A2 June 2024 Paper 2 Q1
1.
| Scheme | Marks | AO |
|---|---|---|
| \(4\sinh^3 x + 3\sinh x \equiv 4\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^3 + 3\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\equiv 4\left(\dfrac{\mathrm{e}^{3x} - 3\mathrm{e}^{x} + 3\mathrm{e}^{-x} - \mathrm{e}^{-3x}}{8}\right) + 3\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) | M1 | 2.1 |
| \(\equiv \dfrac{\mathrm{e}^{3x} - \mathrm{e}^{-3x}}{2} \equiv \sinh 3x\,*\) | A1* | 1.1b |
| (2) |
Notes
M1: Begins the proof by expressing \(\sinh x\) correctly in terms of exponentials, substitutes and makes progress in cubing the bracket.
Award for obtaining an expression of the form \(A\mathrm{e}^{3x} + B\mathrm{e}^{x} + C\mathrm{e}^{-x} + D\mathrm{e}^{-3x}\) but terms do not need to be collected.
Note that \((\mathrm{e}^{2x} - 2 + \mathrm{e}^{-2x})(\mathrm{e}^x - \mathrm{e}^{-x}) = \mathrm{e}^{3x} - \mathrm{e}^{x} - 2\mathrm{e}^{x} + 2\mathrm{e}^{-x} + \mathrm{e}^{-x} - \mathrm{e}^{-3x}\)
A1*: Fully correct proof with no errors. Must see \(= \sinh 3x\) or e.g. LHS = RHS
| Scheme | Marks | AO |
|---|---|---|
| \(\sinh 3x = 19\sinh x \Rightarrow 4\sinh^3 x + 3\sinh x = 19\sinh x\) \(\sinh 3x = 19\sinh x \Rightarrow 4\sinh^3 x - 16\sinh x = 0\) \(4\sinh x\left(\sinh^2 x - 4\right) = 0\) | M1 | 3.1a |
| \(\sinh x = 0 \Rightarrow x = 0\) | B1 | 2.2a |
| \(\sinh^2 x = 4 \Rightarrow \sinh x = \pm 2\) \(\Rightarrow x = \ln\left(\pm 2 + \sqrt{(\pm 2)^2 + 1}\right)\) | M1 | 1.1b |
| \(x = \ln\left(2 + \sqrt{5}\right)\) or \(x = \ln\left(-2 + \sqrt{5}\right)\) oe e.g. \(x = -\ln\left(2 + \sqrt{5}\right)\) | A1 | 1.1b |
| \(x = \ln\left(2 + \sqrt{5}\right)\) and \(x = \ln\left(-2 + \sqrt{5}\right)\) Alternatively, \(x = \ln\left(\sqrt{5} \pm 2\right)\) oe e.g. \(x = \pm\ln\left(2 + \sqrt{5}\right)\) or \(\dfrac{1}{2}\ln\left(9 \pm 4\sqrt{5}\right)\) | A1 | 1.1b |
| (5) | ||
| (7 marks) |
Notes
M1: Uses part (a), collects terms and attempts to factorise or cancel \(\sinh x\).
This can be implied if they go straight from their cubic to writing all correct answers for \(\sinh x\) including zero from their calculator.
B1: Deduces the root \(x = 0\), allow ln1 but not \(\ln\left[0 + \ln\sqrt{0 + 1}\right]\)
M1: Proceeds to \(\sinh x = \alpha\) and uses the correct logarithmic form of arsinh to obtain at least one exact value for \(x\)
Alternatively, candidates proceed from \(\sinh x = \alpha\) to substitute the exponential form and then solve a 3TQ in \(\mathrm{e}^x\) to obtain at least one exact value for \(x\)
A1: One correct non-zero solution
A1: Both correct non-zero solutions and no incorrect other solutions, but isw if they go on to evaluate these answers incorrectly.
Allow \(x = \ln\left(\sqrt{5} \pm 2\right)\) for listing both solutions
Alternative if candidates substitute the exponential form at the start:
M1: Candidates substitutes the exponential form for each term and proceeds to find a four term cubic equation in \(\mathrm{e}^{2x} = 0\), which may not be correct.
B1: Deduces the root \(x = 0\), allow ln1 but not \(\ln\left[0 + \ln\sqrt{0 + 1}\right]\)
M1: They factorise their cubic, proceed to obtain exact values for \(\mathrm{e}^{2x}\), then take logs to obtain at least one exact value for \(x\)
If they go directly to decimal answers this will usually score M0 unless they recover to exact form.
A1: One correct non-zero solution
A1: Both correct non-zero solutions and no incorrect other solutions, but isw if they go on to evaluate these answers incorrectly.
This is how this would be awarded:
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{(\mathrm{e}^{3x} - \mathrm{e}^{-3x})}{2} = 19\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)\) \(\mathrm{e}^{3x} - \mathrm{e}^{-3x} - 19\mathrm{e}^{x} + 19\mathrm{e}^{-x} = 0\) \(\mathrm{e}^{6x} - 1 - 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} = 0\) | M1 (for cubic in \(\mathrm{e}^{2x} = 0\)) | |
| \(\mathrm{e}^{6x} - 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} - 1 = 0\) \(\Rightarrow (\mathrm{e}^{2x} - 1)(\mathrm{e}^{4x} - 18\mathrm{e}^{2x} + 1) = 0\) \(\mathrm{e}^{2x} = 1,\ 9 \pm 4\sqrt{5}\) \(2x = \ln 1,\ 2x = \ln\left(9 \pm 4\sqrt{5}\right)\) \(x = 0,\ x = \dfrac{1}{2}\ln\left(9 \pm 4\sqrt{5}\right)\) | B1, M1, A1, A1 |
(corrected from the printed mark scheme: the line \(\mathrm{e}^{6x} - 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} - 1 = 0\) is printed as \(\mathrm{e}^{6x} + 19\mathrm{e}^{4x} + 19\mathrm{e}^{2x} - 1 = 0\))