JEE Main 2024 — Mathematics Algebra

Given,

$$a{R}_{1}b\Leftrightarrow {a}^{2}+{b}^{2}=1;a,b\in R$$aR1​b⇔a2+b2=1;a,b∈R

${R}_{1}$R1​ is not reflexive as $(a,a)\notin a{R}_{1}b$(a,a)∈/aR1​b

So, it is not equivalence.

Now, solving

$$(a,b){R}_{2}(c,d)\Leftrightarrow a+d=b+c;(a,b),(c,d)\in N$$(a,b)R2​(c,d)⇔a+d=b+c;(a,b),(c,d)∈N

Reflexive: $a+b=b+a\rightarrow \mathrm{True}$a+b=b+a→True

Symmetric: $(a,b){R}_{2}(c,d)$(a,b)R2​(c,d)

$$\Rightarrow a+d=b+c$$⇒a+d=b+c

$$\Rightarrow d+a=c+b$$⇒d+a=c+b

$$\Rightarrow c+b=d+a$$⇒c+b=d+a

$$\Rightarrow (c,d){R}_{2}(a,b)$$⇒(c,d)R2​(a,b)

Transitive: $(a,b){R}_{2}(c,d)\Rightarrow a+d=b+c...(i)$(a,b)R2​(c,d)⇒a+d=b+c...(i)

$$(c,d){R}_{2}(e,f)\Rightarrow c+f=d+e...(ii)$$(c,d)R2​(e,f)⇒c+f=d+e...(ii)

Now, adding above equation we get,

$$\Rightarrow a+f=b+e$$⇒a+f=b+e

$$\Rightarrow (a,b){R}_{2}(e,f)$$⇒(a,b)R2​(e,f)

So, ${R}_{2}$R2​ is reflexive, symmetric and transitive

Hence only ${R}_{2}$R2​ is equivalence relation.