AS June 2024 Paper 1 Q1
1. The cubic equation
\[2x^3 - 3x^2 + 5x + 7 = 0\]has roots \(\alpha\), \(\beta\) and \(\gamma\).
Without solving the equation, determine the exact value of
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \gamma = \dfrac{3}{2},\ \alpha\beta + \alpha\gamma + \beta\gamma = \dfrac{5}{2}\) | B1 | 3.1a |
| \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = \left(\dfrac{3}{2}\right)^2 - 2\left(\dfrac{5}{2}\right) = \ldots\) | M1 | 1.1b |
| \(= -\dfrac{11}{4} = -2.75\) cso | A1 | 1.1b |
| (3) |
Notes
B1: Correct sum and pair sum, they may be seen anywhere in the candidates working.
M1: Uses a correct identity and substitutes in their sum and pair sum to find a value.
A1: Correct value following B1, if uses \(\alpha + \beta + \gamma = -\dfrac{3}{2}\) this can score B0 M1 A0 cso
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha\beta\gamma = -\dfrac{7}{2}\) or \(x = \dfrac{3}{w}\) used in the equation | B1 | 2.2a |
| \(\dfrac{3}{\alpha} + \dfrac{3}{\beta} + \dfrac{3}{\gamma} = \dfrac{3(\alpha\beta + \alpha\gamma + \beta\gamma)}{\alpha\beta\gamma} = \dfrac{3\left(\frac{5}{2}\right)}{\left(-\frac{7}{2}\right)}\) or \(2\left(\dfrac{3}{w}\right)^3 - 3\left(\dfrac{3}{w}\right)^2 + 5\left(\dfrac{3}{w}\right) + 7 = 0 \Rightarrow 7w^3 + 15w^2 - 27w + 54\{= 0\}\) \(\Rightarrow -\dfrac{\text{‘}15\text{’}}{\text{‘}7\text{’}}\) | M1 | 1.1b |
| \(= -\dfrac{15}{7}\) cso | A1 | 1.1b |
| (3) |
Notes
B1: Correct value for the product (may be seen anywhere in the candidates working) or for using \(x = \dfrac{3}{w}\) in the given equation.
M1: Uses a correct identity and substitutes in their pair sum and product to obtain a value or multiplies through by \(w^3\) to identify at least the required terms and finds their new sum.
A1: Correct value from correct pair sum and product cso
| Scheme | Marks | AO |
|---|---|---|
| \((5 - \alpha)(5 - \beta)(5 - \gamma) = A \pm B(\alpha + \beta + \gamma) \pm C(\alpha\beta + \alpha\gamma + \beta\gamma) \pm (\alpha\beta\gamma)\) \(= \left\{5^3 - 5^2(\alpha + \beta + \gamma) + 5(\alpha\beta + \alpha\gamma + \beta\gamma) - \alpha\beta\gamma\right\}\) or \(2(5 - w)^3 - 3(5 - w)^2 + 5(5 - w) + 7\{= 0\}\) or \(\mathrm{f}(x) = A(x - \alpha)(x - \beta)(x - \gamma) \Rightarrow A = 2\) | M1 | 3.1a |
| \((5 - \alpha)(5 - \beta)(5 - \gamma) = 125 - 25\left(\dfrac{3}{2}\right) + 5\left(\dfrac{5}{2}\right) + \dfrac{7}{2}\) or \((5 - \alpha)(5 - \beta)(5 - \gamma) = -\left(\dfrac{2 \times 125 - 3 \times 25 + 25 + 7}{-2}\right)\) Or \(-2w^3 + 27w^2 - 125w + 207\{= 0\} \Rightarrow -\dfrac{\text{‘}207\text{’}}{\text{‘}-2\text{’}}\) Or \(\mathrm{f}(5) = 2(5 - \alpha)(5 - \beta)(5 - \gamma)\) \(\Rightarrow (5 - \alpha)(5 - \beta)(5 - \gamma) = \dfrac{\mathrm{f}(5)}{2}\) | M1 | 1.1b |
| \(= \dfrac{207}{2} = 103.5\) cso | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
M1: Correct strategy for obtaining the required value by expanding, must reach an expression for the form \(A \pm B(\alpha + \beta + \gamma) \pm C(\alpha\beta + \alpha\gamma + \beta\gamma) \pm (\alpha\beta\gamma)\) may not be factorised for example.
\(A \pm B\alpha \pm B\beta \pm B\gamma \pm C\alpha\beta \pm C\alpha\gamma \pm C\beta\gamma \pm (\alpha\beta\gamma)\)
or
Attempts the correct linear transformation of the given equation and expands.
or
Uses \(\mathrm{f}(x) = A(x - \alpha)(x - \beta)(x - \gamma)\) to find a value for \(A\)
M1: Uses their sum, pair sum and product to obtain a value. Allow recovery from a sign slip as long as substituting into an expression of the form \(A \pm B(\alpha + \beta + \gamma) \pm C(\alpha\beta + \alpha\gamma + \beta\gamma) \pm (\alpha\beta\gamma)\).
This would be A0 even if the correct answer is achieved.
or
Simplifies to obtain at least the required terms to find a value for the new product. Ignore the other terms whether correct or not.
Or
Uses \(\dfrac{\mathrm{f}(5)}{2}\)
A1: Correct value with no errors seen cso