QPaperGen

We are given \(A = \{-3,-2,-1,0,1,2,3\}\) and
\[
xRy \iff 0 \leq x^2 + 2y \leq 4.
\]

\(\textbf{Step 1: Find valid pairs:}\)

Check for each \(y \in A\):
\[\]
- For \(y = -3\): condition \(\;0 \leq x^2 - 6 \leq 4 \;\Rightarrow\; 6 \leq x^2 \leq 10\).
Thus \(x = \pm 3\).
Pairs: \((3,-3), (-3,-3)\).
\[\]
- For \(y = -2\): \(0 \leq x^2 - 4 \leq 4 \;\Rightarrow\; 4 \leq x^2 \leq 8\).
Thus \(x = \pm 2\).
Pairs: \((-2,-2), (2,-2)\).
\[\]
- For \(y = -1\): \(0 \leq x^2 - 2 \leq 4 \;\Rightarrow\; 2 \leq x^2 \leq 6\).
Thus \(x = \pm 2\).
Pairs: \((-2,-1), (2,-1)\).
\[\]
- For \(y = 0\): \(0 \leq x^2 \leq 4\).
Thus \(x = -2,-1,0,1,2\).
Pairs: \((-2,0), (-1,0), (0,0), (1,0), (2,0)\).
\[\]
- For \(y = 1\): \(0 \leq x^2 + 2 \leq 4 \;\Rightarrow\; -2 \leq x^2 \leq 2\).
Thus \(x^2 \leq 2 \;\Rightarrow\; x = -1,0,1\).
Pairs: \((-1,1), (0,1), (1,1)\).
\[\]
- For \(y = 2\): \(0 \leq x^2 + 4 \leq 4 \;\Rightarrow\; x^2 = 0\).
Thus \(x=0\).
Pair: \((0,2)\).

\[\]
\(\textbf{Step 2: Count elements of \(R\)}\)

So the relation is
\(
R = \{(3,-3), (-3,-3), (-2,-2), (2,-2), (-2,-1), (2,-1), (-2,0), (-1,0), (0,0), (1,0), (2,0), (-1,1), (0,1), (1,1), (0,2)\}.
\)

Hence \(l = 15\).

\[\]

\(\textbf{Step 3: Make \(R\) reflexive:}\)

A relation is reflexive if \((a,a) \in R\) for all \(a \in A\).

Currently in \(R\):
\((-3,-3), (-2,-2), (0,0), (1,1)\) are present.

Missing: \((-1,-1), (2,2), (3,3)\).

So we must add \(m = 3\) elements.

\[\]

\(\textbf{Step 4: Final Answer}\)

\[
l+m = 15+3 = 18.
\]

\[
\therefore \quad \boxed{18}
\]