Let A_{1}={(x, y):|x| leq y^{2},|x|+2 y leq 8 | vedclass

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(A) The region $A_{1}$ is defined by $|x| \leq y^{2}$ and $|x|+2y \leq 8$. Since both inequalities are symmetric about the $y$-axis,the area is $2 	i

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$6$

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(A) The region $A_{1}$ is defined by $|x| \leq y^{2}$ and $|x|+2y \leq 8$. Since both inequalities are symmetric about the $y$-axis,the area is $2 	imes$ the area in the first quadrant $(x \geq 0)$.In the first quadrant,the region is bounded by $x = y^{2}$ and $x = 8 - 2y$.To find the intersection point,set $y^{2} = 8 - 2y \implies y^{2} + 2y - 8 = 0 \implies (y+4)(y-2) = 0$. Since $y \geq 0$,we have $y = 2$.Area $(A_{1}) = 2 \left[ \int_{0}^{2} y^{2} dy + \int_{2}^{4} (8-2y) dy ight]$$= 2 \left[ \left( \frac{y^{3}}{3} ight)_{0}^{2} + \left( 8y - y^{2} ight)_{2}^{4} ight]$$= 2 \left[ \frac{8}{3} + (32 - 16) - (16 - 4) ight] = 2 \left[ \frac{8}{3} + 16 - 12 ight] = 2 \left[ \frac{8}{3} + 4 ight] = 2 \left( \frac{20}{3} ight) = \frac{40}{3}$.The region $A_{2}$ is $|x|+|y| \leq k$,which is a square with vertices at $(\pm k, 0)$ and $(0, \pm k)$. The area of this square is $2k^{2}$.Given $27 	imes 	ext{Area}(A_{1}) = 5 	imes 	ext{Area}(A_{2})$:$27 	imes \frac{40}{3} = 5 	imes 2k^{2}$$9 	imes 40 = 10k^{2}$$360 = 10k^{2} \implies k^{2} = 36 \implies k = 6$.

Let $A_{1}=\{(x, y):|x| \leq y^{2},|x|+2 y \leq 8\}$ and $A_{2}=\{(x, y):|x|+|y| \leq k\}$. If $27 \times \text{Area}(A_{1}) = 5 \times \text{Area}(A_{2})$,then $k$ is equal to

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