A2 June 2019 Paper 1 Q8
8 In this question you must show detailed reasoning.
The roots of the equation \(x^3 - x^2 + kx - 2 = 0\) are \(\alpha\), \(\dfrac{1}{\alpha}\) and \(\beta\).
(a) Evaluate, in exact form, the roots of the equation. [6]
(b) Find \(k\). [2]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\alpha.\dfrac{1}{\alpha}.\beta = 2 \Rightarrow \beta = 2\) | M1 A1 | 3.1a 1.1b |
| \(\alpha + \dfrac{1}{\alpha} + \beta = 1\) | M1 | 1.1b |
| \(\Rightarrow \alpha^2 + \alpha + 1 = 0\) | A1 | 1.1b |
| \(\Rightarrow \alpha = \dfrac{-1 \pm \sqrt{-3}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}\mathrm{i}\) | M1 | 1.1b |
| roots are [2], \(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i},\ -\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\) | A1 | 1.1b |
| [6] |
Notes
M1: product of roots used
A1: \(\beta = 2\)
M1: sum of roots used; or \((x - 2)(x^2 + x + 1) = 0\) M1A1
A1: or equivalent quadratic (with \(\beta = 2\))
M1: solving their quadratic; \(x = \dfrac{-1 \pm \sqrt{-3}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}\mathrm{i}\)
| Scheme | Marks | AO |
|---|---|---|
| \(k = \alpha.\dfrac{1}{\alpha} + \alpha\beta + \dfrac{1}{\alpha}\beta\) | M1 | 1.1a |
| \(= 1 + 2\left(\alpha + \dfrac{1}{\alpha}\right) = 1 - 2 = -1\) | A1 | 1.1b |
| [2] |
Notes
M1: \(k\) = product of root pairs; or \((x - 2)(x^2 + x + 1) \Rightarrow k = -1\)
A1: or by direct substitution; or by factor theorem