Key Concepts and Formulas

• Relation: A relation $R$ on a set $A$ is a subset of the Cartesian product $A \times A$. An ordered pair $(x, y) \in R$ means $x$ is related to $y$.
• Reflexive Relation: A relation $R$ on a set $A$ is reflexive if for every element $x \in A$, the ordered pair $(x, x)$ is in $R$.
• Set $A$: $A = \{-3, -2, -1, 0, 1, 2, 3\}$.
• Condition for relation $R$: $x \mathrm{R} y$ if and only if $0 \leq x^2 + 2y \leq 4$.

---

Step-by-Step Solution

Step 1: Determine the elements of the relation $R$ and find $l$ (the number of elements in $R$).

The relation $R$ consists of ordered pairs $(x, y)$ where $x, y \in A$ and $0 \leq x^2 + 2y \leq 4$. We can rewrite the inequality as $x^2 \leq 4 - 2y$. Since $x^2 \geq 0$, we also need $4 - 2y \geq 0$, which implies $2y \leq 4$, or $y \leq 2$. This means we only need to consider $y \in \{-3, -2, -1, 0, 1, 2\}$.

We systematically check each possible value of $y \in A$:

• For $y = -3$: The condition becomes $0 \leq x^2 + 2(-3) \leq 4 \implies 0 \leq x^2 - 6 \leq 4$. This implies $6 \leq x^2 \leq 10$. For $x \in A$, the possible values for $x^2$ are $\{0, 1, 4, 9\}$. The only value in this set satisfying $6 \leq x^2 \leq 10$ is $x^2 = 9$. If $x^2 = 9$, then $x = 3$ or $x = -3$. Both $3, -3 \in A$. The pairs are $(-3, -3)$ and $(3, -3)$. (2 pairs)

• For $y = -2$: The condition becomes $0 \leq x^2 + 2(-2) \leq 4 \implies 0 \leq x^2 - 4 \leq 4$. This implies $4 \leq x^2 \leq 8$. For $x \in A$, the only value satisfying $4 \leq x^2 \leq 8$ is $x^2 = 4$. If $x^2 = 4$, then $x = 2$ or $x = -2$. Both $2, -2 \in A$. The pairs are $(-2, -2)$ and $(2, -2)$. (2 pairs)

• For $y = -1$: The condition becomes $0 \leq x^2 + 2(-1) \leq 4 \implies 0 \leq x^2 - 2 \leq 4$. This implies $2 \leq x^2 \leq 6$. For $x \in A$, the only value satisfying $2 \leq x^2 \leq 6$ is $x^2 = 4$. If $x^2 = 4$, then $x = 2$ or $x = -2$. Both $2, -2 \in A$. The pairs are $(-2, -1)$ and $(2, -1)$. (2 pairs)

• For $y = 0$: The condition becomes $0 \leq x^2 + 2(0) \leq 4 \implies 0 \leq x^2 \leq 4$. For $x \in A$, the values satisfying $0 \leq x^2 \leq 4$ are $x^2 \in \{0, 1, 4\}$. If $x^2 = 0$, then $x = 0$. If $x^2 = 1$, then $x = 1$ or $x = -1$. If $x^2 = 4$, then $x = 2$ or $x = -2$. All these $x$ values ($0, \pm 1, \pm 2$) are in $A$. The pairs are $(-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0)$. (5 pairs)

• For $y = 1$: The condition becomes $0 \leq x^2 + 2(1) \leq 4 \implies 0 \leq x^2 + 2 \leq 4$. This implies $-2 \leq x^2 \leq 2$. Since $x^2 \geq 0$, we have $0 \leq x^2 \leq 2$. For $x \in A$, the values satisfying $0 \leq x^2 \leq 2$ are $x^2 \in \{0, 1\}$. If $x^2 = 0$, then $x = 0$. If $x^2 = 1$, then $x = 1$ or $x = -1$. All these $x$ values ($0, \pm 1$) are in $A$. The pairs are $(-1, 1), (0, 1), (1, 1)$. (3 pairs)

• For $y = 2$: The condition becomes $0 \leq x^2 + 2(2) \leq 4 \implies 0 \leq x^2 + 4 \leq 4$. This implies $-4 \leq x^2 \leq 0$. Since $x^2 \geq 0$, we must have $x^2 = 0$. If $x^2 = 0$, then $x = 0$. This $x$ value is in $A$. The pair is $(0, 2)$. (1 pair)

• For $y = 3$: The condition becomes $0 \leq x^2 + 2(3) \leq 4 \implies 0 \leq x^2 + 6 \leq 4$. This implies $-6 \leq x^2 \leq -2$. There are no real values of $x^2$ that satisfy this. No pairs for $y=3$. (0 pairs)

The total number of elements in $R$ is the sum of the counts from each case: $l = 2 + 2 + 2 + 5 + 3 + 1 + 0 = 15$.

Step 2: Determine $m$, the minimum number of elements required to be added to $R$ to make it a reflexive relation.

For a relation $R$ on set $A$ to be reflexive, every element $x \in A$ must be related to itself. This means the pairs $(x, x)$ must be in $R$ for all $x \in A$. The set $A = \{-3, -2, -1, 0, 1, 2, 3\}$ has 7 elements. The required reflexive pairs are: $(-3, -3), (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2), (3, 3)$.

We check which of these pairs are already present in the relation $R$ we found in Step 1:

1. $(-3, -3)$: Found in the case $y = -3$. (Present)
2. $(-2, -2)$: Found in the case $y = -2$. (Present)
3. $(-1, -1)$: Check if $0 \leq (-1)^2 + 2(-1) \leq 4 \implies 0 \leq 1 - 2 \leq 4 \implies 0 \leq -1 \leq 4$. This is false. $(-1, -1) \notin R$. (Missing)
4. $(0, 0)$: Found in the case $y = 0$. (Present)
5. $(1, 1)$: Found in the case $y = 1$. (Present)
6. $(2, 2)$: Check if $0 \leq (2)^2 + 2(2) \leq 4 \implies 0 \leq 4 + 4 \leq 4 \implies 0 \leq 8 \leq 4$. This is false. $(2, 2) \notin R$. (Missing)
7. $(3, 3)$: Check if $0 \leq (3)^2 + 2(3) \leq 4 \implies 0 \leq 9 + 6 \leq 4 \implies 0 \leq 15 \leq 4$. This is false. $(3, 3) \notin R$. (Missing)

The missing pairs required for reflexivity are $(-1, -1), (2, 2), (3, 3)$. Therefore, the minimum number of elements to be added is $m = 3$.

Step 3: Calculate $l+m$.

We found $l=15$ and $m=3$. $l + m = 15 + 3 = 18$.

---

Common Mistakes & Tips

• Careful with inequalities: Ensure all parts of the inequality are considered, especially when $x^2$ is involved (it must be non-negative).
• Check membership in A: Always verify that the $x$ values derived from the inequality are actually elements of the set $A$.
• Systematic approach: Iterate through all possible values of $y$ (or $x$) to ensure no pairs are missed. For reflexivity, ensure all elements of $A$ are checked for self-relation.

---

Summary

We first determined the elements of the relation $R$ by systematically checking all pairs $(x, y)$ from the set $A$ against the given condition $0 \leq x^2 + 2y \leq 4$. This process yielded $l=15$ elements in $R$. To find $m$, the minimum number of elements to add for reflexivity, we identified the pairs $(x, x)$ that are required for all $x \in A$. By checking which of these required pairs were not already in $R$, we found that $m=3$ elements needed to be added. Finally, the sum $l+m$ was calculated as $15+3=18$.

The final answer is $\boxed{\text{18}}$.