A2 October 2020 Paper 1 Q4
4 The roots of the equation \(2x^3 - 5x + 7 = 0\) are \(\alpha\), \(\beta\) and \(\gamma\).
(a) Find \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\). [4]
(b) Find an equation with integer coefficients whose roots are \(2\alpha - 1\), \(2\beta - 1\) and \(2\gamma - 1\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(2x^3 - 5x + 7 = 0\) \([\alpha + \beta + \gamma = 0],\ \beta\gamma + \alpha\gamma + \alpha\beta = -\frac{5}{2},\ \alpha\beta\gamma = -\frac{7}{2}\) | B1 | 1.1a |
| \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\) | B1 | 3.1a |
| \(= \dfrac{-5/2}{-7/2}\) | M1 | 1.1 |
| \(= \dfrac{5}{7}\) | A1 | 1.1 |
| [4] |
| Scheme | Marks | AO |
|---|---|---|
| \(2x^3 - 5x + 7 = 0\) let \(y = 2x - 1,\ x = \frac{1}{2}(y + 1)\) | M1 | 1.1a |
| \(\Rightarrow \dfrac{1}{4}(y + 1)^3 - \dfrac{5}{2}(y + 1) + 7 = 0\) | M1 | 1.1 |
| \(\Rightarrow y^3 + 3y^2 - 7y + 19 = 0\) | A2,1,0 | 1.1,1.1 |
| [4] |
Notes
M1: substituting for \(x\) (not \(2y - 1\))
Alternative method
| Scheme | Marks |
|---|---|
| sum of roots \(= 2(\alpha + \beta + \gamma) - 3 = -3\) | B1 |
| \((2\alpha - 1)(2\beta - 1) + (2\beta - 1)(2\gamma - 1) + (2\gamma - 1)(2\alpha - 1)\) \(= 4(\alpha\beta + \beta\gamma + \gamma\alpha) - 4(\alpha + \beta + \gamma) + 3 = -10 + 3 = -7\) | B1 |
| \((2\alpha - 1)(2\beta - 1)(2\gamma - 1)\) \(= 8\alpha\beta\gamma - 4(\alpha\beta + \beta\gamma + \gamma\alpha) + 2(\alpha + \beta + \gamma) - 1 = -19\) | B1 |
| \(\Rightarrow y^3 + 3y^2 - 7y + 19 = 0\) | B1ft |
| [4] |
B1: or by expanding \((x - 2\alpha + 1)(x - 2\beta + 1)(x - 2\gamma + 1)\); must expand fully for first B1
B1ft: must be an equation