CBSE Class 12 · 2018 · Relations And Functions XIILong Answer

Reflexive: $|a-a|=0$ is divisible by $4$, so $(a,a)\in R$ for all $a\in A$.
Symmetric: $|a-b|=|b-a|$, so $4\mid |a-b|\Rightarrow 4\mid |b-a|$; hence $(a,b)\in R\Rightarrow (b,a)\in R$.
Transitive: if $4\mid |a-b|$ and $4\mid |b-c|$, write $a-b=4m,b-c=4n$. Adding, $a-c=4(m+n)$, so $4\mid |a-c|$ and $(a,c)\in R$.
Hence $R$ is an equivalence relation. The class of an element collects all members of $A$ leaving the same remainder mod $4$.
Elements related to $1$ (remainder $1$): $\{1,5,9\}$.
Equivalence class $[2]$ (remainder $2$): $\{2,6,10\}$.