A2 October 2021 Paper 1 Q3
3. The cubic equation
\[ax^3 + bx^2 - 19x - b = 0\]where \(a\) and \(b\) are constants, has roots \(\alpha\), \(\beta\) and \(\gamma\)
The cubic equation
\[w^3 - 9w^2 - 97w + c = 0\]where \(c\) is a constant, has roots \((4\alpha - 1)\), \((4\beta - 1)\) and \((4\gamma - 1)\)
Without solving either cubic equation, determine the value of \(a\), the value of \(b\) and the value of \(c\). (6)
| Scheme | Marks | AO |
|---|---|---|
| \(w = 4x - 1 \Rightarrow x = \dfrac{w + 1}{4}\) | B1 | 3.1a |
| \(a\left(\dfrac{w + 1}{4}\right)^3 + b\left(\dfrac{w + 1}{4}\right)^2 - 19\left(\dfrac{w + 1}{4}\right) - b\ (= 0)\) or \((4x - 1)^3 - 9(4x - 1)^2 - 97(4x - 1) + c\ (= 0)\) | M1 | 3.1a |
| \(aw^3 + (3a + 4b)w^2 + (3a + 8b - 304)w + (a - 60b - 304) = 0\) or \(64x^3 - 192x^2 - 304x + 87 + c = 0\) | M1 | 1.1b |
| Divides by \(a\) and equates the coefficients of \(w^2\) and \(w\) \(\dfrac{3a + 4b}{a} = -9 \qquad \dfrac{3a + 8b - 304}{a} = -97\) and solves simultaneously to find a value for \(a\) or a value for \(b\) Note: \(12a + 4b = 0\) and \(100a + 8b = 304\) or Divides through by ‘16’ leading to values of \(a\) and \(b\) \(4x^3 - 12x^2 - 19x + \dfrac{87 + c}{16} = 0\) | M1 | 3.1a |
| \(c = \dfrac{a - 60b - 304}{a} = \ldots\) or \(\dfrac{87 + c}{16} = 12 \Rightarrow c = \ldots\) | M1 | 1.1b |
| \(a = 4 \quad b = -12 \quad c = 105\) | A1 | 1.1b |
| (6) | ||
| (6 marks) |
Notes
(Corrected from the printed mark scheme: the printed scheme has \(\dfrac{87 + c}{19}\) in both places; dividing through by 16 gives \(\dfrac{87 + c}{16}\).)
B1: Selects the method of making a connection between \(x\) and \(w\) by writing \(w = 4x - 1\) or \(x = \dfrac{w + 1}{4}\)
M1: Applies the process of substituting their \(x = \dfrac{w + 1}{4}\) into \(ax^3 + bx^2 - 19x - b = 0\) or \(w = 4x - 1\) into \(w^3 - 9w^2 - 97w + c = 0\). Must be substitution of the correct variable into the opposing equation but may be scored if the initial linear equation is incorrect (e.g. \(x = 4w - 1\) into the first equation). Note that the “ = 0 ” can be missing for this mark.
M1: Expands the brackets and collects terms in their equation (in \(x\) or \(w\)). Note that the “ = 0 ” can be missing for this mark.
M1: A complete method for finding a value for \(a\) or \(b\). See scheme, it involves dividing through by an appropriate factor for their equation to balance the \(w^3\) or \(-19x\) terms, then equating other coefficients and solving equations if necessary.
M1: A complete method for finding a value for \(c\). They must have divided through by an appropriate factor as per the previous M before attempting to compare the constant coefficient (and use their \(a\) and \(b\) if appropriate).
A1: \(a = 4 \quad b = -12 \quad c = 105\)
Alternative
| Scheme | Marks | AO |
|---|---|---|
| At least two of \(\alpha + \beta + \gamma = -\dfrac{b}{a} \qquad \alpha\beta + \alpha\gamma + \beta\gamma = -\dfrac{19}{a} \qquad \alpha\beta\gamma = \dfrac{b}{a}\) | B1 | 3.1a |
| New sum \(= 4(\alpha + \beta + \gamma) - 3 = 9 \Rightarrow 4\left(-\dfrac{b}{a}\right) - 3 = 9 \Rightarrow b = -3a\) | M1 | 3.1a |
| New pair sum \(= 16(\alpha\beta + \alpha\gamma + \beta\gamma) - 8(\alpha + \beta + \gamma) + 3 = -97\) \(\Rightarrow 16\left(-\dfrac{19}{a}\right) - 8\left(-\dfrac{b}{a}\right) + 3 = -97\) | M1 | 1.1b |
| \(\Rightarrow 16\left(-\dfrac{19}{a}\right) - 8(3) + 3 = -97 \Rightarrow a = \ldots\) | M1 | 3.1a |
| New product \(64(\alpha\beta\gamma) - 16(\alpha\beta + \alpha\gamma + \beta\gamma) + 4(\alpha + \beta + \gamma) - 1 = -c\) \(\Rightarrow 64\left(\dfrac{b}{a}\right) - 16\left(-\dfrac{19}{a}\right) + 4(3) - 1 = -c \Rightarrow c = \ldots\) | M1 | 1.1b |
| \(a = 4 \quad b = -12 \quad c = 105\) | A1 | 1.1b |
| (6) |
B1: Selects the method of giving at least two correct equations containing \(\alpha\), \(\beta\) and \(\gamma\)
M1: Applies the process of finding the new sum to generate an equation in \(a\) and \(b\). Must be substituting in the correct places.
M1: Attempts the new pair sum to generate another equation connecting \(a\) and \(b\). Must be substituting in the correct places.
M1: Solves their equations to find a value for \(a\) or \(b\).
M1: Uses the new product with their values to find values for \(a\), \(b\) and \(c\)
A1: \(a = 4 \quad b = -12 \quad c = 105\)