CAIE FP1 2013 June — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSubstitution to find new equation
DifficultyChallenging +1.2 This is a Further Maths question involving polynomial root relationships and recurrence relations. While it requires understanding of symmetric functions and the substitution y=x², the method is fairly standard for FP1: square the original equation, rearrange to eliminate odd powers, then use Newton's recurrence. The final part uses the new equation with S_n notation systematically. More challenging than typical A-level due to algebraic manipulation length, but follows a well-established technique taught in Further Maths courses.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

2 The roots of the equation \(x ^ { 4 } - 4 x ^ { 2 } + 3 x - 2 = 0\) are \(\alpha , \beta , \gamma\) and \(\delta\); the sum \(\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } + \delta ^ { n }\) is denoted by \(S _ { n }\). By using the relation \(y = x ^ { 2 }\), or otherwise, show that \(\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }\) and \(\delta ^ { 2 }\) are the roots of the equation $$y ^ { 4 } - 8 y ^ { 3 } + 12 y ^ { 2 } + 7 y + 4 = 0$$ State the value of \(S _ { 2 }\) and hence show that $$S _ { 8 } = 8 S _ { 6 } - 12 S _ { 4 } - 72 .$$

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y^2 - 4y + 3y^{\frac{1}{2}} - 2 = 0\)M1 Makes substitution
\(\Rightarrow 9y = 4 + y^4 + 16y^2 - 4y^2 + 16y - 8y^3\)M1 Squares (must see both terms in \(y^2\))
\(\Rightarrow y^4 - 8y^3 + 12y^2 + 7y + 4 = 0\) (AG)A1 3 marks
\(S_2 = 0^2 - 2\times(-4) = 8\)B1
\(S_8 = 8S_6 - 12S_4 - 7S_2 - 16\)M1
\(\Rightarrow S_8 = 8S_6 - 12S_4 - 56 - 16 = 8S_6 - 12S_4 - 72\) (AG)A1 3 marks; total [6]
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^2 - 4y + 3y^{\frac{1}{2}} - 2 = 0$ | M1 | Makes substitution |
| $\Rightarrow 9y = 4 + y^4 + 16y^2 - 4y^2 + 16y - 8y^3$ | M1 | Squares (must see both terms in $y^2$) |
| $\Rightarrow y^4 - 8y^3 + 12y^2 + 7y + 4 = 0$ (AG) | A1 | 3 marks |
| $S_2 = 0^2 - 2\times(-4) = 8$ | B1 | |
| $S_8 = 8S_6 - 12S_4 - 7S_2 - 16$ | M1 | |
| $\Rightarrow S_8 = 8S_6 - 12S_4 - 56 - 16 = 8S_6 - 12S_4 - 72$ (AG) | A1 | 3 marks; total [6] |

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2 The roots of the equation $x ^ { 4 } - 4 x ^ { 2 } + 3 x - 2 = 0$ are $\alpha , \beta , \gamma$ and $\delta$; the sum $\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } + \delta ^ { n }$ is denoted by $S _ { n }$. By using the relation $y = x ^ { 2 }$, or otherwise, show that $\alpha ^ { 2 } , \beta ^ { 2 } , \gamma ^ { 2 }$ and $\delta ^ { 2 }$ are the roots of the equation

$$y ^ { 4 } - 8 y ^ { 3 } + 12 y ^ { 2 } + 7 y + 4 = 0$$

State the value of $S _ { 2 }$ and hence show that

$$S _ { 8 } = 8 S _ { 6 } - 12 S _ { 4 } - 72 .$$

\hfill \mbox{\textit{CAIE FP1 2013 Q2 [6]}}