| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Volume of revolution with hyperbolics |
| Difficulty | Challenging +1.2 This is a Further Maths question involving hyperbolic functions, which places it above average difficulty on an absolute scale. However, it follows standard techniques: (a) solving a hyperbolic equation using exponential definitions, (b) expanding and using double-angle identities (routine algebraic manipulation), and (c) applying volume of revolution formula with the identity from part (b). The question is well-scaffolded with part (b) providing the key simplification needed for part (c). While requiring knowledge of Further Maths content, the problem-solving is largely procedural rather than requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\cosh x - 6\sinh x = 0 \Rightarrow 5\!\left(\frac{e^x+e^{-x}}{2}\right) - 6\!\left(\frac{e^x - e^{-x}}{2}\right) = 0\) | M1 | Substitutes correct exponential forms; allow the "2"s to be missing |
| \(e^{2x} = 11\) | A1 | Correct equation |
| \(x = \ln\sqrt{11}\) | A1 | Correct value (or e.g. \(\frac{1}{2}\ln 11\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\cosh x - 6\sinh x = 0 \Rightarrow \tanh x = \frac{5}{6}\) | M1 | Rearranges to \(\tanh x = \ldots\) |
| \(x = \text{artanh}\!\left(\frac{5}{6}\right)\) | A1 | Correct equation |
| \(x = \ln\sqrt{11}\) | A1 | Correct value (or e.g. \(\frac{1}{2}\ln 11\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\cosh x - 6\sinh x = 0 \Rightarrow 25\cosh^2 x = 36\sinh^2 x\), leading to \(\sinh^2 x = \frac{25}{11}\) or \(\cosh^2 x = \frac{36}{11}\) | M1 | Rearranges to \(\sinh^2 x = \ldots\) or \(\cosh^2 x = \ldots\) |
| \(\sinh x = (\pm)\frac{5}{\sqrt{11}}\) or \(\cosh x = (\pm)\frac{6}{\sqrt{11}}\) | A1 | Correct equation (allow \(\pm\)) |
| \(x = \ln\sqrt{11}\) | A1 | Correct value (or e.g. \(\frac{1}{2}\ln 11\)) |
# Question 2:
## Part (a): $y = 5\cosh x - 6\sinh x$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\cosh x - 6\sinh x = 0 \Rightarrow 5\!\left(\frac{e^x+e^{-x}}{2}\right) - 6\!\left(\frac{e^x - e^{-x}}{2}\right) = 0$ | M1 | Substitutes correct exponential forms; allow the "2"s to be missing |
| $e^{2x} = 11$ | A1 | Correct equation |
| $x = \ln\sqrt{11}$ | A1 | Correct value (or e.g. $\frac{1}{2}\ln 11$) |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\cosh x - 6\sinh x = 0 \Rightarrow \tanh x = \frac{5}{6}$ | M1 | Rearranges to $\tanh x = \ldots$ |
| $x = \text{artanh}\!\left(\frac{5}{6}\right)$ | A1 | Correct equation |
| $x = \ln\sqrt{11}$ | A1 | Correct value (or e.g. $\frac{1}{2}\ln 11$) |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\cosh x - 6\sinh x = 0 \Rightarrow 25\cosh^2 x = 36\sinh^2 x$, leading to $\sinh^2 x = \frac{25}{11}$ or $\cosh^2 x = \frac{36}{11}$ | M1 | Rearranges to $\sinh^2 x = \ldots$ or $\cosh^2 x = \ldots$ |
| $\sinh x = (\pm)\frac{5}{\sqrt{11}}$ or $\cosh x = (\pm)\frac{6}{\sqrt{11}}$ | A1 | Correct equation (allow $\pm$) |
| $x = \ln\sqrt{11}$ | A1 | Correct value (or e.g. $\frac{1}{2}\ln 11$) |
> Note: This is not a proof so allow "h"s to be lost along the way as long as the intention is clear.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{38487750-8c0f-4c3d-a019-5213ed2866eb-04_616_764_246_584}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$y = 5 \cosh x - 6 \sinh x$$
The curve crosses the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the $x$ coordinate of the point $A$, giving your answer as a natural logarithm.
\item Show that
$$( 5 \cosh x - 6 \sinh x ) ^ { 2 } \equiv a \cosh 2 x + b \sinh 2 x + c$$
where $a , b$ and $c$ are constants to be found.
The finite region $R$, bounded by the curve and the coordinate axes, is shown shaded in Figure 1.
The region $R$ is rotated through $2 \pi$ radians about the $x$-axis.
\item Use calculus to find the volume of the solid generated, giving your answer as an exact multiple of $\pi$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2018 Q2 [10]}}