| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on polynomial roots with clear guidance. The substitution is given explicitly, and finding power sums using recurrence relations is a standard FP1 technique. While it requires multiple steps and familiarity with symmetric functions, the path is well-signposted and the methods are routine for this syllabus level. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks |
|---|---|
| Obtains an equation in \(y\) not involving radicals, e.g., \(y(y+1)^2 = 1\) | M1 |
| \(\Rightarrow ... \Rightarrow y^3 + 2y^2 + y - 1 = 0\) (AG) | A1 |
| [2] | |
| (i) \(S_2 = -2\) | B1 |
| \(S_4 = 4 - 2 = 2\) | M1A1 |
| [3] | |
| (ii) \(S_6 = -2S_4 - S_2 + 3 = 1\) | M1A1 |
| Answer | Marks |
|---|---|
| \(\Sigma\alpha^2 = -2, \Sigma\alpha^2\beta^2 = 1, \alpha^2\beta\gamma^2 = 1\) | |
| \(S_6 = (\Sigma\alpha^2)^3 - 3\Sigma\alpha^2\Sigma\alpha^2\beta^2 + 3\alpha^2\beta^2\gamma^2\) | M1 |
| \(= (-2)^3 - 3 \times (-2) \times 1 + 3\) | |
| \(= -8 + 6 + 3\) | |
| \(= 1\) | A1 |
| \(S_8 = -2S_6 - S_4 + S_2 = -6\) | M1A1 |
| [4] |
Obtains an equation in $y$ not involving radicals, e.g., $y(y+1)^2 = 1$ | M1 |
$\Rightarrow ... \Rightarrow y^3 + 2y^2 + y - 1 = 0$ (AG) | A1 |
| [2] |
**(i)** $S_2 = -2$ | B1 |
$S_4 = 4 - 2 = 2$ | M1A1 |
| [3] |
**(ii)** $S_6 = -2S_4 - S_2 + 3 = 1$ | M1A1 |
**OR**
$\Sigma\alpha^2 = -2, \Sigma\alpha^2\beta^2 = 1, \alpha^2\beta\gamma^2 = 1$ |
$S_6 = (\Sigma\alpha^2)^3 - 3\Sigma\alpha^2\Sigma\alpha^2\beta^2 + 3\alpha^2\beta^2\gamma^2$ | M1 |
$= (-2)^3 - 3 \times (-2) \times 1 + 3$ |
$= -8 + 6 + 3$ |
$= 1$ | A1 |
$S_8 = -2S_6 - S_4 + S_2 = -6$ | M1A1 |
| [4] |6 The equation
$$x ^ { 3 } + x - 1 = 0$$
has roots $\alpha , \beta , \gamma$. Use the relation $x = \sqrt { } y$ to show that the equation
$$y ^ { 3 } + 2 y ^ { 2 } + y - 1 = 0$$
has roots $\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }$.
Let $S _ { n } = \alpha ^ { n } + \beta ^ { n } + \gamma ^ { n }$.\\
(i) Write down the value of $S _ { 2 }$ and show that $S _ { 4 } = 2$.\\
(ii) Find the values of $S _ { 6 }$ and $S _ { 8 }$.
\hfill \mbox{\textit{CAIE FP1 2010 Q6 [9]}}