JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Sets Relations and Functions

Reflexivity holds because $1+a^2>0$ always, and symmetry holds because $1+ab=1+ba$. Transitivity fails: $(-2,0)$ and $(0,2)$ both satisfy $1+ab>0$, yet $1+(-2)(2)=-3<0$, so $(-2,2)\notin R$; hence $R$ is not an equivalence relation and Statement II is false. For the count, among the $25$ ordered pairs the condition $1+ab>0$ fails only when $ab\le -1$, namely $(1,-1),(-1,1),(1,-2),(-1,2),(2,-1),(-2,1),(2,-2),(-2,2)$ — eight pairs. This leaves $25-8=17$ pairs in $R$, so Statement I is true.