| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring substitution using the identity cosh²x = 1 + sinh²x to form a quadratic in sinh x, then solving and applying inverse hyperbolic functions. While it's Further Maths content, the technique is routine and directly follows standard methods with clear signposting ('Given that' and 'Hence'), making it slightly easier than average overall. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
1
\begin{enumerate}[label=(\alph*)]
\item Given that
$$4 \cosh ^ { 2 } x = 7 \sinh x + 1$$
find the two possible values of $\sinh x$.
\item Hence obtain the two possible values of $x$, giving your answers in the form $\ln p$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q1 [7]}}