AS June 2018 Paper 1 Q2
2 In this question you must show detailed reasoning.
The cubic equation \(2x^3 + 3x^2 - 5x + 4 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\). By making an appropriate substitution, or otherwise, find a cubic equation with integer coefficients whose roots are \(\dfrac{1}{\alpha}\), \(\dfrac{1}{\beta}\) and \(\dfrac{1}{\gamma}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(u = \dfrac{1}{x}\) | B1 | 2.2a |
| \(u^3\left(2\left(\dfrac{1}{u}\right)^3 + 3\left(\dfrac{1}{u}\right)^2 - 5\left(\dfrac{1}{u}\right) + 4\right) = 0\) | M1 | 1.1 |
| \(4u^3 - 5u^2 + 3u + 2 = 0\) | A1 | 1.1 |
| [3] |
Notes
B1: SOI
Other letters can be used as the variable. If “\(x\)” used allow B1 for sight of \(2\left(\frac{1}{x}\right)^3 + 3\left(\frac{1}{x}\right)^2 - 5\left(\frac{1}{x}\right) + 4\ (= 0)\)
M1: For substituting \(\dfrac{1}{u}\) into the given equation and attempting to multiply by \(u^3\)
If no \(u^3\) outside brackets then need to see at least two terms multiplied by \(u^3\).
A1: Or multiple of this (with integer coefficients). Condone \(x\) as variable. Must have “\(= 0\)”
Alternative
| Scheme | Marks |
|---|---|
| \(\alpha + \beta + \gamma = -\tfrac{3}{2}\); \(\alpha\beta + \beta\gamma + \gamma\alpha = -\tfrac{5}{2}\); \(\alpha\beta\gamma = -2\) | M1 |
| \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \left(\dfrac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}\right) = \tfrac{5}{4}\) \(\dfrac{1}{\alpha\beta} + \dfrac{1}{\beta\gamma} + \dfrac{1}{\gamma\alpha} = \left(\dfrac{\alpha + \beta + \gamma}{\alpha\beta\gamma}\right) = \tfrac{3}{4}\) \(\dfrac{1}{\alpha\beta\gamma} = -\tfrac{1}{2}\) | M1 |
| \(4x^3 - 5x^2 + 3x + 2 = 0\) | A1 |
M1: (1st) Allow one sign slip
M1: (2nd) At least 2 correct
A1: oe Must be integer coefficients
Must have “= 0”