Let A = {-3, -2, -1, 0, 1, 2, 3}. Let R be a | vedclass

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(D) Given $A = \{-3, -2, -1, 0, 1, 2, 3\}$. The condition is $0 \leq x^2 + 2y \leq 4$,which implies $-x^2 \leq 2y \leq 4 - x^2$,or $-\frac{x^2}{2} \le

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$18$

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(D) Given $A = \{-3, -2, -1, 0, 1, 2, 3\}$. The condition is $0 \leq x^2 + 2y \leq 4$,which implies $-x^2 \leq 2y \leq 4 - x^2$,or $-\frac{x^2}{2} \leq y \leq 2 - \frac{x^2}{2}$.For each $x \in A$,we find $y \in A$ satisfying the condition:If $x = -3, x^2 = 9$: $-4.5 \leq y \leq -2.5 \Rightarrow y = -3$. Pairs: $(-3, -3)$.If $x = -2, x^2 = 4$: $-2 \leq y \leq 0 \Rightarrow y = -2, -1, 0$. Pairs: $(-2, -2), (-2, -1), (-2, 0)$.If $x = -1, x^2 = 1$: $-0.5 \leq y \leq 1.5 \Rightarrow y = 0, 1$. Pairs: $(-1, 0), (-1, 1)$.If $x = 0, x^2 = 0$: $0 \leq y \leq 2 \Rightarrow y = 0, 1, 2$. Pairs: $(0, 0), (0, 1), (0, 2)$.If $x = 1, x^2 = 1$: $-0.5 \leq y \leq 1.5 \Rightarrow y = 0, 1$. Pairs: $(1, 0), (1, 1)$.If $x = 2, x^2 = 4$: $-2 \leq y \leq 0 \Rightarrow y = -2, -1, 0$. Pairs: $(2, -2), (2, -1), (2, 0)$.If $x = 3, x^2 = 9$: $-4.5 \leq y \leq -2.5 \Rightarrow y = -3$. Pairs: $(3, -3)$.The set $R = \{(-3, -3), (-2, -2), (-2, -1), (-2, 0), (-1, 0), (-1, 1), (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, -2), (2, -1), (2, 0), (3, -3)\}$.Counting the elements,$l = 15$.For $R$ to be reflexive,it must contain $(a, a)$ for all $a \in A$. The missing elements are $(-3, -3)$ (present),$(-2, -2)$ (present),$(-1, -1)$ (missing),$(0, 0)$ (present),$(1, 1)$ (present),$(2, 2)$ (missing),$(3, 3)$ (missing).Thus,$m = 3$ elements are required to be added.Therefore,$l + m = 15 + 3 = 18$.

Let $A = \{-3, -2, -1, 0, 1, 2, 3\}$. Let $R$ be a relation on $A$ defined by $x R y$ if and only if $0 \leq x^2 + 2y \leq 4$. Let $l$ be the number of elements in $R$ and $m$ be the minimum number of elements required to be added to $R$ to make it a reflexive relation. Then $l+m$ is equal to

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