JEE Main 2023 — Sets and Relations Question with Solution

Given,

Set $A=\left\{1,2,3\right\}$

Now Cartesian product $A\times A=\left\{\left(1,1\right),\left(2,1\right),\left(1,2\right),........,\left(3,3\right)\right\}$

Now, given the relation is reflexive,

So, $\left(1, 1\right), \left(2, 2\right), \left(3, 3\right) \in R$

Also given $\left(1, 2\right), \left(2, 3\right)\in R, \left(1, 3\right)$ must $\in R$

Now finding, Possible cases :

Case-1: All of $\left(2, 1\right), \left(3, 2\right), \left(3, 1\right) \notin R\to 1$ relation.

Case-2: Only one of $\left(2, 1\right), \left(3, 2\right), \left(3, 1\right)\in R\to 3$ relations.

For example if relation is $\left\{\left(1, 1\right), \left(2, 2\right), \left(3, 3\right),\left(2,1\right)\right\}$ then it is reflexive as well as transitive as $\left(2,2\right),\left(2,1\right)\to \left(2,1\right) \text{is present in relation}$

Note that exactly two of $\left(2, 1\right), \left(3, 2\right), \left(3, 1\right)\in R$ is not possible because if two of these $\in R$, third must $\in R$  to make relation transitive.

Total number of relations $=4$