JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 26)
Let the area of the region $\left\{(x, y): x-2 y+4 \geqslant 0, x+2 y^2 \geqslant 0, x+4 y^2 \leq 8, y \geqslant 0\right\}$ be $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to _____________.
Answer
119
Explanation
_27th_January_Morning_Shift_en_26_1.png)
$$\begin{aligned} & A=\int_\limits0^1\left[\left(8-4 y^2\right)-\left(-2 y^2\right)\right] d y+ \\ & \int_\limits1^{3 / 2}\left[\left(8-4 y^2\right)-(2 y-4)\right] d y \\ & =\left[8 y-\frac{2 y^3}{3}\right]_0^1+\left[12 y-y^2-\frac{4 y^3}{3}\right]_1^{3 / 2}=\frac{107}{12}=\frac{m}{n} \\ & \therefore m+n=119 \end{aligned}$$
Comments (0)
