JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
Let $A=\{-3,-2,-1,0,1,2,3\}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $0\le x^{2}+2y\le4$. Let $l$ be the number of elements in $R$ and $m$ be the minimum number of elements required to be added in $R$ to make it a reflexive relation. Then $l+m$ is equal to
Sets, Relations and Function
Solution
Pairs $(x,y)$ with $0\le x^2+2y\le4$: $x=0$: $y\in\{0,1,2\}$ (3); $x=\pm1$: $-1\le 2y\le3$, $y\in\{0,1\}$ (2 each, 4); $x=\pm2$: $-4\le2y\le0$, $y\in\{-2,-1,0\}$ (6); $x=\pm3$: $-9\le 2y\le-5$, $y\in\{-3\}$ (2). Hence $l=15$. Reflexive pairs present need $x^2+2x\in[0,4]$: $x=0\,(0),\,1\,(3),\,-2\,(0),\,-3\,(3)$ qualify; $x=-1\,(-1),\,2\,(8),\,3\,(15)$ do not. So $3$ pairs must be added, $m=3$. $l+m=18$. Option C.