UPSC NDA I 2019 Mathematics Solved Paper

Concept:A relation on a set is reflexive if every element is related to itself, symmetric if for every ordered pair its reverse is also present, and transitive if whenever $(a,b)$ and $(b,c)$ are present, $(a,c)$ is also present.Explanation:Given $X = \{1, 2, 3, 4\}$ and $R = \{(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)\}$.Check reflexivity:For $R$ to be reflexive, $(a,a) \in R$ for every $a \in X$.Here $(4,4) \notin R$, so $R$ is not reflexive.Check symmetry:For each $(a,b) \in R$, check if $(b,a) \in R$.All pairs satisfy this: e.g., $(1,2) \to (2,1)$, $(2,3) \to (3,2)$, $(1,3) \to (3,1)$, and all self-pairs are symmetric.Thus $R$ is symmetric.Check transitivity:We need: if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.Verify all combinations:• $(1,2)$ and $(2,1)$ give $(1,1)$ – present.• $(1,2)$ and $(2,3)$ give $(1,3)$ – present.• $(2,1)$ and $(1,2)$ give $(2,2)$ – present.• $(2,1)$ and $(1,3)$ give $(2,3)$ – present.• $(2,3)$ and $(3,2)$ give $(2,2)$ – present.• $(3,2)$ and $(2,1)$ give $(3,1)$ – present.• $(1,3)$ and $(3,2)$ give $(1,2)$ – present.• $(3,1)$ and $(1,2)$ give $(3,2)$ – present.• Also all cases with self-pairs hold.No counterexample exists, so $R$ is transitive.Thus $R$ is symmetric and transitive, but not reflexive.Answer:Option B is correct: R is symmetric and transitive, but not reflexive.