1. **Problem Statement:** Given sets $A = B = \{1,2,3,4\}$ and relations: $$R_1 = \{(x,y) \mid x \in A, y \in B, x = y^2 + 1\}$$ $$R_2 = \{(x,y) \mid x \in A, y \in B, x \text{ is even and } y \text{ is odd}\}$$ Find: (i) The ordered pairs for $R_1$ and $R_2$. (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 $x,y \in \{1,2,3,4\}$. Check each $y$: - $y=1$: $x = 1^2 + 1 = 2$ (in $A$) - $y=2$: $x = 2^2 + 1 = 5$ (not in $A$) - $y=3$: $x = 3^2 + 1 = 10$ (not in $A$) - $y=4$: $x = 4^2 + 1 = 17$ (not in $A$) So $R_1 = \{(2,1)\}$. - 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 pairs $= \{2\}$ - $R_1$ range: all $y$ values in pairs $= \{1\}$ - $R_2$ domain: $\{2,4\}$ - $R_2$ 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. - $R_1$: Domain $\{2\}$ maps to $1$ only, so $R_1$ is a function. - $R_2$: Domain $2$ maps to $1$ and $3$ (two outputs), so $R_2$ is not a function. **Final answers:** (i) $R_1 = \{(2,1)\}$, $R_2 = \{(2,1),(2,3),(4,1),(4,3)\}$ (ii) $\text{Domain}(R_1) = \{2\}$, $\text{Range}(R_1) = \{1\}$ $\text{Domain}(R_2) = \{2,4\}$, $\text{Range}(R_2) = \{1,3\}$ (iii) $R_1$ is a function; $R_2$ is not a function.