| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| 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 Further Maths question requiring systematic application of Vieta's formulas and algebraic manipulation to find transformed roots. Part (a) is routine, part (b) is standard, but part (c) requires finding sum and product of asymmetric cubic transformations (α³β+1 and αβ³+1), which demands careful algebraic work beyond typical A-level questions. The multi-step nature and need to manipulate higher powers of roots places this moderately above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha + \beta = -2\) | B1 | |
| \(\alpha\beta = -5\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-2)^2 - 2(-5)\) | M1 | Using correct identity for \(\alpha^2+\beta^2\) with ft or correct substitution |
| \(= 14\) | A1 | CSO. A0 if \(\alpha+\beta\) has wrong sign |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^3\beta + \alpha\beta^3 = \alpha\beta(\alpha^2+\beta^2)\) | M1 | PI. OE e.g. \(\alpha^3\beta+\alpha\beta^3 = \alpha\beta[(\alpha+\beta)^2-2\alpha\beta]\) |
| \(S(\text{sum}) = \alpha^3\beta+\alpha\beta^3+2 = (-5)(14)+2 = -68\) | A1F | Correct or ft c's \(\alpha\beta \times\) c's [answer (b)] \(+2\) |
| \(P(\text{product}) = (\alpha\beta)^4 + \alpha^3\beta+\alpha\beta^3+1 = (-5)^4+(-5)(14)+1 = 556\) | A1F | Correct or ft \([\text{c's } \alpha\beta]^4 + \text{c's } \alpha\beta \times \text{c's [answer (b)]}+1\) |
| \(x^2 - Sx + P = 0\) | M1 | Using correct general form of LHS with ft substitution of c's \(S\) and \(P\) |
| Equation: \(x^2 + 68x + 556 = 0\) | A1 | CSO ACF |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha + \beta = -2$ | B1 | |
| $\alpha\beta = -5$ | B1 | |
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = (-2)^2 - 2(-5)$ | M1 | Using correct identity for $\alpha^2+\beta^2$ with ft or correct substitution |
| $= 14$ | A1 | CSO. A0 if $\alpha+\beta$ has wrong sign |
## Question 5(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^3\beta + \alpha\beta^3 = \alpha\beta(\alpha^2+\beta^2)$ | M1 | PI. OE e.g. $\alpha^3\beta+\alpha\beta^3 = \alpha\beta[(\alpha+\beta)^2-2\alpha\beta]$ |
| $S(\text{sum}) = \alpha^3\beta+\alpha\beta^3+2 = (-5)(14)+2 = -68$ | A1F | Correct or ft c's $\alpha\beta \times$ c's [answer (b)] $+2$ |
| $P(\text{product}) = (\alpha\beta)^4 + \alpha^3\beta+\alpha\beta^3+1 = (-5)^4+(-5)(14)+1 = 556$ | A1F | Correct or ft $[\text{c's } \alpha\beta]^4 + \text{c's } \alpha\beta \times \text{c's [answer (b)]}+1$ |
| $x^2 - Sx + P = 0$ | M1 | Using correct general form of LHS with ft substitution of c's $S$ and $P$ |
| Equation: $x^2 + 68x + 556 = 0$ | A1 | CSO ACF |5 The roots of the quadratic equation
$$x ^ { 2 } + 2 x - 5 = 0$$
are $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $\alpha + \beta$ and the value of $\alpha \beta$.
\item Calculate the value of $\alpha ^ { 2 } + \beta ^ { 2 }$.
\item Find a quadratic equation which has roots $\alpha ^ { 3 } \beta + 1$ and $\alpha \beta ^ { 3 } + 1$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2013 Q5 [9]}}