| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) uses the standard substitution y=1/x³, parts (b) and (c) build on part (a) using symmetric function techniques. While it requires multiple steps and careful algebra, the methods are well-practiced in Further Maths syllabi with no novel insight needed. |
| Spec | 4.05a 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 |
| \(\Rightarrow 2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0\) leading to \(5y^{-\frac{2}{3}} = 6 - 2y^{-1} \Rightarrow 125y = (6y-2)^3\) | M1 | Cubes to eliminate radical |
| \(216y^3 - 216y^2 - 53y - 8 = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2x^3 - 6)^3 = (-5x^2)^3\) | M1 | \(8x^9 - 72x^6 + 125x^6 + 216x^3 - 216 = 0\) |
| \(y = x^{-3}\) leading to \(x^3 = y^{-1}\) | B1 | Substitutes |
| \(216y^3 - 216y^2 - 53y - 8 = 0\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1\) and \(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\dfrac{53}{216}\) | B1 FT | Using their answer to part (a) |
| \(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\dfrac{53}{216}\right)\) | M1 | \(\alpha^{-6}+\beta^{-6}+\gamma^{-6} = (\alpha^{-3}+\beta^{-3}+\gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3}+\beta^{-3}\gamma^{-3}+\gamma^{-3}\alpha^{-3})\) |
| \(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \dfrac{161}{108}\) | A1 | |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(216S_{-9} = 216S_{-6} + 53S_{-3} + 24\) | M1 | Using their \(216\alpha^{-9} - 216\alpha^{-6} - 53\alpha^{-3} - 8 = 0\) |
| \(S_{-9} = \dfrac{399}{216} = \dfrac{133}{72}\) | A1 | |
| Total: 2 |
## Question 4:
### Part 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^{-3} \Rightarrow x = y^{-\frac{1}{3}}$ | B1 | Substitutes |
| $\Rightarrow 2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0$ leading to $5y^{-\frac{2}{3}} = 6 - 2y^{-1} \Rightarrow 125y = (6y-2)^3$ | M1 | Cubes to eliminate radical |
| $216y^3 - 216y^2 - 53y - 8 = 0$ | A1 | |
**Alternative method 4(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2x^3 - 6)^3 = (-5x^2)^3$ | M1 | $8x^9 - 72x^6 + 125x^6 + 216x^3 - 216 = 0$ |
| $y = x^{-3}$ leading to $x^3 = y^{-1}$ | B1 | Substitutes |
| $216y^3 - 216y^2 - 53y - 8 = 0$ | A1 | |
| **Total: 3** | | |
---
### Part 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1$ and $\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\dfrac{53}{216}$ | B1 FT | Using their answer to part (a) |
| $\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\dfrac{53}{216}\right)$ | M1 | $\alpha^{-6}+\beta^{-6}+\gamma^{-6} = (\alpha^{-3}+\beta^{-3}+\gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3}+\beta^{-3}\gamma^{-3}+\gamma^{-3}\alpha^{-3})$ |
| $\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \dfrac{161}{108}$ | A1 | |
| **Total: 3** | | |
---
### Part 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $216S_{-9} = 216S_{-6} + 53S_{-3} + 24$ | M1 | Using their $216\alpha^{-9} - 216\alpha^{-6} - 53\alpha^{-3} - 8 = 0$ |
| $S_{-9} = \dfrac{399}{216} = \dfrac{133}{72}$ | A1 | |
| **Total: 2** | | |
---4 The cubic equation $2 x ^ { 3 } + 5 x ^ { 2 } - 6 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation whose roots are $\frac { 1 } { \alpha ^ { 3 } } , \frac { 1 } { \beta ^ { 3 } } , \frac { 1 } { \gamma ^ { 3 } }$.
\item Find the value of $\frac { 1 } { \alpha ^ { 6 } } + \frac { 1 } { \beta ^ { 6 } } + \frac { 1 } { \gamma ^ { 6 } }$.
\item Find also the value of $\frac { 1 } { \alpha ^ { 9 } } + \frac { 1 } { \beta ^ { 9 } } + \frac { 1 } { \gamma ^ { 9 } }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q4 [8]}}