FP3 June 2010 Q5
5. Given that \(y = \left(\mathrm{arcosh}\,3x\right)^2\), where \(3x > 1\), show that
(a) \((9x^2 - 1)\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 36y\), (5)
(b) \((9x^2 - 1)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 9x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 18\). (4)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\,\mathrm{arcosh}\left(3x\right) \times \dfrac{3}{\sqrt{9x^2 - 1}}\) | M1A1A1 |
| \(\sqrt{9x^2 - 1}\,\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6\,\mathrm{arcosh}\left(3x\right)\) \(\left(9x^2 - 1\right)\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 36\left(\mathrm{arcosh}\left(3x\right)\right)^2\) | dM1 |
| \(\left(9x^2 - 1\right)\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 36y\) * | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\left\{18x\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + \left(9x^2 - 1\right) \times 2\dfrac{\mathrm{d}y}{\mathrm{d}x} \times \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\right\} = 36\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 {A1} A1 |
| \(\left(9x^2 - 1\right)\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} + 9x\dfrac{\mathrm{d}y}{\mathrm{d}x} = 18\) * | A1 |
| (4) | |
| (9 marks) |