Let the area of the region \ (x, y): x-2 y+4 geq 0 , .x+2 y 2 geq 0, x+4 y 2 leq 8, y geq 0 \ be m/n , where m and n are coprime numbers. Then m + n is equal to

Mathematics · Area under the Curves

JEE Main 2024 — 27 January, Shift 1 — Question 22

Let the area of the region $\{(x, y): x-2 y+4 \geq 0$ , $\left.x+2 y^{2} \geq 0, x+4 y^{2} \leq 8, y \geq 0\right\}$ be $\frac{m}{n}$ , where $m$ and n are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to

Answer: 119

Numerical answer — enter this value.

Step-by-step solution

$A=\int_{0}^{1}\left[\left(8-4 y^{2}\right)-\left(-2 y^{2}\right)\right] d y+$

$\int_{1}^{3 / 2}\left[\left(8-4 y^{2}\right)-(2 y-4)\right] d y$

$=\left[8 y-\frac{2 y^{3}}{3}\right]_{0}^{1}+\left[12 y-y^{2}-\frac{4 y^{3}}{3}\right]_{1}^{3 / 2}=\frac{107}{12}=\frac{m}{n}$

$\therefore \mathrm{m}+\mathrm{n}=119$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves