A2 October 2020 Paper 1 Q13
13
(a) Using exponentials, prove that \(\sinh 2x = 2\cosh x\sinh x\). [2]
(b) Hence show that if \(\mathrm{f}(x) = \sinh^2 x\), then \(\mathrm{f}''(x) = 2\cosh 2x\). [2]
(c) Explain why the coefficients of odd powers in the Maclaurin series for \(\sinh^2 x\) are all zero. [2]
(d) Find the coefficient of \(x^n\) in this series when \(n\) is a positive even number. [3]
| Scheme | Marks | AO |
|---|---|---|
| \(2\cosh x\sinh x = 2\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) | M1 | 2.1 |
| \(= \dfrac{(\mathrm{e}^x + \mathrm{e}^{-x})(\mathrm{e}^x - \mathrm{e}^{-x})}{2} = \dfrac{(\mathrm{e}^{2x} - \mathrm{e}^{-2x})}{2}\) \(= \sinh 2x\) | A1 | 2.2a |
| [2] |
Notes
M1: substituting
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}(x) = \sinh^2 x\) \(\mathrm{f}'(x) = 2\sinh x\cosh x\ [= \sinh 2x]\) | B1 | 2.1 |
| \(\Rightarrow \mathrm{f}''(x) = 2\cosh 2x\)* | B1 | 2.2a |
| [2] |
Notes
SC B1 if other methods used
NB AG
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}'''(x) = 4\sinh 2x,\ \mathrm{f}^{(5)}(x) = 16\sinh 2x, \ldots\) so all odd derivatives are multiples of \(\sinh 2x\) | M1 | 2.1 |
| so \(\mathrm{f}'''(0) = \mathrm{f}^{(5)}(0) = \mathrm{f}^{(7)}(0) = \ldots = 0\) | E1 | 2.4 |
| [2] |
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}''(x) = 2\cosh 2x,\ \mathrm{f}^{(4)}(x) = 8\cosh 2x, \ldots\) | M1 | 2.1 |
| \(\mathrm{f}^{(n)}(0) = 2^{n-1}\) [\(n\) even] | A1 | 2.1 |
| coefft of \(x^n = 2^{n-1}/n!\) [\(n\) even] | A1 | 2.2a |
| [3] |
Notes
A1: accept \(f^{(n)}(x) = 2^{n-1}\cosh(2x)\)