$$\begin{gathered} Let \left|ܽa\right|=1. What is the general solution of \sin 2\mathcal{x}=a ܵ? \\ \left(A\right)\mathcal{x}=n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{\sin }{} \end{gathered}$$

$Sketch the graph of \mathcal{Y}=2{\sin }^{-1}3\mathcal{x} showing clearly the domain and range of the function as well as any intercepts.$

$$\begin{gathered} The sketch shows the graph of the curve \mathcal{y}=f\left(\mathcal{x}\right) where f\left(\mathcal{x}\right)=2{\cos }^{-1}\frac{\mathcal{x}}{3} . \\ The area under the curve for 0\le \mathcal{x}\le 3 is shaded. \\ \left(i\right) Find the \mathcal{y} intercept. \\ \left(ii\right) Find the domain and range of f\left(\mathcal{x}\right)=2{\cos }^{-1}\frac{\mathcal{x}}{3} \\ \left(iii\right) Calculate the area of the shaded region. \end{gathered}$$

$Prove {\tan }^{-1}\frac{2}{3}+{\cos }^{-4}\frac{2}{\sqrt{5}}={\tan }^{-1}\frac{7}{4}.$

$$\begin{gathered} what is the value of {\int }_0^{\frac{\mathrm{\pi }}{4}}\left(se{c}^2\mathcal{x}-\mathcal{x}\right)d\mathcal{x}? \\ \left(A\right) 1-\frac{{\mathrm{\pi }}^2}{32} \\ \left(B\right) 1-\frac{{\mathrm{\pi }}^2}{16} \\ \left(C\right) 1-\frac{\mathrm{\pi }}{8} \\ \left(D\right) 1-\frac{\mathrm{\pi }}{4} \end{gathered}$$

Shade the following regions bounded by the curves:

$\mathcal{y}<\sqrt{4-{\left(\mathcal{x}-2\right)}^2} and \mathcal{y}>\frac{{\mathcal{x}}^2}{2}$

In the triangle ABC, ∠ACB = 30, ∠ABC = 50 and BC = 10 cm. The foot of the perpendicular from A to BC is D. 

i) Use the sine rule to find an expression for the length of AB. 

ii) Hence or otherwise, find the length of AD. 

Answer correct to two decimal places

In the diagram above AF is a straight line, ∠ABC = 74 , ∠DEC = 59 , ∠BCF =137 and ∠BCE =15.

Prove that AB||DE

$$\begin{gathered} What is the value of f\text{'}\left(\mathcal{x}\right) if f \left(\mathcal{x}\right) =2{\mathcal{x}}^2{\cos }^{-1}2\mathcal{x} ? \\ \left(A\right) \frac{-8\mathcal{x}}{\sqrt{1-2{\mathcal{x}}^2}} \\ \left(B\right)\frac{-8\mathcal{x}}{\sqrt{1-4{\mathcal{x}}^2}} \\ \left(C\right) \frac{-4{\mathcal{x}}^2}{\sqrt{1-2{\mathcal{x}}^2}}+4\mathcal{x}{\cos }^{-1}2\mathcal{x} \\ \left(D\right)\frac{-4{\mathcal{x}}^2}{\sqrt{1-4{\mathcal{x}}^2}}+4\mathcal{x}{\cos }^{-1}2\mathcal{x} \end{gathered}$$

$$\begin{gathered} Which of the following expressions is correct? \\ \left(A\right) {\tan }^{-1}\mathcal{x}={\cos }^{-1}\frac{1}{\sqrt{1-{\mathcal{x}}^2}} \\ \left(B\right) {\tan }^{-1}\mathcal{x}={\cos }^{-1}\frac{1}{\sqrt{1+{\mathcal{x}}^2}} \\ \left(C\right) {\tan }^{-1}\mathcal{x}={\cos }^{-1}\frac{\mathcal{x}}{\sqrt{1-{\mathcal{x}}^2}} \\ \left(D\right) {\tan }^{-1}\mathcal{x}={\cos }^{-1}\frac{\mathcal{x}}{\sqrt{1+{\mathcal{x}}^2}} \end{gathered}$$