JEE Main 2025 — Sets And Relations Question with Solution

The relation $R$ is defined for the set $\mathrm{A}=\{-3,-2,-1,0,1,2,3\}$ with the condition $0\le x^2+2y\le 4$. Let's determine the pairs $(x,y)$ that satisfy this condition.

For $y=-3$:

Solving $x^2+2(-3)\le 4$, we find $x=\{3,-3\}$.

For $y=-2$:

Solving $x^2+2(-2)\le 4$, we find $x=\{-2,2\}$.

For $y=-1$:

Solving $x^2+2(-1)\le 4$, we find $x=\{-2,2\}$.

For $y=0$:

Solving $x^2+0\le 4$, we find $x=\{-2,-1,0,1,2\}$.

For $y=1$:

Solving $x^2+2(1)\le 4$, we find $x=\{-1,0,1\}$.

For $y=2$:

Solving $x^2+2(2)\le 4$, we find $x=\{0\}$.

The relation $R$ consists of the following pairs:

$R=\{(3,-3),(-3,-3),(-2,-2),(2,-2),(-2,-1),(2,-1),(-2,0),(-1,0),(0,0),(1,0),(2,0),(-1,1),(0,1),(1,1),(0,2)\}$

Currently, $R$ has $l=15$ elements. To make $R$ reflexive, it must include all pairs $(x,x)$ for every $x\in \mathrm{A}$. We identify the missing reflexive pairs $(-1,-1)$, $(2,2)$, and $(3,3)$, which are required to satisfy reflexivity.

Thus, $m=3$ more elements are needed. Therefore, the total $l+m=15+3=18$.