Jee Main 2024 29 01 2024 Shift 1 - Question9

Answer: (2)

Solution:

Reflexive: for all $(a, a) R(a, a)$

$\Rightarrow a b-a b=0$ is divisible by 5 .

So $(a, b) R(a, b) \forall a, b \in \mathbb{Z}$

$\therefore \quad R$ is reflexive

Symmetric :

For $(a, b) \in R(c, d)$

If $ad - bc$ is divisible by 5.

Then $b c-a d$ is also divisible by 5 .

$\Rightarrow \quad (c, d) R(a, b) \forall a, b, c, d \in \mathbb{Z}$

$\therefore \quad R$ is symmetric

Transitive :

If $(a, b) R(c, d) \Rightarrow a d-b c$ divisible by 5

and $(c, d) R(e, f) \Rightarrow c f-d e$ divisible by 5 $a d-b c=5 k _1 \quad k _1$ and $k _2$ are integers

$c f-d e=5 k_2$

afd $-b c f=5 k _1 f$

$b c f - b d e = 5 k _2 b$

$a f d-b d e=5\left(k _1 f+k _2 b\right)$

$d(a f-b e)=5\left(k _1 f+k _2 b\right)$

$\Rightarrow a f-b e$ is not divisible by 5 for every $a, b, e, f$,$

$e, f \in Z$.

$\therefore \quad R$ is not transitive

For e.g., take $a=1, b=2, c=5, d=5, e=2, f=2$