QPaperGen

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



\(\textbf{Step 1: Find \(R\)}\)

\[
\begin{aligned}
x = -2 &\implies y = \max(-2,1) = 1 \quad \Rightarrow (-2,1) \in R, \\
x = -1 &\implies y = 1 \quad \Rightarrow (-1,1) \in R, \\
x = 0 &\implies y = 1 \quad \Rightarrow (0,1) \in R, \\
x = 1 &\implies y = 1 \quad \Rightarrow (1,1) \in R, \\
x = 2 &\implies y = 2 \quad \Rightarrow (2,2) \in R, \\
x = 3 &\implies y = 3 \quad \Rightarrow (3,3) \in R.
\end{aligned}
\]

Thus,
\[
R = \{(-2,1), (-1,1), (0,1), (1,1), (2,2), (3,3)\}, \quad l = 6.
\]


\(\textbf{Step 2: Make \(R\) reflexive:}\)

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

Currently, we already have \((1,1), (2,2), (3,3)\).

Missing pairs: \((-2,-2), (-1,-1), (0,0)\).

Thus, \[m = 3\].



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

A relation is symmetric if \((x,y) \in R \implies (y,x) \in R\).

Check:
\[\begin{aligned}
(-2,1) \implies (1,-2) \qquad & \text{missing.}\\
(-1,1) \implies (1,-1)\qquad & \text{missing.}\\
(0,1) \implies (1,0)\qquad & \text{missing.}\\
(1,1), (2,2), (3,3) \qquad & \text{are already symmetric.}\\
\end{aligned}
\]
So, we need to add \(3\) pairs. Hence \[n=3\]

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

\[
l+m+n = 6+3+3 = 12.
\]

\[
\therefore \quad \text{The required value is } \boxed{12}.
\]