AS June 2024 Paper 1 Q4
4 In this question you must show detailed reasoning.
The roots of the cubic equation \(x^3 - 3x^2 + 19x - 17 = 0\) are \(\alpha\), \(\beta\) and \(\gamma\).
(a) Find 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]
(b) Hence or otherwise solve the equation \(x^3 - 3x^2 + 19x - 17 = 0\). [3]
| Scheme | Marks | AO |
|---|---|---|
| DR let \(y = \tfrac{1}{2}(x - 1) \Rightarrow x = 2y + 1\) | M1 | 1.1 |
| \(\Rightarrow (2y + 1)^3 - 3(2y + 1)^2 + 19(2y + 1) - 17\ [= 0]\) | M1 | 1.1 |
| \(\Rightarrow 8y^3 + 12y^2 + 6y + 1 - 12y^2 - 12y - 3 + 38y + 19 - 17 = 0\) | A1 | 1.1 |
| \(\Rightarrow 8y^3 + 32y = 0\) or \(y^3 + 4y = 0\) | A1 | 1.1 |
| [4] |
Notes
M1: condone rearrangement slips
M1: substituting for \(x\) in the cubic
A1: expanding \((2y + 1)^3\) correctly
A1: must be an equation
Alternative solution
| Scheme | Marks |
|---|---|
| Sum of new roots \(= \tfrac{1}{2}(\Sigma\alpha - 3) = \tfrac{1}{2}(3 - 3) = 0\) | B1 |
| Sum of pairs \(= \tfrac{1}{4}(\Sigma\alpha\beta - 2\Sigma\alpha + 3) = \tfrac{1}{4}(19 - 6 + 3) = 4\) | M1 |
| Product \(= \tfrac{1}{8}(\alpha\beta\gamma - \Sigma\alpha\beta + \Sigma\alpha - 1) = \tfrac{1}{8}(17 - 19 + 3 - 1) = 0\) | A1 |
| Equation is \(y^3 + 4y = 0\) | A1 |
B1: sum of new roots = 0
M1: attempt to find sum of pairs or product of new roots
A1: either correct (from correct working)
A1: must be an equation
| Scheme | Marks | AO |
|---|---|---|
| DR \(y = 0, 2\mathrm{i}, -2\mathrm{i}\) | B1ft | 1.1 |
| \(x = 2y + 1\) | M1 | 1.1 |
| \(\Rightarrow\) roots of original cubic are \(1\), \(1 + 4\mathrm{i}\), \(1 - 4\mathrm{i}\) | A1 | 1.1 |
| [3] |
Notes
B1ft: ft their cubic in \(y\) (but not solved BC)
(Corrected from the printed mark scheme: the list of roots is printed as “1. 1 + 4i, 1 − 4i”; the full stop is a comma.)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1) = 0 \Rightarrow x - 1\) is a factor | B1 |
| \(x^3 - 3x^2 + 19x - 17 = (x - 1)(x^2 - 2x + 17)\) | M1 |
| \(\Rightarrow x = 1, 1 + 4\mathrm{i}\) or \(1 - 4\mathrm{i}\) | A1 |
M1: factorising by inspection or long division