Q73 / 155 · WBJEE 2018 · Mathematics
On the set $R$ of real numbers, the relation $\rho$ is defined by $x \rho y,(x, y) \in R$
- AIf $|x-y| < 2$, then $\rho$ is reflexive but neither symmetric nor transitive.
- BII $x-y < 2$, then $\rho$ is reflexeve and symmetric bu not transitive.
- C$If(x \mid \geq y,$ then $\rho$ is reflexive and transitive but not symmetric.
- Correct answerDIf $x>|y|$, then $p$ is transitive but neither reflexve nor symmetric.
Solution
On the set $R$ of real numbers
For reflexive,
$x \rho x \Rightarrow(x, x) \in R$
$\Rightarrow x > |x|$ which is not true. $\Rightarrow \rho$ is not reflexive.
For symmetric,
$(x, y) \in R \Rightarrow x > |y|$
and $(y, x) \in R \Rightarrow y > |x|$
So, $x > |y| \neq y > |x|$ $\Rightarrow \rho$ is not symmetric.
For transitive, $(x, y) \in R \Rightarrow x > |y|(y, z) \in R \Rightarrow y > |z|$
$\Rightarrow x > |z| \Rightarrow(x, z) \in R$
$\Rightarrow \rho$ is transitive.
For reflexive,
$x \rho x \Rightarrow(x, x) \in R$
$\Rightarrow x > |x|$ which is not true. $\Rightarrow \rho$ is not reflexive.
For symmetric,
$(x, y) \in R \Rightarrow x > |y|$
and $(y, x) \in R \Rightarrow y > |x|$
So, $x > |y| \neq y > |x|$ $\Rightarrow \rho$ is not symmetric.
For transitive, $(x, y) \in R \Rightarrow x > |y|(y, z) \in R \Rightarrow y > |z|$
$\Rightarrow x > |z| \Rightarrow(x, z) \in R$
$\Rightarrow \rho$ is transitive.
Hence, the correct answer is option: 'D'.