$$\begin{gathered} What is the acute angle between the asymptotes of the hyperbola \frac{{\mathcal{x}}^2}{3}-{\mathcal{y}}^2=1? \\ \left(A\right)\frac{\mathrm{\pi }}{3} \\ \left(B\right)\frac{\mathrm{\pi }}{4} \\ \left(C\right)\frac{\mathrm{\pi }}{6} \\ \left(D\right)\frac{\mathrm{\pi }}{2} \end{gathered}$$

$$\begin{gathered} Which of the following is the range of the function f\left(\mathcal{x}\right)={\sin }^{-1}\mathcal{x}+{\tan }^{-1}\mathcal{x}? \\ \left(A\right)-\mathrm{\pi }<\mathcal{y}<\mathrm{\pi } \\ \left(B\right)-\mathrm{\pi }\le \mathcal{y}\le \mathrm{\pi } \\ \left(C\right)\frac{-3\mathrm{\pi }}{4}\le \mathcal{y}\le \frac{3\mathrm{\pi }}{4} \\ \left(D\right)\frac{-\mathrm{\pi }}{2}\le \mathcal{y}\le \frac{\mathrm{\pi }}{2} \end{gathered}$$

$$\begin{gathered} U\sin g the substitution \mathcal{x}=\mathrm{\pi }-\mathcal{y}, the definite integral {\int }_0^{\mathrm{\pi }} will simplify to: \\ \left(A\right) 0 \\ \left(B\right) {\int }_0^{\mathrm{\pi }}\sin \mathcal{x}d\mathcal{x} \\ \left(C\right) \frac{\pi }{2}{\int }_0^{\mathrm{\pi }}\sin \mathcal{x}d\mathcal{x} \\ \left(D\right) \frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$

$$\begin{gathered} If \mathcal{z}=\frac{\cos \theta -i\sin \theta }{\cos \theta +i\sin \theta } then \mathcal{z} can be expressed as: \\ A) \cos 2\theta +i\sin 2\theta \\ B) \cos 2\theta +i\sin 2\theta \\ C) \sin 2\theta +i\cos 2\theta \\ D) \sin 2\theta -i\cos 2\theta \end{gathered}$$

$$\begin{gathered} Let u = 3 - i and w = 4 + 3i. \\ i) Find Im(uw). \\ ii) Find -iw. \\ iii) Evaluate |u + w{|}^2 \\ iv) Express \frac{u}{w} in the form a + ib, where a and b are real numbers. \\ i) Sketch the region in the complex plane which simul\tan eously satisfies \\ Im\left(\mathcal{z}\right)\ge 1 and arg\left(\frac{\mathcal{z}+2i}{\mathcal{z}-2i}\right)=\pm \frac{\mathrm{\pi }}{2} \\ ii) Find the particular z in part (i) that gives the maximum value of arg(\mathcal{z}) , given \\ -\mathrm{\pi }<\arg (\mathcal{z})\le \mathrm{\pi }. \\ \mathrm{i}) \mathrm{Let} {\mathcal{z}}_1 = \mathrm{a} + \mathrm{ib} \mathrm{and} {\mathcal{z}}_2 = \mathcal{x} + \mathrm{i}\mathcal{y} \mathrm{Prove} \mathrm{that} {\overline{{\mathcal{z}}_1\mathcal{z}}}_2 =\overline{ {\mathcal{z}}_1{\mathcal{z}}_2.} \\ \mathrm{ii}\left) \\ \right(\mathrm{\alpha }) \mathrm{Express} (8 + 7\mathrm{i}\left)\right(5 + 4\mathrm{i}) \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{a} + \mathrm{ib}. \\ (\mathrm{\beta }) \mathrm{Use} \mathrm{part} (\mathrm{i}) \mathrm{to} \mathrm{write} \mathrm{down} (8 - 7\mathrm{i}\left)\right(5 - 4\mathrm{i}) \mathrm{in} \mathrm{the} \mathrm{form} \mathrm{a} + \mathrm{ib} . \\ \mathrm{iii}) \mathrm{Hence} \mathrm{find} \mathrm{the} \mathrm{prime} \mathrm{factorisation} \mathrm{of} {12}^2 + {67}^2. \end{gathered}$$

