| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Standard +0.3 This is a straightforward application of standard symmetric function formulas (sum, product of roots) and complex conjugate root properties. Part (i) and (iii) are direct recall, part (ii) uses the fact that complex roots come in conjugate pairs so their sum is real, and part (iv) requires substituting into the product formula. All steps are routine for FP1 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\alpha + \beta + \gamma = 9\) | B1 | 1 mark |
| (ii) \(p = \frac{9-\alpha}{2}\) | B1 M1 A1 A1 | State or use other root is \(p - iq\); Substitute into (i); Obtain \(2p + \alpha = 9\); Obtain correct answer a.e.f. |
| (iii) \(\alpha\beta\gamma = 29\) | B1 | 1 mark |
| (iv) \(\alpha(p^2 + q^2) = 29\) | M1 A1ft | Substitute into (iii); Obtain unsimplified expression with no i's |
| \(q = \sqrt{\frac{29}{\alpha} - \frac{(9-\alpha)^2}{4}}\) | M1 M1 A1 | Rearrange to obtain q or \(q^2\); Substitute their expression for p a.e.f.; Obtain correct answer a.e.f. |
| Answer | Marks | Guidance |
|---|---|---|
| \(2p\alpha + p^2 + q^2 = 27\) | M1 A1 | Substitute into \(\alpha\beta + \beta\gamma + \gamma\alpha = 27\); Obtain unsimplified expression with no i's |
| \(q = \sqrt{27 - \frac{(9-\alpha)^2}{4} - \alpha(9-\alpha)}\) | M1 M1 A1 | Rearrange to obtain q or \(q^2\); Substitute their expression for p a.e.f.; Obtain correct answer a.e.f. |
(i) $\alpha + \beta + \gamma = 9$ | B1 | 1 mark
(ii) $p = \frac{9-\alpha}{2}$ | B1 M1 A1 A1 | State or use other root is $p - iq$; Substitute into (i); Obtain $2p + \alpha = 9$; Obtain correct answer a.e.f.
(iii) $\alpha\beta\gamma = 29$ | B1 | 1 mark
(iv) $\alpha(p^2 + q^2) = 29$ | M1 A1ft | Substitute into (iii); Obtain unsimplified expression with no i's
$q = \sqrt{\frac{29}{\alpha} - \frac{(9-\alpha)^2}{4}}$ | M1 M1 A1 | Rearrange to obtain q or $q^2$; Substitute their expression for p a.e.f.; Obtain correct answer a.e.f.
**Total: 11 marks**
(iv) Alternative method:
$2p\alpha + p^2 + q^2 = 27$ | M1 A1 | Substitute into $\alpha\beta + \beta\gamma + \gamma\alpha = 27$; Obtain unsimplified expression with no i's
$q = \sqrt{27 - \frac{(9-\alpha)^2}{4} - \alpha(9-\alpha)}$ | M1 M1 A1 | Rearrange to obtain q or $q^2$; Substitute their expression for p a.e.f.; Obtain correct answer a.e.f.10 The roots of the equation
$$x ^ { 3 } - 9 x ^ { 2 } + 27 x - 29 = 0$$
are denoted by $\alpha , \beta$ and $\gamma$, where $\alpha$ is real and $\beta$ and $\gamma$ are complex.\\
(i) Write down the value of $\alpha + \beta + \gamma$.\\
(ii) It is given that $\beta = p + \mathrm { i } q$, where $q > 0$. Find the value of $p$, in terms of $\alpha$.\\
(iii) Write down the value of $\alpha \beta \gamma$.\\
(iv) Find the value of $q$, in terms of $\alpha$ only.
\hfill \mbox{\textit{OCR FP1 2006 Q10 [11]}}