| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Standard +0.3 This is a standard Further Maths question on symmetric functions of roots requiring straightforward application of Newton's identities or algebraic manipulation. Part (i) uses the identity (α+β+γ)² - 2(αβ+βγ+γα) with Vieta's formulas, while part (ii) applies the result that each root satisfies the cubic equation. Both parts are textbook exercises with well-established methods, making this easier than average even for Further Maths. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| 1(i) | α+β+γ=5, αβ+αγ+βγ=13 | B1 |
| α2 +β2 +γ2 =52 −2 ( 13 ) | M1 | =( )2 ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| = ‒1 | A1 | www |
| Answer | Marks | Guidance |
|---|---|---|
| 1(ii) | ( ) | |
| α3 +β3 +γ3 =5 α2 +β2 +γ2 −13 ( α+β+γ )+12 | M1 | Uses α3 =5α2 −13α+4. |
| =5 ( −1 ) −13 ( 5 )+12=−58 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 1:
--- 1(i) ---
1(i) | α+β+γ=5, αβ+αγ+βγ=13 | B1 | Sum of roots and αβ+αγ+βγ. SOI
α2 +β2 +γ2 =52 −2 ( 13 ) | M1 | =( )2 ( )
Uses ∑α2 ∑α −2 ∑αβ
= ‒1 | A1 | www
3
--- 1(ii) ---
1(ii) | ( )
α3 +β3 +γ3 =5 α2 +β2 +γ2 −13 ( α+β+γ )+12 | M1 | Uses α3 =5α2 −13α+4.
=5 ( −1 ) −13 ( 5 )+12=−58 | A1
( )
Alt method: Use formula e.g. ∑α3= ( ∑α ) ∑α2 − ∑αβ +3αβγ
( ) ( )
Or ∑α 3 – 3 ∑α (∑αβ) +3α βγ
2
Question | Answer | Marks | GuidanceThe roots of the cubic equation
$$x^3 - 5x^2 + 13x - 4 = 0$$
are $\alpha, \beta, \gamma$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $\alpha^2 + \beta^2 + \gamma^2$. [3]
\item Find the value of $\alpha^3 + \beta^3 + \gamma^3$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP1 2018 Q1 [5]}}