1. Problem 1a: Write the set notations for shaded regions. ai) The shaded region covers sets R and Q but not the third set, so the union of R and Q minus the third set: $$R \cup Q\setminus C$$ assuming the unlabeled set is C. aii) The entire universal set $n(U)$ excluding sets A and B means the complement of $A \cup B$: $$n(U) \setminus (A \cup B)$$. ii) Only set A shaded inside $n(U)$ means just set A: $$A$$. aiv) The shaded region is intersection of A and C: $$A \cap C$$. 2. Problem 1bi: Express $(2 - \sqrt{3})^2$ in the form $a + b\sqrt{c}$ where $a,b,c \in \mathbb{Z}$. Step 1: Expand using formula: $$(2 - \sqrt{3})^2 = 2^2 - 2 \times 2 \times \sqrt{3} + (\sqrt{3})^2$$ Step 2: Calculate terms: $$= 4 - 4\sqrt{3} + 3$$ Step 3: Combine like terms: $$= 7 - 4\sqrt{3}$$ So, $a=7$, $b=-4$, $c=3$. 3. Problem 1bii: Given $$x + \sqrt{7} = \frac{5}{2} - \frac{1}{2}\sqrt{7}$$ find the positive value of $x$ and express $$(x - 1) + \sqrt{7}$$. Step 1: Equate the rational and irrational parts: - Rational part: $$x = \frac{5}{2}$$ - Irrational part agents balance as $\sqrt{7} = -\frac{1}{2} \sqrt{7}$ doesn't match, so user might mean the coefficients equal. Step 2: Find $x$: Given in equation, $x = \frac{5}{2}$. Step 3: Compute $$(x - 1) + \sqrt{7} = \left(\frac{5}{2} - 1\right) + \sqrt{7} = \frac{3}{2} + \sqrt{7}$$ 4. Problem 2a: Write the first three terms of the expansion of $\bigl(1 + \frac{1}{4}x\bigr)^4$. Step 1: Use Binomial Theorem: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ Step 2: Here, $a=1$, $b=\frac{1}{4}x$, $n=4$. Step 3: First three terms: - $k=0$: $\binom{4}{0} 1^{4} \left(\frac{1}{4}x\right)^0 = 1$ - $k=1$: $\binom{4}{1} 1^{3} \left(\frac{1}{4}x\right)^1 = 4 \times \frac{1}{4}x = x$ - $k=2$: $\binom{4}{2} 1^{2} \left(\frac{1}{4}x\right)^2 = 6 \times \frac{1}{16}x^2 = \frac{3}{8} x^2$ So expansion first 3 terms: $$1 + x + \frac{3}{8} x^2$$ 5. Problem 4bi: Using expansion, find $(1.025)^4$ correct to three decimal places. Step 1: Substitute $x=0.1$ in $1 + x + \frac{3}{8} x^2$ for approximation of $(1 + 0.1)^4$ is not directly applicable to 1.025. Step 2: Instead, write $(1.025) = 1 + 0.025$, so $x=0.025$. Step 3: Use first three terms: $$1 + 0.025 + \frac{3}{8} (0.025)^2 = 1 + 0.025 + \frac{3}{8} \times 0.000625 = 1 + 0.025 + 0.000234375 = 1.025234375$$ Step 4: To 3 decimal places: $$1.025$$ 6. Problem 4bii: Solve the equation $$5^{2x+1} - 126 \cdot 5^{x-1} + 1 = 0$$. Step 1: Let $y = 5^x$; rewrite exponents: $$5^{2x+1} = 5^{2x} \times 5^1 = 5 \times (5^x)^2 = 5y^2$$ $$126 \cdot 5^{x-1} = 126 \times 5^{-1} \times 5^x = 126 \times \frac{1}{5} y = 25.2 y$$ Step 2: Substitute: $$5 y^2 - 25.2 y +1 =0$$ Step 3: Divide entire equation by 5: $$y^2 - 5.04 y + 0.2 =0$$ Step 4: Use quadratic formula: $$y = \frac{5.04 \pm \sqrt{5.04^2 - 4 \times 0.2}}{2}$$ Calculate discriminant: $$5.04^2 = 25.4016$$ $$4 \times 0.2 = 0.8$$ $$\sqrt{25.4016 - 0.8} = \sqrt{24.6016} \approx 4.959$$ Step 5: Solutions: $$y_1 = \frac{5.04 + 4.959}{2} = \frac{10}{2} = 5$$ $$y_2 = \frac{5.04 - 4.959}{2} = \frac{0.081}{2} = 0.0405$$ Step 6: Recall $y=5^x$: - For $y=5$, $5^x=5 \Rightarrow x=1$ - For $y=0.0405$, $x=\log_5(0.0405)$ Step 7: Calculate $x = \log_5(0.0405)$: Using change of base: $$x = \frac{\log(0.0405)}{\log(5)} \approx \frac{-1.3927}{0.69897} = -1.992$$ Answer: $$x \approx 1 \text{ or } x \approx -1.992$$ 7. Problem 3a: Population modeled by $$p(t)=500 \times 3^{-0.1 t}$$. i. Initial population when $t=0$: $$p(0) = 500 \times 3^{0} = 500 \times 1 = 500$$ ii. Population after 10 mins: $$p(10) = 500 \times 3^{-0.1 \times 10} = 500 \times 3^{-1} = 500 \times \frac{1}{3} \approx 166.67$$ iii. Find time $t$ when population decreases to 50: $$50 = 500 \times 3^{-0.1 t}$$ Divide both sides by 500: $$\frac{50}{500} = 3^{-0.1 t} \Rightarrow 0.1 = 3^{-0.1 t}$$ Take natural logarithm: $$\ln(0.1) = \ln(3^{-0.1 t}) = -0.1 t \ln 3$$ Solve for $t$: $$t = - \frac{\ln(0.1)}{0.1 \ln 3} = - \frac{-2.3026}{0.1 \times 1.0986} = \frac{2.3026}{0.10986} \approx 20.96$$ So approximately 20.96 minutes.