| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.3 This is a standard FP1 question on roots of polynomials requiring routine application of Vieta's formulas and algebraic manipulation. Part (i) involves direct recall and the identity α²+β²+γ²=(α+β+γ)²-2(αβ+βγ+γα). Part (ii) uses a straightforward substitution x=1/u and applies Vieta's formulas again. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average for Further Maths content. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) \(\alpha + \beta + \gamma = 3, \alpha\beta + \beta\gamma + \gamma\alpha = 2\) | B1 B1 | State correct values. |
| (b) \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 9 - 4 = 5\) | M1 A1 ft | State or imply the result and use their values. Obtain correct answer. |
| \(\frac{3}{u} - \frac{9}{u} + \frac{6}{u} + 2 = 0\) | ||
| (ii)(a) \(2u^3 + 6u^2 - 9u + 3 = 0\) | M1 | |
| \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -3\) | A1 | Use given substitution to obtain an equation. Obtain correct answer. |
| (b) | M1 A1ft | |
| 2 + 2 + 2 + 2 = 8 | Required expression is related to new cubic stated or implied \(-\)(their "b" / their "a"). |
(i)(a) $\alpha + \beta + \gamma = 3, \alpha\beta + \beta\gamma + \gamma\alpha = 2$ | B1 B1 | State correct values. |
(b) $\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 9 - 4 = 5$ | M1 A1 ft | State or imply the result and use their values. Obtain correct answer. |
$\frac{3}{u} - \frac{9}{u} + \frac{6}{u} + 2 = 0$ | | |
(ii)(a) $2u^3 + 6u^2 - 9u + 3 = 0$ | M1 | |
$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = -3$ | A1 | Use given substitution to obtain an equation. Obtain correct answer. |
(b) | M1 A1ft | |
| | 2 + 2 + 2 + 2 = 8 | Required expression is related to new cubic stated or implied $-$(their "b" / their "a"). |
**Total: 8 marks**6 The cubic equation $3 x ^ { 3 } - 9 x ^ { 2 } + 6 x + 2 = 0$ has roots $\alpha , \beta$ and $\gamma$.
\begin{enumerate}[label=(\roman*)]
\item (a) Write down the values of $\alpha + \beta + \gamma$ and $\alpha \beta + \beta \gamma + \gamma \alpha$.\\
(b) Find the value of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$.
\item (a) Use the substitution $x = \frac { 1 } { u }$ to find a cubic equation in $u$ with integer coefficients.\\
(b) Use your answer to part (ii) (a) to find the value of $\frac { 1 } { \alpha } + \frac { 1 } { \beta } + \frac { 1 } { \gamma }$.
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 2007 Q6 [8]}}