| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Standard +0.8 This is a Further Maths question requiring insight that the special relationship among roots (one root equals the sum of the other two) can be exploited using Vieta's formulas. Students must recognize that α + β + (α + β) = 4 gives 2(α + β) = 4, then use the product of roots systematically. While methodical, it requires more algebraic manipulation and strategic thinking than standard A-level questions, placing it moderately above average difficulty. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (a) | 𝛼+𝛽+𝛼+𝛽 = 4 [⇒𝛼+𝛽 = 2] |
| Answer | Marks |
|---|---|
| so roots are 1+i√2, 1−i√2 and 2 | B1 |
| Answer | Marks |
|---|---|
| [6] | 3.1a |
| Answer | Marks |
|---|---|
| 2.2a | or 𝛼𝛽+2𝛼+2𝛽 = 7 or 𝛼𝛽+2(𝛼+𝛽)= 7 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | (b) | [𝑐 =]−6 |
| [1] | 2.2a |
Question 10:
10 | (a) | 𝛼+𝛽+𝛼+𝛽 = 4 [⇒𝛼+𝛽 = 2]
𝛼𝛽+(𝛼+𝛽)𝛼+(𝛼+𝛽)𝛽 = 7
⇒ 𝛼(cid:2870)+𝛽(cid:2870)+3𝛼𝛽 = 7
𝑥+ (cid:2871) = 2 or 3𝑥(2−𝑥)+(2−𝑥)(cid:2870)+𝑥(cid:2870) = 7
(cid:3051)
⇒ 𝑥(cid:2870) – 2𝑥 + 3 = 0
2±√−8
⇒ 𝑥 = = 1±i√2
2
so roots are 1+i√2, 1−i√2 and 2 | B1
B1
M1
A1
M1
A1
[6] | 3.1a
1.1
3.1a
1.1
1.1
2.2a | or 𝛼𝛽+2𝛼+2𝛽 = 7 or 𝛼𝛽+2(𝛼+𝛽)= 7
substitution which could lead to a quadratic in 𝛼 or 𝛽 or 𝑥
could be in 𝛼 or 𝛽
a method to solve their quadratic
10 | (b) | [𝑐 =]−6 | B1
[1] | 2.2a10 The equation $\mathrm { x } ^ { 3 } - 4 \mathrm { x } ^ { 2 } + 7 \mathrm { x } + \mathrm { c } = 0$, where $c$ is a constant, has roots $\alpha , \beta$ and $\alpha + \beta$.
\begin{enumerate}[label=(\alph*)]
\item Determine the roots of the equation.
\item Find c.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2023 Q10 [7]}}