A2 June 2023 Paper 2 Q2
2.
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{e}^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}\) or \(\mathrm{e}^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\) | B1 | 1.1b |
| (1) |
Notes
B1: Correct series (ignore terms beyond \(x^3\)).
| Scheme | Marks | AO |
|---|---|---|
| Version 1 \(\mathrm{e}^{\left(\mathrm{e}^x - 1\right)} = \mathrm{e}^{1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots - 1}\) or Version 2 \(\mathrm{e}^{\left(\mathrm{e}^x - 1\right)} = 1 + \left(\mathrm{e}^x - 1\right) + \dfrac{\left(\mathrm{e}^x - 1\right)^2}{2!} + \dfrac{\left(\mathrm{e}^x - 1\right)^3}{3!} + \ldots\) | M1 | 1.1b |
| Version 1.1 \(= 1 + \left(x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!}\right) + \dfrac{1}{2}\left(x + \dfrac{x^2}{2!} + \ldots\right)^2 + \dfrac{1}{6}(x + \ldots)^3 + \ldots\) Or Version 1.2 \(\mathrm{e}^{x + \frac{x^2}{2} + \frac{x^3}{6}} = \mathrm{e}^x \times \mathrm{e}^{\frac{x^2}{2}} \times \mathrm{e}^{\frac{x^3}{6}}\) \(= \left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right)\left(1 + \dfrac{x^2}{2} + \ldots\right)\left(1 + \dfrac{x^3}{6} + \ldots\right)\) Or Version 2.1 \(= 1 + \left(\mathrm{e}^x - 1\right) + \dfrac{\left(\mathrm{e}^{2x} - 2\mathrm{e}^x + 1\right)}{2} + \dfrac{\left(\mathrm{e}^{3x} - 3\mathrm{e}^{2x} + 3\mathrm{e}^x - 1\right)}{6} + \ldots\) \(= \dfrac{1}{3} + \dfrac{1}{6}\mathrm{e}^{3x} + \dfrac{1}{2}\mathrm{e}^x\) \(= \dfrac{1}{3} + \dfrac{1}{6}\left(1 + 3x + \dfrac{(3x)^2}{2} + \dfrac{(3x)^3}{6}\right) + \dfrac{1}{2}\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right)\) Or Version 2.2 \(= 1 + \left(\mathrm{e}^x - 1\right) + \dfrac{\left(\mathrm{e}^{2x} - 2\mathrm{e}^x + 1\right)}{2} + \dfrac{\left(\mathrm{e}^{3x} - 3\mathrm{e}^{2x} + 3\mathrm{e}^x - 1\right)}{6} + \ldots\) \(= 1 + \left(x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right) + \dfrac{1}{2}\left[\left(1 + 2x + \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{6}\right) - 2\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right) + 1\right]\) \(+ \dfrac{1}{6}\left[\left(1 + 3x + \dfrac{(3x)^2}{2} + \dfrac{(3x)^3}{6}\right) - 3\left(1 + 2x + \dfrac{(2x)^2}{2} + \dfrac{(2x)^3}{6}\right) + 3\left(1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}\right) - 1\right]\) | M1 | 3.1a |
| \(= 1 + x + \left(\dfrac{1}{2} + \dfrac{1}{2}\right)x^2 + \left(\dfrac{1}{6} + \dfrac{1}{2} \times 2 \times \dfrac{1}{2} + \dfrac{1}{6}\right)x^3 + \ldots\) | dM1 | 1.1b |
| \(= 1 + x + x^2 + \dfrac{5}{6}x^3 + \ldots\) | A1 A1 | 1.1b 2.1 |
| (5) | ||
| (6 marks) |
Notes
M1: Correctly applies the exponential Maclaurin expansion at least once, either to the base exponent or in the index. Allow 2 for 2! and 6 for 3! in the cube term. Follow through on their series seen in (a)
M1: A complete attempt to use the exponential Maclaurin series to produce a cubic expression in terms of \(x\) only. Allow if the 3! is incorrect for this mark, but a polynomial in \(x\) must have been achieved. Condone a slip with one term. Follow through on their series seen in (a)
dM1: Dependent on previous method mark only. Expands the brackets and gathers terms (not necessarily fully simplified, but should have a single term for each power).
A1: Any two correct from coefficients of \(x\), \(x^2\) and \(x^3\), need not be simplified.
A1: Fully correct answer with simplified terms.
NB: Question instructs to use standard Maclaurin series, so use of differentiation scores no mark.
Special case: Using \(\mathrm{e}^{\left(\mathrm{e}^x - 1\right)} = \mathrm{e}^{\mathrm{e}^x} \times \mathrm{e}^{-1}\) can score M1M1M0A0A0 for using Maclaurin series \(\mathrm{e}^{\mathrm{e}^x}\) and then on \(\mathrm{e}^x, \mathrm{e}^{2x}, \mathrm{e}^{3x}\)