CAIE FP1 2014 June — Question 1 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with special relationships
DifficultyStandard +0.8 This is a Further Maths question requiring insight into root relationships and symmetric functions. Students must recognize that if α and 1/α are roots, then using Vieta's formulas with the third root β leads to the product of roots being β = q, then cleverly manipulating the sum of products to derive the required identity. It requires more algebraic sophistication than standard A-level but is a well-defined proof with clear structure.
Spec1.01a Proof: structure of mathematical proof and logical steps4.05a Roots and coefficients: symmetric functions

1 The equation \(x ^ { 3 } + p x + q = 0\), where \(p\) and \(q\) are constants, with \(q \neq 0\), has one root which is the reciprocal of another root. Prove that \(p + q ^ { 2 } = 1\).

Question 1:
AnswerMarks Guidance
Working/AnswerMark Guidance
Let roots be \(\alpha\), \(\alpha^{-1}\) and \(\beta \Rightarrow \alpha + \alpha^{-1} + \beta = 0\)M1 Any three for M1
Product of roots: \(\beta = -q\)A1 A1 for another
Sum of products in pairs: \(1 + \beta(\alpha + \alpha^{-1}) = p\)A1 A1 for a third
From (1) and (3): \(1 - \beta^2 = p\)M1 Wrong sign in (2) scores M1A0A1M1A0
Using (2): \(1 - q^2 = p\) or \(p + q^2 = 1\)A1
Total: [5]
## Question 1:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Let roots be $\alpha$, $\alpha^{-1}$ and $\beta \Rightarrow \alpha + \alpha^{-1} + \beta = 0$ | M1 | Any three for M1 |
| Product of roots: $\beta = -q$ | A1 | A1 for another |
| Sum of products in pairs: $1 + \beta(\alpha + \alpha^{-1}) = p$ | A1 | A1 for a third |
| From (1) and (3): $1 - \beta^2 = p$ | M1 | Wrong sign in (2) scores M1A0A1M1A0 |
| Using (2): $1 - q^2 = p$ or $p + q^2 = 1$ | A1 | |

**Total: [5]**

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1 The equation $x ^ { 3 } + p x + q = 0$, where $p$ and $q$ are constants, with $q \neq 0$, has one root which is the reciprocal of another root. Prove that $p + q ^ { 2 } = 1$.

\hfill \mbox{\textit{CAIE FP1 2014 Q1 [5]}}