Q98 / 155 · WBJEE 2018 · Mathematics
Let $\rho$ be a relation defined on $N$, the set of natural numbers, as $\rho=\{(x, y) \in N \times N: 2 x+y=41\} .$ Then
- A$\rho$ is an equivalence relation
- B$\rho$ is only reflexive relation
- C$\rho$ is only symmetric relation
- Correct answerD$\rho$ is not transitive
Solution
We have, $\rho=\{(x, y) \in N \times N: 2 x+y=41\}$
For reflexive,
$\begin{array}{l}<br />x p x \Rightarrow 2 x+x=41 \\<br />\Rightarrow 3 x=41<br />\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.
For reflexive,
$\begin{array}{l}<br />x p x \Rightarrow 2 x+x=41 \\<br />\Rightarrow 3 x=41<br />\end{array}$
$\Rightarrow x=\frac{41}{3} \notin N$
So, $\rho$ is not reflexive.
For symmetric,
$x \rho y \Rightarrow 2 x+y=41$
and $y \rho x \Rightarrow 2 y+x=41$
$\Rightarrow x p y \neq y \rho x$
So, $\rho$ is not symmetric.
For transitive,
$x p y \Rightarrow 2 x+y=41$
and $y \rho z \Rightarrow 2 y+z=41$
$\Rightarrow x p z$
$\Rightarrow \rho$ is not transitive.
Hence, the correct answer is option: 'D'.