1. **Problem 1:** (a) Check if set A is a subset of set B. - Set A = {1, 2, 3, 4, 5, ...} (all positive integers). - Set B = {3, 6, 9, 12, ...} (all positive multiples of 3). - Since 1 \in A but 1 \notin B, A is not a subset of B. (b) Find A \cap B. - Since every element of B is a positive multiple of 3 and all multiples of 3 are positive integers, B \subseteq A. - Therefore, A \cap B = B = {3, 6, 9, 12, ...}. 2. **Problem 2:** Evaluate $\left| \frac{5i}{3 - i} \right|$. - Multiply numerator and denominator by the conjugate of the denominator: $$\frac{5i}{3 - i} \times \frac{3 + i}{3 + i} = \frac{5i(3 + i)}{3^2 + 1^2} = \frac{5i(3 + i)}{10}$$ - Expand numerator: $$5i \times 3 = 15i, \quad 5i \times i = 5i^2 = 5(-1) = -5$$ - So numerator is $15i - 5$. - Fraction becomes: $$\frac{15i - 5}{10} = -\frac{1}{2} + \frac{3}{2}i$$ - Modulus is: $$\left| -\frac{1}{2} + \frac{3}{2}i \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{9}{4}} = \sqrt{\frac{10}{4}} = \frac{\sqrt{10}}{2}$$ 3. **Problem 3:** Determine domain and range of $f(x) = \frac{1}{x^2 - 4}$. - Domain: Denominator $x^2 - 4 \neq 0$ implies $x \neq \pm 2$. - So domain is $\mathbb{R} \setminus \{ -2, 2 \}$. - Let $t = x^2 - 4$; since $x^2 \geq 0$, $t \geq -4$ and $t \neq 0$. - Range of $f(x) = \frac{1}{t}$: - For $t \in (-4, 0)$, $f(x) \in (-\infty, -\frac{1}{4})$. - At $t = -4$, $f(x) = -\frac{1}{4}$. - For $t \in (0, \infty)$, $f(x) \in (0, \infty)$. - Therefore, range is $(-\infty, -\frac{1}{4}] \cup (0, \infty)$. 4. **Problem 4:** Find $(u \circ v)(x)$ and $(v \circ u)(x)$ where $u(x) = 5x + 7$ and $v(x) = 5x + 7$. - Compute $u(v(x))$: $$u(5x + 7) = 5(5x + 7) + 7 = 25x + 35 + 7 = 25x + 42$$ - Compute $v(u(x))$: $$v(5x + 7) = 5(5x + 7) + 7 = 25x + 42$$ - So, $(u \circ v)(x) = (v \circ u)(x) = 25x + 42$. **Final answers:** 1. (a) $A \not\subseteq B$ because $1 \in A$ but $1 \notin B$. (b) $A \cap B = B = \{3, 6, 9, 12, ...\}$. 2. $\left| \frac{5i}{3 - i} \right| = \frac{\sqrt{10}}{2}$. 3. Domain: $\mathbb{R} \setminus \{ -2, 2 \}$. Range: $(-\infty, -\frac{1}{4}] \cup (0, \infty)$. 4. $(u \circ v)(x) = (v \circ u)(x) = 25x + 42$.