2 The matrix \(\mathbf { A }\) is given by
$$\mathbf { A } = \left( \begin{array} { r r r }
- 1 & 2 & 12
0 & 1 & 0
0 & 0 & 3
\end{array} \right) .$$
Use the characteristic equation of \(\mathbf { A }\) to show that
$$\mathbf { A } ^ { 4 } = p \mathbf { A } ^ { 2 } + q \mathbf { l }$$
where \(p\) and \(q\) are integers to be determined.
3 The curve \(C\) has equation
$$x y ^ { 3 } - 4 x ^ { 3 } y = 3$$
Show that, at the point \(( - 1,1 )\) on \(C , \frac { \mathrm { dy } } { \mathrm { dx } } = 11\).
Find the value of \(\frac { d ^ { 2 } y } { d x ^ { 2 } }\) at the point \(( - 1,1 )\).
\includegraphics[max width=\textwidth, alt={}, center]{37db1c60-0f94-413f-b29b-5872975eee9e-06_535_1584_276_276}
The diagram shows the curve with equation \(\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } ^ { 2 } }\) for \(x \geqslant 2\), together with a set of \(( N - 2 )\) rectangles
of unit width.
By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 1 } ^ { N } \frac { \ln r } { r ^ { 2 } } < \frac { 2 + 3 \ln 2 } { 4 } - \frac { 1 + \ln N } { N }$$
Use a similar method to find, in terms of \(N\), a lower bound for \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { N } } \frac { \ln \mathrm { r } } { \mathrm { r } ^ { 2 } }\).
5 Find the particular solution of the differential equation
$$\frac { d ^ { 2 } y } { d x ^ { 2 } } - 2 \frac { d y } { d x } + y = 4 \cos x$$
given that, when \(x = 0 , y = - 4\) and \(\frac { d y } { d x } = 3\).
Hence obtain the roots of the equation
$$x ^ { 5 } - 10 x ^ { 4 } + 40 x ^ { 2 } - 32 = 0$$
in the form \(\operatorname { cosec } ( q \pi )\), where \(q\) is rational.
Show that an appropriate integrating factor for
$$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$
is \(x + \sqrt { x ^ { 2 } - 1 }\).
Hence find the solution of the differential equation
$$\sqrt { x ^ { 2 } - 1 } \frac { d y } { d x } + y = x ^ { 2 } - x \sqrt { x ^ { 2 } - 1 }$$
for which \(y = 1\) when \(x = \frac { 5 } { 4 }\). Give your answer in the form \(y = f ( x )\).
Starting from the definition of cosh in terms of exponentials, prove that
$$2 \cosh ^ { 2 } A = \cosh 2 A + 1$$
The curve \(C\) has parametric equations
$$\mathrm { x } = 2 \cosh 2 \mathrm { t } + 3 \mathrm { t } , \quad \mathrm { y } = \frac { 3 } { 2 } \cosh 2 \mathrm { t } - 4 \mathrm { t } , \quad \text { for } - \frac { 1 } { 2 } \leqslant t \leqslant \frac { 1 } { 2 }$$
The area of the surface generated when \(C\) is rotated through \(2 \pi\) radians about the \(y\)-axis is denoted by \(A\).
Show that \(A = 10 \pi \int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } ( 2 \cosh 2 t + 3 t ) \cosh 2 t d t\).
Hence find \(A\) in terms of \(\pi\) and e.
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