| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using sech/tanh identities |
| Difficulty | Standard +0.3 This is a structured Further Maths question testing standard hyperbolic function techniques: sketching tanh, deriving the inverse formula (routine manipulation of exponentials), using the sech²x = 1 - tanh²x identity to form a quadratic, and solving it. While it requires multiple steps and is from FP2, each part follows predictable patterns with clear signposting, making it slightly easier than average overall. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Sketch, approximately correct shape | B1 | |
| Asymptotes at \(y = \pm 1\) | B1 | 2 marks |
| (b) Use of \(u = \frac{\sinh x}{\cosh x}\) | M1 | |
| \(= \frac{e^x - e^{-x}}{e^x + e^{-x}}\) or \(\frac{e^{2x} - 1}{e^{2x} + 1}\) | A1 | |
| \(u\left(e^x + e^{-x}\right) = e^x - e^{-x}\) | M1 | M1 for multiplying up |
| \(e^{-x}(1 + u) = e^x(1 - u)\) | A1 | A1 for factorizing out \(e\)'s or M1 for attempt at \(1 + u\) and \(1 - u\) in terms of \(e^x\) |
| \(e^{2x} = \frac{1 + u}{1 - u}\) | m1 | |
| \(x = \frac{1}{2}\ln\left(\frac{1 + u}{1 - u}\right)\) | A1 | 6 marks |
| (c)(i) Use of \(\tanh^2 x = 1 - \operatorname{sech}^2 x\) | M1 | |
| Printed answer | A1 | 2 marks |
| (ii) \((3\tanh x - 1)(\tanh x - 2) = 0\) | M1 | Attempt to factorise. Accept \(\tanh x \neq 2\) written down but not ignored or just crossed out |
| \(\tanh x = \frac{1}{3}\) | A1 | |
| \(x = \frac{1}{2}\ln 2\) | M1, A1F | 5 marks |
**(a)** Sketch, approximately correct shape | B1 |
Asymptotes at $y = \pm 1$ | B1 | 2 marks | B0 if curve touches asymptotes. Lines of answer booklet could be used for asymptotes. be strict with sketch
**(b)** Use of $u = \frac{\sinh x}{\cosh x}$ | M1 |
$= \frac{e^x - e^{-x}}{e^x + e^{-x}}$ or $\frac{e^{2x} - 1}{e^{2x} + 1}$ | A1 |
$u\left(e^x + e^{-x}\right) = e^x - e^{-x}$ | M1 | M1 for multiplying up
$e^{-x}(1 + u) = e^x(1 - u)$ | A1 | A1 for factorizing out $e$'s or M1 for attempt at $1 + u$ and $1 - u$ in terms of $e^x$
$e^{2x} = \frac{1 + u}{1 - u}$ | m1 |
$x = \frac{1}{2}\ln\left(\frac{1 + u}{1 - u}\right)$ | A1 | 6 marks | AG
**(c)(i)** Use of $\tanh^2 x = 1 - \operatorname{sech}^2 x$ | M1 |
Printed answer | A1 | 2 marks
**(ii)** $(3\tanh x - 1)(\tanh x - 2) = 0$ | M1 | Attempt to factorise. Accept $\tanh x \neq 2$ written down but not ignored or just crossed out
$\tanh x = \frac{1}{3}$ | A1 |
$x = \frac{1}{2}\ln 2$ | M1, A1F | 5 marks | ft
**Total: 15 marks**4
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = \tanh x$.
\item Given that $u = \tanh x$, use the definitions of $\sinh x$ and $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$ to show that
$$x = \frac { 1 } { 2 } \ln \left( \frac { 1 + u } { 1 - u } \right)$$
\item \begin{enumerate}[label=(\roman*)]
\item Show that the equation
$$3 \operatorname { sech } ^ { 2 } x + 7 \tanh x = 5$$
can be written as
$$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$
\item Show that the equation
$$3 \tanh ^ { 2 } x - 7 \tanh x + 2 = 0$$
has only one solution for $x$.\\
Find this solution in the form $\frac { 1 } { 2 } \ln a$, where $a$ is an integer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q4 [15]}}