$$\begin{gathered} Question Nine What is the range of g\left(\mathcal{x}\right)=\tan \left(\frac{1}{2}{\cos }^{-1}\mathcal{x}\right)? \\ \left(A\right) \mathcal{Y}\le 0 \\ \left(B\right) \mathcal{Y}\in R \\ \left(C\right) 0\le \mathcal{Y}<1 \\ \left(D\right) \mathcal{Y}\ge 0 \end{gathered}$$

$$\begin{gathered} Given that the Argand Digram for |\mathcal{z}-2|+|\mathcal{z}-4|=10 is an ellipse, \\ Find the co-ordinates of the centre of ellipse and the lengths of the major and \\ minor axis \\ On an Argand Digram, showa the region for which \mathcal{z} satisfies the inqualities \\ \mathcal{z}+ \overline{\mathcal{z}}\le 6 and |\mathcal{z}-2|+| \mathcal{z}-4|\le 10 \\ Find the parameter of the shape in the Argand Digram describes by \\ |\mathcal{z}-1|\le 1 and 0\le arg \mathcal{z}\ge \frac{\mathrm{\pi }}{6} \end{gathered}$$

$$\begin{gathered} Find \int \frac{d\mathcal{x}}{{\mathcal{x}}^2+2\mathcal{x}+5} \\ Use integration by parts to evaluate {\int }_0^{\frac{\mathrm{\pi }}{2}} {\tan }^{-1} \mathcal{x}d\mathcal{x} \\ Use the substitution t = \tan \frac{\mathcal{x}}{2} to evaluate {\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{d\mathcal{x}}{1+\sin \mathcal{x}+\cos \mathcal{x}} \end{gathered}$$

$$\begin{gathered} Find the following indefinite integrals. \\ \left(i\right) \int \frac{1}{\sqrt{9{\mathcal{x}}^2+2}}d\mathcal{x} (use the substitution u=3\mathcal{x}) \\ \left(ii\right) \int \frac{1}{\mathcal{x}+\sqrt{\mathcal{x}}}d\mathcal{x} (use the substitution u=\sqrt{\mathcal{x}}) \\ Evaluate \\ \left(i\right) {\int }_0^1\frac{1+\mathcal{x}}{\sqrt{1-{\mathcal{x}}^2}}d\mathcal{x} \\ \left(ii\right) {\int }_0^1\frac{{\mathcal{x}}^2-1}{{\mathcal{x}}^2+1}d\mathcal{x} \\ Use integration by part to find \int \mathcal{x}\ln \mathcal{x} d\mathcal{x} \\ \left(i\right) Express \frac{4\mathcal{x}+2}{(\mathcal{x}+1)({\mathcal{x}}^2+1)} in the form \frac{a}{\mathcal{x}+1}+\frac{b\mathcal{x}+c}{{\mathcal{x}}^2+1} \\ \left(ii\right) Hence, or otherwise evaluate {\int }_0^{\sqrt{3}}\frac{4\mathcal{x}+2}{({\mathcal{x}}^2+1)({\mathcal{x}}^2+1)}d\mathcal{x} \end{gathered}$$

$Find the equation of the \tan gent to the curve {\mathcal{x}}^3+{\mathcal{Y}}^3-5\mathcal{Y}-3=0 at the point(1,-2)$

$$\begin{gathered} A curve has parametric equations \\ \mathcal{x}=\sin \theta \\ \mathcal{y}=\tan \theta \\ where 0<\theta <\mathrm{\pi } \mathrm{and} \mathrm{\theta }\ne \frac{\mathrm{\pi }}{2}. \\ \mathrm{Show} \mathrm{that} \frac{\mathrm{d}\mathcal{y}}{\mathrm{d}\mathcal{x}}={\sec }^2\mathrm{\theta } \mathrm{and} \frac{{\mathrm{d}}^2\mathcal{y}}{\mathrm{d}{\mathcal{x}}^2}=3{\mathrm{tan\theta sec}}^4\mathrm{\theta }. \\ \mathrm{Using} \mathrm{the} \mathrm{substitution} \mathcal{x}=6 \cos u, determine \int \frac{1}{{\mathcal{x}}^2\sqrt{36-{\mathcal{x}}^2}}d\mathcal{x}. \end{gathered}$$

$Use integration by parts determine \int {\mathcal{x}}^2\cos \mathcal{x} d\mathcal{x}.$

$$\begin{gathered} \left(i\right) Show that \frac{d}{du} {\log }_e (u+\sqrt{a^2+u^2)}= \frac{1}{\sqrt{a^2+u^2}} , where a is a cons\tan t. \\ (ii) Hence evaluate{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos \mathcal{x}}{\sqrt{4+{\sin }^2 \mathcal{x}}}d\mathcal{x}. \\ Use the substitution t = \tan \frac{\mathcal{x}}{2} to find \int \frac{1}{1+\cos \mathcal{x}+\sin \mathcal{x}} d\mathcal{x}. \\ A complex number satisfies \left|\mathcal{z}-4\right| \le 2 and Im(\mathcal{z})\le 0. \\ (i) Sketch the locus of \mathcal{z}. \\ (ii) Show that - \frac{\mathrm{\pi }}{6} \le arg \mathcal{z}\le 0. \\ By rationali\sin g the numerator of the integrand, evaluate {\int }_{\frac{1}{2}}^1 \sqrt{\frac{\mathcal{x}}{2-\mathcal{x}}} d\mathcal{x}. \\ Use a suitable substitution to evaluate {\int }_3^9 \frac{3}{(9+\mathcal{x})\sqrt{\mathcal{x}}} d\mathcal{x}. \end{gathered}$$

$$\begin{gathered} \left(i\right) Find a general solution to the equation \cos 3\mathcal{x}= \sin 2\mathcal{x}. \\ \left(ii\right) Hence, or otherwise, find the smallest positive solution of the equation \\ \cos 3\mathcal{x}= \sin 2\mathcal{x}. \end{gathered}$$

$Solve the quadratic equation{\mathcal{z}}^2 - 2i\mathcal{z} + 3 = 0.$