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JEE Main 202126 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Area Under Curves

JEE Main 2021 · 26 August · Morning Shift · Mathematics · Area Under Curves

The area of the region S = \ ( x, y ) : 3 x ^ 2 ≤ 4y ≤ 6x + 24\...

The area of the region

S = \left\{( x, y ) : 3{x}^{2} \le 4y \le 6x + 24\right\}

is______. $The area of the region S = \ ( x, y ) : 3 x ^ 2 ≤ 4y ≤ 6x + 24\ is______.$

Correct answer

0

Step-by-step solution

We have,

$S =$ \left\{( x, y ) : 3{x}^{2} \le 4y \le 6x + 24\right\} $</p><p>Then,</p><p>$ y = \dfrac{3{x}^{2}}{4} $</p><p>$ 4y = 6x + 24 $</p><p>Solving both equations, we get \left(- 2, 3\right)$ and \left(4, 12\right) $.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/0a8114da-f21b-4f81-bb81-6398b075b00b-image.png" style="width: 405px; height: 265px;" /></p><p>Required area</p><p>$ A = {∫}_{- 2}^{4}\left(\dfrac{6x + 24}{4} - \dfrac{3{x}^{2}}{4}\right)dx $</p><p>$ ⇒ A = \dfrac{1}{4}{∫}_{- 2}^{4}\left(6x + 24 - 3{x}^{2}\right)dx $</p><p>$ ⇒ A = {\left(\dfrac{3}{4}{x}^{2} + 6x - \dfrac{{x}^{3}}{4}\right)}_{- 2}^{4} $</p><p>$ ⇒ A = \left(12 + 24 - 16 - 3 + 12 - 2\right) $</p><p>$ ⇒ A = 27sq . units

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