Let $A=\{-2,-1,0,1,2,3\}$. Let R be a relation on $A$ defined by $x \mathrm{R} y$ if and only if $y=\max \{x, 1\}$. Let $l$ be the number of elements in R . Let $m$ and $n$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then $l+m+n$ is equal to

<p>To solve the problem, we start by defining the set $A = \{-2, -1, 0, 1, 2, 3\}$ and the relation $R$ on set $A$, where an element $x$ is related to $y$ (written as $x \, R \, y$) if and only if $y = \max\{x, 1\}$.</p> <p>This leads us to the following pairs in the relation $R$:</p> <p><p>For $x = -2$, $y = \max\{-2, 1\} = 1$, so $(-2, 1)$ is in $R$.</p></p> <p><p>For $x = -1$, $y = \max\{-1, 1\} = 1$, so $(-1, 1)$ is in $R$.</p></p> <p><p>For $x = 0$, $y = \max\{0, 1\} = 1$, so $(0, 1)$ is in $R$.</p></p> <p><p>For $x = 1$, $y = \max\{1, 1\} = 1$, so $(1, 1)$ is in $R$.</p></p> <p><p>For $x = 2$, $y = \max\{2, 1\} = 2$, so $(2, 2)$ is in $R$.</p></p> <p><p>For $x = 3$, $y = \max\{3, 1\} = 3$, so $(3, 3)$ is in $R$.</p></p> <p>Thus, the relation $R$ consists of the pairs: $\{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$, and there are $l = 6$ elements in $R$.</p> <h3>Making the Relation Reflexive</h3> <p>A relation is reflexive if every element in the set $A$ relates to itself. Therefore, the missing reflexive pairs are:</p> <p><p>$(-2, -2)$</p></p> <p><p>$(-1, -1)$</p></p> <p><p>$(0, 0)$</p></p> <p>Adding these three pairs will make the relation reflexive, so $m = 3$.</p> <h3>Making the Relation Symmetric</h3> <p>A relation is symmetric if whenever $(x, y)$ is in $R$, $(y, x)$ must also be in $R$. Therefore, the missing symmetric pairs are:</p> <p><p>$(1, -2)$</p></p> <p><p>$(1, -1)$</p></p> <p><p>$(1, 0)$</p></p> <p>Thus, we need to add these three pairs for symmetry, so $n = 3$.</p> <p>Finally, we calculate the sum $l + m + n = 6 + 3 + 3 = 12$.</p>