| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths question requiring hyperbolic identities, differentiation, and surface area of revolution. Part (a) requires manipulation of multiple-angle hyperbolic identities (non-trivial). Part (b) is a standard differentiation with identity verification. Part (c) combines these results in a surface area integral with specific limits requiring careful algebraic manipulation to reach the given form. The question is structured to guide students through steps, but requires solid technique with hyperbolics and integration—significantly harder than typical A-level but standard for FP2. |
| Spec | 1.07l Derivative of ln(x): and related functions4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08e Mean value of function: using integral |
6
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { 1 } { 4 } ( \cosh 4 x + 2 \cosh 2 x + 1 ) = \cosh ^ { 2 } x \cosh 2 x$$
\item Show that, if $y = \cosh ^ { 2 } x$, then
$$1 + \left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) ^ { 2 } = \cosh ^ { 2 } 2 x$$
\item The arc of the curve $y = \cosh ^ { 2 } x$ between the points where $x = 0$ and $x = \ln 2$ is rotated through $2 \pi$ radians about the $x$-axis. Show that the area $S$ of the curved surface formed is given by
$$S = \frac { \pi } { 256 } ( a \ln 2 + b )$$
where $a$ and $b$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2012 Q6 [13]}}