CAIE FP1 2009 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSubstitution to find new equation
DifficultyStandard +0.8 This is a Further Maths question requiring insight to use the substitution y = x³ to transform the original equation into one relating x⁶ terms, then apply symmetric function techniques. While the algebraic manipulation is manageable once the approach is identified, recognizing the method and executing the multi-step solution elevates this above standard A-level difficulty.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

1 The equation $$x ^ { 4 } - x ^ { 3 } - 1 = 0$$ has roots \(\alpha , \beta , \gamma , \delta\). By using the substitution \(y = x ^ { 3 }\), or by any other method, find the exact value of \(\alpha ^ { 6 } + \beta ^ { 6 } + \gamma ^ { 6 } + \delta ^ { 6 }\).

AnswerMarks
\(x = y^{1/2} \Rightarrow y^4 = (1+y)^3\)M1A1
\(\Rightarrow y^4 - y^3 - 3y^2 - 3y - 1 = 0\)A1
\(\sum \alpha^6 = 1 - 2 \times (-3) = 7\)M1A1
or \(y = x^2 \Rightarrow y^4 - y^3 - 2y^2 + 1 = 0\) has roots \(\alpha^2, \beta^2, \gamma^2\)M1
\(\sum \alpha^6 = (\sum \alpha^2)(\sum \alpha^4) - (\sum \alpha^2)(\sum \alpha^2\beta^2) + 3(\sum \alpha^2\beta^2\gamma^2)\)M1A2
\(= 5 - (-2) + 0 = 7\)A1
or For last 2 marks:
AnswerMarks
\(z = y^2 \Rightarrow z^4 - 7z^3 + z^2 - 3z + 1 = 0\)M1
\(\sum \alpha^6 = \frac{(-7)}{1} = 7\)A1
or Use of \(S_{N+4} = S_N + S_{N+3}\)M1
\(S_{-1} = \frac{0}{1} = 0\), \(S_2 = 1^2 - 2 \times 0 = 1\)(both) M1A1
\(S_3 = 0 + 1 = 1\), \(S_4 = 1 + 4 = 5\)(both) A1
\(S_5 = 5 + 1 = 6\), \(S_6 = 6 + 1 = 7\)(both) A1
$x = y^{1/2} \Rightarrow y^4 = (1+y)^3$ | M1A1 | 

$\Rightarrow y^4 - y^3 - 3y^2 - 3y - 1 = 0$ | A1 |

$\sum \alpha^6 = 1 - 2 \times (-3) = 7$ | M1A1 |

or $y = x^2 \Rightarrow y^4 - y^3 - 2y^2 + 1 = 0$ has roots $\alpha^2, \beta^2, \gamma^2$ | M1 |

$\sum \alpha^6 = (\sum \alpha^2)(\sum \alpha^4) - (\sum \alpha^2)(\sum \alpha^2\beta^2) + 3(\sum \alpha^2\beta^2\gamma^2)$ | M1A2 |

$= 5 - (-2) + 0 = 7$ | A1 |

**or For last 2 marks:**

$z = y^2 \Rightarrow z^4 - 7z^3 + z^2 - 3z + 1 = 0$ | M1 |

$\sum \alpha^6 = \frac{(-7)}{1} = 7$ | A1 |

**or Use of $S_{N+4} = S_N + S_{N+3}$** | M1 |

$S_{-1} = \frac{0}{1} = 0$, $S_2 = 1^2 - 2 \times 0 = 1$ | (both) M1A1 |

$S_3 = 0 + 1 = 1$, $S_4 = 1 + 4 = 5$ | (both) A1 |

$S_5 = 5 + 1 = 6$, $S_6 = 6 + 1 = 7$ | (both) A1 |
1 The equation

$$x ^ { 4 } - x ^ { 3 } - 1 = 0$$

has roots $\alpha , \beta , \gamma , \delta$. By using the substitution $y = x ^ { 3 }$, or by any other method, find the exact value of $\alpha ^ { 6 } + \beta ^ { 6 } + \gamma ^ { 6 } + \delta ^ { 6 }$.

\hfill \mbox{\textit{CAIE FP1 2009 Q1 [5]}}