| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Standard +0.3 This is a straightforward application of relationships between roots and coefficients (Vieta's formulas) with symmetric roots. The structure p-q, p, p+q makes the algebra particularly clean: sum gives 3p directly, product simplifies nicely, and finding k is routine substitution. While it requires understanding of root relationships (FP1 content), the execution involves minimal algebraic manipulation and no problem-solving insight beyond recognizing the given structure. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \((p-q) + p + (p+q) = 6 \Rightarrow p = 2\) | M1, A1 | For use of \(\Sigma q = -b/a\) / For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(4 - q^2 = -5 \Rightarrow q = 3\) | B1, M1, A1 | For use of \(\alpha\beta\gamma = -d/a\) / For expanding and solving for \(q^2\) / For correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Hence \(k = 3\) | B1, M1, A1 | For stating or using three numerical roots / For use of \(\Sigma \alpha\beta = c/a\) / For correct answer from their roots |
| Answer | Marks |
|---|---|
| B1, M1, A1 | For stating or using three numerical roots / For stating and expanding factorised form / For correct answer from their roots |
**(i)** $(p-q) + p + (p+q) = 6 \Rightarrow p = 2$ | M1, A1 | For use of $\Sigma q = -b/a$ / For correct answer
**Total: 2 marks**
**(ii)** $2(2-q)(2+q) = -10$
Hence $4 - q^2 = -5 \Rightarrow q = 3$ | B1, M1, A1 | For use of $\alpha\beta\gamma = -d/a$ / For expanding and solving for $q^2$ / For correct answer
**Total: 3 marks**
**(iii)** **EITHER:** Roots are $-1, 2, 5$
$-1 \times 2 + 2 \times 5 + -1 \times 5 = k$
i.e. $k = 3$
**OR:** Roots are $-1, 2, 5$
Equation is $(x+1)(x-2)(x-5) = 0$
Hence $k = 3$ | B1, M1, A1 | For stating or using three numerical roots / For use of $\Sigma \alpha\beta = c/a$ / For correct answer from their roots
OR
B1, M1, A1 | For stating or using three numerical roots / For stating and expanding factorised form / For correct answer from their roots
**Total: 3 marks**
**Question 2 Total: 8 marks**
---2 The cubic equation $x ^ { 3 } - 6 x ^ { 2 } + k x + 10 = 0$ has roots $p - q , p$ and $p + q$, where $q$ is positive.\\
(i) By considering the sum of the roots, find $p$.\\
(ii) Hence, by considering the product of the roots, find $q$.\\
(iii) Find the value of $k$.
\hfill \mbox{\textit{OCR FP1 Q2 [8]}}