| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.3 This is a standard FP1 question on transformed roots requiring application of Vieta's formulas and algebraic manipulation. While it involves multiple steps (finding sum and product of transformed roots), the techniques are routine for Further Maths students and follow a well-established method with no novel insight required. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha + \beta = \frac{1}{2}, \quad \alpha\beta = \frac{3}{2}\) | B1 | State or use both correct results in (i) or (ii) |
| \(\alpha + \beta + \frac{\alpha+\beta}{\alpha\beta}\) or \(\alpha+\beta+\frac{2}{3}(\alpha+\beta)\) | M1 | Express sum of new roots in terms of \(\alpha+\beta\) and \(\alpha\beta\) |
| M1 | Substitute their values into their expression | |
| \(p = \frac{5}{6}\) | A1 4 | Obtain given answer correctly |
| *Or* | ||
| \(3u^2 - u + 2(= 0)\) | B1 | Substitute \(x = \frac{1}{u}\) and obtain correct quadratic |
| M1 | Use sum of roots of new equation | |
| M1 | Substitute their values into their expression | |
| \(p = \frac{5}{6}\) | A1 | Obtain given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha'\beta' = \alpha\beta + \frac{1}{\alpha\beta} + \frac{\beta}{\alpha} + \frac{\alpha}{\beta}\) | B1 | Correct expansion |
| \(\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}\) | M1 | Show how to deal with \(\alpha^2 + \beta^2\) |
| A1 | Obtain correct expression | |
| M1 | Substitute their values into \(\alpha'\beta'\) | |
| \(q = \frac{1}{3}\) | A1 5 | Obtain correct answer a.e.f. |
## Question 8:
**Part (i) — Either method**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha + \beta = \frac{1}{2}, \quad \alpha\beta = \frac{3}{2}$ | B1 | State or use both correct results in (i) or (ii) |
| $\alpha + \beta + \frac{\alpha+\beta}{\alpha\beta}$ or $\alpha+\beta+\frac{2}{3}(\alpha+\beta)$ | M1 | Express sum of new roots in terms of $\alpha+\beta$ and $\alpha\beta$ |
| | M1 | Substitute their values into their expression |
| $p = \frac{5}{6}$ | A1 **4** | Obtain given answer correctly |
| *Or* | | |
| $3u^2 - u + 2(= 0)$ | B1 | Substitute $x = \frac{1}{u}$ and obtain correct quadratic |
| | M1 | Use sum of roots of new equation |
| | M1 | Substitute their values into their expression |
| $p = \frac{5}{6}$ | A1 | Obtain given answer correctly |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha'\beta' = \alpha\beta + \frac{1}{\alpha\beta} + \frac{\beta}{\alpha} + \frac{\alpha}{\beta}$ | B1 | Correct expansion |
| $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$ | M1 | Show how to deal with $\alpha^2 + \beta^2$ |
| | A1 | Obtain correct expression |
| | M1 | Substitute their values into $\alpha'\beta'$ |
| $q = \frac{1}{3}$ | A1 **5** | Obtain correct answer a.e.f. |
---8 The quadratic equation $2 x ^ { 2 } - x + 3 = 0$ has roots $\alpha$ and $\beta$, and the quadratic equation $x ^ { 2 } - p x + q = 0$ has roots $\alpha + \frac { 1 } { \alpha }$ and $\beta + \frac { 1 } { \beta }$.\\
(i) Show that $p = \frac { 5 } { 6 }$.\\
(ii) Find the value of $q$.
\hfill \mbox{\textit{OCR FP1 2011 Q8 [9]}}