JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Sets Relations and Functions

Let $d=a-b$, so the requirement is $2d^2+3d=d(2d+3)\in \{0,1,2,3,4\}$. Testing integer values: $d=0$ gives $0$ and $d=-2$ gives $(-2)(-1)=2$, both in $B$; while $d=-1$ gives $-1$, $d=1$ gives $5$, $d=2$ gives $14$, and $|d|\ge 3$ gives still larger values, none in $B$. For $d=0$ (that is $a=b$) there are $10$ ordered pairs $(1,1),\dots ,(10,10)$. For $d=-2$ (that is $b=a+2$) the pairs are $(1,3),(2,4),\dots ,(8,10)$, giving $8$ pairs. Therefore the number of elements in $R$ is $10+8=18$.