| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with given sum conditions |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question involving complex polynomial roots with Vieta's formulas and a constraint condition. While parts (a)-(c) follow systematically from the given condition α=β+γ using standard techniques, part (d) requires solving a complex quadratic and identifying which root is real. The complexity lies in careful algebraic manipulation with complex numbers across multiple connected parts rather than deep conceptual insight, placing it moderately above average difficulty. |
| Spec | 4.02i Quadratic equations: with complex roots4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(\alpha + \beta + \gamma = 4i\) | B1 | (1 mark) |
| (ii) \(\alpha\beta\gamma = 4 - 2i\) | B1 | (1 mark) |
| (b)(i) \(\alpha + \alpha = 4i, \alpha = 2i\) | B1 | AG (1 mark) |
| (ii) \(\beta\gamma = \frac{4-2i}{2i} = -2i - 1\) | M1, A1 | Some method must be shown, eg \(\frac{2}{i} - 1\); AG (2 marks) |
| (iii) \(q = \alpha\beta + \beta\gamma + \gamma\alpha = \alpha(\beta + \gamma) + \beta\gamma = 2i.2i - 2i - 1 = -2i - 5\) | M1, M1, A1 | Or \(\alpha^2 + \beta\gamma\), ie suitable grouping; AG (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(z^2 - 2iz - (1+2i) = 0\) | M1, A1 | Elimination of say \(\gamma\) to arrive at \(\beta^2 - 2i\beta - (1+2i) = 0\); M1A0 unless also some reference to \(\gamma\) being a root; AG (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\beta = -1, \gamma = 1 + 2i\) | M1, A1A1 | For any correct method; A1 for each answer (3 marks) |
**(a)(i)** $\alpha + \beta + \gamma = 4i$ | B1 | (1 mark)
**(ii)** $\alpha\beta\gamma = 4 - 2i$ | B1 | (1 mark)
**(b)(i)** $\alpha + \alpha = 4i, \alpha = 2i$ | B1 | AG (1 mark)
**(ii)** $\beta\gamma = \frac{4-2i}{2i} = -2i - 1$ | M1, A1 | Some method must be shown, eg $\frac{2}{i} - 1$; AG (2 marks)
**(iii)** $q = \alpha\beta + \beta\gamma + \gamma\alpha = \alpha(\beta + \gamma) + \beta\gamma = 2i.2i - 2i - 1 = -2i - 5$ | M1, M1, A1 | Or $\alpha^2 + \beta\gamma$, ie suitable grouping; AG (3 marks)
**(c)** Use of $\beta + \gamma = 2i$ and $\beta\gamma = -2i - 1$
$z^2 - 2iz - (1+2i) = 0$ | M1, A1 | Elimination of say $\gamma$ to arrive at $\beta^2 - 2i\beta - (1+2i) = 0$; M1A0 unless also some reference to $\gamma$ being a root; AG (2 marks)
**(d)** $f(-1) = 1 + 2i - 1 - 2i = 0$
$\beta = -1, \gamma = 1 + 2i$ | M1, A1A1 | For any correct method; A1 for each answer (3 marks)
**Total: 13 marks**
---5 The cubic equation
$$z ^ { 3 } - 4 \mathrm { i } z ^ { 2 } + q z - ( 4 - 2 \mathrm { i } ) = 0$$
where $q$ is a complex number, has roots $\alpha , \beta$ and $\gamma$.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of:
\begin{enumerate}[label=(\roman*)]
\item $\alpha + \beta + \gamma$;
\item $\alpha \beta \gamma$.
\end{enumerate}\item Given that $\alpha = \beta + \gamma$, show that:
\begin{enumerate}[label=(\roman*)]
\item $\alpha = 2 \mathrm { i }$;
\item $\quad \beta \gamma = - ( 1 + 2 \mathrm { i } )$;
\item $\quad q = - ( 5 + 2 \mathrm { i } )$.
\end{enumerate}\item Show that $\beta$ and $\gamma$ are the roots of the equation
$$z ^ { 2 } - 2 \mathrm { i } z - ( 1 + 2 \mathrm { i } ) = 0$$
\item Given that $\beta$ is real, find $\beta$ and $\gamma$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q5 [13]}}