1. **Problem Q1:** Express each of the following as an entire surd and write them in ascending order: 3\sqrt{11}, 2\sqrt{23}, 7\sqrt{2}, 10, 4\sqrt{6}. - Convert each to a single square root: - $3\sqrt{11} = \sqrt{9 \times 11} = \sqrt{99}$ - $2\sqrt{23} = \sqrt{4 \times 23} = \sqrt{92}$ - $7\sqrt{2} = \sqrt{49 \times 2} = \sqrt{98}$ - $10 = \sqrt{100}$ - $4\sqrt{6} = \sqrt{16 \times 6} = \sqrt{96}$ - Now order: $\sqrt{92} (2\sqrt{23}) < \sqrt{96} (4\sqrt{6}) < \sqrt{98} (7\sqrt{2}) < \sqrt{99} (3\sqrt{11}) < \sqrt{100} (10)$. 2. **Problem Q2a:** Simplify $\sqrt{63} - \sqrt{7} + \sqrt{28}$. - Simplify each radical: - $\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt{7}$ - $\sqrt{7}$ stays as is - $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$ - Substitute and combine: - $3\sqrt{7} - \sqrt{7} + 2\sqrt{7} = (3 - 1 + 2)\sqrt{7} = 4\sqrt{7}$ 3. **Problem Q2b:** Simplify $(2\sqrt{3} - 1)(3\sqrt{3} - 1)$. - Expand the product: - $2\sqrt{3} \times 3\sqrt{3} = 6 \times 3 = 18$ - $2\sqrt{3} \times (-1) = -2\sqrt{3}$ - $-1 \times 3\sqrt{3} = -3\sqrt{3}$ - $-1 \times -1 = 1$ - Combine terms: - $18 + 1 - 2\sqrt{3} - 3\sqrt{3} = 19 - 5\sqrt{3}$ 4. **Problem Q3:** Express $\dfrac{5}{\sqrt{3} - 1} - \dfrac{1}{2 + \sqrt{3}}$ with rational denominators. - Rationalize each term: - First term: Multiply numerator and denominator by $\sqrt{3} + 1$: $$\frac{5}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{5(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{5(\sqrt{3} + 1)}{3 - 1} = \frac{5(\sqrt{3} + 1)}{2}$$ - Second term: Multiply numerator and denominator by $2 - \sqrt{3}$: $$\frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$$ - Subtract the terms: $$\frac{5(\sqrt{3} + 1)}{2} - (2 - \sqrt{3}) = \frac{5\sqrt{3} + 5}{2} - 2 + \sqrt{3} = \frac{5\sqrt{3} + 5}{2} + \sqrt{3} - 2$$ - Combine $\sqrt{3}$ terms: $$\sqrt{3} = \frac{2\sqrt{3}}{2}$$ - Sum: $$\frac{5\sqrt{3} + 5 + 2\sqrt{3} - 4}{2} = \frac{7\sqrt{3} + 1}{2}$$ Final expression: $\boxed{\frac{7\sqrt{3} + 1}{2}}$. 5. **Problem Q4a:** Solve $2x - 11 = |3x - 4|$. - Case 1: $3x - 4 \geq 0 \Rightarrow x \geq \frac{4}{3}$ $$2x - 11 = 3x - 4 \Rightarrow -11 + 4 = 3x - 2x \Rightarrow -7 = x \Rightarrow x = -7$$ Not valid since $-7 < \frac{4}{3}$. - Case 2: $3x - 4 < 0 \Rightarrow x < \frac{4}{3}$ $$2x - 11 = -(3x - 4) = -3x + 4 \Rightarrow 2x + 3x = 4 + 11 \Rightarrow 5x = 15 \Rightarrow x = 3$$ Valid since $3 \geq \frac{4}{3}$? No, contradicts case condition. - So check for $x=3$ in original: $2(3) - 11 = 6 - 11 = -5$, $|3(3) - 4| = |9 - 4| = 5$; LHS $\neq$ RHS - Check $x=-7$: Not in domain, discard. - Re-examining. Case 1: $x \geq \frac{4}{3}$ Equation: $2x - 11 = 3x - 4 \Rightarrow x = -7$ invalid. Case 2: $x < \frac{4}{3}$ Equation: $2x - 11 = -3x + 4 \Rightarrow 5x = 15 \Rightarrow x=3$ invalid. No solution from above. But check if $x=\frac{4}{3}$ satisfies: $2(\frac{4}{3})-11 = \frac{8}{3} -11 = \frac{8 - 33}{3} = \frac{-25}{3}$, $|3(\frac{4}{3}) - 4| = |4 - 4| = 0$ no equality. Hence **no solutions**. 6. **Problem Q4b:** Solve $|5 - 3x| = |x + 3|$. - Square both sides: $$(5 - 3x)^2 = (x + 3)^2$$ - Expand: $$25 - 30x + 9x^2 = x^2 + 6x + 9$$ - Bring all to one side: $$9x^2 - 30x + 25 - x^2 - 6x - 9 = 0 \Rightarrow 8x^2 - 36x + 16 = 0$$ - Divide by 4: $$2x^2 - 9x + 4 = 0$$ - Use quadratic formula: $$x = \frac{9 \pm \sqrt{81 - 32}}{4} = \frac{9 \pm \sqrt{49}}{4} = \frac{9 \pm 7}{4}$$ - Solutions: - $x = \frac{9 + 7}{4} = 4$ - $x = \frac{9 - 7}{4} = \frac{2}{4} = 0.5$ - Check absolute values equality: - For $x=4$: $|5 - 12| = | -7| = 7$ and $|4 + 3| = 7$ valid. - For $x=0.5$: $|5 - 1.5| = 3.5$, $|0.5 + 3| = 3.5$ valid. 