| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Standard +0.8 This is a Further Maths question requiring systematic application of Vieta's formulas to roots in geometric progression, followed by algebraic manipulation to eliminate α and derive relationships between coefficients. While the techniques are standard, the multi-step algebraic reasoning and the need to strategically combine equations to eliminate variables elevates this above routine exercises. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | α+2α+4α=−p | B1 |
| 2α2 +4α2 +8α2 =q | B1 | Sum of products in pairs. |
| Answer | Marks | Guidance |
|---|---|---|
| 7α p | M1 | Combines equations. |
| ⇒2pα+q=0 | A1 | Verifies result (AG). |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | 8α3 =−r | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| p3 | B1 | Verifies result (AG). |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 2:
--- 2(i) ---
2(i) | α+2α+4α=−p | B1 | Sum of roots.
2α2 +4α2 +8α2 =q | B1 | Sum of products in pairs.
14α2 q
=−
7α p | M1 | Combines equations.
⇒2pα+q=0 | A1 | Verifies result (AG).
4
--- 2(ii) ---
2(ii) | 8α3 =−r | B1 | Product of roots.
q3
⇒r = ⇒ p3r−q3 =0
p3 | B1 | Verifies result (AG).
2
Question | Answer | Marks | GuidanceThe roots of the equation
$$x^3 + px^2 + qx + r = 0$$
are $\alpha$, $2\alpha$, $4\alpha$, where $p$, $q$, $r$ and $\alpha$ are non-zero real constants.
\begin{enumerate}[label=(\roman*)]
\item Show that
$$2p\alpha + q = 0.$$ [4]
\item Show that
$$p^3 r - q^3 = 0.$$ [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2018 Q2 [6]}}