| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Intersection points of hyperbolic curves |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring conversion of hyperbolic functions to exponential form, solving a resulting quadratic in e^x, then integration of hyperbolic functions. While the techniques are standard for FP3, it requires multiple steps, careful algebraic manipulation, and exact answers. The integration part is routine for this level, but the initial algebraic work with exponentials adds moderate challenge. Slightly above average difficulty even for Further Maths students. |
| Spec | 1.08f Area between two curves: using integration4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Put \(6\cosh x = 9 - 2\sinh x\) | M1 | |
| \(6 \times \frac{1}{2}(e^x + e^{-x}) = 9 - 2 \times \frac{1}{2}(e^x - e^{-x})\) | M1 | Replaces \(\cosh x\) and \(\sinh x\) by the correct exponential forms |
| \(4e^x + 2e^{-x} - 9 = 0 \Rightarrow 4e^{2x} - 9e^x + 2 = 0\) | M1 A1 | M1: Multiplies by \(e^x\); A1: Correct quadratic in \(e^x\), terms collected |
| \(e^x = \frac{1}{4}\) or \(2\), and \(x = \ln 2\) or \(\ln\frac{1}{4}\) | M1 A1 | M1: Solves quadratic in \(e^x\); A1: Correct values of \(x\) (any correct equivalent form) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \int(9 - 2\sinh x - 6\cosh x)\,dx\) | M1 | Or \(\int(6\cosh x - (9-2\sinh x))\,dx\) or equivalent exponential form |
| \(\pm(9x - 2\cosh x - 6\sinh x)\) or \(\pm(9x - 4e^x + 2e^{-x})\) | M1 A1 | M1: Attempt to integrate; A1: Correct integration |
| \(\pm\bigl([9\ln 2 - 2\cosh\ln 2 - 6\sinh\ln 2] - [9\ln\tfrac{1}{4} - 2\cosh\ln\tfrac{1}{4} - 6\sinh\ln\tfrac{1}{4}]\bigr)\) | dM1 | Complete substitution of limits from part (a); depends on both previous M's |
| Combines logs correctly and uses \(\cosh\) and \(\sinh\) of \(\ln\) correctly at least once | M1 | |
| \(9\ln 8 - 14\) or \(27\ln 2 - 14\) (any correct equivalent) | A1cao |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Put $6\cosh x = 9 - 2\sinh x$ | M1 | |
| $6 \times \frac{1}{2}(e^x + e^{-x}) = 9 - 2 \times \frac{1}{2}(e^x - e^{-x})$ | M1 | Replaces $\cosh x$ and $\sinh x$ by the correct exponential forms |
| $4e^x + 2e^{-x} - 9 = 0 \Rightarrow 4e^{2x} - 9e^x + 2 = 0$ | M1 A1 | M1: Multiplies by $e^x$; A1: Correct quadratic in $e^x$, terms collected |
| $e^x = \frac{1}{4}$ or $2$, and $x = \ln 2$ or $\ln\frac{1}{4}$ | M1 A1 | M1: Solves quadratic in $e^x$; A1: Correct values of $x$ (any correct equivalent form) |
**(6 marks)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \int(9 - 2\sinh x - 6\cosh x)\,dx$ | M1 | Or $\int(6\cosh x - (9-2\sinh x))\,dx$ or equivalent exponential form |
| $\pm(9x - 2\cosh x - 6\sinh x)$ or $\pm(9x - 4e^x + 2e^{-x})$ | M1 A1 | M1: Attempt to integrate; A1: Correct integration |
| $\pm\bigl([9\ln 2 - 2\cosh\ln 2 - 6\sinh\ln 2] - [9\ln\tfrac{1}{4} - 2\cosh\ln\tfrac{1}{4} - 6\sinh\ln\tfrac{1}{4}]\bigr)$ | dM1 | Complete substitution of limits from part (a); depends on both previous M's |
| Combines logs correctly and uses $\cosh$ and $\sinh$ of $\ln$ correctly at least once | M1 | |
| $9\ln 8 - 14$ or $27\ln 2 - 14$ (any correct equivalent) | A1cao | |
**Note: Subtracting the wrong way round could score 5/6 max. If they use $4e^{2x} - 9e^x + 2$ in (b) to find the area — no marks.**
**(6 marks) — Total 12**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{094b3c91-1460-44a2-b9d6-4de90d3adfa0-13_593_1292_118_328}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curves shown in Figure 1 have equations
$$y = 6 \cosh x \text { and } y = 9 - 2 \sinh x$$
\begin{enumerate}[label=(\alph*)]
\item Using the definitions of $\sinh x$ and $\cosh x$ in terms of $\mathrm { e } ^ { x }$, find exact values for the $x$-coordinates of the two points where the curves intersect.
The finite region between the two curves is shown shaded in Figure 1.
\item Using calculus, find the area of the shaded region, giving your answer in the form $a \ln b + c$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2013 Q7 [12]}}