OCR Further Pure Core AS 2024 June — Question 7 6 marks

Exam BoardOCR
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyChallenging +1.8 This is a Further Maths question requiring systematic application of Vieta's formulas and algebraic manipulation to find relationships between transformed roots. While the technique is standard for FM students (finding sum, sum of products, and product of new roots), it requires careful multi-step reasoning with expressions like (αβ+βγ+γα)² and handling fractions of roots. The 'show detailed reasoning' requirement and the need to construct the final equation from scratch elevate this above routine practice, but it follows a well-established method without requiring novel insight.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

7 In this question you must show detailed reasoning.
The roots of the equation \(2 x ^ { 3 } - 3 x ^ { 2 } - 3 x + 5 = 0\) are \(\alpha , \beta\) and \(\gamma\).
By considering \(( \alpha + \beta + \gamma ) ^ { 2 }\) and \(( \alpha \beta + \beta \gamma + \gamma \alpha ) ^ { 2 }\), determine a cubic equation with integer coefficients whose roots are \(\frac { \alpha \beta } { \gamma } , \frac { \beta \gamma } { \alpha }\) and \(\frac { \gamma \alpha } { \beta }\).

Question 7:
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha\beta\gamma = -\frac{5}{2},\ \sum\alpha\beta = -\frac{3}{2},\ \sum\alpha = \frac{3}{2}\)M1 Attempt to find three Vieta's formulae for original equation. This mark awarded if 2 of 3 are correct and 3rd has been attempted, or if all correct but for signs.
\(\alpha'\beta'\gamma' = \frac{\alpha\beta}{\gamma}\times\frac{\beta\gamma}{\alpha}\times\frac{\gamma\alpha}{\beta} = \alpha\beta\gamma\left(= -\frac{5}{2}\right)\)B1 Showing that the new product of roots is the same as the original.
\((\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)\) \((\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)\)M1 Both considered and at least one correctly written in desired form using Vieta's expressions. May be embedded in expressions for \(\sum\alpha'\beta'\) or \(\sum\alpha'\).
\(\sum\alpha'\beta' = \frac{\alpha\beta}{\gamma}\cdot\frac{\beta\gamma}{\alpha}+\frac{\beta\gamma}{\alpha}\cdot\frac{\gamma\alpha}{\beta}+\frac{\gamma\alpha}{\beta}\cdot\frac{\alpha\beta}{\gamma}\) \(= \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)\) \(= \left(\frac{3}{2}\right)^2 - 2\times\left(-\frac{3}{2}\right) = \frac{9}{4}+\frac{12}{4} = \frac{21}{4}\)B1 For correct \(\sum\alpha'\beta'\) in terms of Vieta's expressions i.e. \(\sum\alpha'\beta' = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\). Can be awarded before numerical value found. Correct numerical value implies correct \(\sum\alpha'\beta'\) (but must be identified as \(\sum\alpha'\beta'\)).
\(\sum\alpha' = \frac{\alpha\beta}{\gamma}+\frac{\beta\gamma}{\alpha}+\frac{\gamma\alpha}{\beta} = \frac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{\alpha\beta\gamma}\) \(= \frac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)}{\alpha\beta\gamma}\) \(= \frac{\left(-\frac{3}{2}\right)^2 - 2\times\left(-\frac{5}{2}\right)\times\frac{3}{2}}{-\frac{5}{2}} = -\frac{9}{4}\times\frac{2}{5} - \frac{30}{10} = -\frac{39}{10}\)B1 For correct \(\sum\alpha'\) in terms of Vieta's expressions. Can be awarded before numerical value found. Correct numerical value implies correct \(\sum\alpha'\) (but must be identified as \(\sum\alpha'\)).
