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(D) We are given the sets $A = \{(x, y) \in \mathbb{R} 	imes \mathbb{R} : |x + y| \geq 3\}$ and $B = \{(x, y) \in \mathbb{R} 	imes \mathbb{R} : |x|

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) We are given the sets $A = \{(x, y) \in \mathbb{R} 	imes \mathbb{R} : |x + y| \geq 3\}$ and $B = \{(x, y) \in \mathbb{R} 	imes \mathbb{R} : |x| + |y| \leq 3\}$.We need to find the set $C = \{(x, y) \in A \cap B : x = 0 	ext{ or } y = 0\}$.Case $1$: If $x = 0$,then $(0, y) \in A \cap B$.From $B$,$|0| + |y| \leq 3 \implies |y| \leq 3 \implies -3 \leq y \leq 3$.From $A$,$|0 + y| \geq 3 \implies |y| \geq 3$.Combining these,$|y| = 3$,so $y = 3$ or $y = -3$. Thus,$(0, 3)$ and $(0, -3)$ are in $C$.Case $2$: If $y = 0$,then $(x, 0) \in A \cap B$.From $B$,$|x| + |0| \leq 3 \implies |x| \leq 3 \implies -3 \leq x \leq 3$.From $A$,$|x + 0| \geq 3 \implies |x| \geq 3$.Combining these,$|x| = 3$,so $x = 3$ or $x = -3$. Thus,$(3, 0)$ and $(-3, 0)$ are in $C$.Therefore,$C = \{(3, 0), (-3, 0), (0, 3), (0, -3)\}$.Now,we calculate $\sum_{(x, y) \in C} |x + y| = |3 + 0| + |-3 + 0| + |0 + 3| + |0 - 3| = 3 + 3 + 3 + 3 = 12$.

Let $A = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + y| \geq 3\}$ and $B = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| + |y| \leq 3\}$. If $C = \{(x, y) \in A \cap B : x = 0 \text{ or } y = 0\}$,then $\sum_{(x, y) \in C} |x + y|$ is :

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