| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring (a) algebraic manipulation of standard hyperbolic definitions to prove an identity, and (b) solving an equation using the proven identity and substitution. While it involves multiple steps and is from FP2, the techniques are routine for students at this level—no novel insight required, just systematic application of definitions and algebraic manipulation. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| - \(\text{LHS} = 1 + \frac{1}{2}\left(e^{2\theta} - 2 + e^{-2\theta}\right)\) | M1 | Expansion of \(\frac{1}{2}\left(e^{\theta} - e^{-\theta}\right)^2\) correctly |
| - \(= \frac{1}{2}\left(e^{2\theta} + e^{-2\theta}\right) = \cosh 2\theta\) | A1, A1 | Any form; AG |
| Answer | Marks | Guidance |
|---|---|---|
| - \(3 + 6\sinh^2\theta = 2\sinh\theta + 11\) | M1 | |
| - \(3\sinh^2\theta - \sinh\theta - 4 = 0\) | A1 | OE |
| - \((3\sinh\theta - 4)(\sinh\theta + 1) = 0\) | M1 | Attempt to factorise or formula |
| - \(\sinh\theta = \frac{4}{3}\) or \(-1\) | A1F | ft if factorise or real roots found |
| - \(\theta = \ln 3\) | A1F | |
| - \(\theta = \ln\left(\sqrt{2} - 1\right)\) | A1F |
**(a)**
- $\text{LHS} = 1 + \frac{1}{2}\left(e^{2\theta} - 2 + e^{-2\theta}\right)$ | M1 | Expansion of $\frac{1}{2}\left(e^{\theta} - e^{-\theta}\right)^2$ correctly
- $= \frac{1}{2}\left(e^{2\theta} + e^{-2\theta}\right) = \cosh 2\theta$ | A1, A1 | Any form; AG
- **Total: 3 marks**
**(b)**
- $3 + 6\sinh^2\theta = 2\sinh\theta + 11$ | M1 |
- $3\sinh^2\theta - \sinh\theta - 4 = 0$ | A1 | OE
- $(3\sinh\theta - 4)(\sinh\theta + 1) = 0$ | M1 | Attempt to factorise or formula
- $\sinh\theta = \frac{4}{3}$ or $-1$ | A1F | ft if factorise or real roots found
- $\theta = \ln 3$ | A1F |
- $\theta = \ln\left(\sqrt{2} - 1\right)$ | A1F |
- **Total: 6 marks**
**Question 1 Total: 9 marks**
---1
\begin{enumerate}[label=(\alph*)]
\item Use the definitions $\sinh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } - \mathrm { e } ^ { - \theta } \right)$ and $\cosh \theta = \frac { 1 } { 2 } \left( \mathrm { e } ^ { \theta } + \mathrm { e } ^ { - \theta } \right)$ to show that
$$1 + 2 \sinh ^ { 2 } \theta = \cosh 2 \theta$$
\item Solve the equation
$$3 \cosh 2 \theta = 2 \sinh \theta + 11$$
giving each of your answers in the form $\ln p$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q1 [9]}}