| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.8 This is a standard Further Maths transformed roots question requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) is routine recall, part (b) uses the standard identity α²+β²=(α+β)²-2αβ, but part (c) requires finding sum and product of transformed roots (3α-β and 3β-α) which involves multiple algebraic steps and careful manipulation. While methodical rather than requiring deep insight, it's more demanding than typical A-level pure questions due to the algebraic complexity and being from FP1. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha + \beta = -\frac{3}{2}\), \(\alpha\beta = \frac{3}{4}\) | B1B1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^2 + \beta^2 = \left(-\frac{3}{2}\right)^2 - 2\left(\frac{3}{4}\right) = \frac{3}{4}\) | M1A1 | AG; A0 if \(\alpha + \beta\) has wrong sign |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sum \(= 2(\alpha + \beta) = -3\) | B1F | ft wrong value for \(\alpha + \beta\) |
| Product \(= 10\alpha\beta - 3(\alpha^2 + \beta^2) = \frac{21}{4}\) | M1A1F | ft wrong values |
| \(x^2 - Sx + P (= 0)\) | M1 | Signs must be correct for the M1 |
| Eqn is \(4x^2 + 12x + 21 = 0\) | A1 | Integer coeffs and \('= 0'\) needed |
| Total | 5 |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha + \beta = -\frac{3}{2}$, $\alpha\beta = \frac{3}{4}$ | B1B1 | |
| **Total** | **2** | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^2 + \beta^2 = \left(-\frac{3}{2}\right)^2 - 2\left(\frac{3}{4}\right) = \frac{3}{4}$ | M1A1 | AG; A0 if $\alpha + \beta$ has wrong sign |
| **Total** | **2** | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sum $= 2(\alpha + \beta) = -3$ | B1F | ft wrong value for $\alpha + \beta$ |
| Product $= 10\alpha\beta - 3(\alpha^2 + \beta^2) = \frac{21}{4}$ | M1A1F | ft wrong values |
| $x^2 - Sx + P (= 0)$ | M1 | Signs must be correct for the M1 |
| Eqn is $4x^2 + 12x + 21 = 0$ | A1 | Integer coeffs and $'= 0'$ needed |
| **Total** | **5** | |
---2 The equation
$$4 x ^ { 2 } + 6 x + 3 = 0$$
has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the values of $\alpha + \beta$ and $\alpha \beta$.
\item Show that $\alpha ^ { 2 } + \beta ^ { 2 } = \frac { 3 } { 4 }$.
\item Find an equation, with integer coefficients, which has roots
$$3 \alpha - \beta \text { and } 3 \beta - \alpha$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2011 Q2 [9]}}