WBJEE 2024 Math Paper

To analyze the relation $p$ given by $x p y$ iff $x-y+\sqrt{2}$ is an irrational number, let's evaluate whether the relation is reflexive, symmetric, transitive, or an equivalence relation.Reflexive : A relation is reflexive if every element is related to itself. For $x p x$ to hold, we need $x-x+\sqrt{2}$ to be irrational. Simplifying, $x-x=0$, and $0+\sqrt{2}=\sqrt{2}$, which is indeed irrational. Therefore, the relation $p$ is reflexive.Reflexive: YesSymmetric : Consider $(\sqrt{2}, 1) \in p \Rightarrow \sqrt{2}-1+\sqrt{2}=2\sqrt{2}-1$ is irrational number. but $(1, \sqrt{2}) p$ as $1-\sqrt{2}+\sqrt{2}=1$ is not an irrational number.Thus $p$ is not symmetric.Symmetric : No Transitive :$(\sqrt{2}, 1) \in p$ as $\sqrt{2}-1+\sqrt{2}$ is irrational.$(1,2\sqrt{2}) \in \rho$ as $1-2\sqrt{2}+\sqrt{2}=1-\sqrt{2}$ is irrational.Now $\sqrt{2}-2\sqrt{2}+\sqrt{2}=0$ is not irrational.$\Rightarrow (\sqrt{2}, 2\sqrt{2}) \notin p$$\Rightarrow p$ is not transitive relation.Transitive: No Equivalence Relation :A relation is an equivalence relation if it is reflexive, symmetric, and transitive. We've established that the relation $p$ is reflexive but not symmetric not transitive. Therefore, the relation $p$ is not an equivalence relation.Equivalence Relation : NoThus, the correct answer is :Option A : Reflexive