$$\begin{gathered} Indicate on an Argand diagram the region in which z lies given both \\ |\mathcal{z}-(3+\mathcal{i}\left)\right|\le 3 and \frac{\mathrm{\pi }}{4}arg[\mathcal{z}-(1+\mathcal{i}\left)\right]\le \frac{\mathrm{\pi }}{2} \\ \mathrm{Find} \mathrm{the} \mathrm{locus} \mathrm{in} \mathrm{the} \mathrm{Argand} \mathrm{diagram} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{P} \mathrm{which} \mathrm{represent} \mathrm{the} \mathrm{complex} \mathrm{number} \mathrm{z} \\ \mathrm{where} \mathrm{z}\overline{\mathrm{z}}-4(\mathrm{z}+\overline{\mathrm{z}})=9 \\ \mathrm{Show} \mathrm{by} \mathrm{geometrical} \mathrm{consideration} \mathrm{or} \mathrm{otherwise} \mathrm{that} \mathrm{if} \mathrm{complex} \mathrm{number} \\ {\mathrm{z}}_1 \mathrm{and} {\mathrm{z}}_2 \mathrm{are} \mathrm{such} \mathrm{that} \left|{\mathrm{z}}_1\right|=\left|{\mathrm{z}}_2\right| \mathrm{then} \frac{{\mathrm{z}}_1+{\mathrm{z}}_2}{{\mathrm{z}}_1-{\mathrm{z}}_2} \mathrm{is} \mathrm{purely} \mathrm{imaginary}. \end{gathered}$$

$$\begin{gathered} The expression \frac{9\mathcal{x}}{(\mathcal{x}+2){(\mathcal{x}-1)}^2} can be expressed in partial fraction as: \\ A) \frac{2}{\mathcal{x}-1}-\frac{2}{\mathcal{x}+2}+\frac{3}{{(\mathcal{x}-1)}^2} \\ B) \frac{2}{\mathcal{x}+2}-\frac{3}{{(\mathcal{x}-1)}^2} \\ C) \frac{2}{\mathcal{x}+2}+\frac{3}{{(\mathcal{x}-1)}^2} \\ D) \frac{2}{\mathcal{x}-1}+\frac{2}{\mathcal{x}+2}+\frac{3}{{(\mathcal{x}-1)}^2} \end{gathered}$$

$$\begin{gathered} Suppose \mathcal{z} and \mathcal{w} are the roots of the quadratic equation 3{\mathcal{x}}^2+(2-i)\mathcal{x}+(4+i)=0. \\ Without solving the quadratic find the value \overline{\mathcal{z}} + \overline{\mathcal{w}}. \\ A) -\frac{2+i}{3} \\ B) \frac{2+i}{3} \\ C) -\frac{4+i}{3} \\ D) \frac{4+i}{3} \end{gathered}$$

$$\begin{gathered} Four medical tests A, B, C and D are carried out within 14 days on a patient. A must precede B and B must precede C and D. On the days when A and B are carried out the patient must not undergo any other test, but C and D can be carried out on the same day or on different days in any order. \\ In how many ways can the days for the test be chosen? \end{gathered}$$

$$\begin{gathered} \left(i\right) Show that {(k+1)}^2(k+4)=k^3+6k^2+9k+4 \\ \left(ii\right) Use mathematical induction to prove that for all integers n\ge 1 \\ \frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+...+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)} \\ \left(iii\right) Hence find \\ \underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{r(r+1)(r+2)} \end{gathered}$$

$$\begin{gathered} A particle is fired vertically with initial velocity of u metres per second, and is subject both to gravity, g and air resis\tan ce, which is proportional to the square of the speed v. \\ Show that the equation of motion is given by \ddot{\mathcal{x}}=-g-k{v}^2, where k is a constant. \\ By taking \ddot{\mathcal{x}}=v\frac{dv}{d\mathcal{x}} and integrating, show that the greatest height H reached by the particle is given by H=\frac{1}{2k}\ln \frac{g+k{u}^2}{g} \\ The particle returns to the point of projection. By considering a suitable equation of motion, show that the velocity w ,with which it returns to the point of projection is given by \\ {w}^2=\frac{g}{k}\left(1-e^{-2kH}\right) \end{gathered}$$

