Step 1: Identify the set and the condition

The set is $A = \{-3, -2, -1, 0, 1, 2, 3\}$. The relation $R$ is defined by $0 \le x^2 + 2y \le 4$. We need to find all pairs $(x, y)$ such that $x, y \in A$ satisfying this inequality.

Step 2: Evaluate pairs for each x

For each $x \in A$, we find $y \in A$ such that $-x^2 \le 2y \le 4 - x^2$, which implies $-\frac{x^2}{2} \le y \le 2 - \frac{x^2}{2}$.

• If $x = 0$: $0 \le y \le 2 \implies y \in \{0, 1, 2\}$. (3 pairs)
• If $x = \pm 1$: $-0.5 \le y \le 1.5 \implies y \in \{0, 1\}$. (2 values for $x=1$, 2 for $x=-1$, total 4 pairs)
• If $x = \pm 2$: $-2 \le y \le 0 \implies y \in \{-2, -1, 0\}$. (3 values for $x=2$, 3 for $x=-2$, total 6 pairs)
• If $x = \pm 3$: $-4.5 \le y \le -2.5 \implies y \in \{-3\}$. (1 value for $x=3$, 1 for $x=-3$, total 2 pairs)

Step 3: Calculate l and m

Total elements $l = 3 + 4 + 6 + 2 = 15$. For $R$ to be reflexive, $(x, x)$ must be in $R$ for all $x \in A$. Checking $(x, x)$: $x=0: (0,0) \in R$ (Yes). $x=\pm 1: (1,1) \in R, (-1,-1) \in R$ (Yes). $x=\pm 2: (2,2) \notin R$ (since $4+4=8 > 4$), $(-2,-2) \notin R$ (since $4-4=0 \in [0,4]$ - wait, $(-2,-2)$ is in $R$). $x=\pm 3: (3,3) \notin R, (-3,-3) \notin R$. Missing reflexive pairs: $(2,2), (3,3), (-3,-3)$. Thus $m = 3$.

Step 4: Final Calculation

The value $l + m = 15 + 3 = 18$.