$$\begin{gathered} \left(1\right)by u\sin g t=\tan \frac{\mathcal{x}}{2}show that \frac{\cos \mathcal{x} }{1+\sin \mathcal{x}}=\frac{1-t^2}{{(1-t)}^2} \\ \left(2\right)hence or otherwise solve \frac{\cos \mathcal{x} }{1+\sin \mathcal{x}}=1for 0\le \mathcal{x}\le 2\mathcal{x} \\ Evaluate {\int }_{\frac{1}{4}}^{\frac{1}{2}}\frac{-2}{\sqrt{1-4{\mathcal{x}}^2}}d\mathcal{x} \end{gathered}$$

$Show that F\left(\mathcal{x}\right) = \frac{\sqrt{1-{\mathcal{x}}^2}}{{\sin }^{-1}\mathcal{x}} is an odd function.$

$$\begin{gathered} U\sin g the substitution u=1-2\mathcal{x} , the integral {\int }_0^1\frac{v}{\sqrt{1-2\mathcal{x}}}d\mathcal{x} can be expressed as: \\ \left(A\right)\frac{1}{4}{\int }_{-1}^1\frac{1-u}{\sqrt{u}}du \\ \left(B\right) \frac{1}{4}{\int }_{11}^{-1}\frac{1-u}{\sqrt{u}}du \\ \left(C\right) \frac{1}{2}{\int }_{-1}^1\frac{1-u}{\sqrt{u}}du \\ \left(D\right)\frac{1}{2}{\int }_1^{-1}\frac{1-u}{\sqrt{u}}du \end{gathered}$$

$$\begin{gathered} Let f\left(\mathcal{x}\right) =2\mathcal{x}-{\mathcal{x}}^2 for \mathcal{x}\le 1. This function has an inverse f^{-1}\left(\mathcal{x}\right) . \\ Sketch the graphs of \mathcal{y} =f\left(\mathcal{x}\right) and \mathcal{y}=f^{-1} \left(\mathcal{x}\right) on the same diagram. \\ Find an expression for f^{-1} \left(\mathcal{x}\right) . \\ Evaluate f^{-1}\left(\frac{3}{4}\right) \end{gathered}$$

$$\begin{gathered} Alarge indoor stadium is 300 metres long. A golf ball is to be hit from flopr level from one end of the stadium to the other without hitting the roof. The roof is just high enough for the ball to reach a maximum height of 25 metres. \\ Show that if the ball is just going to miss the roof then \theta ={\tan }^{-1}\left(\frac{1}{3}\right). \\ U\sin g this value of \theta , and u\sin g g=9.8, find the value of V so that the balljust lands at the end of the stadium. \end{gathered}$$

$$\begin{gathered} Write the expression \mathcal{y}=\sin \mathcal{x}+\sqrt{3}\cos \mathcal{x} in the form R\sin \left(\mathcal{x}+\alpha \right). \\ A) 2\sin \left(\mathcal{x}+\frac{\mathrm{\pi }}{4}\right) \\ B) \sin \left(\mathcal{x}+\frac{\mathrm{\pi }}{3}\right) \\ C) 2\sin \left(\mathcal{x}+\frac{\mathrm{\pi }}{3}\right) \\ D) 2\sin \left(\mathcal{x}-\frac{\mathrm{\pi }}{3}\right) \end{gathered}$$

$$\begin{gathered} if \mathcal{y}={\sin }^{-1}\frac{a}{\mathcal{x}} then \frac{d\mathcal{y}}{d\mathcal{x}} equals: \\ A\left)\frac{-a}{{\mathcal{x}}^2\sqrt{{\mathcal{x}}^2-a^2}} \\ B\right)\frac{\mathcal{x}}{\sqrt{{\mathcal{x}}^2-a^2}} \\ C\left)\frac{-\mathcal{x}}{\sqrt{{\mathcal{x}}^2-a^2}} \\ D\right)\frac{-a}{\mathcal{x}\sqrt{{\mathcal{x}}^2-a^2}} \end{gathered}$$

$$\begin{gathered} A ball is projected from leve ground at an angle of \theta and a velocity of V m/s. You may \\ assume the equations of motion are x=Vt\cos \theta and y=-\frac{g{r}^2}{2}+Vt\sin \theta \\ \left(i\right) Show that the maximum height achieved by the ball is \frac{V^2{\sin }^2\theta }{2g} \\ \left(ii\right) The ball just clears a wall of height h metres that is d metres from the point of projection. \end{gathered}$$

$Show that the greatest height reached is \frac{d^2{\tan }^2\theta }{4(d\tan \theta -h)}$

$Differentiate {\sin }^{-1} \left(\frac{x}{2}\right) with respect to x.$

$$\begin{gathered} Consider the function\int \left(\mathcal{x}\right)=\frac{2\mathcal{x}}{\mathcal{x}+1} and its inverse function \int -1\left(\mathcal{x}\right). Evaluate \int -1\left(3\right). \\ \left(A\right) ‐3 \\ \left(B\right)\frac{2}{3} \\ \left(C\right)\frac{3}{2} \\ \left(D\right) 3 \end{gathered}$$