13
- 1
4
\end{array} \right) + \mu \left( \begin{array} { r }
5
1
- 3
\end{array} \right)$$
where \(\lambda\) and \(\mu\) are scalar parameters.
- Show that \(l _ { 1 }\) and \(l _ { 2 }\) meet and find the position vector of their point of intersection \(A\).
- Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to one decimal place.
A circle with centre \(A\) and radius 35 cuts the line \(l _ { 1 }\) at the points \(P\) and \(Q\). Given that the \(x\) coordinate of \(P\) is greater than the \(x\) coordinate of \(Q\),
- find the coordinates of \(P\) and the coordinates of \(Q\).
6. Use integration by parts to show that
$$\int \mathrm { e } ^ { 2 x } \cos 3 x \mathrm {~d} x = p \mathrm { e } ^ { 2 x } \sin 3 x + q \mathrm { e } ^ { 2 x } \cos 3 x + k$$
where \(p\) and \(q\) are rational numbers to be found and \(k\) is an arbitrary constant.\\
(6)\\
7. Water is flowing into a large container and is leaking from a hole at the base of the container.
At time \(t\) seconds after the water starts to flow, the volume, \(V \mathrm {~cm} ^ { 3 }\), of water in the container is modelled by the differential equation
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = 300 - k V$$
where \(k\) is a constant.
- Solve the differential equation to show that, according to the model,
$$V = \frac { 300 } { k } + A \mathrm { e } ^ { - k t }$$
where \(A\) is a constant.\\
(5)
Given that the container is initially empty and that when \(t = 10\), the volume of water is increasing at a rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
- find the exact value of \(k\).
- Hence find, according to the model, the time taken for the volume of water in the container to reach 6 litres. Give your answer to the nearest second.\\
8. Use proof by contradiction to prove that, for all positive real numbers \(x\) and \(y\),
$$\frac { 9 x } { y } + \frac { y } { x } \geqslant 6$$
9.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{594542dd-ee2d-49b6-9fab-77b2d1a44f8c-24_632_734_214_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Figure 1 shows a sketch of a closed curve with parametric equations
$$x = 5 \cos \theta \quad y = 3 \sin \theta - \sin 2 \theta \quad 0 \leqslant \theta < 2 \pi$$
The region enclosed by the curve is rotated through \(\pi\) radians about the \(x\)-axis to form a solid of revolution. - Show that the volume, \(V\), of the solid of revolution is given by
$$V = 5 \pi \int _ { \alpha } ^ { \beta } \sin ^ { 3 } \theta ( 3 - 2 \cos \theta ) ^ { 2 } \mathrm {~d} \theta$$
where \(\alpha\) and \(\beta\) are constants to be found.
- Use the substitution \(u = \cos \theta\) and algebraic integration to show that \(V = k \pi\) where \(k\) is a rational number to be found.
\includegraphics[max width=\textwidth, alt={}, center]{594542dd-ee2d-49b6-9fab-77b2d1a44f8c-28_2649_1889_109_178}