OCR MEI Further Pure Core AS 2020 November — Question 3 7 marks

Exam BoardOCR MEI
ModuleFurther Pure Core AS (Further Pure Core AS)
Year2020
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Argand & Loci
TypeRoots of polynomial equations
DifficultyStandard +0.3 This is a straightforward Further Maths question requiring standard techniques: solving a quadratic to get complex roots, converting to modulus-argument form using basic formulas, then using De Moivre's theorem to show they satisfy a cubic equation. While it involves multiple steps and Further Maths content, each step is routine application of well-practiced methods with no novel insight required. Slightly above average difficulty due to the multi-part nature and Further Maths context, but still a standard textbook-style exercise.
Spec4.02b Express complex numbers: cartesian and modulus-argument forms4.02g Conjugate pairs: real coefficient polynomials4.02i Quadratic equations: with complex roots4.02q De Moivre's theorem: multiple angle formulae
3 In this question you must show detailed reasoning.
The roots of the equation \(x ^ { 2 } - 2 x + 4 = 0\) are \(\alpha\) and \(\beta\).
  1. Find \(\alpha\) and \(\beta\) in modulus-argument form.
  2. Hence or otherwise show that \(\alpha\) and \(\beta\) are both roots of \(x ^ { 3 } + \lambda = 0\), where \(\lambda\) is a real constant to be determined.
Question 3:
AnswerMarks Guidance
3(a) DR
2± −12
x= =1± 3i
2
π π
= 2(cos +isin )
3 3
π π
2(cos(− )+isin(− ))
AnswerMarks
3 3M1
A1
B1
B1
AnswerMarks
[4]1.1
1.1
1.1
AnswerMarks
1.1or completing the square
5π 5π
or 2(cos +isin )
AnswerMarks Guidance
3 3no working M0, allow 1 slip
3(b) DR
α3 = 8(cos π + i sin π) = −8
β3 = 8(cos (−π) + i sin (−π)) = −8
AnswerMarks
so λ = 8B1
B1
AnswerMarks
B11.1
1.1
1.1
AnswerMarks
Alternative solutionM1
A1
AnswerMarks
A1no working M1A0A0
α3 = (1+√3 i)(1+√3 i)(1+√3 i)attempt to substitute into
= (−2+2√3 i)(1+√3 i) = −8x3 + λ = 0
β3 = (1−√3 i)(1−√3 i)(1−√3 i)
AnswerMarks
= (−2−2√3 i)(1−√3 i) = −8oe (eg β a root as complex
so λ = 8conjugate)
Alternative solutioncondone αβγ = λ
caocondone no check of coefft
of x
AnswerMarks Guidance
α + β + γ = 0 ⇒ γ = −2B1
αβγ = (1+√3 i)(1−√3 i)(−2) = −8M1 condone αβγ = λ
so λ = 8A1 cao
[3]
Question 3:
3 | (a) | DR
2± −12
x= =1± 3i
2
π π
= 2(cos +isin )
3 3
π π
2(cos(− )+isin(− ))
3 3 | M1
A1
B1
B1
[4] | 1.1
1.1
1.1
1.1 | or completing the square
5π 5π
or 2(cos +isin )
3 3 | no working M0, allow 1 slip
3 | (b) | DR
α3 = 8(cos π + i sin π) = −8
β3 = 8(cos (−π) + i sin (−π)) = −8
so λ = 8 | B1
B1
B1 | 1.1
1.1
1.1
Alternative solution | M1
A1
A1 | no working M1A0A0
α3 = (1+√3 i)(1+√3 i)(1+√3 i) | attempt to substitute into
= (−2+2√3 i)(1+√3 i) = −8 | x3 + λ = 0
β3 = (1−√3 i)(1−√3 i)(1−√3 i)
= (−2−2√3 i)(1−√3 i) = −8 | oe (eg β a root as complex
so λ = 8 | conjugate)
Alternative solution | condone αβγ = λ
cao | condone no check of coefft
of x
α + β + γ = 0 ⇒ γ = −2 | B1
αβγ = (1+√3 i)(1−√3 i)(−2) = −8 | M1 | condone αβγ = λ
so λ = 8 | A1 | cao
[3]
3 In this question you must show detailed reasoning.\\
The roots of the equation $x ^ { 2 } - 2 x + 4 = 0$ are $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Find $\alpha$ and $\beta$ in modulus-argument form.
\item Hence or otherwise show that $\alpha$ and $\beta$ are both roots of $x ^ { 3 } + \lambda = 0$, where $\lambda$ is a real constant to be determined.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2020 Q3 [7]}}
This paper (10 questions)
View full paper
Q1 3 Q2 4 Q3 7 Q4 4 Q5 6 Q6 8 Q7 7 Q8 7 Q9 7 Q10 7