| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Sum of powers of roots |
| Difficulty | Challenging +1.2 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and recurrence relations. Part (a) involves routine substitution y=3x+1, parts (b-c) use Newton's sums with the derived equation. While multi-step, each component follows well-established techniques without requiring novel insight. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=3x+1 \Rightarrow x=\frac{1}{3}(y-1)\); \(\Rightarrow 27\left(\frac{y-1}{3}\right)^3+18\left(\frac{y-1}{3}\right)^2+6\left(\frac{y-1}{3}\right)-1=0\) | B1 | Substitutes |
| \(\Rightarrow (y-1)^3+2(y-1)^2+2(y-1)-1=0\); \(\Rightarrow y^3-3y^2+3y-1+2y^2-4y+2+2y-2-1=0\) | M1 | Expands |
| \(y^3-y^2+y-2=0\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_2 = 1^2 - 2(1) = -1\) | M1 A1 | Uses formula for sum of squares, AG |
| \(S_3 = (3\alpha+1)^3+(3\beta+1)^3+(3\gamma+1)^3 = -1-(1)+6\) | M1 | Uses \(y^3 = y^2-y+2\) or expands and uses original equation |
| \(4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{-1} = \frac{(3\alpha+1)(3\beta+1)+(3\beta+1)(3\gamma+1)+(3\gamma+1)(3\alpha+1)}{(3\alpha+1)(3\beta+1)(3\gamma+1)} = \frac{1}{2}\) | B1 | |
| \(2S_{-2} = S_1 - 3 + S_{-1} = 1-3+\frac{1}{2}\) | M1 | Uses \(2y^{-2} = y-1+y^{-1}\) |
| \(S_{-2} = -\frac{3}{4}\) | A1 | CAO |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=3x+1 \Rightarrow x=\frac{1}{3}(y-1)$; $\Rightarrow 27\left(\frac{y-1}{3}\right)^3+18\left(\frac{y-1}{3}\right)^2+6\left(\frac{y-1}{3}\right)-1=0$ | B1 | Substitutes |
| $\Rightarrow (y-1)^3+2(y-1)^2+2(y-1)-1=0$; $\Rightarrow y^3-3y^2+3y-1+2y^2-4y+2+2y-2-1=0$ | M1 | Expands |
| $y^3-y^2+y-2=0$ | A1 | AG |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_2 = 1^2 - 2(1) = -1$ | M1 A1 | Uses formula for sum of squares, AG |
| $S_3 = (3\alpha+1)^3+(3\beta+1)^3+(3\gamma+1)^3 = -1-(1)+6$ | M1 | Uses $y^3 = y^2-y+2$ or expands and uses original equation |
| $4$ | A1 | |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{-1} = \frac{(3\alpha+1)(3\beta+1)+(3\beta+1)(3\gamma+1)+(3\gamma+1)(3\alpha+1)}{(3\alpha+1)(3\beta+1)(3\gamma+1)} = \frac{1}{2}$ | B1 | |
| $2S_{-2} = S_1 - 3 + S_{-1} = 1-3+\frac{1}{2}$ | M1 | Uses $2y^{-2} = y-1+y^{-1}$ |
| $S_{-2} = -\frac{3}{4}$ | A1 | CAO |
---4 The cubic equation $27 x ^ { 3 } + 18 x ^ { 2 } + 6 x - 1 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Show that a cubic equation with roots $3 \alpha + 1,3 \beta + 1,3 \gamma + 1$ is
$$y ^ { 3 } - y ^ { 2 } + y - 2 = 0$$
The sum $( 3 \alpha + 1 ) ^ { n } + ( 3 \beta + 1 ) ^ { n } + ( 3 \gamma + 1 ) ^ { n }$ is denoted by $\mathrm { S } _ { \mathrm { n } }$.
\item Find the values of $S _ { 2 }$ and $S _ { 3 }$.
\item Find the values of $S _ { - 1 }$ and $S _ { - 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q4 [10]}}