| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Standard +0.8 This is a Further Maths question requiring systematic application of Vieta's formulas and transformation of roots. Part (a) needs manipulation of symmetric functions (finding sum of products of squared roots), while part (b) requires constructing a new equation with transformed roots—a multi-step process involving relationships between elementary symmetric polynomials. More demanding than standard A-level but routine for Further Maths students who know the techniques. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha\beta\gamma = -\dfrac{1}{5}\), \(\sum\alpha\beta = \dfrac{3}{5}\), \(\sum\alpha = \dfrac{2}{5}\) | M1, A1 | 1.1a, 1.1 — Any one of \(\alpha\beta\gamma\), \(\alpha\beta+\beta\gamma+\gamma\alpha\), \(\alpha+\beta+\gamma\). All three correct. DR: Detailed reasoning required |
| \((\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + \sum 2\alpha\beta\gamma\alpha\) | M1 | 3.1a — Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have missing 2 |
| \((\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + 2\alpha\beta\gamma\sum\alpha\) | A1 | 1.1 — Correct, useful form |
| \(\sum\alpha^2\beta^2 = \dfrac{13}{25}\) | A1 [5] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitution \(x = \sqrt{u}\) or \(x^2 = u\) used | B1 | Soi by correct substitution — must be used |
| \(5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0\) | M1* | Substitution and dealing with \((\sqrt{u})^3\) term |
| \(((5u+3)\sqrt{u})^2 = (2u-1)^2\) | M1* | Rearrangement with \(\sqrt{u}\) terms collected on one side and squaring. LHS does not need to be factorised before squaring |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 | Correct expansion and rearranging. Must be an equation. Must see correct entire cubic |
| \(\sum\alpha^2\beta^2 = \dfrac{13}{25}\) | A1ft dep(*) | Correct coefficient ratio identified. Must gain method marks and have a cubic |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \dfrac{1}{25}\) | B1ft | 2.2a |
| \((\alpha + \beta + \gamma)^2 = \sum\alpha^2 + 2\sum\alpha\beta\) | M1 | 3.1a — Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in \(\alpha\), \(\beta\) and \(\gamma\). Might have 2 missing |
| \(\sum\alpha^2 = -\dfrac{26}{25}\) | A1 | 1.1 — Seen or implied. \(\sum\alpha^2 = \left(\frac{2}{5}\right)^2 - 2\times\frac{3}{5}\) |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 [4] | 1.1 — Or any non-zero integer multiple. Must be \(= 0\). CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitution \(x = \sqrt{u}\) or \(x^2 = u\) used | B1 | Soi by correct substitution — must be used |
| \(5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0\) | M1 | |
| \(((5u+3)\sqrt{u})^2 = (2u-1)^2\) | M1 | |
| \(25x^3 + 26x^2 + 13x - 1 = 0\) | A1 [4] | CAO |
# Question 5:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha\beta\gamma = -\dfrac{1}{5}$, $\sum\alpha\beta = \dfrac{3}{5}$, $\sum\alpha = \dfrac{2}{5}$ | M1, A1 | 1.1a, 1.1 — Any one of $\alpha\beta\gamma$, $\alpha\beta+\beta\gamma+\gamma\alpha$, $\alpha+\beta+\gamma$. All three correct. DR: Detailed reasoning required |
| $(\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + \sum 2\alpha\beta\gamma\alpha$ | M1 | 3.1a — Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in $\alpha$, $\beta$ and $\gamma$. Might have missing 2 |
| $(\alpha\beta + \beta\gamma + \gamma\alpha)^2 = \sum\alpha^2\beta^2 + 2\alpha\beta\gamma\sum\alpha$ | A1 | 1.1 — Correct, useful form |
| $\sum\alpha^2\beta^2 = \dfrac{13}{25}$ | A1 [5] | 1.1 |
### Part (a) Alternate Method
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitution $x = \sqrt{u}$ or $x^2 = u$ used | B1 | Soi by correct substitution — must be used |
| $5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0$ | M1* | Substitution and dealing with $(\sqrt{u})^3$ term |
| $((5u+3)\sqrt{u})^2 = (2u-1)^2$ | M1* | Rearrangement with $\sqrt{u}$ terms collected on one side and squaring. LHS does not need to be factorised before squaring |
| $25x^3 + 26x^2 + 13x - 1 = 0$ | A1 | Correct expansion and rearranging. Must be an equation. Must see correct entire cubic |
| $\sum\alpha^2\beta^2 = \dfrac{13}{25}$ | A1ft dep(*) | Correct coefficient ratio identified. Must gain method marks and have a cubic |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \dfrac{1}{25}$ | B1ft | 2.2a |
| $(\alpha + \beta + \gamma)^2 = \sum\alpha^2 + 2\sum\alpha\beta$ | M1 | 3.1a — Expansion showing or implying 6 terms including squares and cross-terms, symmetrical in $\alpha$, $\beta$ and $\gamma$. Might have 2 missing |
| $\sum\alpha^2 = -\dfrac{26}{25}$ | A1 | 1.1 — Seen or implied. $\sum\alpha^2 = \left(\frac{2}{5}\right)^2 - 2\times\frac{3}{5}$ |
| $25x^3 + 26x^2 + 13x - 1 = 0$ | A1 [4] | 1.1 — Or any non-zero integer multiple. Must be $= 0$. **CAO** |
### Part (b) Alternate Method
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitution $x = \sqrt{u}$ or $x^2 = u$ used | B1 | Soi by correct substitution — must be used |
| $5u\sqrt{u} - 2u + 3\sqrt{u} + 1 = 0$ | M1 | |
| $((5u+3)\sqrt{u})^2 = (2u-1)^2$ | M1 | |
| $25x^3 + 26x^2 + 13x - 1 = 0$ | A1 [4] | **CAO** |
---5 In this question you must show detailed reasoning.
You are given that $\alpha , \beta$ and $\gamma$ are the roots of the equation $5 x ^ { 3 } - 2 x ^ { 2 } + 3 x + 1 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha ^ { 2 } \beta ^ { 2 } + \beta ^ { 2 } \gamma ^ { 2 } + \gamma ^ { 2 } \alpha ^ { 2 }$.
\item Find a cubic equation whose roots are $\alpha ^ { 2 } , \beta ^ { 2 }$ and $\gamma ^ { 2 }$ giving your answer in the form $a x ^ { 3 } + b x ^ { 2 } + c x + d = 0$ where $a , b , c$ and $d$ are integers.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2019 Q5 [9]}}