JEE Main 2025 — Sets And Relations Question with Solution

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 $xRy$) if and only if $y=\max \{x,1\}$.

This leads us to the following pairs in the relation $R$:

For $x=-2$, $y=\max \{-2,1\}=1$, so $(-2,1)$ is in $R$.

For $x=-1$, $y=\max \{-1,1\}=1$, so $(-1,1)$ is in $R$.

For $x=0$, $y=\max \{0,1\}=1$, so $(0,1)$ is in $R$.

For $x=1$, $y=\max \{1,1\}=1$, so $(1,1)$ is in $R$.

For $x=2$, $y=\max \{2,1\}=2$, so $(2,2)$ is in $R$.

For $x=3$, $y=\max \{3,1\}=3$, so $(3,3)$ is in $R$.

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 $R$.

Making the Relation Reflexive

A relation is reflexive if every element in the set $A$ relates to itself. Therefore, the missing reflexive pairs are:

$(-2,-2)$

$(-1,-1)$

$(0,0)$

Adding these three pairs will make the relation reflexive, so $m=3$.

Making the Relation Symmetric

A relation is symmetric if whenever $(x,y)$ is in $R$, $(y,x)$ must also be in $R$. Therefore, the missing symmetric pairs are:

$(1,-2)$

$(1,-1)$

$(1,0)$

Thus, we need to add these three pairs for symmetry, so $n=3$.

Finally, we calculate the sum $l+m+n=6+3+3=12$.