| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve mixed sinh/cosh linear combinations |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard hyperbolic function definitions and algebraic manipulation. Part (a) is direct substitution of sinh x = (e^x - e^{-x})/2 and cosh x = (e^x + e^{-x})/2, while part (b) involves solving a quadratic in e^x. The techniques are routine for FP2 students with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| \(9\sinh x - \cosh x = 9 \cdot \frac{e^x - e^{-x}}{2} - \frac{e^x + e^{-x}}{2}\) | M1 | Using definitions of sinh and cosh |
| \(= \frac{9e^x - 9e^{-x} - e^x - e^{-x}}{2} = \frac{8e^x - 10e^{-x}}{2} = 4e^x - 5e^{-x}\) | A1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| \(4e^x - 5e^{-x} = 8\) | M1 | Using result from (a) |
| \(4e^{2x} - 8e^x - 5 = 0\) | M1 | Multiplying through by \(e^x\) |
| \((2e^x - 5)(2e^x + 1) = 0\) | M1 | Factorising |
| \(e^x = \frac{5}{2}\) (only valid solution since \(e^x > 0\)) | A1 | Correct value with reason |
| \(\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\frac{5}{2} - \frac{2}{5}}{2} = \frac{21}{20}\) | M1 | Finding sinh x or cosh x |
| \(\cosh x = \frac{e^x + e^{-x}}{2} = \frac{\frac{5}{2} + \frac{2}{5}}{2} = \frac{29}{20}\) | A1 | Correct values |
| \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{21}{29}\) | A1 | Correct final answer |
# Question 1:
## Part (a):
| $9\sinh x - \cosh x = 9 \cdot \frac{e^x - e^{-x}}{2} - \frac{e^x + e^{-x}}{2}$ | M1 | Using definitions of sinh and cosh |
|---|---|---|
| $= \frac{9e^x - 9e^{-x} - e^x - e^{-x}}{2} = \frac{8e^x - 10e^{-x}}{2} = 4e^x - 5e^{-x}$ | A1 | Correct completion |
## Part (b):
| $4e^x - 5e^{-x} = 8$ | M1 | Using result from (a) |
|---|---|---|
| $4e^{2x} - 8e^x - 5 = 0$ | M1 | Multiplying through by $e^x$ |
| $(2e^x - 5)(2e^x + 1) = 0$ | M1 | Factorising |
| $e^x = \frac{5}{2}$ (only valid solution since $e^x > 0$) | A1 | Correct value with reason |
| $\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\frac{5}{2} - \frac{2}{5}}{2} = \frac{21}{20}$ | M1 | Finding sinh x or cosh x |
| $\cosh x = \frac{e^x + e^{-x}}{2} = \frac{\frac{5}{2} + \frac{2}{5}}{2} = \frac{29}{20}$ | A1 | Correct values |
| $\tanh x = \frac{\sinh x}{\cosh x} = \frac{21}{29}$ | A1 | Correct final answer |
---1 (a) Show that
$$9 \sinh x - \cosh x = 4 \mathrm { e } ^ { x } - 5 \mathrm { e } ^ { - x }$$
(b) Given that
$$9 \sinh x - \cosh x = 8$$
find the exact value of $\tanh x$.
\hfill \mbox{\textit{AQA FP2 2010 Q1 [9]}}