| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (i) involves recognizing that u = Σα - 2α and forming a new equation by substitution, while part (ii) uses the reciprocal product transformation. Both parts follow well-established techniques taught in FP1, though the multi-step nature and need for careful algebraic manipulation place it slightly above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = -\alpha + \beta + \gamma \Rightarrow u + 2\alpha = \alpha + \beta + \gamma = 1\) | M1A1 | Deduces initial result |
| \(\Rightarrow \alpha = \left(\frac{1-u}{2}\right)\) | M1 | Substitutes into cubic equation |
| \(\Rightarrow \left(\frac{1-u}{2}\right)^3 - \left(\frac{1-u}{2}\right)^2 - 3\left(\frac{1-u}{2}\right) - 10 = 0\) | A1 | Deduces new cubic equation |
| \(\Rightarrow u^3 - u^2 - 13u + 93 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(-b = \sum\alpha = 1 \Rightarrow b = -1\); \(c = 4\sum\alpha\beta - (\sum\alpha)^2 = 4\times(-3) - 1^2 = -13\); \(-d = 4\sum\alpha\sum\alpha\beta - (\sum\alpha)^3 - 8\alpha\beta\gamma = 4\times1\times(-3) - 1^3 - 8\times10 = -93\) | M1, A1, A1 | Award M1 for attempt at formulae for all three coefficients; A1 for any two correct; A1 for completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha\beta\gamma = 10\) | B1 | Deduces initial result |
| \(\Rightarrow v = \frac{1}{\beta\gamma} \Rightarrow \frac{v}{\alpha} = \frac{1}{\alpha\beta\gamma} = \frac{1}{10} \Rightarrow \alpha = 10v\) | M1A1 | |
| \((10v)^3 - (10v)^2 - 3(10v) - 10 = 0\) | M1 | Substitutes into cubic equation |
| \(\Rightarrow 100v^3 - 10v^2 - 3v - 1 = 0\) | A1 | Deduces new cubic equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(-b = \frac{\Sigma\alpha}{\alpha\beta\gamma} = \frac{1}{10} \Rightarrow b = -\frac{1}{10}\); \(c = \frac{\Sigma\alpha\beta}{(\alpha\beta\gamma)^2} = \frac{-3}{10^2} = -\frac{3}{100}\); \(-d = \frac{1}{(\alpha\beta\gamma)^2} = \frac{1}{10^2} \Rightarrow d = -\frac{1}{100}\) | M1A1, A1, A1 | Award M1 for attempt at formulae; A1 any two correct; A1 second correct; A1 completion |
| So \(100v^3 - 10v^2 - 3v - 1 = 0\) |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = -\alpha + \beta + \gamma \Rightarrow u + 2\alpha = \alpha + \beta + \gamma = 1$ | M1A1 | Deduces initial result |
| $\Rightarrow \alpha = \left(\frac{1-u}{2}\right)$ | M1 | Substitutes into cubic equation |
| $\Rightarrow \left(\frac{1-u}{2}\right)^3 - \left(\frac{1-u}{2}\right)^2 - 3\left(\frac{1-u}{2}\right) - 10 = 0$ | A1 | Deduces new cubic equation |
| $\Rightarrow u^3 - u^2 - 13u + 93 = 0$ | A1 | |
**Alternative method (final 3 marks):** Let equation be $u^3 + bu^2 + cu + d = 0$
| $-b = \sum\alpha = 1 \Rightarrow b = -1$; $c = 4\sum\alpha\beta - (\sum\alpha)^2 = 4\times(-3) - 1^2 = -13$; $-d = 4\sum\alpha\sum\alpha\beta - (\sum\alpha)^3 - 8\alpha\beta\gamma = 4\times1\times(-3) - 1^3 - 8\times10 = -93$ | M1, A1, A1 | Award M1 for attempt at formulae for all three coefficients; A1 for any two correct; A1 for completion |
---
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha\beta\gamma = 10$ | B1 | Deduces initial result |
| $\Rightarrow v = \frac{1}{\beta\gamma} \Rightarrow \frac{v}{\alpha} = \frac{1}{\alpha\beta\gamma} = \frac{1}{10} \Rightarrow \alpha = 10v$ | M1A1 | |
| $(10v)^3 - (10v)^2 - 3(10v) - 10 = 0$ | M1 | Substitutes into cubic equation |
| $\Rightarrow 100v^3 - 10v^2 - 3v - 1 = 0$ | A1 | Deduces new cubic equation |
**Alternative (final 4 marks):** Let equation be $v^3 + bv^2 + cv + d = 0$
| $-b = \frac{\Sigma\alpha}{\alpha\beta\gamma} = \frac{1}{10} \Rightarrow b = -\frac{1}{10}$; $c = \frac{\Sigma\alpha\beta}{(\alpha\beta\gamma)^2} = \frac{-3}{10^2} = -\frac{3}{100}$; $-d = \frac{1}{(\alpha\beta\gamma)^2} = \frac{1}{10^2} \Rightarrow d = -\frac{1}{100}$ | M1A1, A1, A1 | Award M1 for attempt at formulae; A1 any two correct; A1 second correct; A1 completion |
| So $100v^3 - 10v^2 - 3v - 1 = 0$ | | |
---8 The cubic equation $x ^ { 3 } - x ^ { 2 } - 3 x - 10 = 0$ has roots $\alpha , \beta , \gamma$.\\
(i) Let $u = - \alpha + \beta + \gamma$. Show that $u + 2 \alpha = 1$, and hence find a cubic equation having roots $- \alpha + \beta + \gamma$, $\alpha - \beta + \gamma , \alpha + \beta - \gamma$.\\
(ii) State the value of $\alpha \beta \gamma$ and hence find a cubic equation having roots $\frac { 1 } { \beta \gamma } , \frac { 1 } { \gamma \alpha } , \frac { 1 } { \alpha \beta }$.
\hfill \mbox{\textit{CAIE FP1 2012 Q8 [10]}}