A2 June 2025 Paper 1 Q2
2.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Determine the exact values of \(x\) for which
\[\sinh 2x = 3\sinh x\](5)
| Scheme | Marks | AO |
|---|---|---|
| Uses the identity \(\sinh 2x = 2\sinh x\cosh x\) to find a value for \(\sinh x\) or \(\cosh x\) \(2\sinh x\cosh x = 3\sinh x \Rightarrow 2\sinh x\cosh x - 3\sinh x = 0\) \(\sinh x(2\cosh x - 3) = 0 \Rightarrow \sinh x = \ldots\) or \(\cosh x = \ldots\) | M1 | 3.1a |
| \(\sinh x = 0 \Rightarrow x = 0\) | B1 | 1.1b |
| \(\cosh x = \text{``}\dfrac{3}{2}\text{''} \Rightarrow x = \ln\left[\text{``}\dfrac{3}{2}\text{''} + \sqrt{\left(\text{``}\dfrac{3}{2}\text{''}\right)^2 - 1}\right]\) Alternatively \(\cosh x = \dfrac{3}{2} \Rightarrow \dfrac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right) \Rightarrow \mathrm{e}^{2x} - 3\mathrm{e}^x + 1 = 0\) \(\Rightarrow \mathrm{e}^x = \dfrac{3 + \sqrt{5}}{2}\) or \(\dfrac{3 - \sqrt{5}}{2} \Rightarrow x = \ln\ldots\) | dM1 | 1.1b |
| \(x = \ln\left[\dfrac{3}{2} + \sqrt{\left(\dfrac{3}{2}\right)^2 - 1}\right]\) | A1 | 1.1b |
| \(x = \pm\ln\left(\dfrac{3 + \sqrt{5}}{2}\right)\) or \(x = \ln\left(\dfrac{3 \pm \sqrt{5}}{2}\right)\) | A1 | 2.2a |
| (5) | ||
| (5 marks) |
Notes
M1: Uses the identity \(\sinh 2x = 2\sinh x\cosh x\) and proceeds to find a value for \(\sinh x\) or \(\cosh x\)
B1: \(x = 0\)
dM1: Uses the correct formula for \(\operatorname{arcosh} x\) with their value of \(\cosh x\) to find a value for \(x\) as a natural logarithm or alternatively gives awrt to 3sf. e.g. 0.962, \(-0.962\). You may need to check their workings.
Alternatively uses the exponential definition for \(\cosh x\), forms and solves a quadratic for \(\mathrm{e}^x\) leading to find a value for \(x\) as a natural logarithm. Usual rules apply for solving a quadratic by any means including using a calculator. You may need to check their workings. If correct answers are given without workings, you may award this mark.
A1: Deduces one correct exact value for \(x\). Accept exact unsimplified equivalents.
A1: Deduces both correct exact values for \(x\). Accept exact simplified equivalents. isw
Accept, for example \(x = \pm\ln\left(\dfrac{3}{2} + \dfrac{\sqrt{5}}{2}\right)\) and \(x = \pm\ln\left(\dfrac{3}{2} + \sqrt{\dfrac{5}{4}}\right)\)
Also accept equivalent answers which you may need to check; their may be several alternatives.
e.g. Accept \(x = \ln\left(\dfrac{2}{3 + \sqrt{5}}\right)\) for \(x = \ln\left(\dfrac{3 - \sqrt{5}}{2}\right)\)
Alternative method use of exponentials
| Scheme | Marks | AO |
|---|---|---|
| Uses the identities \(\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) and \(\sinh 2x = \dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2}\) \(\dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} = 3\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) leading to a quartic equation for \(\mathrm{e}^x\) (Note correct equation is \(\mathrm{e}^{4x} - 3\mathrm{e}^{3x} + 3\mathrm{e}^x - 1 = 0\)) | M1 | 3.1a |
| \(\mathrm{e}^x = 1 \Rightarrow x = 0\) | B1 | 1.1b |
| Solves quartic to find a value for \(\mathrm{e}^x\) leading to \(x = \ln\ldots\) | dM1 | 1.1b |
| \(x = \ln\left[\dfrac{3}{2} + \sqrt{\left(\dfrac{3}{2}\right)^2 - 1}\right]\) | A1 | 1.1b |
| \(x = \pm\ln\left(\dfrac{3 + \sqrt{5}}{2}\right)\) or \(x = \ln\left(\dfrac{3 \pm \sqrt{5}}{2}\right)\) | A1 | 2.2a |
| (5) |
M1: Uses correct identities such as \(\sinh x = \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) and \(\sinh 2x = \dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2}\) to form a quartic equation for \(\mathrm{e}^x\)
B1: \(x = 0\)
dM1: Solve their quartic equation to find a value for \(x\) (other than \(x = 0\)) via a correct method. If correct answers are given from their quartic equation without workings, you may award this mark. You may need to check their workings.
A1: Deduces one correct exact value for \(x\), which may be unsimplified.
A1: Deduces both correct exact values for \(x\) isw
Note: Inexact answers for the quartic equation \(\mathrm{e}^x = 2.618,\ 0.3819\) leading to awrt 0.962, \(-0.962\) scores maximum M1 B1 dM1 A0 A0