| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| 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 sum of powers of roots. Part (a) requires the routine technique of using α²+β²+γ² = (α+β+γ)² - 2(αβ+βγ+γα) with Vieta's formulas. While it's a Further Maths topic, the method is well-established and commonly practiced, making it slightly easier than average overall but standard for this syllabus. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha\beta + \beta\gamma + \gamma\alpha = 10\) | B1 | SOI |
| \(\alpha^2 + \beta^2 + \gamma^2 = 5^2 - 20 = 5\) | M1 A1 | Uses \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\) |
| Alternative: Put \(y = x^2\) and form cubic not involving \(x\) | M1 A1 | \(y(y+10)^2 = 25y^2 - 20y + 4\), \(y^3 - 5y^2 + 120y - 4 = 0\) |
| \(\alpha^2 + \beta^2 + \gamma^2 = [-(-5)] = 5\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix}1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1\end{vmatrix} = \begin{vmatrix}1 & \gamma \\ \gamma & 1\end{vmatrix} - \alpha\begin{vmatrix}\alpha & \gamma \\ \beta & 1\end{vmatrix} + \beta\begin{vmatrix}\alpha & 1 \\ \beta & \gamma\end{vmatrix} = 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\) | M1 A1 | Finds determinant |
| \(1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 1 - 5 + 2(2) = 0\) | M1 A1 | Substitutes and uses \(\alpha\beta\gamma = 2\) |
| Total: 4 |
## Question 2:
**Part 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha\beta + \beta\gamma + \gamma\alpha = 10$ | B1 | SOI |
| $\alpha^2 + \beta^2 + \gamma^2 = 5^2 - 20 = 5$ | M1 A1 | Uses $\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)$ |
| **Alternative:** Put $y = x^2$ and form cubic not involving $x$ | M1 A1 | $y(y+10)^2 = 25y^2 - 20y + 4$, $y^3 - 5y^2 + 120y - 4 = 0$ |
| $\alpha^2 + \beta^2 + \gamma^2 = [-(-5)] = 5$ | A1 | |
| **Total: 3** | | |
**Part 2(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix}1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1\end{vmatrix} = \begin{vmatrix}1 & \gamma \\ \gamma & 1\end{vmatrix} - \alpha\begin{vmatrix}\alpha & \gamma \\ \beta & 1\end{vmatrix} + \beta\begin{vmatrix}\alpha & 1 \\ \beta & \gamma\end{vmatrix} = 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma$ | M1 A1 | Finds determinant |
| $1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 1 - 5 + 2(2) = 0$ | M1 A1 | Substitutes and uses $\alpha\beta\gamma = 2$ |
| **Total: 4** | | |
---2 The cubic equation $x ^ { 3 } + 5 x ^ { 2 } + 10 x - 2 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$.\\
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\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_65_1566_573_328}\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_65_1570_662_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_65_1570_751_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_67_1570_840_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_68_1570_931_324}\\
\includegraphics[max width=\textwidth, alt={}]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_63_1570_1023_324} .......................................................................................................................................... . .......................................................................................................................................... ......................................................................................................................................... .\\
\includegraphics[max width=\textwidth, alt={}]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_72_1570_1379_324} ........................................................................................................................................\\
\includegraphics[max width=\textwidth, alt={}]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_80_1570_1556_324} .........................................................................................................................................\\
\includegraphics[max width=\textwidth, alt={}]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_72_1572_1740_322} ....................................................................................................................................... .\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_74_1572_1921_322}\\
\includegraphics[max width=\textwidth, alt={}, center]{e34e64b7-d3bb-4614-bf37-5ef90ad56157-04_70_1570_2012_324}
\item Show that the matrix $\left( \begin{array} { c c c } 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{array} \right)$ is singular.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q2 [7]}}