1. **Rationalise the following:** a. $\frac{1 + \sqrt{3}}{2 + \sqrt{3}}$ Multiply numerator and denominator by the conjugate of denominator $2 - \sqrt{3}$: $$\frac{1 + \sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{(1+\sqrt{3})(2 - \sqrt{3})}{(2)^2 - (\sqrt{3})^2}=\frac{2 - \sqrt{3} + 2\sqrt{3} - 3}{4-3} = \frac{-1 + \sqrt{3}}{1} = -1 + \sqrt{3}$$ b. $\frac{4 + \sqrt{2}}{3 + 2\sqrt{3}}$ Multiply numerator and denominator by conjugate $3 - 2\sqrt{3}$: $$\frac{4 + \sqrt{2}}{3 + 2\sqrt{3}} \times \frac{3 - 2\sqrt{3}}{3 - 2\sqrt{3}} = \frac{(4+\sqrt{2})(3 - 2\sqrt{3})}{9 - 12} = \frac{12 -8\sqrt{3} + 3\sqrt{2} - 2\sqrt{6}}{-3} = -4 + \frac{8\sqrt{3}}{3} - \sqrt{2} + \frac{2\sqrt{6}}{3}$$ c. $\frac{3 - \sqrt{5}}{6 - \sqrt{3}}$ Multiply numerator and denominator by conjugate $6 + \sqrt{3}$: $$\frac{3 - \sqrt{5}}{6 - \sqrt{3}} \times \frac{6 + \sqrt{3}}{6 + \sqrt{3}} = \frac{(3 - \sqrt{5})(6 + \sqrt{3})}{36 - 3} = \frac{18 + 3\sqrt{3} - 6\sqrt{5} - \sqrt{15}}{33} = \frac{18 + 3\sqrt{3} - 6\sqrt{5} - \sqrt{15}}{33}$$ d. $\frac{2 - 3\sqrt{3}}{5 - 5\sqrt{6}}$ Multiply numerator and denominator by conjugate $5 + 5\sqrt{6}$: $$\frac{2 - 3\sqrt{3}}{5 - 5\sqrt{6}} \times \frac{5 + 5\sqrt{6}}{5 + 5\sqrt{6}} = \frac{(2 - 3\sqrt{3})(5 + 5\sqrt{6})}{25 - 150} = \frac{10 + 10\sqrt{6} - 15\sqrt{3} - 15\sqrt{18}}{-125}$$ Simplify $\sqrt{18} = 3\sqrt{2}$: $$= \frac{10 + 10\sqrt{6} - 15\sqrt{3} - 45\sqrt{2}}{-125} = -\frac{2}{25} - \frac{2\sqrt{6}}{25} + \frac{3\sqrt{3}}{25} + \frac{9\sqrt{2}}{25}$$ 2. **Simplify the following:** a. $5\sqrt{3} \times 13\sqrt{3} = 65 \times 3 = 195$ b. $10\sqrt{11}(3\sqrt{2} - 2\sqrt{11}) = 10\sqrt{11} \times 3\sqrt{2} - 10\sqrt{11} \times 2\sqrt{11} = 30\sqrt{22} - 20 \times 11 = 30\sqrt{22} - 220$ c. $(2 + 3\sqrt{3})(4 - 5\sqrt{2}) = 2 \times 4 + 2 \times (-5\sqrt{2}) + 3\sqrt{3} \times 4 - 3\sqrt{3} \times 5\sqrt{2} = 8 - 10\sqrt{2} + 12\sqrt{3} - 15\sqrt{6}$ d. $(3 - 4\sqrt{3})^{2} = (3)^2 - 2 \times 3 \times 4\sqrt{3} + (4\sqrt{3})^{2} = 9 - 24\sqrt{3} + 16 \times 3 = 9 - 24\sqrt{3} + 48 = 57 - 24\sqrt{3}$ 3. **Simplify the following:** a. $(-6 + 2\sqrt{5})(7 - 5\sqrt{5}) = -42 + 30\sqrt{5} +14\sqrt{5} - 10 \times 5 = -42 + 44\sqrt{5} - 50 = -92 + 44\sqrt{5}$ b. $4\sqrt{x}(5\sqrt{x} - 10\sqrt{y}) = 20x - 40\sqrt{xy}$ c. $(4 + 2\sqrt{3})^{2} - (4 - 2\sqrt{3})^{2}$ By the difference of squares expansion: $(a+b)^{2} - (a-b)^{2} = 4ab$ Here $a=4$, $b=2\sqrt{3}$, so: $$4 \times 4 \times 2\sqrt{3} = 32\sqrt{3}$$ 4. **Survey Questions:** a. Total graduates = 100, choosing at least one force. Let $|A|=55$, $|P|=50$, $|I|=30$, $|A \cup P \cup I|=100$ Given: number who chose exactly two = 44, let number who chose all three forces = $x$. Formula for union: $$|A \cup P \cup I| = |A| + |P| + |I| - \text{sum of exactly two} + |A \cap P \cap I|$$ Since exactly two chosen count includes all pairs except the triple counted thrice, $$100 = 55 + 50 + 30 - 44 + x \, \Rightarrow \, 100 = 135 - 44 + x = 91 + x$$ So, $$x = 100 - 91 = 9$$ b. Venn diagrams description: (i) All students who study Additional Mathematics ($A$) also study Mathematics ($M$), so $A \subseteq M$. Some students in $M$ do not study $A$, so $A$ is a proper subset of $M$. $U$ is the universal set containing all students. (ii) Every student studying $M$ studies $A$, so $M \subseteq A$. Not all students study $A$, so $A$ is a subset but not equal to $U$. Thus, the Venn diagrams show proper subset relations as explained. **Final answers:** 1.a $-1 + \sqrt{3}$ 1.b $-4 + \frac{8\sqrt{3}}{3} - \sqrt{2} + \frac{2\sqrt{6}}{3}$ 1.c $\frac{18 + 3\sqrt{3} - 6\sqrt{5} - \sqrt{15}}{33}$ 1.d $-\frac{2}{25} - \frac{2\sqrt{6}}{25} + \frac{3\sqrt{3}}{25} + \frac{9\sqrt{2}}{25}$ 2.a $195$ 2.b $30\sqrt{22} - 220$ 2.c $8 - 10\sqrt{2} + 12\sqrt{3} - 15\sqrt{6}$ 2.d $57 - 24\sqrt{3}$ 3.a $-92 + 44\sqrt{5}$ 3.b $20x - 40\sqrt{xy}$ 3.c $32\sqrt{3}$ 4.a Number who chose all three = 9 4.b Explained set relations in Venn diagrams.