CAIE FP1 2012 November — Question 11 EITHER

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeSum of powers of roots
DifficultyChallenging +1.3 This is a standard Further Maths question on Newton-Sums/recurrence relations for power sums of roots. The recurrence relation derivation is routine (multiply the polynomial equation by x^n and sum over roots), computing S_2, S_3, S_4, S_5 using the recurrence is mechanical, and the final expression simplifies to S_2·S_3 using symmetry. While it requires familiarity with FM techniques and careful algebra, it follows a well-established template without requiring novel insight.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots4.06b Method of differences: telescoping series
The roots of the equation \(x ^ { 4 } - 3 x ^ { 2 } + 5 x - 2 = 0\) are \(\alpha , \beta , \gamma , \delta\), and \(\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } + \delta ^ { n }\) is denoted by \(S _ { n }\). Show that $$S _ { n + 4 } - 3 S _ { n + 2 } + 5 S _ { n + 1 } - 2 S _ { n } = 0$$ Find the values of
  1. \(S _ { 2 }\) and \(S _ { 4 }\),
  2. \(S _ { 3 }\) and \(S _ { 5 }\). Hence find the value of $$\alpha ^ { 2 } \left( \beta ^ { 3 } + \gamma ^ { 3 } + \delta ^ { 3 } \right) + \beta ^ { 2 } \left( \gamma ^ { 3 } + \delta ^ { 3 } + \alpha ^ { 3 } \right) + \gamma ^ { 2 } \left( \delta ^ { 3 } + \alpha ^ { 3 } + \beta ^ { 3 } \right) + \delta ^ { 2 } \left( \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } \right) .$$
Question 11:
Recurrence relation (2 marks):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\alpha\) is a root \(\Rightarrow \alpha^4-3\alpha^2+5\alpha-2=0\)
\(\Rightarrow \alpha^{n+4}-3\alpha^{n+2}+5\alpha^{n+1}-2\alpha^n=0\)M1 Substitutes \(\alpha\) into equation, multiplies by \(\alpha^n\)
Repeat for \(\beta,\gamma,\delta\) and sum \(\Rightarrow S_{n+4}-3S_{n+2}+5S_{n+1}-2S_n=0\) (AG)A1 2
Part (i) — Finding \(S_4\) (3 marks):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S_2=0-2\times(-3)=6\)B1 Uses \(\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta\)
\(S_4=3\times6-5\times0+2\times4=26\)M1A1 3
Part (ii) — Finding \(\sum\alpha^2\beta^3\) (6 marks):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}=\frac{-5}{-2}=\frac{5}{2}\)M1A1 Uses \(S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}\)
\(S_3=3\times0-5\times4+2\times\frac{5}{2}=-15\)M1A1 Finds \(S_3\) from formula
\(S_5=3\times(-15)-5\times6+2\times0=-75\)M1A1 6
Answer/WorkingMarks Guidance
\(\sum\alpha^2\beta^3=S_2S_3-S_5\)M1
\(=6\times(-15)-(-75)=-15\)M1A1 3
Question 11 (OR):
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
Reduces M to echelon form: \(\begin{pmatrix} 2 & 1 & -1 & 4 \\ 3 & 4 & 6 & 1 \\ -1 & 2 & 8 & -7 \end{pmatrix} \rightarrow \ldots \rightarrow \begin{pmatrix} 2 & 1 & -1 & 4 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & 0 \end{pmatrix}\)M1A1
\(\text{Dim}(\mathbf{M}) = 4 - 2 = 2\)A1 Uses dimension theorem
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
Basis for \(R\) is \(\left\{ \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \right\}\)B1 OE
\(x = 2\lambda + \mu\), \(y = 3\lambda + 4\mu\), \(z = -\lambda + 2\mu \Rightarrow 2x - y + z = 0\)M1A1 Finds Cartesian equation for \(R\)
Part (iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(2x + y - z + 4t = 0\), \(y + 3z - 2t = 0\), \(t = \lambda\) and \(z = \mu\), \(\Rightarrow y = 2\lambda - 3\mu\) and \(x = -3\lambda + 2\mu\)M1 Finds basis for null space
\(\Rightarrow\) Basis of null space is \(\left\{ \begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix} \right\}\)A1A1 OE
\(2 \times 8 - 7 + k = 0 \Rightarrow k = -9\)B1 Evaluates \(k\)
\(5\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -9 \end{pmatrix}\)M1A1 OE, via equations — finds particular solution
\(\mathbf{x} = \begin{pmatrix} 5 \\ -2 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix}\)M1A1 Finds general solution
Total: [14]
# Question 11:

