CAIE Further Paper 1 2024 November — Question 3 10 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSymmetric functions of roots
DifficultyStandard +0.8 This is a standard Further Maths question on symmetric functions and Newton's identities, requiring systematic application of known formulas (e.g., s₁² = s₂ + 2p₂) and recurrence relations for power sums. While it involves multiple steps and careful algebraic manipulation, the techniques are well-established in the Further Pure syllabus with no novel insight required. It's moderately above average difficulty due to the length and potential for arithmetic errors, but remains a textbook-style exercise.
Spec4.05a Roots and coefficients: symmetric functions

3 It is given that $$\begin{aligned} & \alpha + \beta + \gamma + \delta = 2 \\ & \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = 3 \\ & \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } + \delta ^ { 3 } = 4 \end{aligned}$$
  1. Find the value of \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta\).
  2. Find the value of \(\alpha ^ { 2 } \beta + \alpha ^ { 2 } \gamma + \alpha ^ { 2 } \delta + \beta ^ { 2 } \alpha + \beta ^ { 2 } \gamma + \beta ^ { 2 } \delta + \gamma ^ { 2 } \alpha + \gamma ^ { 2 } \beta + \gamma ^ { 2 } \delta + \delta ^ { 2 } \alpha + \delta ^ { 2 } \beta + \delta ^ { 2 } \gamma\). \includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-06_2717_33_109_2014} \includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-07_2723_33_99_22}
  3. It is given that \(\alpha , \beta , \gamma , \delta\) are the roots of the equation $$6 x ^ { 4 } - 12 x ^ { 3 } + 3 x ^ { 2 } + 2 x + 6 = 0 .$$
    1. Find the value of \(\alpha ^ { 4 } + \beta ^ { 4 } + \gamma ^ { 4 } + \delta ^ { 4 }\).
    2. Find the value of \(\alpha ^ { 5 } + \beta ^ { 5 } + \gamma ^ { 5 } + \delta ^ { 5 }\).

Question 3(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)\); \(3=2^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)\)M1 Substitutes into formula for sum of squares
\(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{1}{2}\)A1
Question 3(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha^2(\beta+\gamma+\delta)+\beta^2(\alpha+\gamma+\delta)+\gamma^2(\alpha+\beta+\delta)+\delta^2(\alpha+\beta+\gamma)=\alpha^2(2-\alpha)+\beta^2(2-\beta)+\gamma^2(2-\gamma)+\delta^2(2-\delta)\)M1 A1 Factorises and substitutes
\(2(3)-4=2\)A1
Question 3(c)(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(6S_4-12S_3+3S_2+2S_1+24=0 \Rightarrow 6S_4-12(4)+3(3)+2(2)+24=0\)M1 Sums and substitutes
A1
\(S_4=\frac{11}{6}\)A1
Question 3(c)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(6S_5-12S_4+3S_3+2S_2+6S_1=0 \Rightarrow 6S_5-12\left(\frac{11}{6}\right)+3(4)+2(3)+12=0\)M1 Multiplies equation through by \(x\), sums and substitutes; \(S_5-2S_4+\frac{1}{2}S_3+\frac{1}{3}S_2+S_1=0 \Rightarrow S_5-2\left(\frac{11}{6}\right)+\frac{1}{2}(4)+\frac{1}{3}(3)+2=0\)
\(S_5=-\frac{4}{3}\)A1
## Question 3(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^2+\beta^2+\gamma^2+\delta^2=(\alpha+\beta+\gamma+\delta)^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)$; $3=2^2-2(\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta)$ | M1 | Substitutes into formula for sum of squares |
| $\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=\frac{1}{2}$ | A1 | |

## Question 3(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^2(\beta+\gamma+\delta)+\beta^2(\alpha+\gamma+\delta)+\gamma^2(\alpha+\beta+\delta)+\delta^2(\alpha+\beta+\gamma)=\alpha^2(2-\alpha)+\beta^2(2-\beta)+\gamma^2(2-\gamma)+\delta^2(2-\delta)$ | M1 A1 | Factorises and substitutes |
| $2(3)-4=2$ | A1 | |

## Question 3(c)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $6S_4-12S_3+3S_2+2S_1+24=0 \Rightarrow 6S_4-12(4)+3(3)+2(2)+24=0$ | M1 | Sums and substitutes |
| | A1 | |
| $S_4=\frac{11}{6}$ | A1 | |

## Question 3(c)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $6S_5-12S_4+3S_3+2S_2+6S_1=0 \Rightarrow 6S_5-12\left(\frac{11}{6}\right)+3(4)+2(3)+12=0$ | M1 | Multiplies equation through by $x$, sums and substitutes; $S_5-2S_4+\frac{1}{2}S_3+\frac{1}{3}S_2+S_1=0 \Rightarrow S_5-2\left(\frac{11}{6}\right)+\frac{1}{2}(4)+\frac{1}{3}(3)+2=0$ |
| $S_5=-\frac{4}{3}$ | A1 | |
3 It is given that

$$\begin{aligned}
& \alpha + \beta + \gamma + \delta = 2 \\
& \alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } + \delta ^ { 2 } = 3 \\
& \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } + \delta ^ { 3 } = 4
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta$.
\item Find the value of $\alpha ^ { 2 } \beta + \alpha ^ { 2 } \gamma + \alpha ^ { 2 } \delta + \beta ^ { 2 } \alpha + \beta ^ { 2 } \gamma + \beta ^ { 2 } \delta + \gamma ^ { 2 } \alpha + \gamma ^ { 2 } \beta + \gamma ^ { 2 } \delta + \delta ^ { 2 } \alpha + \delta ^ { 2 } \beta + \delta ^ { 2 } \gamma$.\\

\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-06_2717_33_109_2014}\\
\includegraphics[max width=\textwidth, alt={}, center]{beb9c1f1-1676-4432-a42a-c418ff9f45d8-07_2723_33_99_22}
\item It is given that $\alpha , \beta , \gamma , \delta$ are the roots of the equation

$$6 x ^ { 4 } - 12 x ^ { 3 } + 3 x ^ { 2 } + 2 x + 6 = 0 .$$
\begin{enumerate}[label=(\roman*)]
\item Find the value of $\alpha ^ { 4 } + \beta ^ { 4 } + \gamma ^ { 4 } + \delta ^ { 4 }$.
\item Find the value of $\alpha ^ { 5 } + \beta ^ { 5 } + \gamma ^ { 5 } + \delta ^ { 5 }$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q3 [10]}}