A2 June 2021 Paper 1 Q4
4 Show that the solutions to the equation
\[3\tanh^2 x - 2\operatorname{sech} x = 2\]can be expressed in the form
\[x = \pm\ln\left(a + \sqrt{b}\right)\]where \(a\) and \(b\) are integers to be found.
You may use without proof the result \(\cosh^{-1} y = \ln\left(y + \sqrt{y^2 - 1}\right)\) [5 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses appropriate hyperbolic identity or substitutes using exponential form | M1 | 1.1a |
| Solves a quadratic or quartic equation and selects positive root | M1 | 2.2a |
| Obtains correct value(s) of \(\operatorname{sech} x\) or \(\cosh x\) or \(\mathrm{e}^x\) | B1 | 1.1b |
| Expresses \(x\) in logarithmic form which contains one of 3 or \(\sqrt{8}\) or \(2\sqrt{2}\) | M1 | 1.1a |
| Obtains correct values of \(x\) Condone missing \(\pm\) in working prior to final answer Condone \(x = \pm\ln\left(3 + 2\sqrt{2}\right)\) | A1 | 1.1b |
| (5 marks) |
Typical solution
\[3\tanh^2 x - 2\operatorname{sech} x = 2\]\[3 - 3\operatorname{sech}^2 x - 2\operatorname{sech} x - 2 = 0\]\[0 = 3\operatorname{sech}^2 x + 2\operatorname{sech} x - 1\]\[\operatorname{sech} x = \frac{1}{3} \text{ or } -1\]But \(\operatorname{sech} x \gt 0\) \(\therefore \operatorname{sech} x = \dfrac{1}{3}\)
\[\cosh x = 3 \Rightarrow x = \pm\cosh^{-1} 3\]\[x = \pm\ln\left(3 + \sqrt{8}\right)\]