1. **Problem 2:** Given functions $f(x) = a + x^2$ and $L(x) = c$ where $a$, $c$ are constants, and the equation $3f(2) + 3L(x) = 6$ holds for all $x$. We need to find $2f(0) + 2L(7)$. Step 1: Calculate $f(2)$ and express $f(2)$ in terms of $a$. $$f(2) = a + 2^2 = a + 4$$ Step 2: Since $L(x) = c$ (constant), $L(x) = c$ for any $x$. Rewrite the given equation: $$3f(2) + 3L(x) = 3(a + 4) + 3c = 6$$ $$3a + 12 + 3c = 6$$ Step 3: Divide both sides by 3: $$a + 4 + c = 2$$ Step 4: Rearrange to isolate $a + c$: $$a + c = 2 - 4 = -2$$ Step 5: Calculate $2f(0) + 2L(7)$. Since $f(0) = a + 0^2 = a$, and $L(7) = c$, then: $$2f(0) + 2L(7) = 2a + 2c = 2(a + c) = 2 imes (-2) = -4$$ --- 2. **Problem 3:** Given sets $X = \{3, 4, 5\}$, $Y = \{4, 5\}$, $Z = \{3, 5\}$. Find: 3.1) $(X - Z) \times Y$ Step 1: Calculate $X - Z$, the elements in $X$ but not in $Z$: $$X - Z = \{4\}$$ Step 2: Cartesian product $(X - Z) \times Y$: $$\{4\} \times \{4,5\} = \{(4,4), (4,5)\}$$ 3.2) $n((Z \cap Y) \times X)$ Step 1: Find intersection $Z \cap Y$: $$Z \cap Y = \{5\}$$ Step 2: Cartesian product $(Z \cap Y) \times X$: $$\{5\} \times \{3,4,5\} = \{(5,3), (5,4), (5,5)\}$$ Step 3: Find number of elements (cardinality) $n((Z \cap Y) \times X) = 3$ --- 3. **Problem 4:** Given $f(x) = x^2 - \sqrt{2}x$, $g(x) = x+1$. Find: 4.1) $f(3) + 3g(\sqrt{2})$ Step 1: Calculate $f(3)$: $$f(3) = 3^2 - \sqrt{2} \times 3 = 9 - 3\sqrt{2}$$ Step 2: Calculate $g(\sqrt{2})$: $$g(\sqrt{2}) = \sqrt{2} + 1$$ Step 3: Multiply $g(\sqrt{2})$ by 3: $$3g(\sqrt{2}) = 3(\sqrt{2} +1 ) = 3\sqrt{2} + 3$$ Step 4: Add $f(3)$ and $3g(\sqrt{2})$: $$f(3) + 3g(\sqrt{2}) = (9 - 3\sqrt{2}) + (3\sqrt{2} + 3) = 9 + 3 = 12$$ 4.2) Prove $f(\sqrt{2}) = g(-1)$ Step 1: Calculate $f(\sqrt{2})$: $$f(\sqrt{2}) = (\sqrt{2})^2 - \sqrt{2} \times \sqrt{2} = 2 - 2 = 0$$ Step 2: Calculate $g(-1)$: $$g(-1) = -1 + 1 = 0$$ Step 3: Since $f(\sqrt{2}) = 0$ and $g(-1) = 0$, the equality holds, thus proved. --- 4. **Problem 5:** Given a function $f$ from $X$ to $Y$, with $aRb$ meaning "$a$ is a multiple of $b$", where $a \in X$, $b \in Y$, $n(X) = 4$, $n(Y) = 2$, and $X \cup Y = \{4, 8, 9, 27\}$. Find $X$, $Y$, write the relation $R$, and find its range. Step 1: Since $n(Y) = 2$, select 2 elements from the union set as $Y$. Remaining 4 elements total (but only 4 elements given), so $X$ has 4 elements total and $Y$ 2 elements. The union has 4 elements total, so $X$ and $Y$ are subsets partitioning the set. Step 2: To satisfy $aRb$ meaning $a$ is multiple of $b$, $b$ must divide $a$. Looking at the numbers: - $4$, $8$, $9$, $27$ Step 3: Let's try $Y = \{4, 9\}$ and $X = \{4, 8, 9, 27\}$ (since $X$ has to have 4 elements). Step 4: Relation $R$ is pairs $(a,b)$ where $a$ is a multiple of $b$. Check each $a$ in $X$ with $b$ in $Y$: - For $a=4$: multiple of 4? yes; multiple of 9? no - For $a=8$: multiple of 4? yes; multiple of 9? no - For $a=9$: multiple of 4? no; multiple of 9? yes - For $a=27$: multiple of 4? no; multiple of 9? yes (9 × 3 = 27) Relation $R = \{(4,4), (8,4), (9,9), (27,9)\}$ Step 5: Range of $f$ is the set of $b$ values related to some $a$: $$\text{Range} = \{4, 9\}$$