1. **Problem Statement:** Given sets $A = B = \{1,2,3,4\}$ and relations: - $R_1 = \{(x,y) \mid x \in A, y \in B, x = y^2 + 1\}$ - $R_2 = \{(x,y) \mid x \in A, y \in B, x$ is even and $y$ is odd$\}$ Find: (i) The relations $R_1$ and $R_2$ as sets of ordered pairs. (ii) The domain and range of $R_1$ and $R_2$. (iii) Whether $R_1$ and $R_2$ represent functions. 2. **Step (i): Write relations as ordered pairs** - For $R_1$, find all $(x,y)$ with $x = y^2 + 1$ and $x,y \in \{1,2,3,4\}$. Calculate $y^2 + 1$ for each $y$: - $y=1 \Rightarrow 1^2+1=2$ (in $A$) - $y=2 \Rightarrow 4+1=5$ (not in $A$) - $y=3 \Rightarrow 9+1=10$ (not in $A$) - $y=4 \Rightarrow 16+1=17$ (not in $A$) So only $y=1$ gives $x=2$ in $A$. Thus, $R_1 = \{(2,1)\}$. - For $R_2$, $x$ is even and $y$ is odd, both in $\{1,2,3,4\}$. Even $x$: $2,4$. Odd $y$: $1,3$. So $R_2 = \{(2,1),(2,3),(4,1),(4,3)\}$. 3. **Step (ii): Domain and Range** - $R_1$: - Domain: all $x$ values in $R_1$ = $\{2\}$. - Range: all $y$ values in $R_1$ = $\{1\}$. - $R_2$: - Domain: $\{2,4\}$. - Range: $\{1,3\}$. 4. **Step (iii): Check if relations are functions** - A relation is a function if each element in the domain maps to exactly one element in the range. - $R_1$ has domain $\{2\}$ and maps $2 \to 1$ uniquely, so $R_1$ is a function. - $R_2$ has domain $\{2,4\}$ but $2$ maps to both $1$ and $3$, and $4$ maps to both $1$ and $3$. Since elements in domain map to multiple values, $R_2$ is not a function. **Final answers:** (i) $R_1 = \{(2,1)\}$, $R_2 = \{(2,1),(2,3),(4,1),(4,3)\}$. (ii) Domain and range: - $R_1$: Domain $= \{2\}$, Range $= \{1\}$. - $R_2$: Domain $= \{2,4\}$, Range $= \{1,3\}$. (iii) $R_1$ is a function; $R_2$ is not a function.