AS June 2019 Paper 1 Q10
10
(a) Using the definition of \(\cosh x\) and the Maclaurin series expansion of \(\mathrm{e}^x\), find the first three non-zero terms in the Maclaurin series expansion of \(\cosh x\). [3 marks]
(b) Hence find a trigonometric function for which the first three terms of its Maclaurin series are the same as the first three terms of the Maclaurin series for \(\cosh(\mathrm{i}x)\). [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses, or writes, \(\cosh x = \frac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x})\) | B1 | 1.2 |
| Substitutes into their \(\cosh x\) the Maclaurin expansions of \(\mathrm{e}^x\) and \(\mathrm{e}^{-x}\) up to \(x^3\) or beyond. Allow one sign error. Ignore errors beyond \(x^4\) | M1 | 1.1a |
| Correctly simplifies their expression to \(1 + \frac{x^2}{2} + \frac{x^4}{24}\) or equivalent. Accept \(2!\) for 2 and \(4!\) for 24. Ignore terms beyond \(x^4\) NMS can score 3/3 | A1 | 1.1b |
Typical solution
\[\cosh x = \frac{1}{2}\left(\mathrm{e}^x + \mathrm{e}^{-x}\right)\]\[\begin{aligned}\cosh x = {}&\frac{1}{2}\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots\ldots\right) \\ &+ \frac{1}{2}\left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} + \ldots\ldots\right)\end{aligned}\]\[\cosh x = 1 + \frac{x^2}{2} + \frac{x^4}{24}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(\mathrm{i}x\) for \(x\) in their expansion of \(\cosh x\) | M1 | 1.1a |
| Correctly simplifies the powers of \(\mathrm{i}\) to give \(1 - \frac{x^2}{2} + \frac{x^4}{24}\) or equivalent. Accept \(2!\) for 2 and \(4!\) for 24. Ignore terms beyond \(x^4\) Implied by a correct answer of \(\cos x\) | A1 | 1.1b |
| Recognises the Maclaurin expansion of \(\cos x\) NMS can score 3/3 | B1 | 1.2 |
| (6 marks) |