\begin{questions} \Question[6] Simplify the following: \begin{parts} \begin{multicols}{2} \part \((2 \sqrt{3}+\sqrt{2})^{2}\) \begin{solutionordottedlines}[2cm] $(2 \sqrt{3}+\sqrt{2})^{2} = 4 \cdot 3 + 4 \sqrt{3} \sqrt{2} + 2 = 12 + 4 \sqrt{6} + 2 = 14 + 4 \sqrt{6}$ \end{solutionordottedlines} \part \((2 \sqrt{5}+4 \sqrt{3})^{2}\) \begin{solutionordottedlines}[2cm] $(2 \sqrt{5}+4 \sqrt{3})^{2} = 4 \cdot 5 + 16 \cdot 3 + 2 \cdot 8 \sqrt{15} = 20 + 48 + 16 \sqrt{15} = 68 + 16 \sqrt{15}$ \end{solutionordottedlines} \part \((\sqrt{x y}+1)^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{x y}+1)^{2} = xy + 2\sqrt{xy} + 1$ \end{solutionordottedlines} \part \((4 \sqrt{2}-3 \sqrt{7})^{2}\) \begin{solutionordottedlines}[2cm] $(4 \sqrt{2}-3 \sqrt{7})^{2} = 16 \cdot 2 + 9 \cdot 7 - 2 \cdot 12 \sqrt{14} = 32 + 63 - 24 \sqrt{14} = 95 - 24 \sqrt{14}$ \end{solutionordottedlines} \part \((\sqrt{x}-\sqrt{y})^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{x}-\sqrt{y})^{2} = x - 2\sqrt{xy} + y$ \end{solutionordottedlines} \part \((\sqrt{11}-2 \sqrt{22})^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{11}-2 \sqrt{22})^{2} = 11 - 4 \cdot 22 + 4 \cdot 11 = 11 - 88 + 44 = -33 + 44 = 11$ \end{solutionordottedlines} \end{multicols} \part If \(x=\sqrt{2}-1\) and \(y=\sqrt{2}+1\), find: \begin{subparts}\begin{multicols}{2} \subpart \(x y\) \begin{solutionordottedlines}[2cm] $x y = (\sqrt{2}-1)(\sqrt{2}+1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$ \end{solutionordottedlines} \subpart \(x^{2} y\) \begin{solutionordottedlines}[2cm] $x^{2} y = (\sqrt{2}-1)^2(\sqrt{2}+1) = (2 - 2\sqrt{2} + 1)(\sqrt{2}+1) = (3 - 2\sqrt{2})(\sqrt{2}+1) = 3\sqrt{2} + 3 - 2\cdot 2 - 2\sqrt{2} = 3\sqrt{2} + 3 - 4 - 2\sqrt{2} = \sqrt{2} - 1$ \end{solutionordottedlines} \subpart \(y^{2} x\) \begin{solutionordottedlines}[2cm] $y^{2} x = (\sqrt{2}+1)^2(\sqrt{2}-1) = (2 + 2\sqrt{2} + 1)(\sqrt{2}-1) = (3 + 2\sqrt{2})(\sqrt{2}-1) = 3\sqrt{2} - 3 + 2\cdot 2 - 2\sqrt{2} = 3\sqrt{2} - 3 + 4 - 2\sqrt{2} = \sqrt{2} + 1$ \end{solutionordottedlines} \subpart \(\frac{1}{x}+\frac{1}{y}\) \begin{solutionordottedlines}[2cm] $\frac{1}{x}+\frac{1}{y} = \frac{1}{\sqrt{2}-1} + \frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} + \frac{\sqrt{2}-1}{2-1} = \sqrt{2}+1 + \sqrt{2}-1 = 2\sqrt{2}$ \end{solutionordottedlines} \end{multicols}\end{subparts} \part \[(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2}\] \begin{solutionordottedlines}[2cm] $(\sqrt{5}+\sqrt{3})^{2}-(\sqrt{5}-\sqrt{3})^{2} = (5 + 2\sqrt{15} + 3) - (5 - 2\sqrt{15} + 3) = 8 + 2\sqrt{15} 8 + 2\sqrt{15} = 4\sqrt{15}$ \end{solutionordottedlines} \part \[(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2}\] \begin{solutionordottedlines}[2cm] $(\sqrt{3}+\sqrt{2})^{2}-(\sqrt{3}-\sqrt{2})^{2} = (3 + 2\sqrt{6} + 2) - (3 - 2\sqrt{6} + 2) = 5 + 2\sqrt{6} - 5 2\sqrt{6} = 4\sqrt{6}$ \end{solutionordottedlines} \end{parts} \Question[2] Simplify \[(a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d})\]. \begin{solutionordottedlines}[2cm] $(a \sqrt{b}+c \sqrt{d})(a \sqrt{b}-c \sqrt{d}) = (a \sqrt{b})^2 - (c \sqrt{d})^2 = a^2 b - c^2 d$ \end{solutionordottedlines} \Question[9] For each of the figures shown, find: \begin{enumerate}[(i)] \item the value of \(x\) \begin{solutionordottedlines}[2cm] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \item the area of the triangle \begin{solutionordottedlines}[2cm] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \item the perimeter of the triangle \begin{solutionordottedlines}[2cm] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \end{enumerate} \begin{parts}\begin{multicols}{3} \part \begin{center} \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(3)} \end{center} \begin{solutionordottedlines}[3in] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \part \begin{center} \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(1)} \end{center} \begin{solutionordottedlines}[3in] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \part \begin{center} \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(4)} \end{center} \begin{solutionordottedlines}[3in] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \end{multicols}\end{parts} \Question[4] For the 2 figures below, find: \begin{enumerate}[(i)] \item the perimeter \item the area \end{enumerate} \begin{parts}\begin{multicols}{2} \part \begin{center} \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04} \end{center} \begin{solutionordottedlines}[2cm] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \part \begin{center} \includegraphics[width=0.2\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-04(2)} \end{center} \begin{solutionordottedlines}[2cm] % Since no figures are provided, no solution can be given. \end{solutionordottedlines} \end{multicols}\end{parts} \Question[8] Rationalise the following: \begin{parts}\begin{multicols}{2} \part \[\frac{14}{\sqrt{7}}\] \begin{solutionordottedlines}[2cm] $\frac{14}{\sqrt{7}} = \frac{14}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{14\sqrt{7}}{7} = 2\sqrt{7}$ \end{solutionordottedlines} \part \[\frac{\sqrt{14}}{\sqrt{7}}\] \begin{solutionordottedlines}[2cm] $\frac{\sqrt{14}}{\sqrt{7}} = \sqrt{\frac{14}{7}} = \sqrt{2}$ \end{solutionordottedlines} \part \[\frac{\sqrt{5}}{3 \sqrt{7}}\] \begin{solutionordottedlines}[2cm] $\frac{\sqrt{5}}{3 \sqrt{7}} = \frac{\sqrt{5}}{3 \sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} = \frac{\sqrt{35}}{21}$ \end{solutionordottedlines} \part \[\frac{\sqrt{3}}{4 \sqrt{6}}\] \begin{solutionordottedlines}[2cm] $\frac{\sqrt{3}}{4 \sqrt{6}} = \frac{\sqrt{3}}{4 \sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{18}}{24} = \frac{\sqrt{2}}{8}$ \end{solutionordottedlines} \part \[\frac{5}{\sqrt{3}}\] \begin{solutionordottedlines}[2cm] $\frac{5}{\sqrt{3}} = \frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$ \end{solutionordottedlines} \part \[\frac{1}{\sqrt{18}}\] \begin{solutionordottedlines}[2cm] $\frac{1}{\sqrt{18}} = \frac{1}{\sqrt{18}} \cdot \frac{\sqrt{18}}{\sqrt{18}} = \frac{\sqrt{18}}{18} = \frac{\sqrt{2}}{6}$ \end{solutionordottedlines} \part \[\frac{1}{\sqrt{2}}+\sqrt{2}\] \begin{solutionordottedlines}[2cm] $\frac{1}{\sqrt{2}}+\sqrt{2} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2} = \frac{\sqrt{2}}{2} + \sqrt{2} = \frac{3\sqrt{2}}{2}$ \end{solutionordottedlines} \part \[\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}}\] \begin{solutionordottedlines}[2cm] $\frac{\sqrt{72}}{\sqrt{3}}+\frac{3}{\sqrt{2}}-\frac{2}{2 \sqrt{2}} = \sqrt{24} + \frac{3\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 2\sqrt{6} + \frac{2\sqrt{2}}{2} = 2\sqrt{6} + \sqrt{2}$ \end{solutionordottedlines} \part \[\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}}\] \begin{solutionordottedlines}[2cm] $\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{7}{2 \sqrt{3}} = \frac{1+2+\frac{7}{2}}{\sqrt{3}} = \frac{\frac{2+4+7}{2}}{\sqrt{3}} = \frac{\frac{13}{2}}{\sqrt{3}} = \frac{13}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{13\sqrt{3}}{6}$ \end{solutionordottedlines} \end{multicols}\end{parts} \Question[6] If \(x=2 \sqrt{14}\) and \(y=4 \sqrt{2}\), find and rationalise the denominator. \begin{parts}\begin{multicols}{3} \part \(\frac{x}{y}\) \begin{solutionordottedlines}[2cm] $\frac{x}{y} = \frac{2 \sqrt{14}}{4 \sqrt{2}} = \frac{\sqrt{14}}{2 \sqrt{2}} = \frac{\sqrt{14}}{2 \sqrt{2}} \cdots \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{28}}{4} = \frac{2\sqrt{7}}{4} = \frac{\sqrt{7}}{2}$ \end{solutionordottedlines} \part \(\frac{y}{x}\) \begin{solutionordottedlines}[2cm] $\frac{y}{x} = \frac{4 \sqrt{2}}{2 \sqrt{14}} = \frac{2 \sqrt{2}}{\sqrt{14}} = \frac{2 \sqrt{2}}{\sqrt{14}} \cdots \frac{\sqrt{14}}{\sqrt{14}} = \frac{2\sqrt{28}}{14} = \frac{2\cdot 2\sqrt{7}}{14} = \frac{\sqrt{7}}{7}$ \end{solutionordottedlines} \part \(\frac{\sqrt{2} x}{\sqrt{3} y}\) \begin{solutionordottedlines}[2cm] $\frac{\sqrt{2} x}{\sqrt{3} y} = \frac{\sqrt{2} \cdot 2 \sqrt{14}}{\sqrt{3} \cdot 4 \sqrt{2}} = \frac{2\sqrt{28}}{4\sqrt{6}} = \frac{\sqrt{28}}{2\sqrt{6}} = \frac{\sqrt{28}}{2\sqrt{6}} \cdots \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{168}}{12} = \frac{2\sqrt{42}}{12} = \frac{\sqrt{42}}{6}$ \end{solutionordottedlines} \end{multicols}\end{parts} \Question[3] A bowl in the shape of a hemisphere of radius length \(5 \mathrm{~cm}\) is partially filled with water. The surface of the water is a circle of radius \(4 \mathrm{~cm}\) when the rim of the bowl is horizontal. Find the depth of the water. \begin{solutionordottedlines}[2cm] Using the Pythagorean theorem for the right triangle formed by the radius of the water's surface, the radius of the bowl, and the depth of the water: $5^2 = 4^2 + d^2$ $25 = 16 + d^2$ $d^2 = 25 - 16$ $d^2 = 9$ $d = 3 \mathrm{~cm}$ The depth of the water is 3 cm. \end{solutionordottedlines} \Question[3] A bobbin for an industrial knitting machine is in the shape of a truncated cone. The diameter of the top is \(4 \mathrm{~cm}\), t diameter of the base is \(6 \mathrm{~cm}\) and the length of the slant is \(10 \mathrm{~cm}\). Find the height of the bobbin. \begin{center} \includegraphics[width=0.3\textwidth]{img/2023_11_09_1c5231c22d14cc915b41g-09} \end{center} \begin{solutionordottedlines}[2cm] Let the height of the bobbin be \(h\), the radius of the top be \(r_1 = 2 \mathrm{~cm}\), and the radius of the base be \(r_2 = 3 \mathrm{~cm}\). By the Pythagorean theorem, we have: \[l^2 = h^2 + (r_2 - r_1)^2\] Substituting the given values: \[10^2 = h^2 + (3 - 2)^2\] \[100 = h^2 + 1\] \[h^2 = 99\] \[h = \sqrt{99}\] \[h \approx 9.95 \mathrm{~cm}\] The height of the bobbin is approximately \(9.95 \mathrm{~cm}\). \end{solutionordottedlines} \Question[4] Rationalise the following: \begin{parts}\begin{multicols}{2} \part \[\frac{1}{\sqrt{3}+2}\] \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{1}{\sqrt{3}+2} \cdot \frac{\sqrt{3}-2}{\sqrt{3}-2} = \frac{\sqrt{3}-2}{(\sqrt{3}+2)(\sqrt{3}-2)} = \frac{\sqrt{3}-2}{3-4} 2-\sqrt{3}\] \end{solutionordottedlines} \part \[\frac{1}{\sqrt{3}+\sqrt{2}}\] \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{1}{\sqrt{3}+\sqrt{2}} \cdot \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{3-2} = \sqrt{3}-\sqrt{2}\] \end{solutionordottedlines} \end{multicols}\end{parts} \Question[3] Find the integers \(p\) and \(q\) such that \[\frac{\sqrt{5}}{\sqrt{5}-2}=p+q \sqrt{5}\]. \begin{solutionordottedlines}[2cm] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{\sqrt{5}}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} = \frac{5+2\sqrt{5}}{5-4} = \frac{5+2\sqrt{5}}{1} = 