| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Substitution to find new equation |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring polynomial substitution, relationships between roots and coefficients, and manipulation of power sums including negative powers. While the substitution in part (i) is mechanical, parts (ii) and (iii) require understanding of recurrence relations or Newton's identities for power sums, which goes beyond standard A-level and requires problem-solving insight typical of FP1. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitutes \(x = \frac{y+1}{3}\) | M1 | Accept substitution of \(y = 3x-1\) into given equation and derivation of equation in \(x\) |
| Obtains the given result | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_3 = 2S_1 + 7 \times 3\) | M1 | Uses \(y^3 = 2y + 7\). Or uses formula for \(\Sigma(3\alpha-1)^3\) |
| \(= 21\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{-1} = \frac{(3\alpha-1)(3\beta-1)+(3\alpha-1)(3\gamma-1)+(3\beta-1)(3\gamma-1)}{(3\alpha-1)(3\beta-1)(3\gamma-1)} = -\frac{2}{7}\) | M1 A1 | Award M1A1 if \(S_{-1} = -\frac{2}{7}\) written down directly |
| \(7S_{-2} = S_1 - 2S_{-1}\) | M1 | Uses \(7y^{-2} = y - 2y^{-1}\) |
| \(s_{-2} = \frac{4}{49}\) | A1 | |
| Alt method: \(S_{-2} = \sum\frac{1}{(3\alpha-1)^2} = \frac{\Sigma(3\alpha-1)^2(3\beta-1)^2}{(3\alpha-1)^2(3\beta-1)^2(3\gamma-1)^2}\) | M1 A1 | Alt method: Finds cubic with roots \(\frac{1}{3\alpha-1}\) etc. M1; \(7z^3+2z^2-1=0\) A1; Uses \(S_2=(S_1)^2-2x\Sigma\alpha\beta\) M1; \(=\frac{4}{49}\) A1 |
| \(\frac{(\Sigma(3\alpha-1)(3\beta-1))^2 - 2(3\alpha-1)(3\beta-1)(3\gamma-1)(\Sigma(3\alpha-1))}{(3\alpha-1)^2(3\beta-1)^2(3\gamma-1)^2}\) | M1 | |
| \(= \frac{(-2)2 - 2(7)(0)}{7^2} = \frac{4}{49}\) | A1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitutes $x = \frac{y+1}{3}$ | M1 | Accept substitution of $y = 3x-1$ into given equation and derivation of equation in $x$ |
| Obtains the given result | A1 | AG |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_3 = 2S_1 + 7 \times 3$ | M1 | Uses $y^3 = 2y + 7$. Or uses formula for $\Sigma(3\alpha-1)^3$ |
| $= 21$ | A1 | |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{-1} = \frac{(3\alpha-1)(3\beta-1)+(3\alpha-1)(3\gamma-1)+(3\beta-1)(3\gamma-1)}{(3\alpha-1)(3\beta-1)(3\gamma-1)} = -\frac{2}{7}$ | M1 A1 | Award M1A1 if $S_{-1} = -\frac{2}{7}$ written down directly |
| $7S_{-2} = S_1 - 2S_{-1}$ | M1 | Uses $7y^{-2} = y - 2y^{-1}$ |
| $s_{-2} = \frac{4}{49}$ | A1 | |
| Alt method: $S_{-2} = \sum\frac{1}{(3\alpha-1)^2} = \frac{\Sigma(3\alpha-1)^2(3\beta-1)^2}{(3\alpha-1)^2(3\beta-1)^2(3\gamma-1)^2}$ | M1 A1 | Alt method: Finds cubic with roots $\frac{1}{3\alpha-1}$ etc. M1; $7z^3+2z^2-1=0$ A1; Uses $S_2=(S_1)^2-2x\Sigma\alpha\beta$ M1; $=\frac{4}{49}$ A1 |
| $\frac{(\Sigma(3\alpha-1)(3\beta-1))^2 - 2(3\alpha-1)(3\beta-1)(3\gamma-1)(\Sigma(3\alpha-1))}{(3\alpha-1)^2(3\beta-1)^2(3\gamma-1)^2}$ | M1 | |
| $= \frac{(-2)2 - 2(7)(0)}{7^2} = \frac{4}{49}$ | A1 | |
---6 The equation
$$9 x ^ { 3 } - 9 x ^ { 2 } + x - 2 = 0$$
has roots $\alpha , \beta , \gamma$.\\
(i) Use the substitution $y = 3 x - 1$ to show that $3 \alpha - 1,3 \beta - 1,3 \gamma - 1$ are the roots of the equation
$$y ^ { 3 } - 2 y - 7 = 0$$
The sum $( 3 \alpha - 1 ) ^ { n } + ( 3 \beta - 1 ) ^ { n } + ( 3 \gamma - 1 ) ^ { n }$ is denoted by $S _ { n }$.\\
(ii) Find the value of $S _ { 3 }$.\\
(iii) Find the value of $S _ { - 2 }$.\\
\hfill \mbox{\textit{CAIE FP1 2018 Q6 [8]}}