| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Surface area of revolution with hyperbolics |
| Difficulty | Challenging +1.3 Part (a) is routine manipulation using the exponential definition of cosh. Part (b)(i) requires knowing the surface area formula and that d/dx(cosh x) = sinh x, then using cosh²x - sinh²x = 1. Part (b)(ii) involves integrating cosh²x using the double angle result from (a), which is a standard technique. This is a well-structured multi-part question testing standard FP2 techniques with clear scaffolding, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 18.06d |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) \((e^x + e^{-x})^2\) expanded correctly | B1 | \(e^{2x} + 2e^x + e^{-2x}\) is acceptable |
| Result | B1 | 2 marks; AG |
| 5(b)(i) \(\frac{dy}{dx} = \sinh x\) | B1 | |
| \(\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \sinh^2 x} = \cosh x\) | M1 | use of \(\cosh^2 x - \sinh^2 x = 1\) |
| \(S = 2\pi \int_0^{ln a} \cosh x \, dx\) | A1 | 3 marks; AG (clearly derived) |
| 5(b)(ii) Use of \(\cosh^2 x = \frac{1}{2}(1 + \cosh 2x)\) | M1 | allow one slip in formula; M0 if \(\int \cosh^2 x \, dx\) is given as \(\sinh^2 x\) |
| \(S = \pi\left[x + \frac{1}{2}\sinh 2x\right]_0^{ln a}\) | A1 | |
| \(= \pi\left[\ln a + \frac{1}{2}\left(\frac{e^{2\ln a} - e^{-2\ln a}}{2}\right)\right]\) | M1 | |
| \(= \pi\left[\ln a + \frac{1}{4}(a^2 - a^{-2})\right]\) | A1F | |
| \(= \pi\left[\ln a + \frac{1}{4a^2}(a^4 - 1)\right]\) | A1 | 5 marks (10 marks for question); AG |
**5(a)** $(e^x + e^{-x})^2$ expanded correctly | B1 | $e^{2x} + 2e^x + e^{-2x}$ is acceptable
Result | B1 | 2 marks; AG
**5(b)(i)** $\frac{dy}{dx} = \sinh x$ | B1
$\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \sinh^2 x} = \cosh x$ | M1 | use of $\cosh^2 x - \sinh^2 x = 1$
$S = 2\pi \int_0^{ln a} \cosh x \, dx$ | A1 | 3 marks; AG (clearly derived)
**5(b)(ii)** Use of $\cosh^2 x = \frac{1}{2}(1 + \cosh 2x)$ | M1 | allow one slip in formula; M0 if $\int \cosh^2 x \, dx$ is given as $\sinh^2 x$
$S = \pi\left[x + \frac{1}{2}\sinh 2x\right]_0^{ln a}$ | A1
$= \pi\left[\ln a + \frac{1}{2}\left(\frac{e^{2\ln a} - e^{-2\ln a}}{2}\right)\right]$ | M1
$= \pi\left[\ln a + \frac{1}{4}(a^2 - a^{-2})\right]$ | A1F
$= \pi\left[\ln a + \frac{1}{4a^2}(a^4 - 1)\right]$ | A1 | 5 marks (10 marks for question); AG5
\begin{enumerate}[label=(\alph*)]
\item Use the definition $\cosh x = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right)$ to show that $\cosh 2 x = 2 \cosh ^ { 2 } x - 1$.\\
(2 marks)
\item \begin{enumerate}[label=(\roman*)]
\item The arc of the curve $y = \cosh x$ between $x = 0$ and $x = \ln a$ is rotated through $2 \pi$ radians about the $x$-axis. Show that $S$, the surface area generated, is given by
$$S = 2 \pi \int _ { 0 } ^ { \ln a } \cosh ^ { 2 } x \mathrm {~d} x$$
\item Hence show that
$$S = \pi \left( \ln a + \frac { a ^ { 4 } - 1 } { 4 a ^ { 2 } } \right)$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q5 [10]}}