Question

Sol. [Key Concept: $R:\ 1 + ab > 0$ on $\{ - 2, - 1,0,1,2\}$ is reflexive and symmetric but not transitive (e.g. $( - 2,0),(0,2) \in \ R$ but $( - 2,2)notin\ R$); counting the valid pairs gives $17$ elements, so only Statement I is true.]

Reflexive If $(a,a) \in R$

$$\Rightarrow 1 + a^{2} > 0 $$⇒ It is reflexive relation

Symmetric If $1 + ab > 0 \Rightarrow 1 + ba > 0$

i.e. If $(a,b) \in R$ and $(b,a) \in R$

Therefore it is symmetric relation

Transitive $\ \because( - 2,0) \in R\&(0,2) \in R$

But $( - 2,2) \notin R$

Hence it is not transitive

⇒ It is not an equivalence relation.

Check Number of elements in R

$$\because 1 + ab > 0 $$$a\&\text{ }b$ each can be filled by 5 ways

∴ total elements $= 25$ But out of these

$$( - 2, - 1),( - 2,2),( - 1,2),(1, - 2),(2, - 2),(2, - 1) $$$( - 1,1)\&(1, - 1)$ are not in relation.