CAIE FP1 2011 June — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2011
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with special relationships
DifficultyStandard +0.8 This is a Further Maths question requiring systematic application of Vieta's formulas with algebraically related roots. Students must manipulate the sum and product of roots (β/k + β + kβ = -p and β³ = -r) to eliminate k and establish two non-trivial relationships. While the techniques are standard, the algebraic manipulation across multiple steps and the need to strategically combine equations elevates this above routine problems.
Spec4.05a Roots and coefficients: symmetric functions

2 The roots of the equation $$x ^ { 3 } + p x ^ { 2 } + q x + r = 0$$ are \(\frac { \beta } { k } , \beta , k \beta\), where \(p , q , r , k\) and \(\beta\) are non-zero real constants. Show that \(\beta = - \frac { q } { p }\). Deduce that \(r p ^ { 3 } = q ^ { 3 }\).

Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{\beta + \beta k + \beta k^2}{k} = -p\)B1 Sum of roots
\(\frac{\beta^2}{k} + k\beta^2 + \beta^2 = q\)B1 Sum of products in pairs
\(\Rightarrow \beta\left(\frac{k^2+k+1}{k}\right) = -p\) and \(\beta^2\left(\frac{k^2+k+1}{k}\right) = q\)M1 Factorises
\(\Rightarrow \beta = -\frac{q}{p}\) (AG)A1 Part mark: 4
\(\beta^3 = -r\)B1 Product of roots
\(\Rightarrow -\frac{q^3}{p^3} = -r \Rightarrow rp^3 = q^3\) (AG)B1 Part mark: 2
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\beta + \beta k + \beta k^2}{k} = -p$ | B1 | Sum of roots |
| $\frac{\beta^2}{k} + k\beta^2 + \beta^2 = q$ | B1 | Sum of products in pairs |
| $\Rightarrow \beta\left(\frac{k^2+k+1}{k}\right) = -p$ and $\beta^2\left(\frac{k^2+k+1}{k}\right) = q$ | M1 | Factorises |
| $\Rightarrow \beta = -\frac{q}{p}$ (AG) | A1 | Part mark: 4 |
| $\beta^3 = -r$ | B1 | Product of roots |
| $\Rightarrow -\frac{q^3}{p^3} = -r \Rightarrow rp^3 = q^3$ (AG) | B1 | Part mark: 2 |

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2 The roots of the equation

$$x ^ { 3 } + p x ^ { 2 } + q x + r = 0$$

are $\frac { \beta } { k } , \beta , k \beta$, where $p , q , r , k$ and $\beta$ are non-zero real constants. Show that $\beta = - \frac { q } { p }$.

Deduce that $r p ^ { 3 } = q ^ { 3 }$.

\hfill \mbox{\textit{CAIE FP1 2011 Q2 [6]}}