1. Problem: Find $x \in \mathbb{N}$ such that $D = A$ where $D = \{x^2 - 10, 12\}$ and $A$ is given. 2. Since $D = A$, and $A$ is not explicitly given, assume $A = \emptyset$ or $A$ is the empty set as per part a). 3. For $D$ to be empty, both elements must not exist or be equal to elements in $A$. 4. Calculate $x^2 - 10$ and check if it equals any element in $A$. 5. Since $A = \emptyset$, $D$ must be empty, so $x^2 - 10$ and 12 must not be in $A$. 6. But 12 is in $D$, so $D$ cannot be empty unless $A$ contains 12. 7. Therefore, $A$ must be $\{12\}$ and $x^2 - 10 = 12$. 8. Solve $x^2 - 10 = 12$: $$x^2 = 22$$ 9. Since $x \in \mathbb{N}$, $x = \sqrt{22}$ which is not a natural number. 10. No natural number $x$ satisfies this. --- Exercise 4: 1. Represent data with Venn diagram (not shown here). 2. Number playing only football: $$|F| - |F \cap V| - |F \cap B| + |F \cap V \cap B| = 44 - 8 - 21 + 14 = 29$$ 3. Number playing volleyball or basketball but not football: $$|V \cup B| - |F \cap (V \cup B)|$$ Calculate $|V \cup B|$: $$|V| + |B| - |V \cap B| = 42 + 33 - 20 = 55$$ Calculate $|F \cap (V \cup B)|$: $$|F \cap V| + |F \cap B| - |F \cap V \cap B| = 8 + 21 - 14 = 15$$ So, $$55 - 15 = 40$$ 4. Number playing exactly two games: $$|F \cap V| + |F \cap B| + |V \cap B| - 3 \times |F \cap V \cap B| = 8 + 21 + 20 - 3 \times 14 = 49 - 42 = 7$$ 5. Number playing only one game: $$|F| + |V| + |B| - 2(|F \cap V| + |F \cap B| + |V \cap B|) + 3|F \cap V \cap B|$$ $$= 44 + 42 + 33 - 2(8 + 21 + 20) + 3 \times 14 = 119 - 2 \times 49 + 42 = 119 - 98 + 42 = 63$$ --- Exercise 5: 1. Set $A = \{x \in \mathbb{N}^* : |3x + 1| < 10\}$ Solve inequality: $$|3x + 1| < 10 \implies -10 < 3x + 1 < 10$$ Subtract 1: $$-11 < 3x < 9$$ Divide by 3: $$-\frac{11}{3} < x < 3$$ Since $x \in \mathbb{N}^*$ (positive integers), $x = 1, 2$. So, $$A = \{1, 2\}$$ 2. Write sets $B$ and $C$ in comprehension form: - $B = \{x \in \mathbb{N} : x \text{ divides } 24\}$ - $C = \{3k : k \in \mathbb{N}, 0 \leq k \leq 6\}$ Final answers: - $x$ such that $D = A$ with $x \in \mathbb{N}$: no solution. - Only football players: 29 - Volleyball or basketball but not football: 40 - Exactly two games: 7 - Only one game: 63 - $A = \{1, 2\}$ - $B = \{x \in \mathbb{N} : x | 24\}$ - $C = \{3k : k \in \mathbb{N}, 0 \leq k \leq 6\}$