$$\begin{gathered} The equation {\mathcal{x}}^3+ 6{\mathcal{x}}^2- 2\mathcal{x}+ 4 = 0 has roots \alpha , \beta and \gamma . Find the value of: \\ i) \alpha +\beta +\gamma and \alpha \beta +\beta \gamma +\gamma \alpha \\ ii) {\alpha }^2+{\beta }^2+{\gamma }^2 \end{gathered}$$

$$\begin{gathered} In the diagram above O is the centre of the circle. Points A, B and C \\ all lie on the 2 circumference of the circle. If <OAB = \alpha find the size \\ of <ACB. Give reasons for your answer. \end{gathered}$$

$$\begin{gathered} In trying to solve \frac{\mathcal{x}-2}{\mathcal{x}-3}<\frac{4}{\sqrt{\mathcal{x}-2}} over its natural domain in the \\ set of real numbers, three students produce the following inequalities. \\ Student I (\mathcal{x}-2)\sqrt{\mathcal{x}- 2} < 4(\mathcal{x}-3) \\ Student II p \sqrt{{(\mathcal{x}-2)}^3}(\mathcal{x}-3)< 4{(\mathcal{x}-3)}^2 Student III {(\mathcal{x}-2)}^2(\mathcal{x}-3) < 4{(\mathcal{x}-3)}^2 \sqrt{\mathcal{x}- 2} Which students are still on track to obtain the correct solution set? \\ \left(A\right) Just student II \\ \left(B\right) Student III only \\ \left(C\right) Both students II and III but not student I \\ \left(D\right) All three students \end{gathered}$$

$$\begin{gathered} A person on horizontal ground is looking at an aeroplane A through a telescope T. \\ The aeroplane is approaching at a speed of 80m Is at a cons\tan t altitude of 200 meters \\ above the telescope. When the horizontal dis\tan ce of the aeroplane \\ from the telescope is x metres, the angle of elevation of the aeroplane is e radians. \\ \left(i\right) Show that \theta ={\tan }^{-1}\frac{200}{\mathcal{x}} \\ \left(ii\right) Show that\frac{d\theta }{dt}=\frac{16000}{{\mathcal{x}}^2+40000} \\ \left(iii\right) Find the rate at which \theta is changing when \theta =\frac{\pi }{4} , (answer in degrees) \end{gathered}$$

$$\begin{gathered} A group of 4 women and 8 boys include a mother and son. From this group, a team consisting of 2 women and 2 boys is to be chosen. \\ How many ways can the team be chosen if the mother and son cannot be on the team together? \end{gathered}$$

$Solve \frac{3}{\mathcal{x}}<2.$

$Solve \frac{3}{\mathcal{x}}<2.$

$$\begin{gathered} In the diagram above AB is a diameter of the circle, T P is a \tan gent \\ at point T, O is the centre of the circle and <ATP = 111°. Find \\ <BAT giving reasons. \end{gathered}$$

$$\begin{gathered} If the roots of the quadratic equation 8{\mathcal{x}}^2 -5\mathcal{x}+ a = 0 are \sin \theta \\ and \cos 2\theta for some angle \theta , find the possible values of a. \end{gathered}$$

$$\begin{gathered} The rate at which a cool object warms in air is proportional to the difference between i \\ ts temperature T, in degrees Celsius, and the cons\tan t ambient temperature A°C of the \\ surrounding air. This rate can be expressed by the differential equation: \\ \frac{dT}{dt}= k(A - T) \\ where t is time in minutes and k is a positive cons\tan t. The solution of this differential \\ equation is T = A + B{e}^{-kt}, where B is a cons\tan t. (You need NOT show this.) \\ A bottle of baby milk is at 4 3°C when it is removed from a refrigerator and placed \\ on the kitchen bench where the room temperature is 22°C. Five minutes later it has \\ warmed to 12°C. \\ Find the temperature of the milk after a further three minutes sitting on the bench. \\ Give your answer correct to the nearest degree. \end{gathered}$$

$$\begin{gathered} Consider the region enclosed by the upper semicircle \mathcal{y}=\sqrt{a^2-{(\mathcal{x}-a)}^2} and the vertical line \\ \mathcal{x}=b where 0 < b < 2a, shown shaded in the diagram above. \\ A spherical cap is generated by rotating this region around the \mathcal{x}-axis. Show \\ that the volume V, in cubic units, of this solid is given by: \\ V=\frac{\pi b^2}{3}(3a-b). \end{gathered}$$

$$\begin{gathered} In the diagram above, a spherical vase of radius 10 cm is being filled with water \\ at a cons\tan t rate 90 c{m}^3/min. \\ Let h cm be the depth of the water after t minutes. Find the rate at which the depth \\ of the water is ri\sin g at the ins\tan t when the depth is 5 cm. \\ Give your answer in terms of \pi . \end{gathered}$$

$$\begin{gathered} In the diagram above, points A and B are separated by a horizontal dis\tan ce R and \\ point B is located H metres higher than A. Define the angle of inclination of B from \\ A as \alpha . Define the origin for both motions at point A and positive directions as right \\ and up respectively. \\ At the same ins\tan t, identical projectiles are launched from each location directed \\ towards each other. The projectile from A is fired with initial speed U m/s at an \\ angle \theta above the horizontal while the object launched from B has only half as \\ much initial speed but the same angle of elevation above the horizontal. \\ Consequently, the equations of motion for the projectile from A, t seconds after launch, \\ are as follows: \\ {\mathcal{x}}_A= Ut\cos \theta {\mathcal{y}}_A= U t\sin \theta -\frac{1}{2}g{t}^2 . \\ Similarly for the projectile from B: \\ {\mathcal{x}}_B=R-\frac{Ut\cos \theta }{2} , {\mathcal{y}}_B=H+\frac{Ut\sin \theta }{2}-\frac{1}{2}g{t}^2 \\ [Do NOT prove these equations.] \\ Given that the projectiles collide, show that this requires \\ \tan \alpha =\frac{1 }{3}\tan \theta \end{gathered}$$

$$\begin{gathered} The velocity v of a particle moving in a straight line is governed by the equation v = \mathcal{x}-2, \\ where \mathcal{x} is its displacement. The particle started at \mathcal{x}= 5. What is the displacement \\ function of the particle? \\ \left(A\right) \mathcal{x} = 5e^t \\ \left(B\right) \mathcal{x} = 2 +\frac{1}{3}{e}^t \\ \left(C\right) \mathcal{x} = 2 + e^t \\ \left(D\right) \mathcal{x} = 2 + 3e^t \end{gathered}$$