| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Standard +0.3 This is a straightforward substitution problem where u = cosh²x converts the equation to a quadratic in u. Students must solve the quadratic, reject negative solutions (since cosh²x ≥ 1), then use the definition cosh x = (e^x + e^(-x))/2 to find x. While it requires multiple steps and careful algebraic manipulation to reach the ln 2 form, the technique is standard for Core Pure 1 hyperbolic function questions with no novel insight required. |
| Spec | 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Solves quadratic in \(\cosh^2 x\), e.g. \((8\cosh^2 x - 9)(8\cosh^2 x + 1) = 0\) | M1 | 3.1a — any valid method; calculator: accept reaching positive value for \(\cosh^2 x\) |
| \(\cosh^2 x = \dfrac{9}{8}\) | A1 | 1.1b — ignore negative/incorrect extra roots |
| \(\cosh x = \dfrac{3}{4}\sqrt{2} \Rightarrow x = \ln\!\left[\dfrac{3}{4}\sqrt{2} + \sqrt{\left(\dfrac{3}{4}\sqrt{2}\right)^2 - 1}\right]\) | M1 | 1.1b — takes positive square root and uses correct \(\text{arcosh}\) formula; Alt 2: solving quadratic in \(e^{4x}\) |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a — both values, no others; must be in form specified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(64\cosh^2 x(\cosh^2 x -1)-9=0 \Rightarrow 16\sinh^2 2x = \dfrac{9}{16}\) | M1, A1 | 3.1a, 1.1b |
| \(\sinh 2x = \pm\dfrac{3}{4} \Rightarrow x = \dfrac{1}{2}\ln\!\left[\pm\dfrac{3}{4} + \sqrt{\dfrac{9}{16}+1}\right]\) | M1 | 1.1b |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(64\!\left(\dfrac{e^x+e^{-x}}{2}\right)^4 - 64\!\left(\dfrac{e^x+e^{-x}}{2}\right)^2 - 9 = 0\) expanded | M1 | 3.1a |
| \(4e^{4x} - 17 + 4e^{-4x} = 0\) | A1 | 1.1b |
| \((4e^{4x}-1)(1-4e^{-4x})=0 \Rightarrow e^{4x} = \ldots \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a |
# Question 2:
## Main Scheme
| Answer | Mark | Guidance |
|--------|------|----------|
| Solves quadratic in $\cosh^2 x$, e.g. $(8\cosh^2 x - 9)(8\cosh^2 x + 1) = 0$ | M1 | 3.1a — any valid method; calculator: accept reaching positive value for $\cosh^2 x$ |
| $\cosh^2 x = \dfrac{9}{8}$ | A1 | 1.1b — ignore negative/incorrect extra roots |
| $\cosh x = \dfrac{3}{4}\sqrt{2} \Rightarrow x = \ln\!\left[\dfrac{3}{4}\sqrt{2} + \sqrt{\left(\dfrac{3}{4}\sqrt{2}\right)^2 - 1}\right]$ | M1 | 1.1b — takes positive square root and uses correct $\text{arcosh}$ formula; Alt 2: solving quadratic in $e^{4x}$ |
| $x = \pm\dfrac{1}{2}\ln 2$ | A1 | 2.2a — both values, no others; must be in form specified |
## Alt 1
| Answer | Mark | Guidance |
|--------|------|----------|
| $64\cosh^2 x(\cosh^2 x -1)-9=0 \Rightarrow 16\sinh^2 2x = \dfrac{9}{16}$ | M1, A1 | 3.1a, 1.1b |
| $\sinh 2x = \pm\dfrac{3}{4} \Rightarrow x = \dfrac{1}{2}\ln\!\left[\pm\dfrac{3}{4} + \sqrt{\dfrac{9}{16}+1}\right]$ | M1 | 1.1b |
| $x = \pm\dfrac{1}{2}\ln 2$ | A1 | 2.2a |
## Alt 2
| Answer | Mark | Guidance |
|--------|------|----------|
| $64\!\left(\dfrac{e^x+e^{-x}}{2}\right)^4 - 64\!\left(\dfrac{e^x+e^{-x}}{2}\right)^2 - 9 = 0$ expanded | M1 | 3.1a |
| $4e^{4x} - 17 + 4e^{-4x} = 0$ | A1 | 1.1b |
| $(4e^{4x}-1)(1-4e^{-4x})=0 \Rightarrow e^{4x} = \ldots \Rightarrow x = \ldots$ | M1 | 1.1b |
| $x = \pm\dfrac{1}{2}\ln 2$ | A1 | 2.2a |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
Determine the values of $x$ for which
$$64 \cosh ^ { 4 } x - 64 \cosh ^ { 2 } x - 9 = 0$$
Give your answers in the form $q \ln 2$ where $q$ is rational and in simplest form.
\hfill \mbox{\textit{Edexcel CP1 2022 Q2 [4]}}