R={(f,g):f(0)=g(1) and f(1)=g(0)} Reflexive: (f,f)∈R=f(0)=f(1) and f(1)=f(0)→ must hold ⇒ but this is not true for all function
so not reflexive
Symmetric: If (f,g)∈R⇒(g,f)∈R
Now, g(0)=f(1) and g(1)=f(0)→ true
∴ symmetric
Transitive : If(f,g)∈R and (g,h)∈R
⇒(f,h)∈R
Now (f,g)∈R⇒f(0)=g(1) and f(1)=g(0)
(g,h)∈R⇒g(0)=h(1) and g(1)=h(0)
For (f,h)∈R we need f(0)=h(1) and f(1)=h(0)
Now f(0)=g(1)=h(0) and f(1)=g(0)=h(1)
Hence not transitive