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$Let A_{1}, A_{2} and A_{3} be the regions on R^{2} defined by$
$A_{1} = \{(x, y) : x \geq 0, y \geq 0, 2 x + 2 y - x^{2} - y^{2} > 1 > x + y\}$
$A_{2} = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{2} + y^{2}\}$
$A_{3} = \{(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{3} + y^{3}\}$
$Denote by |A_{1}|, |A_{2}| and |A_{3}| the areas of the regions A_{1}, A_{2} and A_{3} respectively. Then$

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$|A_{1}| > |A_{2}| > |A_{3}|$

$|A_{1}| > |A_{3}| > |A_{2}|$

$|A_{1}| = |A_{2}| < |A_{3}|$

$|A_{1}| = |A_{3}| > |A_{2}|$

answer is C.

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Detailed Solution

$T h e a r e a s A_{1}, A_{2}, A_{3} a r e a s b e l o w$

Observing the areas $A_{1}, A_{2}$ are same, because both circles intersect $x + y = 1$ at the same points and symmetric too.

The area $A_{3}$ is more than $A_{2}$

Hence, $A_{1} = A_{2} < A_{3}$

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