| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2017 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with given sum conditions |
| Difficulty | Standard +0.8 This Further Maths question requires systematic application of Vieta's formulas and manipulation of symmetric functions of roots. While the techniques are standard for FP1 (relating sum of roots to coefficients, using the identity α²+β²+γ² = (α+β+γ)² - 2(αβ+αγ+βγ)), part (ii) requires careful algebraic manipulation to isolate individual roots from the given constraint αβ+αγ=36, which is less routine. The multi-step nature and need to work backwards from symmetric functions to individual roots places this above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha + \beta + \gamma = -p = 15 \Rightarrow p = -15\) | 1 | B1 |
| \(2(\alpha\beta + \beta\gamma + \gamma\alpha) = (\alpha+\beta+\gamma)^2 - (\alpha^2+\beta^2+\gamma^2) = 2q\) | 1 | M1 |
| \(\Rightarrow q = \frac{1}{2}(225 - 83) = 71\) | 1 | A1 |
| Subtotal | 3 | |
| \(\frac{36}{\alpha} = 15 - \alpha \quad (= [\beta + \gamma])\) | 1 | M1 |
| \(\Rightarrow \alpha^2 - 15\alpha + 36 = 0 \Rightarrow \alpha = 3\), \(\alpha \neq 12\), e.g. since \(12^2 > 83\) or other reason | 2 | M1A1 |
| \(\beta\gamma = 71 - 36 = 35\) | 1 | B1 |
| \(\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105\) | 1 | A1 |
| Total | 5 |
## Question 5:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha + \beta + \gamma = -p = 15 \Rightarrow p = -15$ | 1 | B1 |
| $2(\alpha\beta + \beta\gamma + \gamma\alpha) = (\alpha+\beta+\gamma)^2 - (\alpha^2+\beta^2+\gamma^2) = 2q$ | 1 | M1 |
| $\Rightarrow q = \frac{1}{2}(225 - 83) = 71$ | 1 | A1 |
| **Subtotal** | **3** | |
| $\frac{36}{\alpha} = 15 - \alpha \quad (= [\beta + \gamma])$ | 1 | M1 |
| $\Rightarrow \alpha^2 - 15\alpha + 36 = 0 \Rightarrow \alpha = 3$, $\alpha \neq 12$, e.g. since $12^2 > 83$ or other reason | 2 | M1A1 |
| $\beta\gamma = 71 - 36 = 35$ | 1 | B1 |
| $\Rightarrow r = -\alpha\beta\gamma = -3 \times 35 = -105$ | 1 | A1 | (extra answer penalised) |
| **Total** | **5** | |5 The cubic equation $x ^ { 3 } + p x ^ { 2 } + q x + r = 0$, where $p , q$ and $r$ are integers, has roots $\alpha , \beta$ and $\gamma$, such that
$$\begin{aligned}
\alpha + \beta + \gamma & = 15 \\
\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } & = 83
\end{aligned}$$
(i) Write down the value of $p$ and find the value of $q$.\\
(ii) Given that $\alpha , \beta$ and $\gamma$ are all real and that $\alpha \beta + \alpha \gamma = 36$, find $\alpha$ and hence find the value of $r$. [5]\\
\hfill \mbox{\textit{CAIE FP1 2017 Q5 [8]}}