Let the area of the region { ( x , y ) : x - 2 y + 4 ? 0 , x + 2 y 2 ? 0 , x + 4 y 2 ? 8 , y ? 0 } be m n , where m and n are coprime numbers. Then m + n is equal to [2024]

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Solution

Q.
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 $\frac{m}{n},$ where $m$ and $n$ are coprime numbers. Then $m + n$ is equal to _____ . [2024]

Ans.
(119)

Given, $x - 2 y + 4 = 0$ ...(i)

$x + 2 y^{2} = 0$ ...(ii)

$x + 4 y^{2} = 8$ ...(iii)

The point intersection of (i) and (iii) are $(- 1, 1.5)$ and $(- 8, - 2) .$

The point of intersection of (i) and (ii) are $(- 2, 1)$ and $(- 8, - 2) .$

$∴$ Required area

$\int_{0}^{1} [(8 - 4 y^{2}) - (- 2 y^{2})] d y + \int_{1}^{3 / 2} [(8 - 4 y^{2}) - (2 y - 4)] d y$

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

$= (8 - \frac{2}{3}) + (12 (\frac{3}{2}) - {(\frac{3}{2})}^{2} - \frac{4}{3} {(\frac{3}{2})}^{3}) - (12 - 1 - \frac{4}{3})$

$= \frac{22}{3} + 18 - \frac{9}{4} - \frac{9}{2} - \frac{29}{3} = \frac{107}{12} = \frac{m}{n}$

$∴ m + n = 119$

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