A2 October 2020 Paper 2 Q1
1. The curve \(C\) has equation
\[y = 31\sinh x - 2\sinh 2x \qquad x \in \mathbb{R}\]Determine, in terms of natural logarithms, the exact \(x\) coordinates of the stationary points of \(C\).
(7)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 31\cosh x - 4\cosh 2x\) | B1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 31\cosh x - 4\left(2\cosh^2 x - 1\right)\) | M1 | 3.1a |
| \(8\cosh^2 x - 31\cosh x - 4 = 0\) | A1 | 1.1b |
| \((8\cosh x + 1)(\cosh x - 4) = 0 \Rightarrow \cosh = \ldots\) | M1 | 1.1b |
| \(\cosh x = 4, \left(-\dfrac{1}{8}\right)\) | A1 | 1.1b |
| \(\cosh x = \alpha \Rightarrow x = \ln\left(\alpha + \sqrt{\alpha^2 - 1}\right)\) or \(\ln\left(\alpha + \sqrt{\alpha^2 - 1}\right)\) or \(-\ln\left(\alpha + \sqrt{\alpha^2 - 1}\right)\) or \(\ln\left(\alpha - \sqrt{\alpha^2 - 1}\right)\) or \(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = 4 \Rightarrow \mathrm{e}^{2x} - 8\mathrm{e}^x + 1 = 0 \Rightarrow \mathrm{e}^x = \ldots \Rightarrow x = \ln(\ldots)\) | M1 | 1.2 |
| \(\pm\ln\left(4 + \sqrt{15}\right)\) or \(\ln\left(4 \pm \sqrt{15}\right)\) | A1 | 2.2a |
| (7) |
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 31\cosh x - 4\cosh 2x\) or \(31\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) - 4\left(\dfrac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2}\right)\) | B1 | 1.1b |
| Using \(\cosh x = \left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)\) and \(\sinh x = \left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) as required \(\Rightarrow 31\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) - 4\left(\dfrac{\mathrm{e}^{2x} + \mathrm{e}^{-2x}}{2}\right) = 0\) leading to \(4\mathrm{e}^{4x} - 31\mathrm{e}^{3x} - 31\mathrm{e}^x + 4 = 0\) o.e. | M1 A1 | 3.1a 1.1b |
| Solves \(4\mathrm{e}^{4x} - 31\mathrm{e}^{3x} - 31\mathrm{e}^x + 4 = 0\) \(\Rightarrow \mathrm{e}^x = \ldots\) | M1 | 1.1b |
| \(\mathrm{e}^x = 4 \pm \sqrt{15}\) or awrt 7.87, 0.13 | A1 | 1.1b |
| \(x = \ln(b)\) where \(b\) is a real exact value | M1 | 1.2 |
| \(\ln\left(4 \pm \sqrt{15}\right)\) | A1 | 2.2a |
| (7) | ||
| (7 marks) |
Notes
B1: Correct differentiation
M1: Identifies a correct approach by using a correct identity to make progress to obtain a quadratic in \(\cosh x\)
A1: Correct 3 term quadratic obtained
M1: Solves their 3TQ
A1: Correct values (may only see 4 here)
M1: Correct process to reach at least one value for \(x\) from their \(\cosh x\)
A1: Deduces the correct 2 values with no incorrect values or work involving \(\cosh x = -\dfrac{1}{8}\)
Alternative
B1: Correct differentiation
M1: Using the exponential form for \(\cosh x\), and \(\sinh x\) if required, and forms a quartic equation for \(\mathrm{e}^x\) with all terms simplified and all on one side
A1: Correct quartic equation for \(\mathrm{e}^x\)
M1: Solves their quartic equation in \(\mathrm{e}^x\)
A1: Correct values to two decimal places or exact values
M1: \(x = \ln(b)\) where \(b\) is a real exact value
A1: Deduces the correct 2 values only
(corrected from the printed mark scheme: \(\mathrm{e}^{2x} - 8\mathrm{e}^x + 1 = 0\) is printed as \(\mathrm{e}^{2x} - 8\mathrm{e}^x + 7 = 0\))