AS June 2018 Paper 1 Q17
17 Find the exact solution to the equation
\[\sinh\theta(\sinh\theta + \cosh\theta) = 1\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Recalls and uses \(\sinh\theta = \frac{1}{2}(\mathrm{e}^\theta - \mathrm{e}^{-\theta})\) and \(\cosh\theta = \frac{1}{2}(\mathrm{e}^\theta + \mathrm{e}^{-\theta})\) PI | B1 | 1.2 |
| Forms equation and rearranges to obtain exactly one exponential term | M1 | 3.1a |
| Takes logarithms of an equation of the form \(\mathrm{e}^{2\theta} = k\) where \(k \gt 0\) | M1 | 1.1a |
| Obtains correct answer in required form | A1 | 1.1b |
Typical solution
\[\frac{1}{2}\left(\mathrm{e}^\theta - \mathrm{e}^{-\theta}\right) \times \left(\frac{1}{2}\left(\mathrm{e}^\theta - \mathrm{e}^{-\theta}\right) + \frac{1}{2}\left(\mathrm{e}^\theta + \mathrm{e}^{-\theta}\right)\right) = 1\]\[\frac{1}{2}\left(\mathrm{e}^\theta - \mathrm{e}^{-\theta}\right) \times \left(\frac{1}{2}\mathrm{e}^\theta + \frac{1}{2}\mathrm{e}^\theta\right) = 1\]\[\mathrm{e}^{2\theta} - \mathrm{e}^0 = 2\]\[\mathrm{e}^{2\theta} = 3\]\[2\theta = \ln 3\]\[\theta = \frac{1}{2}\ln 3\]ALT 17
| Scheme | Marks | AO |
|---|---|---|
| Use of \(\cosh^2\theta - \sinh^2\theta = 1\) PI | M1 | 3.1a |
| Recalls and uses \(\tanh\theta = \frac{\sinh\theta}{\cosh\theta}\) | B1 | 1.2 |
| Solves a three-term quadratic in \(\tanh\theta\) (oe) | M1 | 1.1a |
| Obtains the correct answer. ISW Condone lack of reference to \(\tanh\theta \neq -1\) | A1 | 1.1b |
| (4 marks) |
Typical solution
\[\sinh^2\theta + \sinh\theta\cosh\theta = \cosh^2\theta - \sinh^2\theta\]\[\frac{\sinh^2\theta}{\cosh^2\theta} + \frac{\sinh\theta\cosh\theta}{\cosh^2\theta} = \frac{\cosh^2\theta}{\cosh^2\theta} - \frac{\sinh^2\theta}{\cosh^2\theta}\]\[\tanh^2\theta + \tanh\theta = 1 - \tanh^2\theta\]\[2\tanh^2\theta + \tanh\theta - 1 = 0\]\[(2\tanh\theta - 1)(\tanh\theta + 1) = 0\]\[\tanh\theta = \frac{1}{2} \quad \text{or} \quad \tanh\theta = -1\]but \(\tanh\theta \neq -1\) \(\therefore \tanh\theta = \frac{1}{2}\) only
\[\theta = \operatorname{artanh}\left(\frac{1}{2}\right)\]