| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 7 |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Challenging +1.3 This is a Further Pure 1 question on roots of equations requiring systematic application of Vieta's formulas and algebraic manipulation. Part (i) is straightforward substitution into a given identity. Part (ii) requires forming a new cubic from symmetric functions of transformed roots, which is a standard FP1 technique but demands careful algebraic work across multiple steps. The question is harder than typical A-level pure maths but represents core FP1 material without requiring exceptional insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: DR \(\alpha\beta\gamma = -4\) and \(\alpha + \beta + \gamma = -3\) | B1 | In this question simply generating and using values of \(\alpha\), \(\beta\) and \(\gamma\) should gain no credit. Sufficient working must be shown as evidence that this has not been done. |
| Answer: \(\alpha\beta + \beta\gamma + \gamma\alpha = -2\) | B1 | |
| Answer: \(\alpha^3 + \beta^3 + \gamma^3 \equiv (-3)^3 - 3(-2)(-3) + 3(-4) = -57\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer: DR Understanding how the new roots (\(A\), \(B\) and \(\Gamma\)) relate to the old ones | M1ft | |
| Answer: \(AB\Gamma = \alpha^3\beta^3\gamma^3 = (-4)^3\) | B1 | Seen or implied |
| Answer: \(a(x^3 + 57x^2 + 112x + 64) = 0\) | M1 | |
| Answer: \(x^3 + 57x^2 + 112x + 64 = 0\) | A1 |
## (i)
Answer: DR $\alpha\beta\gamma = -4$ and $\alpha + \beta + \gamma = -3$ | B1 | In this question simply generating and using values of $\alpha$, $\beta$ and $\gamma$ should gain no credit. Sufficient working **must** be shown as evidence that this has not been done.
Answer: $\alpha\beta + \beta\gamma + \gamma\alpha = -2$ | B1 |
Answer: $\alpha^3 + \beta^3 + \gamma^3 \equiv (-3)^3 - 3(-2)(-3) + 3(-4) = -57$ | B1 |
## (ii)
Answer: DR Understanding how the new roots ($A$, $B$ and $\Gamma$) relate to the old ones | M1ft |
Answer: $AB\Gamma = \alpha^3\beta^3\gamma^3 = (-4)^3$ | B1 | Seen or implied
Answer: $a(x^3 + 57x^2 + 112x + 64) = 0$ | M1 |
Answer: $x^3 + 57x^2 + 112x + 64 = 0$ | A1 |
---\textbf{In this question you must show detailed reasoning.}
The equation $x^3 + 3x^2 - 2x + 4 = 0$ has roots $\alpha$, $\beta$ and $\gamma$.
\begin{enumerate}[label=(\roman*)]
\item Using the identity $\alpha^3 + \beta^3 + \gamma^3 \equiv (\alpha + \beta + \gamma)^3 - 3(\alpha\beta + \beta\gamma + \gamma\alpha)(\alpha + \beta + \gamma) + 3\alpha\beta\gamma$ find the value of $\alpha^3 + \beta^3 + \gamma^3$. [3]
\item Given that $\alpha^2\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 112$ find a cubic equation whose roots are $\alpha^2$, $\beta^3$ and $\gamma^3$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 AS 2017 Q5 [7]}}