| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Standard +0.8 Part (a) requires applying Vieta's formulas and the identity (Σα)² = Σα² + 2Σαβ, which is standard Further Maths technique. Part (b) is more demanding: students must expand the summation, apply standard summation formulas from MF19, and manipulate algebraically to reach the required form with constants a and b. The multi-step algebraic manipulation and synthesis of multiple techniques elevates this above routine exercises, though it remains within expected Further Maths scope. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \((-4)^2 - 2(6)\) | M1 | Uses formula for sum of squares |
| \(4\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \((\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2\) | B1 | Expands |
| \(\sum_{r=1}^{n}\left((\alpha+r)^2+(\beta+r)^2+(\gamma+r)^2\right) = \sum_{r=1}^{n}\left(4+2(-4)r+3r^2\right)\) | M1 A1 | Collects like terms and uses \(\alpha+\beta+\lambda=-4\) and *their* result from part (a) |
| \(4n - 4n(n+1) + \frac{1}{2}n(n+1)(2n+1)\) | M1 | Applies formulae from MF19 |
| \(-4n^2 + \frac{1}{2}n(n+1)(2n+1) = n\left(-4n+\frac{1}{2}(2n^2+3n+1)\right) = n\left(n^2-\frac{5}{2}n+\frac{1}{2}\right)\) | M1 A1 | Simplifies |
| Total: 6 |
## Question 2:
### Part 2(a):
| $(-4)^2 - 2(6)$ | M1 | Uses formula for sum of squares |
| $4$ | A1 | |
| **Total: 2** | | |
### Part 2(b):
| $(\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2$ | B1 | Expands |
| $\sum_{r=1}^{n}\left((\alpha+r)^2+(\beta+r)^2+(\gamma+r)^2\right) = \sum_{r=1}^{n}\left(4+2(-4)r+3r^2\right)$ | M1 A1 | Collects like terms and uses $\alpha+\beta+\lambda=-4$ and *their* result from part (a) |
| $4n - 4n(n+1) + \frac{1}{2}n(n+1)(2n+1)$ | M1 | Applies formulae from MF19 |
| $-4n^2 + \frac{1}{2}n(n+1)(2n+1) = n\left(-4n+\frac{1}{2}(2n^2+3n+1)\right) = n\left(n^2-\frac{5}{2}n+\frac{1}{2}\right)$ | M1 A1 | Simplifies |
| **Total: 6** | | |
---2 The cubic equation $x ^ { 3 } + 4 x ^ { 2 } + 6 x + 1 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$.
\item Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { n } \left( ( \alpha + r ) ^ { 2 } + ( \beta + r ) ^ { 2 } + ( \gamma + r ) ^ { 2 } \right) = n \left( n ^ { 2 } + a n + b \right)$$
where $a$ and $b$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q2}}