| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring substitution into a cubic equation followed by application of relationships between roots and coefficients. Part (i) involves algebraic manipulation with fractional powers (moderately technical but routine for FP1). Part (ii) requires recognizing that the new roots relate to 1/α², 1/β², 1/γ² and applying Vieta's formulas to find symmetric functions—this demands conceptual understanding beyond mechanical calculation. Slightly above average difficulty even for Further Maths students. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{u^{\frac{3}{2}}} + \frac{3}{u} + 2 = 0\) | B1 | Use substitution correctly |
| *EITHER:* | M1 | Rearrange |
| M1 | Square | |
| \(\frac{9}{u^2} + \frac{12}{u} + 4 = \frac{1}{u^3}\) | A1 | Obtain correct equation |
| \(4u^3 + 12u^2 + 9u - 1 = 0\) | A1 [5] | Obtain given answer |
| *OR:* \((2u^{\frac{3}{2}} + 3u^{\frac{1}{2}} + 1)(2u^{\frac{3}{2}} + 3u^{\frac{1}{2}} - 1) = 0\) | M2 | Multiply their equation in \(u\) by appropriate related expression |
| A2 | Obtain given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| B1 | Stated or imply that \(u = \frac{1}{x^2}\) | |
| \(-3\) | M1, A1 | Use \(-\frac{b}{a}\); obtain correct answer |
| \(\frac{9}{4}\) | M1, A1 [5] | Use \(\frac{c}{a}\); obtain correct answer |
## Question 10:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{u^{\frac{3}{2}}} + \frac{3}{u} + 2 = 0$ | B1 | Use substitution correctly |
| *EITHER:* | M1 | Rearrange |
| | M1 | Square |
| $\frac{9}{u^2} + \frac{12}{u} + 4 = \frac{1}{u^3}$ | A1 | Obtain correct equation |
| $4u^3 + 12u^2 + 9u - 1 = 0$ | A1 **[5]** | Obtain **given** answer |
| *OR:* $(2u^{\frac{3}{2}} + 3u^{\frac{1}{2}} + 1)(2u^{\frac{3}{2}} + 3u^{\frac{1}{2}} - 1) = 0$ | M2 | Multiply their equation in $u$ by appropriate related expression |
| | A2 | Obtain **given** answer |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | B1 | Stated or imply that $u = \frac{1}{x^2}$ |
| $-3$ | M1, A1 | Use $-\frac{b}{a}$; obtain correct answer |
| $\frac{9}{4}$ | M1, A1 **[5]** | Use $\frac{c}{a}$; obtain correct answer |10 The cubic equation $x ^ { 3 } + 3 x ^ { 2 } + 2 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
(i) Use the substitution $x = \frac { 1 } { \sqrt { u } }$ to show that $4 u ^ { 3 } + 12 u ^ { 2 } + 9 u - 1 = 0$.\\
(ii) Hence find the values of $\frac { 1 } { \alpha ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } }$ and $\frac { 1 } { \alpha ^ { 2 } \beta ^ { 2 } } + \frac { 1 } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { 1 } { \gamma ^ { 2 } \alpha ^ { 2 } }$.
\hfill \mbox{\textit{OCR FP1 2011 Q10 [10]}}