k

is equal to $Let A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x, y): |x| + |y| ≤ k\ . If 27 (Area A _ 1 ) = 5 (Area$ $Let A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x, y): |x| + |y| ≤ k\ . If 27 (Area A _ 1 ) = 5 (Area$

Correct answer

Step-by-step solution

${A}_{1} = \left\{\left(x, y\right): \left|x\right| \le {y}^{2}, \left|x\right| + 2y \le 8\right\}$ and

${A}_{2} = \left\{\left(x, y\right): \left|x\right| + \left|y\right| \le k\right\}$

$Let A _ 1 = \ (x, y): |x| ≤ y ^ 2 , |x| + 2y ≤ 8\ and A _ 2 = \ (x, y): |x| + |y| ≤ k\ . If 27 (Area A _ 1 ) = 5 (Area$

Area $\left({A}_{1}\right) = 2\left[{∫}_{0}^{2}{y}^{2}dy + {∫}_{2}^{4}\left(8 - 2y\right)dy\right]$

$= 2\left[{\left(\dfrac{{y}^{3}}{3}\right)}_{0}^{2} + {\left(8y - {y}^{2}\right)}_{2}^{4}\right]$

Area \left({A}_{1}\right) = 2 \times \dfrac{20}{3} = \dfrac{40}{3} $<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/9b12a2c0-eebf-4cd7-9a4d-dc78cecdd728-image.png" style="width: 200px; height: 181px;" />Area$ \left({A}_{2}\right) = 4 \times \dfrac{1}{2}{k}^{2} $Area \left({A}_{2}\right) = 2{k}^{2}$

Given $27\left(Area{A}_{1}\right) = 5\left(Area{A}_{2}\right)$

$⇒ 27 \times \dfrac{40}{3} = 5 \times 2{k}^{2}$

$⇒ k = 6$

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