$R=\{(f, g): f(0)=g(1) \text { and } f(1)=g(0)\}$
Reflexive: ( $f , f ) \in R$
$= f (0)= f (1)$ and $f (1)= f (0) \rightarrow$ must hold
$\Rightarrow$ but this is not true for all function
so not reflexive
Symmetric: If $( f , g ) \in R \Rightarrow( g , f ) \in R$
Now, $g(0)=f(1)$ and $g(1)=f(0) \rightarrow$ true
$\therefore$ symmetric
Transitive : $\operatorname{If}( f , g ) \in R$ and $( g , h ) \in R$
$\Rightarrow(f, h) \in R$
Now $( f , g ) \in R \Rightarrow f (0)= g (1)$ and $f (1)= g (0)$
$( g , h ) \in R \Rightarrow g (0)= h (1)$ and $g (1)= h (0)$
For $( f , h ) \in R$ 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