JEE Main 2022 — Sets And Relations Question with Solution

$$R=\{(x,y):3x+\alpha y$$ is multiple of 7$$\}$$, now R to be an equivalence relation

(1) R should be reflexive : $$(a,a)\in R\forall a\in N$$

$$\therefore$$ $$3a+a\alpha =7k$$

$$\therefore$$ $$(3+\alpha )a=7k$$

$$\therefore$$ $$3+\alpha =7k_1\Rightarrow \alpha =7k_1-3$$

$$=7k_1+4$$

(2) R should be symmetric : $$aRb\iff bRa$$

$$aRb:3a+(7k-3)b=7m$$

$$\Rightarrow 3(a-b)+7kb=7m$$

$$\Rightarrow 3(b-a)+7ka=7m$$

So, $$aRb\Rightarrow bRa$$

$$\therefore$$ R will be symmetric for $$a=7k_1-3$$

(3) Transitive : Let $$(a,b)\in R,(b,c)\in R$$

$$\Rightarrow 3a+(7k-3)b=7k_1$$ and

$$3b+(7k_2-3)c=7k_3$$

Adding $$3a+7kb+(7k_2-3)c=7(k_1+k_3)$$

$$3a+(7k_2-3)c=7m$$

$$\therefore$$ $$(a,c)\in R$$

$$\therefore$$ R is transitive

$$\therefore$$ $$\alpha =7k-3=7k+4$$