$$\begin{gathered} Take \mathcal{x}=2.5 as a first approximation for a root of {\mathcal{x}}^3-3\mathcal{x}-20=0 . Use one application of \\ Newton\text{'}s Method to find a second approximation correct to 2 decimal places. \end{gathered}$$

$Evaluate \underset{\mathcal{x}\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\mathcal{x}}{2}}\right)}{3\mathcal{x}}$

$$\begin{gathered} Use the substitution u = {\tan }^{-1}\mathcal{x} to evaluate the following \\ .Leave your answer in exact form \\ {\int }_0^1\frac{{\tan }^{-1}\mathcal{x}}{1+{\mathcal{x}}^2} d\mathcal{x} \end{gathered}$$

$$\begin{gathered} The function f\left(\mathcal{x}\right)=\frac{1}{1+3e^{-\mathcal{x}}}- is defined for all real \mathcal{x} and e^{\mathcal{x}} > 0 \\ i) Sketch the curve \mathcal{y}=f(\mathcal{x}) mark in any asymptotes, \mathcal{x}, \mathcal{y} intercepts \\ ii) Explain why an inverse function exists for \mathcal{y}=f\left(\mathcal{x}\right) \\ iii) Find the inverse function \mathcal{y}= f\text{'}(\mathcal{x}) \end{gathered}$$

$$\begin{gathered} A cone with base radius r, vertical height h and slant height s has curved surface area A= \mathrm{\pi }rs and volume V =\frac{1}{3}{\mathrm{\pi r}}^{2}h . \\ \left(i\right) Show that a cone with vertical height \mathcal{x} and base radius \frac{3}{4} has A=\frac{15\mathrm{\pi }{\mathcal{x}}^2}{16}and V=\frac{3\mathrm{\pi }{\mathcal{x}}^3}{16}. \\ \left(ii\right) A similar cone whose base radius is \frac{3}{4} of its vertical height, with its axis vertical and vertex downwards, is \\ being lowered into a container which is overflowing with water. When the vertex is 2 metres below surface of the \\ water (with the cone not fully submerged) the surface area of the cone is being covered with water at a rate of 0.5 m^2{s}^{-1}. \\ Show that the cone is being lowered into the water at a rate of \frac{2}{15\mathrm{\pi }}m{s}^{-1}. \\ \left(iii\right) Find the rate at which the water is overflowing from the container when the vertex of the cone is 2 metres below the surface \end{gathered}$$

$$\begin{gathered} \left(i\right) What is the domain of h\left(\mathcal{x}\right)= {\sin }^{-1}\sqrt{1-{\mathcal{x}}^2}+{\sin }^{-1}\mathcal{x}? \\ \left(ii\right) Find h\text{'}\left(\mathcal{x}\right). \\ \left(iii\right) Hence determine the interval over which h\left(\mathcal{x}\right) is cons\tan t and find this cons\tan t. \end{gathered}$$

$$\begin{gathered} \left(i\right)Sketch the graph of \mathcal{y}= 2 \cos 2 \mathcal{x}, show any intercepts with axes, \\ and the domain and range. \\ \left(ii\right)The region in the first quadrant in the above graph is rotated about \\ the \mathcal{y}axis. \\ \left(\alpha \right)Show that {\mathcal{x}}^2 =\frac{1}{4} {\cos }^2\frac{ \mathcal{y}}{2}? \\ \left(\beta \right) Find the volume of the solid formed (Answer in terms of \mathrm{\pi }) \\ Find \int 2{\mathcal{x}}^2{e}^{4 {\mathcal{x}}^3+2} \\ The acceleration of a particle moving in a straight line is given by \\ \ddot{ \mathcal{x}}=-2e^{- \mathcal{x}} where \mathcal{x} is the displacement from the origin. Initially \\ the object is at the origin with velocity \left(v\right) 2m{s}^{-1} \\ i) Prove that V = 2e^{\frac{- \mathcal{x}}{2}} ? \\ ii)What happens to v as \mathcal{x} increases without bound ? \end{gathered}$$

$$\begin{gathered} The graph below shows the graph of \mathcal{y}={\log }_e\mathcal{x} and the secant joining \\ points P and Q on the curve .P is at \mathcal{x} =1 and Q is at \mathcal{x} =1+\frac{r}{n} \end{gathered}$$

