| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Challenging +1.2 Part (a) is a standard application of Newton's identities/symmetric functions (α²+β²+γ² = (Σα)² - 2Σαβ = 16-12 = 4), requiring only routine manipulation. Part (b) involves summing powers and applying standard summation formulas, which is more involved but still follows a predictable method using the MF19 formula list. This is a typical Further Maths question testing systematic application of known results rather than requiring novel insight. |
| 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 |
| 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. |
## Question 2:
**Part 2(a):**
$(-4)^2 - 2(6)$ | M1 | Uses formula for sum of squares.
$4$ | A1 |
**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.
---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 $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$.\\
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\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.\\
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\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q2 [8]}}