| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on transformed roots requiring standard techniques: algebraic identity verification, application of sum/product of roots formulas, and forming a new equation. While it involves multiple steps, each is routine for FP1 students with no novel insight required, making it slightly easier than average. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Expand brackets | M1 | Expand at least 1 of the brackets |
| Correct derivation | A1 | Derive given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\alpha + \beta = 6k,\ \alpha\beta = k^2\) | B1 B1 | State or use correct values |
| Find \(\alpha - \beta\) | M1 | Find value of \(\alpha - \beta\) using (i) |
| \(\alpha - \beta = (4\sqrt{2})k\) | A1 | Obtain given value correctly (allow if \(-6k\) used) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum\alpha' = 6k\) | B1ft | Sum of new roots stated or used |
| \(\alpha'\beta' = \alpha\beta - (\alpha - \beta) - 1\) | M1 | Express new product in terms of old roots |
| \(\alpha'\beta' = k^2 - (4\sqrt{2})k - 1\) | A1ft | Obtain correct value for new product |
| \(x^2 - 6kx + k^2 - (4\sqrt{2})k - 1 = 0\) | B1ft | Write down correct quadratic equation |
## Question 8:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Expand brackets | M1 | Expand at least 1 of the brackets |
| Correct derivation | A1 | Derive given answer correctly |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha + \beta = 6k,\ \alpha\beta = k^2$ | B1 B1 | State or use correct values |
| Find $\alpha - \beta$ | M1 | Find value of $\alpha - \beta$ using (i) |
| $\alpha - \beta = (4\sqrt{2})k$ | A1 | Obtain given value correctly (allow if $-6k$ used) |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum\alpha' = 6k$ | B1ft | Sum of new roots stated or used |
| $\alpha'\beta' = \alpha\beta - (\alpha - \beta) - 1$ | M1 | Express new product in terms of old roots |
| $\alpha'\beta' = k^2 - (4\sqrt{2})k - 1$ | A1ft | Obtain correct value for new product |
| $x^2 - 6kx + k^2 - (4\sqrt{2})k - 1 = 0$ | B1ft | Write down correct quadratic equation |
**Total: 10 marks**
---8 (i) Show that $( \alpha - \beta ) ^ { 2 } \equiv ( \alpha + \beta ) ^ { 2 } - 4 \alpha \beta$.
The quadratic equation $x ^ { 2 } - 6 k x + k ^ { 2 } = 0$, where $k$ is a positive constant, has roots $\alpha$ and $\beta$, with $\alpha > \beta$.\\
(ii) Show that $\alpha - \beta = 4 \sqrt { 2 } k$.\\
(iii) Hence find a quadratic equation with roots $\alpha + 1$ and $\beta - 1$.
\hfill \mbox{\textit{OCR FP1 2009 Q8 [10]}}