AS June 2025 Paper 1 Q12
12 Use the definitions of \(\sinh x\) and \(\cosh x\) in terms of \(\mathrm{e}^x\) and \(\mathrm{e}^{-x}\) to solve
\[5\sinh x - 3\cosh x = 6\]Give your answer in the form
\[x = \ln\left(a + \sqrt{b}\right)\]where \(a\) and \(b\) are integers.
Fully justify your answer. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Rewrites the given equation in exponential form. | M1 | 1.1a |
| Obtains a correct 3-term exponential equation. PI by a correct value for \(\mathrm{e}^x\) | A1 | 1.1b |
| Solves their 3-term quadratic equation in \(\mathrm{e}^x\) to find a value for \(\mathrm{e}^x\) May be unsimplified. | M1 | 3.1a |
| Obtains \(3 + \sqrt{13}\) or \(3 - \sqrt{13}\) | A1 | 1.1b |
| Gives a correct reason for ignoring their negative value of \(\mathrm{e}^x\) | E1F | 2.4 |
| Obtains \(\ln\left(3 + \sqrt{13}\right)\) only. | A1 | 1.1b |
| (6 marks) |
Typical solution
\[5\sinh x - 3\cosh x = 6\]\[5\left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) - 3\left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) = 6\]\[5\mathrm{e}^x - 5\mathrm{e}^{-x} - 3\mathrm{e}^x - 3\mathrm{e}^{-x} = 12\]\[2\mathrm{e}^x - 8\mathrm{e}^{-x} = 12\]\[\mathrm{e}^x - 6 - 4\mathrm{e}^{-x} = 0\]\[\mathrm{e}^{2x} - 6\mathrm{e}^x - 4 = 0\]\[(\mathrm{e}^x - 3)^2 = 13\]\[\mathrm{e}^x = 3 \pm \sqrt{13}\]but \(\mathrm{e}^x \gt 0\)
\[x = \ln\left(3 + \sqrt{13}\right)\]