CAIE Further Paper 1 2020 June — Question 1 7 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2020
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeEquation with nonlinearly transformed roots
DifficultyStandard +0.3 This is a standard Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) uses the reciprocal root transformation (reverse coefficients), while parts (b) and (c) involve expressing power sums in terms of elementary symmetric functions—techniques that are well-practiced in Further Pure syllabi. The calculations are straightforward with no conceptual surprises, making this slightly easier than average for Further Maths content.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

1 The cubic equation \(7 x ^ { 3 } + 3 x ^ { 2 } + 5 x + 1 = 0\) has roots \(\alpha , \beta , \gamma\).
  1. Find a cubic equation whose roots are \(\alpha ^ { - 1 } , \beta ^ { - 1 } , \gamma ^ { - 1 }\).
  2. Find the value of \(\alpha ^ { - 2 } + \beta ^ { - 2 } + \gamma ^ { - 2 }\).
  3. Find the value of \(\alpha ^ { - 3 } + \beta ^ { - 3 } + \gamma ^ { - 3 }\).

Question 1:
Part (a)
AnswerMarks Guidance
AnswerMarks Guidance
\(y = x^{-1}\)B1
\(7y^{-3} + 3y^{-2} + 5y^{-1} + 1 = 0 \Rightarrow y^3 + 5y^2 + 3y + 7 = 0\)M1 A1
Total3
Part (b)
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha^{-2} + \beta^{-2} + \gamma^{-2} = (-5)^2 - 2(3) = 19\)M1 A1
Part (c)
AnswerMarks Guidance
AnswerMarks Guidance
\(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = -5(19) - 3(-5) - 21 = -101\)M1 A1
Total4
## Question 1:

**Part (a)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^{-1}$ | B1 | |
| $7y^{-3} + 3y^{-2} + 5y^{-1} + 1 = 0 \Rightarrow y^3 + 5y^2 + 3y + 7 = 0$ | M1 A1 | |
| **Total** | **3** | |

**Part (b)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = (-5)^2 - 2(3) = 19$ | M1 A1 | |

**Part (c)**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha^{-3} + \beta^{-3} + \gamma^{-3} = -5(19) - 3(-5) - 21 = -101$ | M1 A1 | |
| **Total** | **4** | |

---
1 The cubic equation $7 x ^ { 3 } + 3 x ^ { 2 } + 5 x + 1 = 0$ has roots $\alpha , \beta , \gamma$.
\begin{enumerate}[label=(\alph*)]
\item Find a cubic equation whose roots are $\alpha ^ { - 1 } , \beta ^ { - 1 } , \gamma ^ { - 1 }$.
\item Find the value of $\alpha ^ { - 2 } + \beta ^ { - 2 } + \gamma ^ { - 2 }$.
\item Find the value of $\alpha ^ { - 3 } + \beta ^ { - 3 } + \gamma ^ { - 3 }$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q1 [7]}}