| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Standard +0.8 This is a standard Further Maths question on sums of powers of roots using Newton's identities and recurrence relations. Parts (i) and (ii) are routine applications of symmetric functions, but part (iii) requires establishing a recurrence relation for S₅, involving multiple algebraic steps. While systematic, it demands careful manipulation beyond typical A-level and is appropriately challenging for FP1. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 - 2(-p) = 2p\) (AG) | B1 | Finds \(S_2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^3+\beta^3+\gamma^3 = p\sum\alpha + 3q = 0 + 3q = 3q\) (AG) | M1A1 | Finds \(S_3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha^5+\beta^5+\gamma^5 = p\sum\alpha^3 + q\sum\alpha^2\) | M1 | Finds \(S_5\) |
| \(= p\cdot 3q + q\cdot 2p = 5pq\) | A1 | |
| \(\Rightarrow 6\sum\alpha^5 = 30pq = 5\sum\alpha^3\sum\alpha^2\) (AG) | A1 |
## Question 2:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 0 - 2(-p) = 2p$ (AG) | B1 | Finds $S_2$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^3+\beta^3+\gamma^3 = p\sum\alpha + 3q = 0 + 3q = 3q$ (AG) | M1A1 | Finds $S_3$ |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^5+\beta^5+\gamma^5 = p\sum\alpha^3 + q\sum\alpha^2$ | M1 | Finds $S_5$ |
| $= p\cdot 3q + q\cdot 2p = 5pq$ | A1 | |
| $\Rightarrow 6\sum\alpha^5 = 30pq = 5\sum\alpha^3\sum\alpha^2$ (AG) | A1 | |
---2 The cubic equation $x ^ { 3 } - p x - q = 0$, where $p$ and $q$ are constants, has roots $\alpha , \beta , \gamma$. Show that\\
(i) $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } = 2 p$,\\
(ii) $\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } = 3 q$,\\
(iii) $6 \left( \alpha ^ { 5 } + \beta ^ { 5 } + \gamma ^ { 5 } \right) = 5 \left( \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } \right) \left( \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } \right)$.
\hfill \mbox{\textit{CAIE FP1 2013 Q2 [6]}}