1. **Problem 1: Multiple Choice with Justifications** 1) Calculate $|\sqrt{5} - 2| + |2 - \sqrt{5}|$. Since $|a| = |-a|$, and $\sqrt{5} \approx 2.236$, we have: $$|\sqrt{5} - 2| = \sqrt{5} - 2 \approx 0.236$$ $$|2 - \sqrt{5}| = 2 - \sqrt{5} = -(\sqrt{5} - 2) \approx 0.236$$ Sum: $$0.236 + 0.236 = 2(\sqrt{5} - 2) = 2\sqrt{5} - 4$$ Answer: B) $2\sqrt{5} - 4$ 2) Find $]-5,7] \cap ]3,9]$. The intersection of intervals is the overlap: $$]-5,7] \cap ]3,9] = ]3,7]$$ Answer: B) $]3,7]$ 3) Solve $|-x + 3| + 10 < 9$. Rewrite: $$|-x + 3| < -1$$ Absolute value is always $\geq 0$, so no solution. Answer: A) $\emptyset$ 4) Identify the set $\{x \in \mathbb{N} \mid (x^2 - 4)(x - 1) = 0\}$. Solve: $$x^2 - 4 = 0 \Rightarrow x = \pm 2$$ Since $x \in \mathbb{N}$ (natural numbers), $x=2$ only. Also, $x - 1 = 0 \Rightarrow x=1$. So set is $\{1,2\}$, which has two elements. Answer: B) Pair 5) The shaded area is $A \cap B^c$ (elements in $A$ but not in $B$). Answer: A) $A \cap B^c$ 2. **Exercise 2: Solve equations and inequalities** a) Solve $|2x - 3| = 3$. Two cases: $$2x - 3 = 3 \Rightarrow 2x = 6 \Rightarrow x = 3$$ $$2x - 3 = -3 \Rightarrow 2x = 0 \Rightarrow x = 0$$ Solution: $\{0,3\}$ b) Solve $|x - 6| > 2$. Inequality splits into: $$x - 6 > 2 \Rightarrow x > 8$$ $$x - 6 < -2 \Rightarrow x < 4$$ Solution: $(-\infty,4) \cup (8,\infty)$ c) Solve $2|1 - 2x| + 1 \leq 9$. Subtract 1: $$2|1 - 2x| \leq 8 \Rightarrow |1 - 2x| \leq 4$$ Split: $$-4 \leq 1 - 2x \leq 4$$ Subtract 1: $$-5 \leq -2x \leq 3$$ Divide by -2 (reverse inequalities): $$\frac{5}{2} \geq x \geq -\frac{3}{2}$$ Rewrite: $$-\frac{3}{2} \leq x \leq \frac{5}{2}$$ Solution: $[-1.5, 2.5]$ 3. **Exercise 3: Set operations** Universal set $E = \{x \in \mathbb{N} \mid 1 < x < 12\} = \{2,3,4,5,6,7,8,9,10,11\}$ Sets: $$A = \{6,10\}, \quad B = \{3,6\}$$ 1) Is $A \subset E$? Since $6,10 \in E$, yes. Is $A \subset B$? No, because $10 \notin B$. 2) Write in extension: a) $A \cup B = \{3,6,10\}$ b) $\overline{A} = E \setminus A = \{2,3,4,5,7,8,9,11\}$ c) $A \cap \overline{B} = A \cap (E \setminus B) = \{6,10\} \cap \{2,4,5,7,8,9,10,11\} = \{10\}$ d) $\overline{A} \cap B = \{2,3,4,5,7,8,9,11\} \cap \{3,6\} = \{3\}$ 3) Card(B) = number of elements in B = 2 Power set $P(B)$ has $2^{2} = 4$ elements: $$P(B) = \{\emptyset, \{3\}, \{6\}, \{3,6\}\}$$ 4) Complete using $\in, \notin, \subset, \notsubset$: - $6 \in A$ - $5 \notin A$ - $A \subset E$ - $A \notsubset B$ **Final answers summarized:** 1) B 2) B 3) A 4) B 5) A Exercise 2: a) $\{0,3\}$ b) $(-\infty,4) \cup (8,\infty)$ c) $[-1.5, 2.5]$ Exercise 3: 1) $A \subset E$ true, $A \subset B$ false 2) a) $\{3,6,10\}$ b) $\{2,3,4,5,7,8,9,11\}$ c) $\{10\}$ d) $\{3\}$ 3) $\text{Card}(B) = 2$, $\text{Card}(P(B)) = 4$, $P(B) = \{\emptyset, \{3\}, \{6\}, \{3,6\}\}$ 4) $6 \in A$, $5 \notin A$, $A \subset E$, $A \notsubset B$