Verified Solution • 2025–26 Session moderate

QJEEM23Ma245a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma245a Solution: Let and . Then the ratio of the area of A to the area of B is

Let A={(x,y)∈R2:y≥0,2x≤y≤4−(x−1)2}A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\} and

B={(x,y)∈R×R:0≤y≤min⁡{2x,4−(x−1)2}}. B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. } .

Then the ratio of the area of A to the area of B is

Multiple Choice Options

  • A ππ+1\frac{\pi}{\pi+1}
  • B π−1π+1\frac{\pi-1}{\pi+1}
  • C ππ−1\frac{\pi}{\pi-1}
  • D π+1π−1\frac{\pi+1}{\pi-1}

Step-by-Step Verified Solution

y2+(x−1)2=4y^{2}+(x-1)^{2}=4

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 49 English Explanation 1

shaded portion == circular (OABC)(\mathrm{OABC})

amp;−Ar⁡(ΔOAB)amp;=π(4)4−12(2)(1)amp;A=(π−1)\begin{aligned}& -\operatorname{Ar}(\Delta \mathrm{OAB}) \\\\& =\frac{\pi(4)}{4}-\frac{1}{2}(2)(1) \\\\& \mathrm{A}=(\pi-1)\end{aligned}

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 49 English Explanation 2
Area B=Ar⁡(△AOB)+B=\operatorname{Ar}(\triangle \mathrm{AOB})+ Area of arc of circle (ABC)(\mathrm{ABC})

=12(1)(2)+π(2)24=π+1=\frac{1}{2}(1)(2)+\frac{\pi(2)^{2}}{4}=\pi+1

AB=π−1π+1\frac{\mathrm{A}}{\mathrm{B}}=\frac{\pi-1}{\pi+1}

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