| Exam Board | OCR |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths roots question requiring a standard substitution followed by applying Vieta's formulas. The substitution is given explicitly, and recognizing that the roots of the new equation are α-2, β-2, γ-2 leads directly to summing reciprocals using sum/product of roots. While it's FP1 content, it's a routine application of standard techniques with no novel insight required. |
| Spec | 4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute and attempt to simplify | M1 | |
| \(2u^3 + 12u^2 + 27u + 25 = 0\) — obtain correct equation, A1 for only 1 error | A2 | Missing \(= 0\) is an error |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Combine 3 terms with correct denominator | M1 | |
| \(\frac{\sum \alpha'\beta'}{\alpha'\beta'\gamma'}\) — obtain correct expression in their notation | A1 | |
| Attempt to use values from (i) correctly | M1 | Must be \(\pm c/\mathbf{a}\) and \(\pm d/\mathbf{a}\) for M1 |
| \(-\frac{27}{25}\) — obtain correct answer with no errors seen | A1ft | ft for their answer in (i) |
| [4] | ||
| *Or*: \(25y^3 + 27y^2 + 12y + 2 = 0\) | M1 | \(y = \frac{1}{u}\) |
| Use correct symmetric function | A1ft | |
| Obtain correct answer | M1 | |
| A1ft | ||
| *Or*: \(\frac{\sum'\alpha'\beta'}{\alpha'\beta'\gamma'}\) — combine 3 terms with correct denominator | M1 | |
| Expand numerator and denominator and use values from original equation correctly | A1, M1 | Condone \(\pm\), but must be "/2" |
| Obtain correct answer with no errors seen | A1 |
# Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute and attempt to simplify | M1 | |
| $2u^3 + 12u^2 + 27u + 25 = 0$ — obtain correct equation, A1 for only 1 error | A2 | Missing $= 0$ is an error |
| **[3]** | | |
---
# Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Combine 3 terms with correct denominator | M1 | |
| $\frac{\sum \alpha'\beta'}{\alpha'\beta'\gamma'}$ — obtain correct expression in their notation | A1 | |
| Attempt to use values from (i) correctly | M1 | Must be $\pm c/\mathbf{a}$ and $\pm d/\mathbf{a}$ for M1 |
| $-\frac{27}{25}$ — obtain correct answer **with no errors seen** | A1ft | ft for their answer in (i) |
| **[4]** | | |
| *Or*: $25y^3 + 27y^2 + 12y + 2 = 0$ | M1 | $y = \frac{1}{u}$ |
| Use correct symmetric function | A1ft | |
| Obtain correct answer | M1 | |
| | A1ft | |
| *Or*: $\frac{\sum'\alpha'\beta'}{\alpha'\beta'\gamma'}$ — combine 3 terms with correct denominator | M1 | |
| Expand numerator and denominator and use values from original equation correctly | A1, M1 | Condone $\pm$, but must be "/2" |
| Obtain correct answer **with no errors seen** | A1 | |
---5 The cubic equation $2 x ^ { 3 } + 3 x + 3 = 0$ has roots $\alpha , \beta$ and $\gamma$.\\
(i) Use the substitution $x = u + 2$ to find a cubic equation in $u$.\\
(ii) Hence find the value of $\frac { 1 } { \alpha - 2 } + \frac { 1 } { \beta - 2 } + \frac { 1 } { \gamma - 2 }$.
\hfill \mbox{\textit{OCR FP1 2014 Q5 [7]}}