AS June 2022 Paper 1 Q4
4 In this question you must show detailed reasoning.
The equation \(z^3 + 2z^2 + kz + 3 = 0\), where \(k\) is a constant, has roots \(\alpha\), \(\dfrac{1}{\alpha}\) and \(\beta\).
Determine the roots in exact form. [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\alpha \cdot \dfrac{1}{\alpha} \cdot \beta = -3\) | M1 | 3.1a |
| \(\Rightarrow \beta = -3\) | A1 | 1.1 |
| \(\alpha + \dfrac{1}{\alpha} + \beta = -2\) | M1 | 1.1 |
| \(\Rightarrow \alpha^2 - \alpha + 1 = 0\) | M1 | 1.1 |
| \(\Rightarrow \alpha = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\) or \(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\) | A1 | 1.1 |
| [as \(\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\right)\left(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\right) = 1\), \(\Rightarrow \alpha = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i} \Rightarrow \dfrac{1}{\alpha} = \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\)] | ||
| so roots are \(-3\), \(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\) and \(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\) | A1 | 3.2a |
| [6] |
Notes
M1: (1st) product of roots = −3 (condone 3 for M1)
M1: (2nd) sum of roots = −2 (condone 2 for M1)
M1: (3rd) getting quadratic in \(\alpha\) or \(k = -2\) found \(\Rightarrow z^2 - z + 1\) by factorising
A1: (2nd) \(\Rightarrow z = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\) or \(\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\)
[ … ] not required for final A1
(corrected from the printed mark scheme: the last line printed only “so roots are \(\frac{1}{2} + \frac{\sqrt{3}}{2}\mathrm{i}\) and \(\frac{1}{2} - \frac{\sqrt{3}}{2}\mathrm{i}\)”, leaving out the third root \(\beta = -3\) found above)