## Recurrence relation (2 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha$ is a root $\Rightarrow \alpha^4-3\alpha^2+5\alpha-2=0$ | | |
| $\Rightarrow \alpha^{n+4}-3\alpha^{n+2}+5\alpha^{n+1}-2\alpha^n=0$ | M1 | Substitutes $\alpha$ into equation, multiplies by $\alpha^n$ |
| Repeat for $\beta,\gamma,\delta$ and sum $\Rightarrow S_{n+4}-3S_{n+2}+5S_{n+1}-2S_n=0$ (AG) | A1 | 2 | |

## Part (i) — Finding $S_4$ (3 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_2=0-2\times(-3)=6$ | B1 | Uses $\sum\alpha^2=\left(\sum\alpha\right)^2-2\sum\alpha\beta$ |
| $S_4=3\times6-5\times0+2\times4=26$ | M1A1 | 3 | Finds $S_4$ from formula |

## Part (ii) — Finding $\sum\alpha^2\beta^3$ (6 marks):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}=\frac{-5}{-2}=\frac{5}{2}$ | M1A1 | Uses $S_{-1}=\frac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$ |
| $S_3=3\times0-5\times4+2\times\frac{5}{2}=-15$ | M1A1 | Finds $S_3$ from formula |
| $S_5=3\times(-15)-5\times6+2\times0=-75$ | M1A1 | 6 | Finds $S_5$ from formula |

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum\alpha^2\beta^3=S_2S_3-S_5$ | M1 | |
| $=6\times(-15)-(-75)=-15$ | M1A1 | 3 | **[14]** |

# Question 11 (OR):

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Reduces **M** to echelon form: $\begin{pmatrix} 2 & 1 & -1 & 4 \\ 3 & 4 & 6 & 1 \\ -1 & 2 & 8 & -7 \end{pmatrix} \rightarrow \ldots \rightarrow \begin{pmatrix} 2 & 1 & -1 & 4 \\ 0 & 1 & 3 & -2 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ | M1A1 | |
| $\text{Dim}(\mathbf{M}) = 4 - 2 = 2$ | A1 | Uses dimension theorem |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Basis for $R$ is $\left\{ \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \right\}$ | B1 | OE |
| $x = 2\lambda + \mu$, $y = 3\lambda + 4\mu$, $z = -\lambda + 2\mu \Rightarrow 2x - y + z = 0$ | M1A1 | Finds Cartesian equation for $R$ |

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + y - z + 4t = 0$, $y + 3z - 2t = 0$, $t = \lambda$ and $z = \mu$, $\Rightarrow y = 2\lambda - 3\mu$ and $x = -3\lambda + 2\mu$ | M1 | Finds basis for null space |
| $\Rightarrow$ Basis of null space is $\left\{ \begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix} \right\}$ | A1A1 | OE |
| $2 \times 8 - 7 + k = 0 \Rightarrow k = -9$ | B1 | Evaluates $k$ |
| $5\begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} - 2\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 7 \\ -9 \end{pmatrix}$ | M1A1 | OE, via equations — finds particular solution |
| $\mathbf{x} = \begin{pmatrix} 5 \\ -2 \\ 0 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} -3 \\ 2 \\ 0 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ -3 \\ 1 \\ 0 \end{pmatrix}$ | M1A1 | Finds general solution |

**Total: [14]**
The roots of the equation $x ^ { 4 } - 3 x ^ { 2 } + 5 x - 2 = 0$ are $\alpha , \beta , \gamma , \delta$, and $\alpha ^ { n } + \beta ^ { n } + \gamma ^ { n } + \delta ^ { n }$ is denoted by $S _ { n }$. Show that

$$S _ { n + 4 } - 3 S _ { n + 2 } + 5 S _ { n + 1 } - 2 S _ { n } = 0$$

Find the values of\\
(i) $S _ { 2 }$ and $S _ { 4 }$,\\
(ii) $S _ { 3 }$ and $S _ { 5 }$.

Hence find the value of

$$\alpha ^ { 2 } \left( \beta ^ { 3 } + \gamma ^ { 3 } + \delta ^ { 3 } \right) + \beta ^ { 2 } \left( \gamma ^ { 3 } + \delta ^ { 3 } + \alpha ^ { 3 } \right) + \gamma ^ { 2 } \left( \delta ^ { 3 } + \alpha ^ { 3 } + \beta ^ { 3 } \right) + \delta ^ { 2 } \left( \alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } \right) .$$

\hfill \mbox{\textit{CAIE FP1 2012 Q11 EITHER}}
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