CAIE FP1 2009 November — Question 5 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. While it involves multiple steps (substitution, verification of roots, computing symmetric functions), the techniques are well-practiced in FP1 and follow predictable patterns without requiring novel insight.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

5 The equation $$x ^ { 3 } + 5 x + 3 = 0$$ has roots \(\alpha , \beta , \gamma\). Use the substitution \(x = - \frac { 3 } { y }\) to find a cubic equation in \(y\) and show that the roots of this equation are \(\beta \gamma , \gamma \alpha , \alpha \beta\). Find the exact values of \(\beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 }\) and \(\beta ^ { 3 } \gamma ^ { 3 } + \gamma ^ { 3 } \alpha ^ { 3 } + \alpha ^ { 3 } \beta ^ { 3 }\).

AnswerMarks
Shows \(y = -3/x \Rightarrow y^3 - 5y^2 - 9 = 0\)B1
\(y = -3/x \Rightarrow y = a\theta/x\)M1
\(\Rightarrow y = [\beta y, \gamma a\beta] \text{ when } x = a, \beta, \gamma, \text{ respectively}\)M1A1
OR for previous 3 marks:
\([\beta y + \gamma a + a\beta = 5\)B1
\(a[^2\beta y + a\beta^2y + a\beta\gamma^2 = a\beta\gamma(a + \beta + \gamma) = a\beta \times 0 = 0\)B1
\(\beta\gamma a a\beta = (a\beta\gamma)^2 = 9\)B1
\(\Sigma a[\beta]^2 = 25 - (2 \times 0) = 25\)M1A1
\(\Sigma a[\beta]^3 - 5\Sigma a^2[\beta]^2 - 27 = 0\)M1A1
\(\Rightarrow ... \Rightarrow \Sigma a[\beta]^3 = 152\)A1
Shows $y = -3/x \Rightarrow y^3 - 5y^2 - 9 = 0$ | B1 |
$y = -3/x \Rightarrow y = a\theta/x$ | M1 |
$\Rightarrow y = [\beta y, \gamma a\beta] \text{ when } x = a, \beta, \gamma, \text{ respectively}$ | M1A1 |

**OR** for previous 3 marks: | |
$[\beta y + \gamma a + a\beta = 5$ | B1 |
$a[^2\beta y + a\beta^2y + a\beta\gamma^2 = a\beta\gamma(a + \beta + \gamma) = a\beta \times 0 = 0$ | B1 |
$\beta\gamma a a\beta = (a\beta\gamma)^2 = 9$ | B1 |
$\Sigma a[\beta]^2 = 25 - (2 \times 0) = 25$ | M1A1 |
$\Sigma a[\beta]^3 - 5\Sigma a^2[\beta]^2 - 27 = 0$ | M1A1 |
$\Rightarrow ... \Rightarrow \Sigma a[\beta]^3 = 152$ | A1 |
5 The equation

$$x ^ { 3 } + 5 x + 3 = 0$$

has roots $\alpha , \beta , \gamma$. Use the substitution $x = - \frac { 3 } { y }$ to find a cubic equation in $y$ and show that the roots of this equation are $\beta \gamma , \gamma \alpha , \alpha \beta$.

Find the exact values of $\beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 } + \alpha ^ { 2 } \beta ^ { 2 }$ and $\beta ^ { 3 } \gamma ^ { 3 } + \gamma ^ { 3 } \alpha ^ { 3 } + \alpha ^ { 3 } \beta ^ { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2009 Q5 [9]}}