CAIE FP1 2005 November — Question 5 7 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2005
SessionNovember
Marks7
PaperDownload PDF ↗
TopicRoots of polynomials
TypeSum of powers of roots
DifficultyChallenging +1.8 This is a Further Maths question requiring systematic application of Vieta's formulas combined with inequality manipulation across three related parts. While the techniques (AM-GM, sum of powers identities) are standard for FP1, the multi-part proof structure with progressively complex expressions and the need to carefully handle inequalities with roots constrained to be >1 elevates this above routine exercises. It requires mathematical maturity but follows a clear logical progression.
Spec4.05a Roots and coefficients: symmetric functions

In the equation $$x^3 + ax^2 + bx + c = 0,$$ the coefficients \(a\), \(b\) and \(c\) are real. It is given that all the roots are real and greater than \(1\).
  1. Prove that \(a < -3\). [1]
  2. By considering the sum of the squares of the roots, prove that \(a^2 > 2b + 3\). [2]
  3. By considering the sum of the cubes of the roots, prove that \(a^3 < -9b - 3c - 3\). [4]

In the equation
$$x^3 + ax^2 + bx + c = 0,$$
the coefficients $a$, $b$ and $c$ are real. It is given that all the roots are real and greater than $1$.

\begin{enumerate}[label=(\roman*)]
\item Prove that $a < -3$. [1]
\item By considering the sum of the squares of the roots, prove that $a^2 > 2b + 3$. [2]
\item By considering the sum of the cubes of the roots, prove that $a^3 < -9b - 3c - 3$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2005 Q5 [7]}}