| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| 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 Pure question on sum of powers of roots using Newton's recurrence relations. Part (a) applies Vieta's formulas directly, part (b) derives and uses the recurrence relation (routine for Further Maths students), part (c) is a standard substitution technique, and part (d) applies Vieta's to the transformed equation. While it requires multiple techniques and is appropriately challenging for Further Maths, all steps follow well-established procedures without requiring novel insight. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_1 = 2\) | B1 | |
| \(S_2 = S_1^2 - 2(0)\) | M1 | Uses formula for sum of squares. |
| \(= 4\) | A1 | Correct answer implies M1A1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n\) | B1 | CAO or as a single fraction. |
| Answer | Marks | Guidance |
|---|---|---|
| \(S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1\) | M1 | Uses their recursive formula from part (i) to find \(S_4\) \(\left[S_3 = \frac{7}{2}\right]\). |
| \(= 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 2 - y\) | B1 | SOI |
| \(2(2-y)^3 - 4(2-y)^2 + 3 = 0\) | M1 | Makes *their* substitution. |
| \(2y^3 - 8y^2 + 8y - 3 = 0\) | A1 | OE but must be an equation. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{\frac{8}{2}}{\frac{-3}{2}}\) OR use \(2S_2 - 8S_1 + 8S_0 - 3S_{-1} = 0\) with substitution of their values | M1 | Uses \(\dfrac{1}{\alpha'} + \dfrac{1}{\beta'} + \dfrac{1}{\gamma'} = \dfrac{\alpha'\beta' + \beta'\gamma' + \gamma'\alpha'}{\alpha'\beta'\gamma'}\) |
| \(= \dfrac{8}{3}\) | A1 FT | FT from 2(c). |
# Question 2:
## Part (a):
$S_1 = 2$ | **B1** |
$S_2 = S_1^2 - 2(0)$ | **M1** | Uses formula for sum of squares.
$= 4$ | **A1** | Correct answer implies M1A1.
## Part (b)(i):
$S_{n+3} = 2S_{n+2} - \frac{3}{2}S_n$ | **B1** | CAO or as a single fraction.
## Part (b)(ii):
$S_4 = 2S_3 - \frac{3}{2}S_1 = 2(2S_2 - \frac{3}{2}S_0) - \frac{3}{2}S_1$ | **M1** | Uses their recursive formula from part (i) to find $S_4$ $\left[S_3 = \frac{7}{2}\right]$.
$= 4$ | **A1** |
## Part (c):
$x = 2 - y$ | **B1** | SOI
$2(2-y)^3 - 4(2-y)^2 + 3 = 0$ | **M1** | Makes *their* substitution.
$2y^3 - 8y^2 + 8y - 3 = 0$ | **A1** | OE but must be an equation.
## Part (d):
$\dfrac{\frac{8}{2}}{\frac{-3}{2}}$ OR use $2S_2 - 8S_1 + 8S_0 - 3S_{-1} = 0$ with substitution of their values | **M1** | Uses $\dfrac{1}{\alpha'} + \dfrac{1}{\beta'} + \dfrac{1}{\gamma'} = \dfrac{\alpha'\beta' + \beta'\gamma' + \gamma'\alpha'}{\alpha'\beta'\gamma'}$
$= \dfrac{8}{3}$ | **A1 FT** | FT from 2(c).
---2 The cubic equation $2 x ^ { 3 } - 4 x ^ { 2 } + 3 = 0$ has roots $\alpha , \beta , \gamma$. Let $\mathrm { S } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } } + \beta ^ { \mathrm { n } } + \gamma ^ { \mathrm { n } }$.
\begin{enumerate}[label=(\alph*)]
\item State the value of $S _ { 1 }$ and find the value of $S _ { 2 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Express $\mathrm { S } _ { \mathrm { n } + 3 }$ in terms of $\mathrm { S } _ { \mathrm { n } + 2 }$ and $\mathrm { S } _ { \mathrm { n } }$.
\item Hence, or otherwise, find the value of $S _ { 4 }$.
\end{enumerate}\item Use the substitution $\mathrm { y } = \mathrm { S } _ { 1 } - \mathrm { x }$, where $S _ { 1 }$ is the numerical value found in part (a), to find and simplify an equation whose roots are $\alpha + \beta , \beta + \gamma , \gamma + \alpha$.
\item Find the value of $\frac { 1 } { \alpha + \beta } + \frac { 1 } { \beta + \gamma } + \frac { 1 } { \gamma + \alpha }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q2 [11]}}