| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. While it involves multiple steps and careful bookkeeping with symmetric functions, the approach is methodical rather than requiring novel insight—students follow a clear template of substitution, using root relationships, and computing symmetric expressions. The difficulty is elevated above average due to the algebraic complexity and being Further Maths content, but remains accessible to well-prepared students. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sqrt{-7y}(-7y) + 2(-7y) + \sqrt{-7y} + 7 = 0 \Rightarrow \sqrt{-7y}(-7y+1) = 14y-7 \Rightarrow -7y(-7y+1)^2 = (14y-7)^2\) | M1 | Uses given substitution and eliminates radical |
| \(\Rightarrow 49y^3 + 14y^2 - 27y + 7 = 0\) | A1 | AG |
| \(y = \frac{x^2}{-7} = \frac{x^2}{\alpha\beta\gamma}\) | M1 | Uses \(\alpha\beta\gamma = -7\) |
| So roots are \(\frac{\alpha^2}{\alpha\beta\gamma} = \frac{\alpha}{\beta\gamma}\), \(\frac{\beta^2}{\alpha\beta\gamma} = \frac{\beta}{\alpha\gamma}\), \(\frac{\gamma^2}{\alpha\beta\gamma} = \frac{\gamma}{\alpha\beta}\) | A1 | AG |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\alpha}{\beta\gamma} + \frac{\beta}{\alpha\gamma} + \frac{\gamma}{\alpha\beta} = -\frac{2}{7}\), \(\frac{1}{\gamma^2} + \frac{1}{\beta^2} + \frac{1}{\alpha^2} = -\frac{27}{49}\) | B1 | States sum of roots and \(\alpha'\beta' + \alpha'\gamma' + \beta'\gamma'\) |
| \(\frac{\alpha^2}{\beta^2\gamma^2} + \frac{\beta^2}{\alpha^2\gamma^2} + \frac{\gamma^2}{\alpha^2\beta^2} = \left(-\frac{2}{7}\right)^2 - 2\left(-\frac{27}{49}\right) = \frac{58}{49}\) | M1 A1 | Uses \(\alpha'^2 + \beta'^2 + \gamma'^2 = (\alpha'+\beta'+\gamma')^2 - 2(\alpha'\beta'+\alpha'\gamma'+\beta'\gamma')\). AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(49\left(\frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\alpha^3\gamma^3} + \frac{\gamma^3}{\alpha^3\beta^3}\right) = -14\left(\frac{58}{49}\right) + 27\left(-\frac{2}{7}\right) - 21\) | M1 | Uses \(49\alpha'^3 = -14\alpha'^2 + 27\alpha' - 7\) |
| \(\Rightarrow \frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\alpha^3\gamma^3} + \frac{\gamma^3}{\alpha^3\beta^3} = -\frac{317}{343}\) | A1 | |
| 2 |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{-7y}(-7y) + 2(-7y) + \sqrt{-7y} + 7 = 0 \Rightarrow \sqrt{-7y}(-7y+1) = 14y-7 \Rightarrow -7y(-7y+1)^2 = (14y-7)^2$ | M1 | Uses given substitution and eliminates radical |
| $\Rightarrow 49y^3 + 14y^2 - 27y + 7 = 0$ | A1 | AG |
| $y = \frac{x^2}{-7} = \frac{x^2}{\alpha\beta\gamma}$ | M1 | Uses $\alpha\beta\gamma = -7$ |
| So roots are $\frac{\alpha^2}{\alpha\beta\gamma} = \frac{\alpha}{\beta\gamma}$, $\frac{\beta^2}{\alpha\beta\gamma} = \frac{\beta}{\alpha\gamma}$, $\frac{\gamma^2}{\alpha\beta\gamma} = \frac{\gamma}{\alpha\beta}$ | A1 | AG |
| | **4** | |
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\alpha}{\beta\gamma} + \frac{\beta}{\alpha\gamma} + \frac{\gamma}{\alpha\beta} = -\frac{2}{7}$, $\frac{1}{\gamma^2} + \frac{1}{\beta^2} + \frac{1}{\alpha^2} = -\frac{27}{49}$ | B1 | States sum of roots and $\alpha'\beta' + \alpha'\gamma' + \beta'\gamma'$ |
| $\frac{\alpha^2}{\beta^2\gamma^2} + \frac{\beta^2}{\alpha^2\gamma^2} + \frac{\gamma^2}{\alpha^2\beta^2} = \left(-\frac{2}{7}\right)^2 - 2\left(-\frac{27}{49}\right) = \frac{58}{49}$ | M1 A1 | Uses $\alpha'^2 + \beta'^2 + \gamma'^2 = (\alpha'+\beta'+\gamma')^2 - 2(\alpha'\beta'+\alpha'\gamma'+\beta'\gamma')$. AG |
| | **3** | |
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $49\left(\frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\alpha^3\gamma^3} + \frac{\gamma^3}{\alpha^3\beta^3}\right) = -14\left(\frac{58}{49}\right) + 27\left(-\frac{2}{7}\right) - 21$ | M1 | Uses $49\alpha'^3 = -14\alpha'^2 + 27\alpha' - 7$ |
| $\Rightarrow \frac{\alpha^3}{\beta^3\gamma^3} + \frac{\beta^3}{\alpha^3\gamma^3} + \frac{\gamma^3}{\alpha^3\beta^3} = -\frac{317}{343}$ | A1 | |
| | **2** | |7 The equation $x ^ { 3 } + 2 x ^ { 2 } + x + 7 = 0$ has roots $\alpha , \beta , \gamma$.\\
(i) Use the relation $x ^ { 2 } = - 7 y$ to show that the equation
$$49 y ^ { 3 } + 14 y ^ { 2 } - 27 y + 7 = 0$$
has roots $\frac { \alpha } { \beta \gamma } , \frac { \beta } { \gamma \alpha } , \frac { \gamma } { \alpha \beta }$.\\
(ii) Show that $\frac { \alpha ^ { 2 } } { \beta ^ { 2 } \gamma ^ { 2 } } + \frac { \beta ^ { 2 } } { \gamma ^ { 2 } \alpha ^ { 2 } } + \frac { \gamma ^ { 2 } } { \alpha ^ { 2 } \beta ^ { 2 } } = \frac { 58 } { 49 }$.\\
(iii) Find the exact value of $\frac { \alpha ^ { 3 } } { \beta ^ { 3 } \gamma ^ { 3 } } + \frac { \beta ^ { 3 } } { \gamma ^ { 3 } \alpha ^ { 3 } } + \frac { \gamma ^ { 3 } } { \alpha ^ { 3 } \beta ^ { 3 } }$.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q7 [9]}}