5+2\sqrt{5}\] Thus, \(p = 5\) and \(q = 2\). \end{solutionordottedlines} \Question[9] Simplify the following: \begin{parts}\begin{multicols}{2} \part \(\frac{3}{\sqrt{5}-2}+\frac{2}{\sqrt{5}+2}\) \begin{solutionordottedlines}[2cm] Multiply each term by the conjugate of its denominator: \[\frac{3}{\sqrt{5}-2} \cdot \frac{\sqrt{5}+2}{\sqrt{5}+2} + \frac{2}{\sqrt{5}+2} \cdot \frac{\sqrt{5}-2}{\sqrt{5}-2}\] \[= \frac{3(\sqrt{5}+2)}{5-4} + \frac{2(\sqrt{5}-2)}{5-4}\] \[= 3\sqrt{5} + 6 + 2\sqrt{5} - 4\] \[= 5\sqrt{5} + 2\] \end{solutionordottedlines} \part \(\frac{5}{(\sqrt{7}-\sqrt{2})^{2}}\) \begin{solutionordottedlines}[2cm] Expand the denominator and simplify: \[\frac{5}{7 - 2\sqrt{14} + 2}\] \[= \frac{5}{9 - 2\sqrt{14}}\] Multiply the numerator and denominator by the conjugate of the denominator: \[\frac{5}{9 - 2\sqrt{14}} \cdot \frac{9 + 2\sqrt{14}}{9 + 2\sqrt{14}}\] \[= \frac{45 + 10\sqrt{14}}{81 - 56}\] \[= \frac{45 + 10\sqrt{14}}{25}\] \[= \frac{9}{5} + 2\sqrt{14}\] \end{solutionordottedlines} \part \(0.0 \mathrm{i} \dot{6}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.0\dot{6}\), then: \[10x = 0.\dot{6}\] \[10x - x = 0.\dot{6} - 0.0\dot{6}\] \[9x = 0.6\] \[x = \frac{0.6}{9}\] \[x = \frac{2}{30}\] \[x = \frac{1}{15}\] \end{solutionordottedlines} \part \(0.3 \dot{2} \dot{4}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.3\dot{2}\dot{4}\), then: \[1000x = 324.\dot{2}\dot{4}\] \[1000x - x = 324.\dot{2}\dot{4} - 0.3\dot{2}\dot{4}\] \[999x = 324\] \[x = \frac{324}{999}\] \[x = \frac{108}{333}\] \[x = \frac{36}{111}\] \[x = \frac{12}{37}\] \end{solutionordottedlines} \part \(0.51 \dot{2} \dot{6}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.51\dot{2}\dot{6}\), then: \[1000x = 512.\dot{2}\dot{6}\] \[1000x - 10x = 512.\dot{2}\dot{6} - 5.1\dot{2}\dot{6}\] \[990x = 507.1\] \[x = \frac{507.1}{990}\] \[x = \frac{5071}{9900}\] \[x = \frac{5071}{9900}\] Simplifying the fraction, we get: \[x = \frac{5071}{9900}\] \[x = \frac{563}{1100}\] \[x = \frac{563}{1100}\] \end{solutionordottedlines} \part \(0.001 \dot{1}\) \begin{solutionordottedlines}[2cm] This is a repeating decimal. Let \(x = 0.001\dot{1}\), then: \[1000x = 1.\dot{1}\] \[1000x - x = 1.\dot{1} - 0.001\dot{1}\] \[999x = 1.1\] \[x = \frac{1.1}{999}\] \[x = \frac{11}{9990}\] \[x = \frac{1}{909}\] \end{solutionordottedlines} \end{multicols}\end{parts} \Question[6] \textbf{Challenge:} \begin{parts} \part Show that \[\frac{138}{19}=7+\frac{1}{3+\frac{1}{1+\frac{1}{4}}}\] \begin{solutionordottedlines}[2cm] Start by evaluating the continued fraction from the innermost fraction outward: \[1 + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}\] \[3 + \frac{1}{\frac{5}{4}} = 3 + \frac{4}{5} = \frac{15}{5} + \frac{4}{5} = \frac{19}{5}\] \[7 + \frac{1}{\frac{19}{5}} = 7 + \frac{5}{19} = \frac{133}{19} + \frac{5}{19} = \frac{138}{19}\] Therefore, the continued fraction equals \(\frac{138}{19}\). \end{solutionordottedlines} \part Express \[\frac{153}{11}\] as a continued fraction with all numerators 1 . \begin{solutionordottedlines}[2cm] Perform the Euclidean algorithm to find the continued fraction: \[\frac{153}{11} = 13 + \frac{10}{11}\] \[\frac{11}{10} = 1 + \frac{1}{10}\] \[\frac{10}{1} = 10\] So, \[\frac{153}{11} = 13 + \frac{1}{1 + \frac{1}{10}}\] Therefore, the continued fraction is \(13 + \frac{1}{1 + \frac{1}{10}}\). \end{solutionordottedlines} \end{parts} \end{questions}