AS June 2023 Paper 1 Q2
2 In this question you must show detailed reasoning.
The equation \(x^2 - kx + 2k = 0\), where \(k\) is a non-zero constant, has roots \(\alpha\) and \(\beta\).
Find \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}\) in terms of \(k\), simplifying your answer. [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\alpha + \beta = k,\ \alpha\beta = 2k\) | B1 | 1.1a |
| \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}\) | M1 | 2.3 |
| \(= \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}\) | M1 | 3.1a |
| \(= \dfrac{k^2 - 4k}{2k} = \dfrac{1}{2}k - 2\) | A1 | 1.1 |
| [4] |
Notes
M1: (1st) for combining fractions correctly
M1: (2nd) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) (soi)
A1: must be simplified (accept \(\frac{k - 4}{2}\))
Alternative solution
| Scheme | Marks |
|---|---|
| \(x = \dfrac{k \pm \sqrt{k^2 - 8k}}{2}\) | B1 |
| \(\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{k + \sqrt{k^2 - 8k}}{k - \sqrt{k^2 - 8k}} + \dfrac{k - \sqrt{k^2 - 8k}}{k + \sqrt{k^2 - 8k}}\) \(= \dfrac{\left(k + \sqrt{k^2 - 8k}\right)^2 + \left(k - \sqrt{k^2 - 8k}\right)^2}{\left(k - \sqrt{k^2 - 8k}\right)\left(k + \sqrt{k^2 - 8k}\right)}\) | M1 |
| \(= \dfrac{2(k^2 + k^2 - 8k)}{k^2 - k^2 + 8k}\) | A1 |
| \(= \dfrac{2(2k^2 - 8k)}{8k} = \dfrac{1}{2}k - 2\) | A1 |
B1: by formula or completing the square
M1: combining fractions
A1: (1st) expanding correctly
A1: (2nd) must be simplified (accept \(\frac{k - 4}{2}\))