Q194 / 270 · NDA 2022 (Phase 1) · Mathematics
Consider the following statements in respect of the relation $R$ in the set IN of natural numbers defined by $x R y$ if $x^2-5 x y+4 y^2=0$ :
1. $R$ is reflexive
2. $R$ is symmetric
3. $R$ is transitive
Which of the above statements is/are correct?
- Correct answerA1 only
- B2 only
- C1 and 2 only
- D1, 2 and 3
Solution
Given that
$\begin{aligned} & R=\left\{(x, y): x^2-5 x y+4 y^2=0, x, y \in N\right\} \\ & x^2-5 x y+4 y^2=0 \\ & \Rightarrow x^2-4 x y-x y+4 y^2=0 \\ & \Rightarrow x(x-4 y)-y(x-4 y)=0 \\ & \Rightarrow(x-y)(x-4 y)=0\end{aligned}$
For reflexive.
$\begin{aligned} & (\mathrm{x}-\mathrm{x})(\mathrm{x}-4 \mathrm{x})=0 \\ & \therefore(\mathrm{x}, \mathrm{x}) \in \mathrm{R}\end{aligned}$
So it is reflexive
For symmetric
Let $(\mathrm{x}, \mathrm{y}) \in \mathrm{R}$
$x^2-5 x y+4 y^2=0$
but
$y^2-5 x y+4 x^2$
may be equal to zero
So it is not symmetric For transitive
Let $(\mathrm{x}, \mathrm{y}) \in \mathrm{R}$
$x^2-5 x y+4 y^2=0$...(i)
and $(y, z) \in R$
$y^2-5 y^2+4 z^2=0$...(ii)
from (i) and (ii)
$\begin{aligned} & \mathrm{x}^2-5 x y+4 y^2=y^2-5 y z+4 z^2 \\ & x^2+3 y^2-4 z^2-5 x y+5 y z=0 \\ & \left(x^2-5 x z+4 z^2\right)+\left(3 y^2-4 z^2+5 x z-5 x y+5 y z\right)=0 \\ & \therefore 3 y^2-4 z^2+5 x z-5 x y+5 y z \neq 0 \\ & \therefore x^2-5 x z+4 z^2 \neq 0\end{aligned}$
$\therefore(\mathrm{x}, \mathrm{z}) \notin \mathrm{R}$
So, it is not transitive
Hence, the correct answer is option: 'A'.