| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Standard +0.8 This is a Further Maths question requiring systematic application of symmetric functions (sum and product of roots) with complex conjugate pairs. Part (i) is straightforward substitution, but parts (ii)-(iii) require careful algebraic manipulation linking real and complex roots through Vieta's formulas. The multi-step nature and need to express complex root components in terms of the real root elevates this above standard A-level, though it follows a predictable pattern for FP1. |
| Spec | 4.02g Conjugate pairs: real coefficient polynomials4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(\alpha\) into equation and rearrange | M1 | \(\alpha^3 - k\alpha^2 - 2 = 0\) |
| \(k = \alpha - \frac{2}{\alpha^2}\) — obtain given answer a.e.f. | A1 | |
| [2] | ||
| *Or*: Substitute for \(k\) and \(x\) in terms of \(\alpha\) and simplify; show simplification leads to consistency | M1, A1 | e.g. "LHS \(= 0\)" |
| *Or*: Eliminate \(\beta\) and \(\gamma\) from symmetric functions; obtain given answer correctly | M1, A1 | Don't penalise sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| State or use \((\gamma) = u - iv\) | B1 | |
| Use sum of roots \(= (\pm)k\); can use \(\sum\alpha\beta\) with \(\alpha\beta\gamma\) | M1* | \(\alpha + u + iv + u - iv = -(-k)\) |
| Rearrange to get \(u\) | DM1 | \(\alpha + 2u = \alpha - \frac{2}{\alpha^2}\) |
| \(u = -\frac{2}{2\alpha^2}\) or better — obtain correct answer | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use product of roots \(= (\pm)2\) or \(\sum\alpha\beta = 0\) | M1* | In terms of \(u\) and \(v\) |
| \(\alpha(u^2 + v^2) = 2\) or \(2\alpha u + u^2 + v^2 = 0\) — obtain correct answer | A1 | |
| Substitute for \(u\) and rearrange to get \(v^2\) | DM1 | |
| \(v^2 = \frac{2}{\alpha} - \frac{1}{\alpha^4}\) — obtain correct answer a.e.f. | A1 | |
| [4] |
# Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $\alpha$ into equation and rearrange | M1 | $\alpha^3 - k\alpha^2 - 2 = 0$ |
| $k = \alpha - \frac{2}{\alpha^2}$ — obtain **given** answer a.e.f. | A1 | |
| **[2]** | | |
| *Or*: Substitute for $k$ and $x$ in terms of $\alpha$ and simplify; show simplification leads to consistency | M1, A1 | e.g. "LHS $= 0$" |
| *Or*: Eliminate $\beta$ and $\gamma$ from symmetric functions; obtain **given** answer correctly | M1, A1 | Don't penalise sign errors |
---
# Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| State or use $(\gamma) = u - iv$ | B1 | |
| Use sum of roots $= (\pm)k$; can use $\sum\alpha\beta$ with $\alpha\beta\gamma$ | M1* | $\alpha + u + iv + u - iv = -(-k)$ |
| Rearrange to get $u$ | DM1 | $\alpha + 2u = \alpha - \frac{2}{\alpha^2}$ |
| $u = -\frac{2}{2\alpha^2}$ or better — obtain correct answer | A1 | |
| **[4]** | | |
---
# Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use product of roots $= (\pm)2$ or $\sum\alpha\beta = 0$ | M1* | In terms of $u$ and $v$ |
| $\alpha(u^2 + v^2) = 2$ or $2\alpha u + u^2 + v^2 = 0$ — obtain correct answer | A1 | |
| Substitute for $u$ and rearrange to get $v^2$ | DM1 | |
| $v^2 = \frac{2}{\alpha} - \frac{1}{\alpha^4}$ — obtain correct answer a.e.f. | A1 | |
| **[4]** | | |
---9 The roots of the equation $x ^ { 3 } - k x ^ { 2 } - 2 = 0$ are $\alpha , \beta$ and $\gamma$, where $\alpha$ is real and $\beta$ and $\gamma$ are complex.\\
(i) Show that $k = \alpha - \frac { 2 } { \alpha ^ { 2 } }$.\\
(ii) Given that $\beta = u + \mathrm { i } v$, where $u$ and $v$ are real, find $u$ in terms of $\alpha$.\\
(iii) Find $v ^ { 2 }$ in terms of $\alpha$.
\hfill \mbox{\textit{OCR FP1 2014 Q9 [10]}}