The correct option is B ρ1∩ρ2 is equivalence relation but ρ1∪ρ2 is not so.
Given ρ1,ρ2 are equivalence relations on S.
⇒ ρ1,ρ2 are reflexive, symmetric and transitive.
Reflexive:
Let x∈S
⇒(x,x)∈ρ1 and (x,x)∈ρ2
⇒(x,x)∈ρ1∩ρ2
⇒ρ1∩ρ2 is reflexive.
Symmetric:
Let (x,y)∈ρ1∩ρ2
We have to show (y,x)∈ρ1∩ρ2
(x,y)∈ρ1∩ρ2
⇒(x,y)∈ρ1 and (x,y)∈ρ2
⇒(y,x)∈ρ1 and (y,x)∈ρ2
⇒(y,x)∈ρ1∩ρ2
⇒ρ1∩ρ2 is symmetric.
Transitive:
Let (x,y),(y,z)∈ρ1∩ρ2
⇒(x,y),(y,z)∈ρ1 and (x,y),(y,z)∈ρ2
⇒(x,z)∈ρ1 and (x,z)∈ρ2
⇒(x,z)∈ρ1∩ρ2
⇒ρ1∩ρ2 is transitive.
Therefore, ρ1∩ρ2 is equivalence relation.
ρ1∪ρ2 is always reflexive and symmetric but not transitive.
e.g. Let S={1,2,3}
ρ1={(1,1),(2,2),(3,3),(1,2),(2,1)}
ρ2={(1,1),(2,2),(3,3),(2,3),(3,2)}
ρ1,ρ2 is equivalence relation.
But ρ1∪ρ2 is not transitive as (1,2),(2,3)∈ρ1∪ρ2 but (1,3)∉ρ1∪ρ2