| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring hyperbolic function differentiation, arc length formula application, and integration of coth x. While it involves multiple steps and FP2 content, each part follows standard techniques: (a) uses chain rule with known hyperbolic identities, (b)(i) applies the arc length formula with the result from (a), and (b)(ii) uses the standard integral of coth x. The 'show that' structure provides clear targets, making this a methodical rather than insightful problem—moderately harder than average A-level due to FP2 content but routine for that syllabus. |
| Spec | 1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{dy}{dx} = \frac{1}{\tanh \frac{x}{2}} \cdots\) | B1 | |
| \(\operatorname{sech}^2 \frac{x}{2} \cdots\) | B1 | |
| \(\frac{1}{2}\) | B1 | |
| \(= \frac{1}{\sinh \frac{x}{2} \cdot \frac{x}{2} \cosh^2 \frac{x}{2}}\) | M1 | OE ie expressing in \(\sinh \frac{x}{2}\) and \(\cosh \frac{x}{2}\) |
| \(= \frac{1}{2\sinh \frac{x}{2} \cosh \frac{x}{2}}\) | M1 | ie use of \(\sinh 2A = 2\sinh A \cosh A\) |
| \(= \frac{1}{\sinh x}\) | M1 | |
| \(= \operatorname{cosech} x\) | A1 | 6 marks; AG |
| Alternative: | ||
| In \(\sinh \frac{x}{2} - \ln \cosh \frac{x}{2}\) | B1 | |
| \(\frac{1}{2} \frac{\cosh \frac{x}{2}}{2} - \frac{1}{2} \frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}\) | B1B1 | |
| \(\frac{\cosh^2 \frac{x}{2} - \sinh^2 \frac{x}{2}}{2\sinh \frac{x}{2} \cosh \frac{x}{2}}\) | M1 | |
| Use of \(\sinh 2A = 2\sinh A \cosh A\) result | M1, A1 | |
| (b)(i) \(s = \int_1^2 \sqrt{1 + \operatorname{cosech}^2 x} \, dx\) | M1 | |
| \(= \int_1^2 \coth x \, dx\) | A1 | 2 marks; AG |
| (ii) \(s = [\ln \sinh x]_1^2\) | M1 | needs to be correct |
| \(= \ln \sinh 2 - \ln \sinh 1\) | A1 | |
| \(= \ln \frac{2\sinh 1\cosh 1}{\sinh 1}\) | A1F | must be seen |
| \(= \ln(2\cosh 1)\) | A1 | 4 marks; AG |
**(a)** $\frac{dy}{dx} = \frac{1}{\tanh \frac{x}{2}} \cdots$ | B1 |
$\operatorname{sech}^2 \frac{x}{2} \cdots$ | B1 |
$\frac{1}{2}$ | B1 |
$= \frac{1}{\sinh \frac{x}{2} \cdot \frac{x}{2} \cosh^2 \frac{x}{2}}$ | M1 | OE ie expressing in $\sinh \frac{x}{2}$ and $\cosh \frac{x}{2}$
$= \frac{1}{2\sinh \frac{x}{2} \cosh \frac{x}{2}}$ | M1 | ie use of $\sinh 2A = 2\sinh A \cosh A$
$= \frac{1}{\sinh x}$ | M1 |
$= \operatorname{cosech} x$ | A1 | 6 marks; AG
**Alternative:** | |
In $\sinh \frac{x}{2} - \ln \cosh \frac{x}{2}$ | B1 |
$\frac{1}{2} \frac{\cosh \frac{x}{2}}{2} - \frac{1}{2} \frac{\sinh \frac{x}{2}}{\cosh \frac{x}{2}}$ | B1B1 |
$\frac{\cosh^2 \frac{x}{2} - \sinh^2 \frac{x}{2}}{2\sinh \frac{x}{2} \cosh \frac{x}{2}}$ | M1 |
Use of $\sinh 2A = 2\sinh A \cosh A$ result | M1, A1 |
**(b)(i)** $s = \int_1^2 \sqrt{1 + \operatorname{cosech}^2 x} \, dx$ | M1 |
$= \int_1^2 \coth x \, dx$ | A1 | 2 marks; AG
**(ii)** $s = [\ln \sinh x]_1^2$ | M1 | needs to be correct
$= \ln \sinh 2 - \ln \sinh 1$ | A1 |
$= \ln \frac{2\sinh 1\cosh 1}{\sinh 1}$ | A1F | must be seen
$= \ln(2\cosh 1)$ | A1 | 4 marks; AG
**Question 7 Total: 12 marks**
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**TOTAL MARKS: 75**7
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \ln \tanh \frac { x } { 2 }$, where $x > 0$, show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \operatorname { cosech } x$$
\item A curve has equation $y = \ln \tanh \frac { x } { 2 }$, where $x > 0$. The length of the arc of the curve between the points where $x = 1$ and $x = 2$ is denoted by $s$.
\begin{enumerate}[label=(\roman*)]
\item Show that
$$s = \int _ { 1 } ^ { 2 } \operatorname { coth } x \mathrm {~d} x$$
\item Hence show that $s = \ln ( 2 \cosh 1 )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q7 [12]}}