A2 June 2022 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 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. (4)
| Scheme | Marks | AO |
|---|---|---|
| Solves the quadratic equation for \(\cosh^2 x\) e.g. \(\left(8\cosh^2 x - 9\right)\left(8\cosh^2 + 1\right) = 0 \Rightarrow \cosh^2 x = \ldots\) | M1 | 3.1a |
| \(\cosh^2 x = \dfrac{9}{8}\ \left\{-\dfrac{1}{8}\right\}\) | A1 | 1.1b |
| \(\cosh x = \dfrac{3}{4}\sqrt{2} \Rightarrow x = \ln\left[\dfrac{3}{4}\sqrt{2} + \sqrt{\left(\dfrac{3}{4}\sqrt{2}\right)^2 - 1}\right]\) Alternatively \(\cosh x = \dfrac{3}{4}\sqrt{2} \Rightarrow \dfrac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right) \Rightarrow \mathrm{e}^{2x} - \dfrac{3}{2}\sqrt{2}\,\mathrm{e}^x + 1 = 0\) \(\Rightarrow \mathrm{e}^x = \sqrt{2}\) or \(\dfrac{\sqrt{2}}{2} \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a |
| (4) | ||
| (4 marks) |
Notes
M1: Solves the quadratic equation for \(\cosh^2 x\) by any valid means. If by calculator accept for reaching the positive value for \(\cosh^2 x\) (negative may be omitted or incorrect) but do not allow for going directly to a value for \(\cosh x\). Alternatively score a correct process leading to a value for \(\sinh 2x\) or its square (Alt 1) or use of correct exponential form for \(\cosh x\) to form and expand to an equation in \(\mathrm{e}^{4x}\) and \(\mathrm{e}^{2x}\) (Alt 2)
A1: Correct value for \(\cosh^2 x\) (ignore negative or incorrect extra roots.). In Alt 1 score for a correct value for \(\sinh^2 2x\) or \(\sinh 2x\). In Alt 2 score for a correct simplified equation in \(\mathrm{e}^{4x}\).
M1: For a correct method to achieve at least one value for \(x\) (from \(\cosh^2 x\)). In the main scheme or Alt 1, takes positive square root (if appropriate) and uses the correct formula for \(\operatorname{arcosh} x\) or \(\operatorname{arsinh} x\) to find a value for \(x\). (No need to see negative square root rejected.) In Alt 2 it is for solving the quadratic in \(\mathrm{e}^{4x}\) and proceeding to find a value for \(x\).
Alternatively uses the exponential definition for \(\cosh x\), forms and solves a quadratic for \(\mathrm{e}^x\) leading to a value for \(x\)
A1: Deduces (both) the correct values for \(x\) and no others. Must be in the form specified.
SC Allow M0A0M1A1 for cases where a calculator was used to get the value for \(\cosh x\) with no evidence if a correct method for find both values is shown.
Alt 1
| Scheme | Marks | AO |
|---|---|---|
| \(64\cosh^2 x\left(\cosh^2 x - 1\right) - 9 = 0 \Rightarrow 64\cosh^2 x\sinh^2 x - 9 = 0\) \(\Rightarrow 16\sinh^2 2x = 9 \Rightarrow \sinh^2 2x = \dfrac{9}{16}\) Or \((8\sinh x\cosh x - 3)(8\sinh x\cosh x + 3) = 0 \Rightarrow \sinh 2x = \pm\dfrac{3}{4}\) | M1 A1 | 3.1a 1.1b |
| \(\sinh 2x = \pm\dfrac{3}{4} \Rightarrow x = \dfrac{1}{2}\ln\left[\pm\dfrac{3}{4} + \sqrt{\dfrac{9}{16} + 1}\right]\) (or use exponentials, or proceed via \(\cosh 4x\)) | M1 | 1.1b |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a |
| (4) |
Alt 2
| Scheme | Marks | AO |
|---|---|---|
| \(64\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^4 - 64\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 - 9 = 0 \Rightarrow\) \(4\left(\mathrm{e}^{4x} + 4\mathrm{e}^{2x} + 6 + 4\mathrm{e}^{-2x} + \mathrm{e}^{-4x}\right) - 16\left(\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}\right) - 9 = 0\) | M1 | 3.1a |
| \(4\mathrm{e}^{4x} - 17 + 4\mathrm{e}^{-4x} = 0\) | A1 | 1.1b |
| \(\left(4\mathrm{e}^{4x} - 1\right)\left(1 - 4\mathrm{e}^{-4x}\right) = 0 \Rightarrow \mathrm{e}^{4x} = \ldots \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(x = \pm\dfrac{1}{2}\ln 2\) | A1 | 2.2a |
| (4) |