Mathematics - Sets and Relations Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{R}=\{(\mathrm{f},\mathrm{g}):\mathrm{f}(0)=\mathrm{g}(1)\text{ and }\mathrm{f}(1)=\mathrm{g}(0)\}\\ & \text{ Reflexive: }(\mathrm{f},\mathrm{f})\in \mathrm{R}\\ & =\mathrm{f}(0)=\mathrm{f}(1)\text{ and }\mathrm{f}(1)=\mathrm{f}(0)\to \text{ must hold }\\ & \Rightarrow \text{ but this is not true for all function }\end{array}$
so not reflexive
Symmetric: If $(\mathrm{f},\mathrm{g})\in \mathrm{R}\Rightarrow (\mathrm{g},\mathrm{f})\in \mathrm{R}$
Now, $g(0)=f(1)$ and $g(1)=f(0)\to$ true
$\therefore$ symmetric
Transitive : $\mathrm{If}(\mathrm{f},\mathrm{g})\in \mathrm{R}$ and $(\mathrm{g},\mathrm{h})\in \mathrm{R}$
$\Rightarrow (\mathrm{f},\mathrm{h})\in \mathrm{R}$
Now $(f,g)\in R\Rightarrow f(0)=g(1)$ and $f(1)=g(0)$
$(\mathrm{g},\mathrm{h})\in \mathrm{R}\Rightarrow \mathrm{g}(0)=\mathrm{h}(1)$ and $\mathrm{g}(1)=\mathrm{h}(0)$
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