| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring multiple advanced techniques: finding equations with transformed roots (cubing), manipulating symmetric functions using Newton's identities or power sum formulas, and connecting singularity conditions to polynomial roots. While each part follows established methods, the multi-step nature, the need to work with sixth powers, and the matrix singularity connection make this substantially harder than typical A-level questions but still within standard Further Maths territory. |
| Spec | 4.03l Singular/non-singular matrices4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^3 \Rightarrow x = y^{\frac{1}{3}}\) | B1 | Substitutes |
| \(y + cy^{\frac{1}{3}} + 1 = 0 \Rightarrow -c^3y = (y+1)^3 = y^3 + 3y^2 + 3y + 1\) | M1 | Correct attempt to eliminate cube root |
| \(y^3 + 3y^2 + (3+c^3)y + 1 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha^3 + \beta^3 + \gamma^3 = -3\), \(\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 3 + c^3\) | B1 FT | Using their answer to (a) |
| \(\alpha^6 + \beta^6 + \gamma^6 = (-3)^2 - 2(3+c^3)\) | M1 | \(\alpha^6+\beta^6+\gamma^6 = (\alpha^3+\beta^3+\gamma^3)^2 - 2(\alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3)\) |
| \(= 3 - 2c^3\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha^3\beta^3\gamma^3 = -1\) | B1 | If using their answer to (a) FT |
| \(\begin{vmatrix} 1 & \alpha^3 & \beta^3 \\ \alpha^3 & 1 & \gamma^3 \\ \beta^3 & \gamma^3 & 1 \end{vmatrix} = 1 - (\alpha^6+\beta^6+\gamma^6) + 2\alpha^3\beta^3\gamma^3 = 2c^3 - 4\) | M1 A1 | Evaluates determinant |
| \(2c^3 - 4 = 0\) | M1 | Sets determinant equal to zero |
| \(c = \sqrt[3]{2}\) | A1 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^3 \Rightarrow x = y^{\frac{1}{3}}$ | B1 | Substitutes |
| $y + cy^{\frac{1}{3}} + 1 = 0 \Rightarrow -c^3y = (y+1)^3 = y^3 + 3y^2 + 3y + 1$ | M1 | Correct attempt to eliminate cube root |
| $y^3 + 3y^2 + (3+c^3)y + 1 = 0$ | A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^3 + \beta^3 + \gamma^3 = -3$, $\alpha^3\beta^3 + \beta^3\gamma^3 + \gamma^3\alpha^3 = 3 + c^3$ | B1 FT | Using their answer to (a) |
| $\alpha^6 + \beta^6 + \gamma^6 = (-3)^2 - 2(3+c^3)$ | M1 | $\alpha^6+\beta^6+\gamma^6 = (\alpha^3+\beta^3+\gamma^3)^2 - 2(\alpha^3\beta^3+\beta^3\gamma^3+\gamma^3\alpha^3)$ |
| $= 3 - 2c^3$ | A1 | AG |
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^3\beta^3\gamma^3 = -1$ | B1 | If using their answer to (a) FT |
| $\begin{vmatrix} 1 & \alpha^3 & \beta^3 \\ \alpha^3 & 1 & \gamma^3 \\ \beta^3 & \gamma^3 & 1 \end{vmatrix} = 1 - (\alpha^6+\beta^6+\gamma^6) + 2\alpha^3\beta^3\gamma^3 = 2c^3 - 4$ | M1 A1 | Evaluates determinant |
| $2c^3 - 4 = 0$ | M1 | Sets determinant equal to zero |
| $c = \sqrt[3]{2}$ | A1 | |
---3 The cubic equation $\mathrm { x } ^ { 3 } + \mathrm { cx } + 1 = 0$, where $c$ is a constant, has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation whose roots are $\alpha ^ { 3 } , \beta ^ { 3 } , \gamma ^ { 3 }$.
\item Show that $\alpha ^ { 6 } + \beta ^ { 6 } + \gamma ^ { 6 } = 3 - 2 c ^ { 3 }$.
\item Find the real value of $c$ for which the matrix $\left( \begin{array} { c c c } 1 & \alpha ^ { 3 } & \beta ^ { 3 } \\ \alpha ^ { 3 } & 1 & \gamma ^ { 3 } \\ \beta ^ { 3 } & \gamma ^ { 3 } & 1 \end{array} \right)$ is singular.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q3 [11]}}