1. **Problem Statement:** Given three sets: - $A = \{ x \in \mathbb{R} : 2 < x \leq 5 \}$ - $B = \{ x \in \mathbb{R} : x \leq 2 \}$ - $C = \{ x \in \mathbb{R} : -2 \leq x < 3 \}$ We need to find and sketch the following sets on the real number line: (i) $A \cup B'$ (ii) $A' \cap B$ (iii) $(A \cup B) \cap C$ --- 2. **Recall important rules:** - The complement of a set $S$, denoted $S'$, is all real numbers not in $S$. - Union $\cup$ means all elements in either set. - Intersection $\cap$ means elements common to both sets. --- 3. **Find $B'$:** Since $B = \{ x : x \leq 2 \}$, its complement is $$B' = \{ x : x > 2 \}$$ --- 4. **Calculate (i) $A \cup B'$:** - $A = \{ x : 2 < x \leq 5 \}$ - $B' = \{ x : x > 2 \}$ Since $B'$ includes all $x > 2$, and $A$ is a subset of $B'$, their union is simply: $$A \cup B' = \{ x : x > 2 \}$$ --- 5. **Calculate (ii) $A' \cap B$:** - $A' = \{ x : x \leq 2 \text{ or } x > 5 \}$ (complement of $A$) - $B = \{ x : x \leq 2 \}$ Intersection is numbers in both $A'$ and $B$: $$A' \cap B = \{ x : x \leq 2 \}$$ --- 6. **Calculate (iii) $(A \cup B) \cap C$:** - $A \cup B = \{ x : x \leq 5 \}$ (since $B$ covers $x \leq 2$ and $A$ covers $2 < x \leq 5$) - $C = \{ x : -2 \leq x < 3 \}$ Intersection: $$ (A \cup B) \cap C = \{ x : -2 \leq x < 3 \}$$ --- **Final answers:** - (i) $A \cup B' = \{ x : x > 2 \}$ - (ii) $A' \cap B = \{ x : x \leq 2 \}$ - (iii) $(A \cup B) \cap C = \{ x : -2 \leq x < 3 \}$ These sets can be sketched on the real number line accordingly.