$$\begin{gathered} Show that the gradient of the secant is \frac{1}{r}{\log }_e{(1+\frac{r}{n})}^n use \frac{d}{d\mathcal{x}}{\log }_e\mathcal{x}=\frac{1}{\mathcal{x}} to show that \underset{n\to \infty }{lim}{(1+\frac{r}{n})}^n=e^r \\ use part 2 to determine an expression for the effective annual rate \\ of interest when an annual rate of 6\% p.a is compounded continually \\ that is , compounded an infinite number of times per year \end{gathered}$$

$$\begin{gathered} What is the value of k such that {\int }_0^{2k}\frac{1}{\sqrt{3-{\mathcal{x}}^2}}d\mathcal{x}=\frac{\mathrm{\pi }}{3}? \\ \left(A\right) \frac{\sqrt{3}}{4} \left(B\right) \frac{\sqrt{3}}{2} \\ \left(C\right)\frac{3}{4} \left(D\right)\frac{3}{2} \end{gathered}$$

$Slove the inequality \frac{4}{1-\mathcal{x}}\le 3 and graph your solution on a number line$

$Find \underset{\mathcal{x}\to 0}{lim} \frac{\sin 3\mathcal{x}}{2\mathcal{x}}.$

$$\begin{gathered} A curve is defined by the parameters \mathcal{x}=2p+\frac{2}{p}, \mathcal{y}=p^2+\frac{1}{p^2}. \\ Which of the following represents this curve in Cartesian form? \\ \left(A\right)\mathcal{y}=\frac{{\mathcal{x}}^2}{4}-2 \\ \left(B\right)\mathcal{y}=\frac{{\mathcal{x}}^2}{2} \\ \left(C\right)\mathcal{x}+{\mathcal{y}}^2=2 \\ \left(D\right)\mathcal{y}={\mathcal{x}}^2-2 \end{gathered}$$

$$\begin{gathered} Differentiate {\sin }^{-1}\left(2\mathcal{x}\right) \\ Find \int \mathcal{x}\sqrt{1+{\mathcal{x}}^2}d\mathcal{x} u\sin g the substitution u=1+{\mathcal{x}}^2 \\ Evaluate {\int }_0^2\frac{d\mathcal{x}}{4+{\mathcal{x}}^2} \\ Evaluate \underset{\mathcal{x}\to 0}{\lim } \frac{\sin 3\mathcal{x}}{5\mathcal{x}} \end{gathered}$$

$$\begin{gathered} Which integral is obtained when the substitution u = \mathcal{x} + 2 is applied to \int \frac{\mathcal{x}}{3}\sqrt{\mathcal{x}+2}d\mathcal{x}? \\ \left(A\right)\frac{1}{3}\int (u^{\frac{1}{2}}+2u^{\frac{1}{3}})du \\ \left(B\right)\frac{1}{3}\int (u^{\frac{1}{2}}-2u^{\frac{3}{2}})du \\ \left(C\right)\frac{1}{3}\int (u^{\frac{3}{2}}+2u^{\frac{1}{2}})du \\ \left(D\right)\frac{1}{3}\int (u^{\frac{3}{2}}-2u^{\frac{1}{2}})du \end{gathered}$$

$$\begin{gathered} Solve \cos \cos \theta +\sqrt{3}\sin \theta =1 (0\le \theta \le 2\mathrm{\pi }) \\ Solve \cos 2\theta =\cos \theta (0\le \theta \le 2\mathrm{\pi }) \end{gathered}$$

$Find the value of \underset{\mathcal{x}\to 0}{lim} \frac{\tan 3\mathcal{x}}{2\mathcal{x}}.$

$Evaluate {\int }_{\frac{\mathrm{\pi }}{3}}^{\frac{\mathrm{\pi }}{2}}\sin \mathcal{x}{\cos }^2\mathcal{x}d\mathcal{x} u\sin g the substitution u=\cos \mathcal{x}$

$Solve \frac{{\mathcal{x}}^2+\mathcal{x}-6}{\mathcal{x}}\ge 2.$