OCR MEI Further Pure Core 2022 June — Question 10 10 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks10
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Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with special relationships
DifficultyChallenging +1.2 This is a Further Maths polynomial roots question requiring systematic use of Vieta's formulas and the given root relationships. While it involves multiple steps (using product of roots to find α, then sum conditions to find β, then back-substituting for coefficients), the approach is methodical rather than requiring novel insight. The special root relationships (α and 2/α, β and 3β) guide the solution path clearly, making this moderately harder than average but still a standard Further Maths exercise.
Spec4.05a Roots and coefficients: symmetric functions
10 The equation \(4 x ^ { 4 } + 16 x ^ { 3 } + a x ^ { 2 } + b x + 6 = 0\),
where \(a\) and \(b\) are real, has roots \(\alpha , \frac { 2 } { \alpha } , \beta\) and \(3 \beta\).
  1. Given that \(\beta < 0\), determine all 4 roots.
  2. Determine the values of \(a\) and \(b\).
Question 10:
AnswerMarks Guidance
10(a) 2 6
𝛼× ×𝛽×3𝛽 =
𝛼 4
1
⇒ 𝛽 = −
2
2 16
𝛼+ +𝛽+3𝛽 = −
𝛼 4
⇒ 𝛼2+2𝛼+2 = 0
−2±√4−8
𝛼 = = −1+𝑖, −1−𝑖
2
1 3
so roots are −1+𝑖, −1−𝑖, − , −
AnswerMarks
2 2M1
A1
M1
A1
M1
A1
AnswerMarks
[6]3.1a
1.1
3.1a
1.1
1.1
AnswerMarks
3.2a`
Solving their quadratic to find 𝛼
1
(−1+𝑖)(−1−𝑖)+(−1+𝑖)(− )+(−1
2
3
+𝑖)(− )
2
1 3 1 3
+(−1−𝑖)(− )+(−1−𝑖)(− )+(− )(− )
2 2 2 2
𝑎
=
4
⇒ 𝑎 = 27
1 3
(−1+𝑖)(−1−𝑖)(− )+(−1+𝑖)(−1−𝑖)(− )
2 2
1 3 1 3
+(−1+𝑖)(− )(− )+(−1−𝑖)(− )(− )
2 2 2 2
𝑏
= −
4
AnswerMarks
⇒ 𝑏 = 22M1
A1
M1
AnswerMarks
A11.1
1.1
1.1
AnswerMarks
1.1Allow in terms of 𝛼 and 𝛽, no missing terms
Allow in terms of 𝛼 and 𝛽, no missing terms
Alternative solution
AnswerMarks
(𝑥+1+𝑖)(𝑥+1−𝑖)(2𝑥+1)(2𝑥+3) = 0M1
⇒ (𝑥2+2𝑥+2)(4𝑥2+8𝑥+3) = 0M1
⇒ 4𝑥4+16𝑥3+27𝑥2+22𝑥+6 = 0A1
⇒ 𝑎 = 27, 𝑏 = 22A1
Alternative solutionAlternative solution
3 81 9𝑎 3𝑏
𝑓(− ) = −54+ − +6 = 0
AnswerMarks
2 4 4 2M1
M1
A1
A1
1 1 𝑎 𝑏
𝑓(− ) = −2+ − +6 = 0
2 4 4 2
𝑎 = 27
𝑏 = 22
[4]
M1
M1
A1
A1
Question 10:
10 | (a) | 2 6
𝛼× ×𝛽×3𝛽 =
𝛼 4
1
⇒ 𝛽 = −
2
2 16
𝛼+ +𝛽+3𝛽 = −
𝛼 4
⇒ 𝛼2+2𝛼+2 = 0
−2±√4−8
𝛼 = = −1+𝑖, −1−𝑖
2
1 3
so roots are −1+𝑖, −1−𝑖, − , −
2 2 | M1
A1
M1
A1
M1
A1
[6] | 3.1a
1.1
3.1a
1.1
1.1
3.2a | `
Solving their quadratic to find 𝛼
1
(−1+𝑖)(−1−𝑖)+(−1+𝑖)(− )+(−1
2
3
+𝑖)(− )
2
1 3 1 3
+(−1−𝑖)(− )+(−1−𝑖)(− )+(− )(− )
2 2 2 2
𝑎
=
4
⇒ 𝑎 = 27
1 3
(−1+𝑖)(−1−𝑖)(− )+(−1+𝑖)(−1−𝑖)(− )
2 2
1 3 1 3
+(−1+𝑖)(− )(− )+(−1−𝑖)(− )(− )
2 2 2 2
𝑏
= −
4
⇒ 𝑏 = 22 | M1
A1
M1
A1 | 1.1
1.1
1.1
1.1 | Allow in terms of 𝛼 and 𝛽, no missing terms
Allow in terms of 𝛼 and 𝛽, no missing terms
Alternative solution
(𝑥+1+𝑖)(𝑥+1−𝑖)(2𝑥+1)(2𝑥+3) = 0 | M1
⇒ (𝑥2+2𝑥+2)(4𝑥2+8𝑥+3) = 0 | M1
⇒ 4𝑥4+16𝑥3+27𝑥2+22𝑥+6 = 0 | A1
⇒ 𝑎 = 27, 𝑏 = 22 | A1
Alternative solution | Alternative solution
3 81 9𝑎 3𝑏
𝑓(− ) = −54+ − +6 = 0
2 4 4 2 | M1
M1
A1
A1
1 1 𝑎 𝑏
𝑓(− ) = −2+ − +6 = 0
2 4 4 2
𝑎 = 27
𝑏 = 22
[4]
M1
M1
A1
A1
10 The equation\\
$4 x ^ { 4 } + 16 x ^ { 3 } + a x ^ { 2 } + b x + 6 = 0$,\\
where $a$ and $b$ are real, has roots $\alpha , \frac { 2 } { \alpha } , \beta$ and $3 \beta$.
\begin{enumerate}[label=(\alph*)]
\item Given that $\beta < 0$, determine all 4 roots.
\item Determine the values of $a$ and $b$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q10 [10]}}
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