UPSC NDA I 2022 Mathematics Solved Paper

Concept:A relation R on a set is reflexive if each element is related to itself, symmetric if $xRy$ implies $yRx$, and transitive if $xRy$ and $yRz$ imply $xRz$.Explanation:Given $R = \{(x,y) \in \mathbb{N} \times \mathbb{N} \mid x^{2} - 5xy + 4y^{2} = 0\}$.Check reflexivity: For any $x \in \mathbb{N}$, substitute $y = x$: $x^{2} - 5x^{2} + 4x^{2} = 0$. So $(x,x) \in R$. Hence R is reflexive.Check symmetry: Take a counterexample: $(4,1)$ satisfies $4^{2} - 5 \cdot 4 \cdot 1 + 4 \cdot 1^{2} = 16 - 20 + 4 = 0$, so $(4,1) \in R$. But $(1,4)$ gives $1 - 20 + 64 = 45 \neq 0$, so $(1,4) \notin R$. Thus R is not symmetric.Check transitivity: Take $(16,4)$ and $(4,1)$. Both satisfy the condition: $(16,4)$: $256 - 320 + 64 = 0$; $(4,1)$: $16 - 20 + 4 = 0$. However, $(16,1)$ gives $256 - 80 + 4 = 180 \neq 0$, so $(16,1) \notin R$. Hence R is not transitive.Answer:Only statement 1 is correct, i.e., R is reflexive only. Therefore, option A (1 only) is correct.