Let A be the set of all functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$ and R be a relation on A such that $\mathrm{R}=\{(\mathrm{f}, \mathrm{g}): f(0)=\mathrm{g}(1)$ and $f(1)=\mathrm{g}(0)\}$. Then R is :

<p>For $R$ to be reflexive, $(f, f)$ must be in $R$.</p> <p>The means $f(0)=f(1)$ and $f(1)=f(0)$ must be true for all $f$.</p> <p>But $f(0) \neq f(1)$ always</p> <p>Therefore, $R$ is not reflexive</p> <p>If $(f, g) \in R$, then $f(0)=g(1)$ and $f(1)=g(0)$</p> <p>$$\begin{aligned} & \because \quad f(0)=g(1) \Rightarrow g(1)=f(0) \\ & \text { and } f(1)=g(0) \Rightarrow g(0)=f(1) \end{aligned}$$</p> <p>$R$ is symmetric</p> <p>If $(f, g) \in R$ and $(g, h) \in \mathrm{R}$, then $f(0)=g(1)$, $f(1)=g(0), g(0)=n(1) \& g(1)=h(0)$</p> <p>Since, $f(0)=g(1)$ and $g(1)=h(0)$, then $f(0)$ is not necessarily equal to $h(0)$.</p> <p>Therefore, $R$ is not transitive.</p> <p>$\therefore \quad$ The relation $R$ is symmetric but not reflexive or transitive.</p>