A2 June 2025 Paper 1 Q5
5 The cubic equation \(2x^3 - 3x + 4 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\).
Determine a cubic equation with integer coefficients whose roots are \(\tfrac{1}{2}(\alpha + 1)\), \(\tfrac{1}{2}(\beta + 1)\) and \(\tfrac{1}{2}(\gamma + 1)\). [4]
| Scheme | Marks | AO |
|---|---|---|
| let \(y = \tfrac{1}{2}(x + 1) \Rightarrow x = 2y - 1\) | M1 | 1.1 |
| \(2(2y - 1)^3 - 3(2y - 1) + 4 = 0\) | M1 | 1.1 |
| \(2(8y^3 - 12y^2 + 6y - 1) - 6y + 3 + 4 = 0\) | B1 | 1.1 |
| \(16y^3 - 24y^2 + 6y + 5 = 0\) | A1 | 1.1 |
| [4] |
Notes
M1: correct substitution chosen and rearranged, condone a rearrangement slip
M1: substitution made, condone one slip only
B1: correctly expanding cubic, soi. Do not ft.
A1: must be an equation with integer coefficients
Alternative method
| Scheme | Marks |
|---|---|
| \(\sum\alpha = 0, \sum\alpha\beta = -\dfrac{3}{2}, \alpha\beta\gamma = -2\) | B1 |
| \(\sum\frac{1}{2}(\alpha + 1) = \frac{3}{2}\), \(\sum\frac{1}{2}(\alpha + 1)\frac{1}{2}(\beta + 1) = \frac{3}{8}\) | M1 |
| \(\frac{1}{2}(\alpha + 1)\frac{1}{2}(\beta + 1)\frac{1}{2}(\gamma + 1) = -\frac{5}{16}\) | A1 |
| \(\Rightarrow 16y^3 - 24y^2 + 6y + 5 = 0\) | A1 |
B1: Soi
M1: using sum and products to correctly find at least one of sum of roots, products of pairs or triple product for transformed equation
A1: two of \(\frac{3}{2}, \frac{3}{8}, -\frac{5}{16}\)
A1: must be an equation with integer coefficients