$$\begin{gathered} Find: \\ \int \mathcal{x}{e}^{\mathcal{x}} d\mathcal{x} \\ \int \frac{1}{\mathcal{x}{(\ln \mathcal{x})}^2} d\mathcal{x} \end{gathered}$$

$$\begin{gathered} Given that \mathcal{z}= 1- i\sqrt{3} : \\ \left(i\right) Express \mathcal{z} in modulus–argument form. 1 \\ \left(ii\right) Find {\mathcal{z}}^6 \end{gathered}$$

$$\begin{gathered} A curve is implicitly defined by {\mathcal{x}}^3+{\mathcal{y}}^3={\mathcal{x}}^2{\mathcal{y}}^2 \\ Find an expression for \frac{d\mathcal{y}}{d\mathcal{x}} in terms of \mathcal{x} and \mathcal{y}. \end{gathered}$$

$$\begin{gathered} You are given that \\ 2\cos A\sin B=\sin (A+B)-\sin (A-B) \\ Let S=1+2\cos \theta +2\cos 2\theta +2\cos 3\theta \\ i) Show that S\times \sin \frac{\theta }{2}=\sin \frac{7\theta }{2} \\ ii) Hence show that 1+2\cos \frac{2\mathrm{\pi }}{7}+2\cos \frac{4\mathrm{\pi }}{7}+2\cos \frac{6\mathrm{\pi }}{7}=0 \\ iii) By writing S as a sum of powers of \cos \theta ,Show that \cos \frac{2\mathrm{\pi }}{7} is a solution \\ of the polynomial equation 8{\mathcal{x}}^3+4{\mathcal{x}}^2-4\mathcal{x}-1=0. \end{gathered}$$

$$\begin{gathered} Sketch the region in the complex plane which simul\tan eously satisfies \\ \frac{\pi }{2} \le arg \left(\mathcal{z}\right) \le \frac{ 3\pi }{4} and \left|\mathcal{z}\right| \le 2. \\ Clearly label the coordinates of any corners of the region, indicating if they are included in the region \end{gathered}$$

$$\begin{gathered} The diagram above shows the region bound by the curve \mathcal{y}= \sqrt{\mathcal{x}}, the \mathcal{x}-axis, \\ and the 3 line \mathcal{x} = 1. This region is rotated about the line \mathcal{x} = 1 \\ to form a solid. Use the method of cylindrical shells to find the volume of the solid. \end{gathered}$$

$$\begin{gathered} The polynomial P\left(\mathcal{x}\right) = {\mathcal{x}}^3 - 9{\mathcal{x}}^2 + 11\mathcal{x} +21 has zeroes \alpha , \beta and \gamma . \\ Find a simplified polynomial with zeroes \alpha + 1, \beta + 1 and \gamma + 1. 2 \\ Hence fully factorise P\left(\mathcal{x}\right). \end{gathered}$$

$$\begin{gathered} The diagram above shows the graph of \mathcal{y}= f\left(\mathcal{x}\right). \\ Copy or trace the graph onto three separate number planes, each one third of a page. \\ Use your diagrams to sketch following graphs, clearly showing any intercepts with \\ axes, turning points, and asymptotes. \\ \mathcal{y}= f \left(\right|\mathcal{x}\left|\right) \\ \mathcal{y}= {\left[f\right(\mathcal{x}\left)\right]}^2 \\ \mathcal{y}= e^{f\left(\mathcal{x}\right)} \end{gathered}$$

$$\begin{gathered} The diagram above shows the graph \mathcal{y} = \frac{b}{a}\sqrt{a^2 - {\mathcal{x}}^2}, that is, \\ the section of the ellipse \frac{{\mathcal{x}}^2}{a^2}+\frac{{\mathcal{y}}^2}{b^2}= 1 where\mathcal{y} \ge 0 \\ Write down the value of{\int }_{-a}^a \sqrt{a^2 - {\mathcal{x}}^2}d\mathcal{x} \\ Deduce that the area of the ellipse\frac{{\mathcal{x}}^2}{a^2} +\frac{{ }^{\mathcal{y}2}}{b^2} = 1 is \pi ab unit{s}^2 \end{gathered}$$