| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.8 This requires differentiating hyperbolic functions (including chain rule for sinh 2x), solving a quadratic equation in cosh x using the identity cosh²x - sinh²x = 1, then applying inverse hyperbolic functions. It's a multi-step problem requiring fluency with hyperbolic identities and algebraic manipulation, placing it moderately above average difficulty but still within standard Further Maths territory. |
| Spec | 4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 31\cosh x - 4\cosh 2x\) | B1 | Correct differentiation |
| \(\frac{dy}{dx} = 31\cosh x - 4(2\cosh^2 x - 1)\) | M1 | Identifies a correct approach by using a correct identity to make progress to obtain a quadratic in \(\cosh x\) |
| \(8\cosh^2 x - 31\cosh x - 4 = 0\) | A1 | Correct 3 term quadratic obtained |
| \((8\cosh x + 1)(\cosh x - 4) = 0 \Rightarrow \cosh x = \ldots\) | M1 | Solves their 3TQ |
| \(\cosh x = 4, \left(-\frac{1}{8}\right)\) | A1 | Correct values (may only see 4 here) |
| \(\cosh x = \alpha \Rightarrow x = \ln\!\left(\alpha + \sqrt{\alpha^2-1}\right)\) or equivalent exponential approach | M1 | Correct process to reach at least one value for \(x\) from their \(\cosh x\) |
| \(\pm\ln\!\left(4 + \sqrt{15}\right)\) or \(\ln\!\left(4 \pm \sqrt{15}\right)\) | A1 | Deduces the correct 2 values with no incorrect values or work involving \(\cosh x = -\frac{1}{8}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 31\cosh x - 4\cosh 2x\) or exponential form | B1 | Correct differentiation |
| Using exponential forms for \(\cosh x\) and \(\sinh x\), forms a quartic equation in \(e^x\) with all terms simplified and all on one side | M1, A1 | Correct quartic equation for \(e^x\) leading to \(4e^{4x} - 31e^{3x} - 31e^x + 4 = 0\) |
| Solves \(4e^{4x} - 31e^{3x} - 31e^x + 4 = 0\), \(\Rightarrow e^x = \ldots\) | M1 | Solves their quartic equation in \(e^x\) |
| \(e^x = 4 \pm \sqrt{15}\) or awrt \(7.87, 0.13\) | A1 | Correct values to two decimal places or exact values |
| \(x = \ln(b)\) where \(b\) is a real exact value | M1 | \(x = \ln(b)\) where \(b\) is a real exact value |
| \(\ln\!\left(4 \pm \sqrt{15}\right)\) | A1 | Deduces the correct 2 values only |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 31\cosh x - 4\cosh 2x$ | B1 | Correct differentiation |
| $\frac{dy}{dx} = 31\cosh x - 4(2\cosh^2 x - 1)$ | M1 | Identifies a correct approach by using a correct identity to make progress to obtain a quadratic in $\cosh x$ |
| $8\cosh^2 x - 31\cosh x - 4 = 0$ | A1 | Correct 3 term quadratic obtained |
| $(8\cosh x + 1)(\cosh x - 4) = 0 \Rightarrow \cosh x = \ldots$ | M1 | Solves their 3TQ |
| $\cosh x = 4, \left(-\frac{1}{8}\right)$ | A1 | Correct values (may only see 4 here) |
| $\cosh x = \alpha \Rightarrow x = \ln\!\left(\alpha + \sqrt{\alpha^2-1}\right)$ or equivalent exponential approach | M1 | Correct process to reach at least one value for $x$ from their $\cosh x$ |
| $\pm\ln\!\left(4 + \sqrt{15}\right)$ or $\ln\!\left(4 \pm \sqrt{15}\right)$ | A1 | Deduces the correct 2 values with no incorrect values or work involving $\cosh x = -\frac{1}{8}$ |
**Total: 7 marks**
---
### Alternative Method:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 31\cosh x - 4\cosh 2x$ or exponential form | B1 | Correct differentiation |
| Using exponential forms for $\cosh x$ and $\sinh x$, forms a quartic equation in $e^x$ with all terms simplified and all on one side | M1, A1 | Correct quartic equation for $e^x$ leading to $4e^{4x} - 31e^{3x} - 31e^x + 4 = 0$ |
| Solves $4e^{4x} - 31e^{3x} - 31e^x + 4 = 0$, $\Rightarrow e^x = \ldots$ | M1 | Solves their quartic equation in $e^x$ |
| $e^x = 4 \pm \sqrt{15}$ or awrt $7.87, 0.13$ | A1 | Correct values to two decimal places or exact values |
| $x = \ln(b)$ where $b$ is a real exact value | M1 | $x = \ln(b)$ where $b$ is a real exact value |
| $\ln\!\left(4 \pm \sqrt{15}\right)$ | A1 | Deduces the correct 2 values only |
**Total: 7 marks**\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = 31 \sinh x - 2 \sinh 2 x \quad x \in \mathbb { R }$$
Determine, in terms of natural logarithms, the exact $x$ coordinates of the stationary points of $C$.
\hfill \mbox{\textit{Edexcel CP2 2020 Q1 [7]}}