| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve mixed sinh/cosh linear combinations |
| Difficulty | Standard +0.3 Part (a) is a standard bookwork proof requiring direct substitution of exponential definitions—routine for Further Maths students. Part (b) involves substituting definitions, forming a quadratic in e^x, and solving for x using logarithms. While it requires multiple steps and algebraic manipulation, this is a standard technique for hyperbolic equations covered in F3, making it slightly above average difficulty overall but well within expected scope. |
| 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 |
| \(\{\frac{1}{2}(e^x+e^{-x})\}^2 - \{\frac{1}{2}(e^x-e^{-x})\}^2 = \{\frac{1}{4}(e^{2x}+2+e^{-2x})\} - \{\frac{1}{4}(e^{2x}-2+e^{-2x})\}\) | M1 | Uses correct exponential forms for cosh and sinh and squares both brackets obtaining 3 terms each time |
| \(= \frac{1}{2}+\frac{1}{2}=1\) | A1 | At least one line of intermediate working (e.g. combines fractions with common denominator) with no errors seen, and concludes \(=1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((e^x-e^{-x})+7\times\frac{1}{2}(e^x+e^{-x})=9 \Rightarrow \frac{9}{2}e^x+\frac{5}{2}e^{-x}-9=0\) | M1A1 | M1: Uses exponential forms and collects terms. A1: Any correct form with terms collected |
| \(\Rightarrow 9e^{2x}-18e^x+5=0\) so \(e^x = \ldots\) | M1 | Solves their three term quadratic in \(e^x\) as far as \(e^x=\ldots\) |
| \(e^x = \frac{1}{3}\) or \(\frac{5}{3}\) | A1 | Both values correct |
| \(x = \ln\frac{1}{3}\) and \(\ln\frac{5}{3}\) | A1 | Both values correct (accept equivalents) |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\{\frac{1}{2}(e^x+e^{-x})\}^2 - \{\frac{1}{2}(e^x-e^{-x})\}^2 = \{\frac{1}{4}(e^{2x}+2+e^{-2x})\} - \{\frac{1}{4}(e^{2x}-2+e^{-2x})\}$ | M1 | Uses correct exponential forms for cosh and sinh and squares both brackets obtaining 3 terms each time |
| $= \frac{1}{2}+\frac{1}{2}=1$ | A1 | At least one line of intermediate working (e.g. combines fractions with common denominator) with no errors seen, and concludes $=1$ |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(e^x-e^{-x})+7\times\frac{1}{2}(e^x+e^{-x})=9 \Rightarrow \frac{9}{2}e^x+\frac{5}{2}e^{-x}-9=0$ | M1A1 | M1: Uses exponential forms **and** collects terms. A1: Any correct form with terms collected |
| $\Rightarrow 9e^{2x}-18e^x+5=0$ so $e^x = \ldots$ | M1 | Solves their three term quadratic in $e^x$ as far as $e^x=\ldots$ |
| $e^x = \frac{1}{3}$ or $\frac{5}{3}$ | A1 | Both values correct |
| $x = \ln\frac{1}{3}$ and $\ln\frac{5}{3}$ | A1 | Both values correct (accept equivalents) |
---3. Using the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials,
\begin{enumerate}[label=(\alph*)]
\item prove that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1$$
\item find algebraically the exact solutions of the equation
$$2 \sinh x + 7 \cosh x = 9$$
giving your answers as natural logarithms.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2014 Q3 [7]}}