Let A = \ (x, y) ∈ ℝ ^ 2 : y ≥ 0, 2x ≤ y ≤ 4 - (x - 1) ^ 2 \ and B = \ (x, y) ∈ ℝ × ℝ: 0 ≤ y ≤ min\ 2x, 4 - (x - 1) ^ 2 \ \ . Then the ratio of the area of A to the area of B is

Correct answer: A. (π - 1)/(π + 1) Given, 2x ≤ y ≤ 4 - (x - 1) ^ 2 : y ≥ 0 ⇒ y ^ 2 + (x - 1) ^ 2 = 4 and y = 2x Now plotting the diagram we get, Area of A : area of shaded region as shown in the above figure, Shaded portion = circular (OABC) - Ar (∆ OAB) = (π(4))/(4) - (1)/(2)(2)(1) Hence, Area of A = (π - 1) Now, plotting the diagram of 0 ≤ y ≤ min\ 2x, 4 - (x - 1) ^ 2 \ we get, Area of B : area of shaded region as shown in the above figure, Area of B = Ar(∆ AOB) + Area of arc of circle (ABC) = (1)/(2)(1)(2) + π (2) ^ 2 4 = π + 1 Hence, (A)/(B) = areaof A areaof B = (π - 1)/(π + 1)

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 A = \ (x, y) ∈ ℝ ^ 2 : y ≥ 0, 2x ≤ y ≤ 4 - (x - 1) ^ 2 \ and B...

Let

A = \left\{\left(x, y\right) \in {\mathbb{R}}^{2}: y \ge 0, 2x \le y \le \sqrt{4 - {\left(x - 1\right)}^{2}}\right\}

and

B = \left\{\left(x, y\right) \in \mathbb{R} \times \mathbb{R}: 0 \le y \le min\left\{2x, \sqrt{4 - {\left(x - 1\right)}^{2}}\right\}\right\}

. Then the ratio of the area of

A

to the area of

B

is $Let A = \ (x, y) ∈ ℝ ^ 2 : y ≥ 0, 2x ≤ y ≤ 4 - (x - 1) ^ 2 \ and B = \ (x, y) ∈ ℝ × ℝ: 0 ≤ y ≤ min\ 2x, 4 - (x - 1) ^ 2$ $Let A = \ (x, y) ∈ ℝ ^ 2 : y ≥ 0, 2x ≤ y ≤ 4 - (x - 1) ^ 2 \ and B = \ (x, y) ∈ ℝ × ℝ: 0 ≤ y ≤ min\ 2x, 4 - (x - 1) ^ 2$

Options

A $\dfrac{\pi - 1}{\pi + 1}$
B $\dfrac{\pi}{\pi - 1}$
C $\dfrac{\pi}{\pi + 1}$
D $\dfrac{\pi + 1}{\pi - 1}$

Correct answer

A. $\dfrac{\pi - 1}{\pi + 1}$

Step-by-step solution

Given,

$2x \le y \le \sqrt{4 - {\left(x - 1\right)}^{2}}: y \ge 0$

$⇒ {y}^{2} + {\left(x - 1\right)}^{2} = 4 and$ y = 2x $Now plotting the diagram we get,<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/26dee46e-dbfb-45c3-8938-02e6f41cac05-9895075re.PNG" style="width: 317px; height: 200px;" />Area of$ A $: area of shaded region as shown in the above figure,Shaded portion = circular \left(OABC\right)$

$- Ar \left(∆ OAB\right)$ = \dfrac{\pi\left(4\right)}{4} - \dfrac{1}{2}\left(2\right)\left(1\right) $Hence, Area of$ A = \left(\pi - 1\right) $Now, plotting the diagram of$ 0 \le y \le min\left\{2x, \sqrt{4 - {\left(x - 1\right)}^{2}}\right\} $we get,<img alt="" src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/5a717159-199c-42aa-a5cc-fe2ce03b6428-9895075_1re.PNG" style="width: 310px; height: 200px;" />Area of$ B $: area of shaded region as shown in the above figure,Area of$ B = Ar\left(∆ AOB\right) + $Area of arc of circle \left(ABC\right)$

$= \dfrac{1}{2}\left(1\right)\left(2\right) + \dfrac{\pi{\left(2\right)}^{2}}{4} = \pi + 1$

Hence, $\dfrac{A}{B} = \dfrac{\text{area}of A}{\text{area}of B} = \dfrac{\pi - 1}{\pi + 1}$

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