Q153 / 270 · NDA 2019 (Phase 1) · Mathematics
$A$ relation $R$ is defined on the set $N$ of natural numbers as $x R y \Rightarrow x^2-4 x y+3 y^2=0$, Then which one of the following is correct?
- AR is reflexive and symmetric, but not transitive
- BR is reflexive and transitive , but not symmetric
- CR is reflexive, symmetric and transitive
- Correct answerDR is reflexive, but neither symmetric nor transitive
Solution
$\begin{aligned} & \text { Given, } x R Y \Rightarrow x^2-4 x y+3 y^2=0 \\ & \Rightarrow x^2-x y-3 x y+3 y^2=0 \\ & \Rightarrow x(x-y)-3 y(x-y)=0 \\ & \Rightarrow(x-y)(x-3 y)=0\end{aligned}$
Reflexive property :
$x R x \Rightarrow(x-x)(-3 x)=0$
So, $R$ is reflexive...(1)
Symmetric property:
Let us check using an example $(1,2)$ and $(2,1)$ for $(1,2) \Rightarrow(1-2)(1-6)=(-1)(-5)=10$ For $(2,1) \Rightarrow(2-1)(2-3)=(1)(-1)=-1$ So, $R$ is not symmetric
...(2)
Transitive property :
For $(9 x, 3 x) \Rightarrow(9 x-3 x)(9 x-9 x)=0$
for $(3 x, x) \Rightarrow(3 x-3 x)(3 x-9 x)=0$
For $(9 x, x) \Rightarrow(9 x-x)(9 x-3 x) \neq 0$
So, $(9 x, 3 x) \in R,(3 x, x) \in R$ but $(9 x, x) \notin R$
So, $R$ is not transitive ...(3)
From (1), (2), (3), $R$ is reflexive, but not symmetric and transitive.
Hence, the correct answer is option: 'D'.