AQA Further AS Paper 1 2018 June — Question 18 4 marks

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with given sum conditions
DifficultyChallenging +1.8 This is a Further Maths inequality proof requiring non-standard manipulation of symmetric functions. Students must recognize that the sum of squared differences is non-negative, expand it using Vieta's formulas (α+β+γ=-m, αβ+βγ+γα=n, αβγ=-2), and algebraically manipulate to reach m²≥3n. While the technique is elegant once seen, it requires insight beyond routine application and careful algebraic work across multiple steps.
Spec4.01a Mathematical induction: construct proofs4.05a Roots and coefficients: symmetric functions

\(\alpha\), \(\beta\) and \(\gamma\) are the real roots of the cubic equation $$x^3 + mx^2 + nx + 2 = 0$$ By considering \((\alpha - \beta)^2 + (\gamma - \alpha)^2 + (\beta - \gamma)^2\), prove that $$m^2 \geq 3n$$ [4 marks]

Question 18:
AnswerMarks
18Writes the expression in terms of and
∑𝛼𝛼 ∑𝛼𝛼𝛽𝛽
Award for correct expansion followed by use of
2 2
AnswerMarks Guidance
∑𝛼𝛼 =(∑𝛼𝛼) −2∑𝛼𝛼𝛽𝛽3.1a M1
(𝛼𝛼−𝛽𝛽) +(𝛾𝛾−𝛼𝛼) +(𝛽𝛽−𝛾𝛾)
2 2 2 2 2 2
= 𝛼𝛼 −2𝛼𝛼𝛽𝛽+𝛽𝛽 +𝛾𝛾 −2𝛾𝛾𝛼𝛼+𝛼𝛼 +𝛽𝛽 −2𝛽𝛽𝛾𝛾+𝛾𝛾
2 2 2
= 2𝛼𝛼 +2𝛽𝛽 +2𝛾𝛾 −2𝛼𝛼𝛽𝛽−2𝛾𝛾𝛼𝛼−2𝛽𝛽𝛾𝛾
2
= 2∑𝛼𝛼 −2∑𝛼𝛼𝛽𝛽
2
= 2((∑𝛼𝛼) −2∑𝛼𝛼𝛽𝛽)−2∑𝛼𝛼𝛽𝛽
2
= 2(∑𝛼𝛼) −6∑𝛼𝛼𝛽𝛽
2 2
= 2(−𝑚𝑚) −6×𝑛𝑛 =2𝑚𝑚 −6𝑛𝑛
But as , and are real then each of and
must be non-negative. 2 2
𝛼𝛼 𝛽𝛽 𝛾𝛾 (𝛼𝛼−𝛽𝛽) , (𝛾𝛾−𝛼𝛼)
2
(𝛽𝛽−𝛾𝛾)
2 2 2
∴ (𝛼𝛼−𝛽𝛽) +(𝛾𝛾−𝛼𝛼) +(𝛽𝛽−𝛾𝛾) ≥ 0
2
∴ 2𝑚𝑚 −6𝑛𝑛 ≥ 0
2
2𝑚𝑚 ≥ 6𝑛𝑛
2
AG
𝑚𝑚 ≥ 3𝑛𝑛
Substitutes for
and for
AnswerMarks Guidance
±𝑚𝑚 ∑𝛼𝛼1.1a M1
±𝑛𝑛 ∑𝛼𝛼𝛽𝛽
Gives a reason for expression
Condone lack of reference to roots being real.
AnswerMarks Guidance
≥ 02.4 E1
Completes fully correct proof to reach the required result.
This mark is only available if all previous marks have been awarded.
Lose this mark for sight of
AnswerMarks Guidance
∑𝛼𝛼 = 𝑚𝑚2.1 R1
Total4
QMarking instructions AO
Question 18:
18 | Writes the expression in terms of and
∑𝛼𝛼 ∑𝛼𝛼𝛽𝛽
Award for correct expansion followed by use of
2 2
∑𝛼𝛼 =(∑𝛼𝛼) −2∑𝛼𝛼𝛽𝛽 | 3.1a | M1 | 2 2 2
(𝛼𝛼−𝛽𝛽) +(𝛾𝛾−𝛼𝛼) +(𝛽𝛽−𝛾𝛾)
2 2 2 2 2 2
= 𝛼𝛼 −2𝛼𝛼𝛽𝛽+𝛽𝛽 +𝛾𝛾 −2𝛾𝛾𝛼𝛼+𝛼𝛼 +𝛽𝛽 −2𝛽𝛽𝛾𝛾+𝛾𝛾
2 2 2
= 2𝛼𝛼 +2𝛽𝛽 +2𝛾𝛾 −2𝛼𝛼𝛽𝛽−2𝛾𝛾𝛼𝛼−2𝛽𝛽𝛾𝛾
2
= 2∑𝛼𝛼 −2∑𝛼𝛼𝛽𝛽
2
= 2((∑𝛼𝛼) −2∑𝛼𝛼𝛽𝛽)−2∑𝛼𝛼𝛽𝛽
2
= 2(∑𝛼𝛼) −6∑𝛼𝛼𝛽𝛽
2 2
= 2(−𝑚𝑚) −6×𝑛𝑛 =2𝑚𝑚 −6𝑛𝑛
But as , and are real then each of and
must be non-negative. 2 2
𝛼𝛼 𝛽𝛽 𝛾𝛾 (𝛼𝛼−𝛽𝛽) , (𝛾𝛾−𝛼𝛼)
2
(𝛽𝛽−𝛾𝛾)
2 2 2
∴ (𝛼𝛼−𝛽𝛽) +(𝛾𝛾−𝛼𝛼) +(𝛽𝛽−𝛾𝛾) ≥ 0
2
∴ 2𝑚𝑚 −6𝑛𝑛 ≥ 0
2
2𝑚𝑚 ≥ 6𝑛𝑛
2
AG
𝑚𝑚 ≥ 3𝑛𝑛
Substitutes for
and for
±𝑚𝑚 ∑𝛼𝛼 | 1.1a | M1
±𝑛𝑛 ∑𝛼𝛼𝛽𝛽
Gives a reason for expression
Condone lack of reference to roots being real.
≥ 0 | 2.4 | E1
Completes fully correct proof to reach the required result.
This mark is only available if all previous marks have been awarded.
Lose this mark for sight of
∑𝛼𝛼 = 𝑚𝑚 | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
$\alpha$, $\beta$ and $\gamma$ are the real roots of the cubic equation
$$x^3 + mx^2 + nx + 2 = 0$$

By considering $(\alpha - \beta)^2 + (\gamma - \alpha)^2 + (\beta - \gamma)^2$, prove that
$$m^2 \geq 3n$$
[4 marks]

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q18 [4]}}