AS June 2025 Paper 1 Q11
11.
In this question you must show detailed reasoning.
\[\mathrm{f}(x) = x^3 - 21x^2 + Ax - 91 \qquad \text{where } A \text{ is a real constant}\]The roots of \(\mathrm{f}(x) = 0\) are
\[\alpha,\ \alpha + 3\beta \text{ and } \alpha + 6\beta\]where \(\alpha\) and \(\beta\) are real constants.
Use algebra to determine the value of each of these roots. (7)
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + (\alpha + 3\beta) + (\alpha + 6\beta) = 21\) and \(\alpha(\alpha + 3\beta)(\alpha + 6\beta) = 91\) | M1 | 1.1b |
| \(3\alpha + 9\beta = 21 \Rightarrow \alpha + 3\beta = 7 \Rightarrow \alpha = 7 - 3\beta\) \((7 - 3\beta)(7 - 3\beta + 3\beta)(7 - 3\beta + 6\beta) = 91\) \((7 - 3\beta)^3 + 9(7 - 3\beta)^2\beta + 18(7 - 3\beta)\beta^2 = 91\) Or \(3\beta = 7 - \alpha\) and \(\alpha^3 + 9\alpha^2\beta + 18\alpha\beta^2 = 91\) leading to \(\alpha^3 + 3\alpha^2(7 - \alpha) + 2\alpha(7 - \alpha)^2 = 91\) | M1 | 3.1a |
| \(343 - 63\beta^2 = 91\) or \(49 - 9\beta^2 = 13\) Or \(7\alpha^2 - 98\alpha + 91 = 0\) | A1 | 1.1b |
| \(49 - 9\beta^2 = 13 \Rightarrow \beta = \ldots\{\pm 2\}\) or \(7\alpha^2 - 98\alpha + 91 = 0 \Rightarrow \alpha = \ldots\{1, 13\}\) | dM1 | 1.1b |
| \(\alpha = 7 - 3(\pm 2) = \ldots\{1, 13\}\) Or \(\beta = \dfrac{7 - \text{‘}\alpha\text{’}}{3}\) | ddM1 | 1.1b |
| Roots = 1, 7, 13 only | dddM1 A1 | 1.1b 1.1b |
| (7) | ||
| (7 marks) |
Notes
M1: Sets the sum of the roots = 21 and the product of the roots = 91
M1: Forms an equation in one variable using their sum of roots and product of roots equations
A1: Correct simplified quadratic equation
dM1: Dependent on the previous method mark, solves their equation to find real roots, may be a cubic if slips earlier
ddM1: Dependent on previous method mark. Finds the value of the other variable
dddM1: Dependent on previous method mark. Uses their values of \(\alpha\) and \(\beta\) to find the value of the roots
A1: Correct roots
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + (\alpha + 3\beta) + (\alpha + 6\beta) = 21\) | M1 | 1.1b |
| \(\alpha + 3\beta = 7\) therefore one root must be 7 | M1 A1 | 3.1a 1.1b |
| \((x - 7)\left(x^2 - 14x + 13\right)\) | dM1 | 1.1b |
| \((x - 7)(x - 1)(x - 13)\) | ddM1 | 1.1b |
| Roots = 1, 13 | dddM1 A1 | 1.1b 1.1b |
M1: Sets the sum of the roots = 21
M1: Simplifies and deduces one root
A1: One root = 7
dM1: Factorises into linear and quadratic, achieves correct first and last term
ddM1: Factorises their quadratic, which must be factorisable
dddM1: Uses their factorised quadratic to find the remining 2 roots
A1: Correct roots 1 and 13