| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 5 |
| Topic | Roots of polynomials |
| Type | Symmetric functions of roots |
| Difficulty | Moderate -0.5 This is a standard Further Pure 1 question on symmetric functions of roots requiring direct application of formulas (sum and product of roots from coefficients, then algebraic manipulation). While it's Further Maths content, the techniques are routine and well-practiced, making it slightly easier than an average A-level question overall. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\alpha + \beta = \frac{7}{3}\) oe | B1 | |
| \(\alpha\beta = \frac{5}{3}\) oe | B1 | If answers not specified then assume order \(\alpha + \beta, \alpha\beta\). |
| [2] | ||
| (ii)(a) \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7}{5}\) | B1 | |
| [1] | ||
| (ii)(b) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) | M1 | Seen or implied |
| \(\frac{19}{9}\) | A1 | |
| [2] |
**(i)** $\alpha + \beta = \frac{7}{3}$ oe | B1 |
$\alpha\beta = \frac{5}{3}$ oe | B1 | If answers not specified then assume order $\alpha + \beta, \alpha\beta$.
| [2] |
**(ii)(a)** $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7}{5}$ | B1 |
| [1] |
**(ii)(b)** $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ | M1 | Seen or implied
$\frac{19}{9}$ | A1 |
| [2] |
---2 In this question you must show detailed reasoning.\\
The quadratic equation $3 x ^ { 2 } - 7 x + 5 = 0$ has roots $\alpha$ and $\beta$.\\
(i) Write down the values of $\alpha + \beta$ and $\alpha \beta$.\\
(ii) Hence find the values of the following expressions.
\begin{enumerate}[label=(\alph*)]
\item $\frac { 1 } { \alpha } + \frac { 1 } { \beta }$
\item $\alpha ^ { 2 } + \beta ^ { 2 }$\\
$3 \quad l _ { 1 }$ and $l _ { 2 }$ are two intersecting straight lines with the following equations.
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = \left( \begin{array} { c }
3 \\
3 \\
- 5
\end{array} \right)
\end{aligned}$$
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 AS 2018 Q2 [5]}}