| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring differentiation of arcosh, solving a hyperbolic equation, and expressing the answer in exact form. While the differentiation is standard (d/dx[arcosh x] = 1/√(x²-1)), solving 40/√(x²-1) = 9 for x and then substituting back to find y in the form r ln 3 + s requires careful algebraic manipulation and knowledge of arcosh in logarithmic form. It's more demanding than typical A-level questions but follows a clear procedure once the derivative is found. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{40}{\sqrt{x^2-1}} - 9\) | M1, A1 | M1: \(\frac{dy}{dx} = \frac{p}{\sqrt{x^2-1}} - q\). A1: Cao |
| Put \(\frac{dy}{dx} = 0\) and obtain \(x^2 = \ldots\) (allow sign errors only) e.g. \(\left(\frac{1681}{81}\right)\) | dM1 | |
| \(x = \frac{41}{9}\) | M1, A1 | M1: square root. A1: \(x=\frac{41}{9}\) or exact equivalent (not \(\pm\frac{41}{9}\)) |
| \(y = 40\ln\!\left\{\frac{41}{9}+\sqrt{\left(\frac{41}{9}\right)^2-1}\right\} - \text{"41"}\) | M1 | Substitutes \(x = "\frac{41}{9}"\) into curve and uses logarithmic form of arcosh |
| So \(y = 80\ln 3 - 41\) | A1 | Cao |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{40}{\sqrt{x^2-1}} - 9$ | M1, A1 | M1: $\frac{dy}{dx} = \frac{p}{\sqrt{x^2-1}} - q$. A1: Cao |
| Put $\frac{dy}{dx} = 0$ and obtain $x^2 = \ldots$ (allow sign errors only) e.g. $\left(\frac{1681}{81}\right)$ | dM1 | |
| $x = \frac{41}{9}$ | M1, A1 | M1: square root. A1: $x=\frac{41}{9}$ or exact equivalent (not $\pm\frac{41}{9}$) |
| $y = 40\ln\!\left\{\frac{41}{9}+\sqrt{\left(\frac{41}{9}\right)^2-1}\right\} - \text{"41"}$ | M1 | Substitutes $x = "\frac{41}{9}"$ into curve and uses logarithmic form of arcosh |
| So $y = 80\ln 3 - 41$ | A1 | Cao |
**Total: 7**4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bd4cd798-61ae-49b6-a297-bb4b9ed15fb1-05_384_1040_226_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows part of the curve with equation
$$y = 40 \operatorname { arcosh } x - 9 x , \quad x \geqslant 1$$
Use calculus to find the exact coordinates of the turning point of the curve, giving your answer in the form $\left( \frac { p } { q } , r \ln 3 + s \right)$, where $p , q , r$ and $s$ are integers.\\
\hfill \mbox{\textit{Edexcel FP3 2013 Q4 [7]}}