A2 October 2021 Paper 2 Q5
5 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR RHS \(= 2\cosh^2 x - 1 = 2\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 - 1\) | M1 | 2.1 |
| \(= 2\left(\dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4}\right) - 1 = \dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2} - 1\) \(= \dfrac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2} = \cosh 2x =\) LHS | A1 | 2.1 |
| [2] |
Notes
M1: Uses correct exponential form in an attempt at proof
A1: AG. Proof must be complete
(Corrected from the printed mark scheme: the second line is printed as \(2\left(\dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{4}\right) - 1 = \dfrac{\mathrm{e}^{2x} + 2 + \mathrm{e}^{-2x}}{2}\), with the \(-1\) missing on the right-hand side, as typed above.)
| Scheme | Marks | AO |
|---|---|---|
| DR \(2\cosh^2 x - 1 = 3\cosh x + 1\) \(\Rightarrow 2\cosh^2 x - 3\cosh x - 2 = 0\) | M1 | 3.1a |
| \((2\cosh x + 1)(\cosh x - 2) = 0\) | M1 | 1.1 |
| \(\cosh x = 2\) or \(-\tfrac{1}{2}\) | A1 | 1.1 |
| \(\cosh x \geqslant 1\) so \(\neq -\tfrac{1}{2}\) | A1 | 2.3 |
| \(x = \cosh^{-1} 2 = \ln\left(2 + \sqrt{3}\right)\) | A1 | 1.1 |
| \(x = \ln\left(2 - \sqrt{3}\right)\) | A1 | 1.1 |
| [6] |
Notes
M1: (first) Use of identity in (a) to leave a three term quadratic equation in just \(\cosh x\)
M1: (second) Attempt to solve eg \(\dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times (-2)}}{2 \times 2}\) or \(2\left(\cosh x - \dfrac{3}{4}\right)^2 - \dfrac{9}{8} - 2 = 0\)
A1: (first) Or solves quadratic BC
A1: (second) Justification must be seen and must contain no incorrect statements
A1: (third) For either correct answer seen
A1: (fourth) Both correct values for \(x\)
Or \(x = -\ln\left(2 + \sqrt{3}\right)\)
Mark final answer