AS October 2020 Paper 1 Q3
3 In this question you must show detailed reasoning.
The roots of the equation \(x^2 - 2x + 4 = 0\) are \(\alpha\) and \(\beta\).
(a) Find \(\alpha\) and \(\beta\) in modulus-argument form. [4]
(b) Hence or otherwise show that \(\alpha\) and \(\beta\) are both roots of \(x^3 + \lambda = 0\), where \(\lambda\) is a real constant to be determined. [3]
| Scheme | Marks | AO |
|---|---|---|
| DR \(x = \dfrac{2 \pm \sqrt{-12}}{2} = 1 \pm \sqrt{3}\,\mathrm{i}\) | M1 A1 | 1.1 1.1 |
| \(= 2\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)\) | B1 | 1.1 |
| \(2\left(\cos\left(-\dfrac{\pi}{3}\right) + \mathrm{i}\sin\left(-\dfrac{\pi}{3}\right)\right)\) | B1 | 1.1 |
| [4] |
Notes
M1: or completing the square
no working M0, allow 1 slip
B1: (2nd) or \(2\left(\cos\dfrac{5\pi}{3} + \mathrm{i}\sin\dfrac{5\pi}{3}\right)\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\alpha^3 = 8(\cos\pi + \mathrm{i}\sin\pi) = -8\) | B1 | 1.1 |
| \(\beta^3 = 8(\cos(-\pi) + \mathrm{i}\sin(-\pi)) = -8\) | B1 | 1.1 |
| so \(\lambda = 8\) | B1 | 1.1 |
| [3] |
Notes
Alternative solution
| Scheme | Marks |
|---|---|
| \(\alpha^3 = (1 + \sqrt{3}\,\mathrm{i})(1 + \sqrt{3}\,\mathrm{i})(1 + \sqrt{3}\,\mathrm{i})\) \(= (-2 + 2\sqrt{3}\,\mathrm{i})(1 + \sqrt{3}\,\mathrm{i}) = -8\) | M1 |
| \(\beta^3 = (1 - \sqrt{3}\,\mathrm{i})(1 - \sqrt{3}\,\mathrm{i})(1 - \sqrt{3}\,\mathrm{i})\) \(= (-2 - 2\sqrt{3}\,\mathrm{i})(1 - \sqrt{3}\,\mathrm{i}) = -8\) | A1 |
| so \(\lambda = 8\) | A1 |
M1: attempt to substitute into \(x^3 + \lambda = 0\)
no working M1A0A0
A1: (1st) oe (eg \(\beta\) a root as complex conjugate)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\alpha + \beta + \gamma = 0 \Rightarrow \gamma = -2\) | B1 |
| \(\alpha\beta\gamma = (1 + \sqrt{3}\,\mathrm{i})(1 - \sqrt{3}\,\mathrm{i})(-2) = -8\) | M1 |
| so \(\lambda = 8\) | A1 |
M1: condone \(\alpha\beta\gamma = \lambda\)
A1: cao
condone no check of coefft of \(x\)