Choosing \(a'=20\) gives \(20x^3+78x^2+105x+50=0\)A1 Any non-zero integer multiple. Must be a cubic *equation* (i.e. must have "\(=0\)") with integer coefficients but can be in any unknown. Correct answer from finding roots on calculator only scores 0/6.
[6]
## Question 7:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha\beta\gamma = -\frac{5}{2},\ \sum\alpha\beta = -\frac{3}{2},\ \sum\alpha = \frac{3}{2}$ | M1 | Attempt to find three Vieta's formulae for original equation. This mark awarded if 2 of 3 are correct and 3rd has been attempted, or if all correct but for signs. |
| $\alpha'\beta'\gamma' = \frac{\alpha\beta}{\gamma}\times\frac{\beta\gamma}{\alpha}\times\frac{\gamma\alpha}{\beta} = \alpha\beta\gamma\left(= -\frac{5}{2}\right)$ | B1 | Showing that the new product of roots is the same as the original. |
| $(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha)$ $(\alpha\beta+\beta\gamma+\gamma\alpha)^2 = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2+2\alpha\beta\gamma(\alpha+\beta+\gamma)$ | M1 | Both considered and at least one correctly written in desired form using Vieta's expressions. May be embedded in expressions for $\sum\alpha'\beta'$ or $\sum\alpha'$. |
| $\sum\alpha'\beta' = \frac{\alpha\beta}{\gamma}\cdot\frac{\beta\gamma}{\alpha}+\frac{\beta\gamma}{\alpha}\cdot\frac{\gamma\alpha}{\beta}+\frac{\gamma\alpha}{\beta}\cdot\frac{\alpha\beta}{\gamma}$ $= \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)$ $= \left(\frac{3}{2}\right)^2 - 2\times\left(-\frac{3}{2}\right) = \frac{9}{4}+\frac{12}{4} = \frac{21}{4}$ | B1 | For correct $\sum\alpha'\beta'$ in terms of Vieta's expressions i.e. $\sum\alpha'\beta' = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$. Can be awarded before numerical value found. Correct numerical value implies correct $\sum\alpha'\beta'$ (but must be identified as $\sum\alpha'\beta'$). |
| $\sum\alpha' = \frac{\alpha\beta}{\gamma}+\frac{\beta\gamma}{\alpha}+\frac{\gamma\alpha}{\beta} = \frac{\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2}{\alpha\beta\gamma}$ $= \frac{(\alpha\beta+\beta\gamma+\gamma\alpha)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma)}{\alpha\beta\gamma}$ $= \frac{\left(-\frac{3}{2}\right)^2 - 2\times\left(-\frac{5}{2}\right)\times\frac{3}{2}}{-\frac{5}{2}} = -\frac{9}{4}\times\frac{2}{5} - \frac{30}{10} = -\frac{39}{10}$ | B1 | For correct $\sum\alpha'$ in terms of Vieta's expressions. Can be awarded before numerical value found. Correct numerical value implies correct $\sum\alpha'$ (but must be identified as $\sum\alpha'$). |
| Choosing $a'=20$ gives $20x^3+78x^2+105x+50=0$ | A1 | Any non-zero integer multiple. Must be a cubic *equation* (i.e. must have "$=0$") with integer coefficients but can be in any unknown. Correct answer from finding roots on calculator only scores 0/6. |
| **[6]** | | |
7 In this question you must show detailed reasoning.\\
The roots of the equation $2 x ^ { 3 } - 3 x ^ { 2 } - 3 x + 5 = 0$ are $\alpha , \beta$ and $\gamma$.\\
By considering $( \alpha + \beta + \gamma ) ^ { 2 }$ and $( \alpha \beta + \beta \gamma + \gamma \alpha ) ^ { 2 }$, determine a cubic equation with integer coefficients whose roots are $\frac { \alpha \beta } { \gamma } , \frac { \beta \gamma } { \alpha }$ and $\frac { \gamma \alpha } { \beta }$.

\hfill \mbox{\textit{OCR Further Pure Core AS 2024 Q7 [6]}}