A2 June 2019 Paper 1 Q6
6
(a) Show that\[\cosh^3 x + \sinh^3 x = \frac{1}{4}\mathrm{e}^{mx} + \frac{3}{4}\mathrm{e}^{nx}\]
where \(m\) and \(n\) are integers. [3 marks]
(b) Hence find \(\cosh^6 x - \sinh^6 x\) in the form\[\frac{a\cosh(kx) + b}{8}\]
where \(a\), \(b\) and \(k\) are integers. [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses correct expressions for \(\cosh x\) and \(\sinh x\), and uses them to simplify LHS. | M1 | 1.1a |
| Finds a correct, unsimplified expansion of the LHS in terms of exponentials. | A1 | 1.1b |
| Completes a rigorous argument to obtain the correct result. Must include clear definitions for \(\cosh x\) & \(\sinh x\). NMS = 0/3 | R1 | 2.1 |
Typical solution
\[\cosh^3 x = \frac{1}{8}\left(\mathrm{e}^{3x} + 3\mathrm{e}^{x} + 3\mathrm{e}^{-x} + \mathrm{e}^{-3x}\right)\]\[\sinh^3 x = \frac{1}{8}\left(\mathrm{e}^{3x} - 3\mathrm{e}^{x} + 3\mathrm{e}^{-x} - \mathrm{e}^{-3x}\right)\]\[\cosh^3 x + \sinh^3 x = \frac{1}{4}\mathrm{e}^{3x} + \frac{3}{4}\mathrm{e}^{-x}\]| Scheme | Marks | AO |
|---|---|---|
| Finds \(\cosh^3 x - \sinh^3 x\) in exponential form PI correct exponential expression. | B1 | 3.1a |
| Uses their expressions to find \(\cosh^6 x - \sinh^6 x\) in exponential form. | M1 | 3.1a |
| Obtains their correct result | A1F | 1.1b |
| Correctly separates out \(\dfrac{3}{8}\cosh 4x\) or equivalent from their expression | M1 | 2.2a |
| Completes a rigorous argument to obtain the correct result. | R1 | 2.1 |
| (8 marks) |