OCR MEI Further Pure Core 2022 June — Question 14 8 marks

Exam BoardOCR MEI
ModuleFurther Pure Core (Further Pure Core)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSum geometric series with complex terms
DifficultyChallenging +1.2 This is a Further Maths question requiring expansion of complex exponentials, use of Euler's formula, and application of geometric series summation. Part (a) is straightforward algebraic manipulation, while part (b) requires recognizing the imaginary part of a geometric series with complex termsβ€”a standard Further Pure technique but more sophisticated than typical A-level. The 'hence' structure provides significant scaffolding, making it moderately above average difficulty.
Spec4.02d Exponential form: re^(i*theta)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.06b Method of differences: telescoping series
14
  1. Find \(\left( 3 - \mathrm { e } ^ { 2 \mathrm { i } \theta } \right) \left( 3 - \mathrm { e } ^ { - 2 \mathrm { i } \theta } \right)\) in terms of \(\cos 2 \theta\).
  2. Hence show that the sum of the infinite series \(\sin \theta + \frac { 1 } { 3 } \sin 3 \theta + \frac { 1 } { 9 } \sin 5 \theta + \frac { 1 } { 27 } \sin 7 \theta + \ldots\) can be expressed as \(\frac { 6 \sin \theta } { 5 - 3 \cos 2 \theta }\).
Question 14:
AnswerMarks Guidance
14(a) (3βˆ’π‘’2π‘–πœƒ)(3βˆ’π‘’βˆ’2π‘–πœƒ) = 9βˆ’3(𝑒2π‘–πœƒ+π‘’βˆ’2π‘–πœƒ)+1 =
= 10βˆ’6cos2πœƒM1
A1
AnswerMarks Guidance
[2]1.1
1.1For expanding correctly
14(b) 1 1 1
let 𝑆 = sinπœƒ+ sin3πœƒ+ sin5πœƒ+ sin7πœƒ+...
3 9 27
1 1 1
and 𝐢 = cosπœƒ+ cos3πœƒ+ cos5πœƒ+ cos7πœƒ+...
3 9 27
1 1 1
𝐢+𝑖𝑆 = eπ‘–πœƒ+ e3π‘–πœƒ+ e5π‘–πœƒ+ e7π‘–πœƒ+...
3 9 27
π‘’π‘–πœƒ
=
1
1βˆ’ e2π‘–πœƒ
3
3eπ‘–πœƒ 3eπ‘–πœƒ(3βˆ’eβˆ’2π‘–πœƒ)
= =
3βˆ’e2π‘–πœƒ 10βˆ’6cos2πœƒ
9(cosπœƒ+𝑖sinπœƒ)βˆ’3(cosπœƒβˆ’π‘–sinπœƒ)
=
10βˆ’6cos2πœƒ
9sinπœƒ+3sinπœƒ 6sinπœƒ
𝑆 = =
AnswerMarks
10βˆ’6cos2πœƒ 5βˆ’3cos2πœƒM1
M1
A1
M1*
M1de
p*
A1
AnswerMarks
[6]2.1
2.1
2.2a
3.1a
2.1
AnswerMarks
2.2aAt least 2 terms of C + iS soi by correct GP formula
sum to infinity of GP formula for their series (which must be
geometric)
oe
multiply numerator and denominator by 3βˆ’eβˆ’2π‘–πœƒ
eπ‘–πœƒ = cosπœƒ+𝑖sinπœƒ used when denominator has been simplified
to a real expression
AG
Question 14:
14 | (a) | (3βˆ’π‘’2π‘–πœƒ)(3βˆ’π‘’βˆ’2π‘–πœƒ) = 9βˆ’3(𝑒2π‘–πœƒ+π‘’βˆ’2π‘–πœƒ)+1 =
= 10βˆ’6cos2πœƒ | M1
A1
[2] | 1.1
1.1 | For expanding correctly
14 | (b) | 1 1 1
let 𝑆 = sinπœƒ+ sin3πœƒ+ sin5πœƒ+ sin7πœƒ+...
3 9 27
1 1 1
and 𝐢 = cosπœƒ+ cos3πœƒ+ cos5πœƒ+ cos7πœƒ+...
3 9 27
1 1 1
𝐢+𝑖𝑆 = eπ‘–πœƒ+ e3π‘–πœƒ+ e5π‘–πœƒ+ e7π‘–πœƒ+...
3 9 27
π‘’π‘–πœƒ
=
1
1βˆ’ e2π‘–πœƒ
3
3eπ‘–πœƒ 3eπ‘–πœƒ(3βˆ’eβˆ’2π‘–πœƒ)
= =
3βˆ’e2π‘–πœƒ 10βˆ’6cos2πœƒ
9(cosπœƒ+𝑖sinπœƒ)βˆ’3(cosπœƒβˆ’π‘–sinπœƒ)
=
10βˆ’6cos2πœƒ
9sinπœƒ+3sinπœƒ 6sinπœƒ
𝑆 = =
10βˆ’6cos2πœƒ 5βˆ’3cos2πœƒ | M1
M1
A1
M1*
M1de
p*
A1
[6] | 2.1
2.1
2.2a
3.1a
2.1
2.2a | At least 2 terms of C + iS soi by correct GP formula
sum to infinity of GP formula for their series (which must be
geometric)
oe
multiply numerator and denominator by 3βˆ’eβˆ’2π‘–πœƒ
eπ‘–πœƒ = cosπœƒ+𝑖sinπœƒ used when denominator has been simplified
to a real expression
AG
14
\begin{enumerate}[label=(\alph*)]
\item Find $\left( 3 - \mathrm { e } ^ { 2 \mathrm { i } \theta } \right) \left( 3 - \mathrm { e } ^ { - 2 \mathrm { i } \theta } \right)$ in terms of $\cos 2 \theta$.
\item Hence show that the sum of the infinite series $\sin \theta + \frac { 1 } { 3 } \sin 3 \theta + \frac { 1 } { 9 } \sin 5 \theta + \frac { 1 } { 27 } \sin 7 \theta + \ldots$ can be expressed as $\frac { 6 \sin \theta } { 5 - 3 \cos 2 \theta }$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q14 [8]}}
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