Let A_1, A_2 and A_3 be the regions on mathbb | vedclass

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(C) The region $A_1$ is defined by $x \geq 0, y \geq 0$ and $2x + 2y - x^2 - y^2 > 1 > x + y$.Rearranging the inequality $2x + 2y - x^2 - y^2 > 1$,we

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$|A_1| = |A_2| < |A_3|$

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(C) The region $A_1$ is defined by $x \geq 0, y \geq 0$ and $2x + 2y - x^2 - y^2 > 1 > x + y$.Rearranging the inequality $2x + 2y - x^2 - y^2 > 1$,we get $x^2 - 2x + y^2 - 2y For $A_2$,the region is $x \geq 0, y \geq 0$ and $x + y > 1 > x^2 + y^2$. This is the region between the line $x + y = 1$ and the circle $x^2 + y^2 = 1$ in the first quadrant.For $A_3$,the region is $x \geq 0, y \geq 0$ and $x + y > 1 > x^3 + y^3$. This is the region between the line $x + y = 1$ and the curve $x^3 + y^3 = 1$ in the first quadrant.By symmetry and geometric transformation,$|A_1| = |A_2|$. Comparing the curves $x^2 + y^2 = 1$ and $x^3 + y^3 = 1$ in the first quadrant,for $0 Therefore,$|A_3| > |A_2| = |A_1|$.Hence,the correct option is $(c)$.

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