| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| 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 the standard identity cosh²x = 1 + sinh²x to convert to a quadratic in sinh x, then solving and using the inverse hyperbolic function formula. While it's Further Maths content, it's a routine textbook exercise with a clear method and no novel insight required. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(2(1+\sinh^2 x) - 3\sinh x = 1\) | M1 | Attempt to use \(\cosh^2 x = 1 + \sinh^2 x\) |
| \(2\sinh^2 x - 3\sinh x + 1 = 0\) | A1 | Correct 3 term quadratic. "\(= 0\)" may be implied |
| \((2\sinh x - 1)(\sinh x - 1) = 0\) | M1 | Attempts to solve their \(3TQ = 0\) leading to \(\sinh x = \ldots\) |
| \(\sinh x = \frac{1}{2}\) or \(1\) | A1 | Both values correct |
| \(x = \ln\frac{1}{2}(1+\sqrt{5}),\ \ln(1+\sqrt{2})\) | A1, A1 (M1A1 on ePEN) | A1: one correct value; A1: both correct values and no other values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(2\left(\frac{e^x+e^{-x}}{2}\right)^2 - 3\left(\frac{e^x-e^{-x}}{2}\right) = 1\) | M1 | Substitutes correct definitions for \(\sinh x\) and \(\cosh x\) in terms of exponentials |
| \(e^{4x} - 3e^{3x} + 3e^x + 1 = 0\) | A1 | Correct quartic in \(e^x\) |
| \((e^{2x}-e^x-1)(e^{2x}-2e^x-1)=0 \Rightarrow e^x = \ldots\) | M1 | Solves quartic; for correct quartic must be recognisable attempt e.g. product of two 3TQs in \(e^x\) |
| \(e^x = \frac{1+\sqrt{5}}{2},\ \frac{2+\sqrt{8}}{2}\) | A1 | Correct values for \(e^x\); allow awrt 1.62, 2.41 but no incorrect values |
| \(x = \ln\frac{1}{2}(1+\sqrt{5}),\ \ln(1+\sqrt{2})\) | A1, A1 (M1A1 on ePEN) | A1: one correct; A1: both correct and no other values; allow awrt 3SF |
## Question 1:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $2(1+\sinh^2 x) - 3\sinh x = 1$ | M1 | Attempt to use $\cosh^2 x = 1 + \sinh^2 x$ |
| $2\sinh^2 x - 3\sinh x + 1 = 0$ | A1 | Correct 3 term quadratic. "$= 0$" may be implied |
| $(2\sinh x - 1)(\sinh x - 1) = 0$ | M1 | Attempts to solve their $3TQ = 0$ leading to $\sinh x = \ldots$ |
| $\sinh x = \frac{1}{2}$ or $1$ | A1 | Both values correct |
| $x = \ln\frac{1}{2}(1+\sqrt{5}),\ \ln(1+\sqrt{2})$ | A1, A1 (M1A1 on ePEN) | A1: one correct value; A1: both correct values and **no other values** |
**Alternative:**
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $2\left(\frac{e^x+e^{-x}}{2}\right)^2 - 3\left(\frac{e^x-e^{-x}}{2}\right) = 1$ | M1 | Substitutes correct definitions for $\sinh x$ and $\cosh x$ in terms of exponentials |
| $e^{4x} - 3e^{3x} + 3e^x + 1 = 0$ | A1 | Correct quartic in $e^x$ |
| $(e^{2x}-e^x-1)(e^{2x}-2e^x-1)=0 \Rightarrow e^x = \ldots$ | M1 | Solves quartic; for correct quartic must be recognisable attempt e.g. product of two 3TQs in $e^x$ |
| $e^x = \frac{1+\sqrt{5}}{2},\ \frac{2+\sqrt{8}}{2}$ | A1 | Correct values for $e^x$; allow awrt 1.62, 2.41 but no incorrect values |
| $x = \ln\frac{1}{2}(1+\sqrt{5}),\ \ln(1+\sqrt{2})$ | A1, A1 (M1A1 on ePEN) | A1: one correct; A1: both correct and **no other values**; allow awrt 3SF |
---\begin{enumerate}
\item Solve the equation
\end{enumerate}
$$2 \cosh ^ { 2 } x - 3 \sinh x = 1$$
giving your answers in terms of natural logarithms.\\
\hfill \mbox{\textit{Edexcel FP3 2015 Q1 [6]}}