$$\begin{gathered} A particle moves in simple harmonic motion such that v^2=-4(\mathcal{x}-5)(\mathcal{x}+1) , where v is velocity in m/s. Maximum acceleration of this particle happens when: \\ \left(A\right) \mathcal{x}=5 and -1 \\ \left(B\right) \mathcal{x}=2 \\ \left(C\right) \mathcal{x}=-5 and 1 \\ \left(D\right) \mathcal{x}=0 \end{gathered}$$

$$\begin{gathered} The acceleration-time graph of a particle is shown. \\ The time\left(s\right) when the particle has a maximum velocity is: \\ \left(A\right) t=2 \\ \left(B\right) t=3 \\ \left(C\right) t=4 \\ \left(D\right) 0<t< 2 \end{gathered}$$

$$\begin{gathered} Let \mathcal{y}=\frac{2-\sin \theta }{\cos \theta } for0\le \theta \le \frac{\mathrm{\pi }}{4} \\ Show that \frac{dy}{d\theta }=se{c}^2\theta (2\sin \theta -1) \\ Hence or otherwise find the minimum value of \frac{2-\sin \theta }{\cos \theta }for 0\le \theta \le \frac{\mathrm{\pi }}{4} \\ Find the maximum value of \frac{2-\sin \theta }{\cos \theta } for -\le \theta \le \frac{\mathrm{\pi }}{4} \end{gathered}$$

$$\begin{gathered} FB is a \tan gent touching a circle at A .CE is the diameter,O is the centre \\ and D lies on the circumference. \angle BAE=36°. \end{gathered}$$

$$\begin{gathered} \left(i\right) Find the size of \angle ACE, giving reasons. \\ \left(ii\right) Find the size of \angle ADC, giving reasons. \end{gathered}$$

$$\begin{gathered} After t minutes the number of bacteria N in a culture is given by N=\frac{900}{1+b{e}^{-ct}} for \\ some cons\tan ts b>0 and c >0 . Initially there are 300 bacteria in the culture and the \\ number of bacteria is initially increa\sin g at a rate of 20 per minute. \\ Show that \frac{DN}{dt}=\frac{cN}{900}(900-N) . \\ Show that b = 2 and c = 0.1. \\ Show that the maximum rate of increase in the number of bacteria occurs when \\ N=450 \end{gathered}$$

$A projectile has the equation of path y=3x^2+2x+4. How far will it have travelled horizentally before it returns to its original height?$

$Find \mathcal{y} if {\left(\frac{d\mathcal{y}}{d\mathcal{x}}\right)}^2 =\frac{9}{9-{\mathcal{x}}^2}and \mathcal{y}=2\mathrm{\pi } when \mathcal{x}-3.$

$Evaluate {\int }_{-1}^1\frac{-1}{\sqrt{2-{\mathcal{x}}^2}}d\mathcal{x}$

$Evaluate {\int }_0^{\frac{\mathrm{\pi }}{12}}2 {\sin }^2 4\mathcal{x}d\mathcal{x}$

$$\begin{gathered} \left(ii\right) Differentiate \mathcal{Y}=\mathcal{x}{\cos }^{-1}\mathcal{x}-\sqrt{1-{\mathcal{x}}^2} \\ \left(iii\right) Hence or otherwise evaluate {\int }_{-1}^1\left({\cos }^{-1}\mathcal{x}\right)-\frac{\mathrm{\pi }}{2}d\mathcal{x} \end{gathered}$$

$$\begin{gathered} A particle moves in such a way that its displacement, \mathcal{x} cm, from the origin at any time is given by the function \mathcal{x}=2+{\cos }^{2}t where t is in seconds. \\ \left(i\right) Show that acceleration is given by \stackrel{..}{\mathcal{x}}=10- 4\mathcal{x} \\ \left(ii\right) Find the centre of the motion. \\ \left(iii\right) Find the amplitude of the motion. \end{gathered}$$