7. **Problem Q4c:** Solve $|2x - 1| > 7$. - Split into two inequalities: - $2x - 1 > 7 \Rightarrow 2x > 8 \Rightarrow x > 4$ - $2x - 1 < -7 \Rightarrow 2x < -6 \Rightarrow x < -3$ - Solution: $x < -3$ or $x > 4$. 8. **Problem Q5a:** Sketch and find key points of $y = (x + 1)(2 - x)$. - Rewrite: $$y = (x + 1)(2 - x) = -x^2 - x + 2$$ - Roots: $$x = -1 \text{ or } x = 2$$ - Vertex: $$x_v = -\frac{b}{2a} = -\frac{-1}{2 \times -1} = -\frac{-1}{-2} = -\frac{1}{2} = -0.5$$ - Find $y_v$: $$y(-0.5) = -( -0.5)^2 - (-0.5) + 2 = -0.25 + 0.5 + 2 = 2.25$$ - Intercepts: - $y$-intercept at $x=0$: $y = (0 +1)(2 - 0) = 1 \times 2 = 2$ 9. **Problem Q5b:** Circle $x^2 + (y-1)^2 = 9$. - Center: $(0,1)$ - Radius: $3$ - Intercepts: - $x$-intercepts: Set $y=0$: $$x^2 + (0 - 1)^2 = 9 \Rightarrow x^2 + 1 =9 \Rightarrow x^2=8 \Rightarrow x=\pm 2\sqrt{2}$$ - $y$-intercepts: Set $x=0$: $$(y -1)^2 = 9 \Rightarrow y -1 = \pm 3 \Rightarrow y = 4 \text{ or } -2$$ 10. **Problem Q5c:** Line $y - 2x - 12=0$ or $y=2x+12$. - Y-intercept: $y=12$ at $x=0$. - X-intercept: $0 = 2x + 12 \Rightarrow x = -6$. 11. **Problem Q5d:** Graph $y=|(x + 1)(x + 3)|$. - The quadratic inside is $x^2 + 4x + 3$ with roots at $x = -3, -1$. - The parabola opens upwards and is negative between $-3$ and $-1$, absolute value reflects this part above x-axis. - Vertex: at $x = -2$, $$y = |-1| = 1$$ - The graph looks like a 'W' shape with minimum value 0 at roots. 12. **Problem Q6a:** Sketch region $x - 3y < -6$. - Rewrite as $3y > x + 6$ or $y > \frac{x}{3} + 2$. - Region is above the line $y = \frac{x}{3} + 2$. 13. **Problem Q6b:** Sketch region $y \leq 2x - x^2$. - Parabola opening down with vertex at $$x_v = \frac{-b}{2a} = \frac{-2}{-2} =1$$ $$y_v = 2(1) - 1^2 = 1$$ - Region includes points under or on this parabola. 14. **Problem Q7a:** Sets from first 20 positive integers $S = \{1,...,20\}$ - $A$ prime numbers in $S$: $$A = \{2,3,5,7,11,13,17,19\}$$ - $B$ factors of 30 less than 20: Factors of 30 are $1,2,3,5,6,10,15,30$ discard 30: $$B = \{1,2,3,5,6,10,15\}$$ - $C$ subset $>1$ and $<9$: $$C = \{2,3,4,5,6,7,8\}$$ 15. **Problem Q7b:** Venn diagram would show overlaps of these sets with intersections like $A \cap B$, $A \cap C$, $B \cap C$, and triple intersection. 16. **Problem Q7c:** Find $(A \cup C)^c \cap B$ - $A \cup C$: $$\{2,3,4,5,6,7,8,11,13,17,19\}$$ - Its complement relative to $S$ is all elements in $S$ not in union: $$ (A \cup C)^c = \{1,9,10,12,14,15,16,18,20\}$$ - Intersection with $B = \{1,2,3,5,6,10,15\}$: $$\{1,10,15\}$$ Answer: $\{1,10,15\}$. 17. **Problem Q7d:** Verify $$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(A \cap C) - n(B \cap C) + n(A \cap B \cap C)$$ - Calculate cardinalities: - $n(A) = 8$ - $n(B) = 7$ - $n(C) = 7$ - Intersections: - $A \cap B = \{2,3,5\}, n=3$ - $A \cap C = \{2,3,5,7\}, n=4$ - $B \cap C = \{2,3,5,6\}, n=4$ - $A \cap B \cap C = \{2,3,5\}, n=3$ - Compute right side: $$8 + 7 + 7 - 3 - 4 - 4 + 3 = 22 - 11 + 3 = 14$$ - Compute $A \cup B \cup C$: $$A \cup B \cup C = \{1,2,3,4,5,6,7,8,10,11,13,15,17,19\}$$ $$n = 14$$ - Equality verified. **Final answers** are shown for each question.