FP3 June 2010 Q3
3.
(a) Starting from the definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials, prove that \[\cosh 2x = 1 + 2\sinh^2 x\] (3)
(b) Solve the equation \[\cosh 2x - 3\sinh x = 15,\] giving your answers as exact logarithms. (5)
| Scheme | Marks |
|---|---|
| \(rhs = 1 + 2\sinh^2 x = 1 + 2\left(\dfrac{e^{x} - e^{-x}}{2}\right)^2\) | M1 |
| \(= \dfrac{2 + e^{2x} - 2 + e^{-2x}}{2}\) | M1 |
| \(= \dfrac{e^{2x} + e^{-2x}}{2} = \cosh 2x = lhs\) * | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(1 + 2\sinh^2 x - 3\sinh x = 15\) \(2\sinh^2 x - 3\sinh x - 14 = 0\) | M1 |
| \(\left(\sinh x + 2\right)\left(2\sinh x - 7\right) = 0\) | M1 |
| \(\sinh x = -2, \dfrac{7}{2}\) | A1 |
| \(x = \ln\left(-2 + \sqrt{(-2)^2 + 1}\right) = \ln\left(-2 + \sqrt{5}\right)\) | M1 |
| \(x = \ln\left(\dfrac{7}{2} + \sqrt{\left(\dfrac{7}{2}\right)^2 + 1}\right) = \ln\left(\dfrac{7 + \sqrt{53}}{2}\right)\) | A1 |
| (5) | |
| (8 marks) |