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

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(A) Given set $A = \{-2, -1, 0, 1, 2, 3\}$.Relation $R$ is defined as $x R y \iff y = \max \{x, 1\}$.Calculating elements of $R$:For $x = -2, y = \max

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$12$

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(A) Given set $A = \{-2, -1, 0, 1, 2, 3\}$.Relation $R$ is defined as $x R y \iff y = \max \{x, 1\}$.Calculating elements of $R$:For $x = -2, y = \max \{-2, 1\} = 1 \implies (-2, 1) \in R$.For $x = -1, y = \max \{-1, 1\} = 1 \implies (-1, 1) \in R$.For $x = 0, y = \max \{0, 1\} = 1 \implies (0, 1) \in R$.For $x = 1, y = \max \{1, 1\} = 1 \implies (1, 1) \in R$.For $x = 2, y = \max \{2, 1\} = 2 \implies (2, 2) \in R$.For $x = 3, y = \max \{3, 1\} = 3 \implies (3, 3) \in R$.So,$R = \{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$.Thus,$l = 6$.For $R$ to be reflexive,we need $(x, x) \in R$ for all $x \in A$. Currently,$R$ contains $(1, 1), (2, 2), (3, 3)$. We need to add $(-2, -2), (-1, -1), (0, 0)$. So,$m = 3$.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. The pairs in $R$ are $(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)$.Symmetry requires $(1, -2), (1, -1), (1, 0)$ to be added. So,$n = 3$.Therefore,$l + m + n = 6 + 3 + 3 = 12$.

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