OCR Further Pure Core 2 2022 June — Question 8 7 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeDe Moivre to derive trigonometric identities
DifficultyChallenging +1.2 This is a standard Further Maths question using De Moivre's theorem to derive trigonometric identities. Part (a) requires expanding binomially and collecting real parts—routine for FM students. Part (b) involves substituting a specific angle and algebraic manipulation, but follows a clear path once part (a) is complete. More mechanical than insightful, though the nested radical form requires care.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae
  1. Show that \(\operatorname { Re } \left( \mathrm { e } ^ { \mathrm { Ai } \theta } \left( \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 4 } \right) = a \cos 4 \theta \cos ^ { 4 } \theta\), where \(a\) is an integer to be determined.
  2. Hence show that \(\cos \frac { 1 } { 12 } \pi = \frac { 1 } { 2 } \sqrt [ 4 ] { \mathrm { b } + \mathrm { c } \sqrt { 3 } }\), where \(b\) and \(c\) are integers to be determined.
Question 8:
AnswerMarks
8DR
5r+2 A B C
= + +
r(r+1)(r+2) r r+1 r+2
1 3 4
= + −
r r + 1 r + 2
98 5r+2 98 1 3 4 
 =  + − 
r(r+1)(r+2) r r+1 r+2
r=k r=k
1 3 4
= + −
k k+1 k+2
1 3 4
+ + −
k+1 k+2 k+3
1 3 4
+ + −
k+2 k+3 k+4
+...
1 3 4
+ + −
96 97 98
1 3 4
+ + −
97 98 99
1 3 4
+ + −
98 99 100
1 4 1 2 4 1 4 1 4
= + − or + − − oe
k k + 1 2 4 7 5 k k + 1 9 9 1 0 0
1 4 1 2 4 2 0 5 3 9 1 4 9
+ − =  + =
k k + 1+ 2+ 4 7 5 3 4 6 5 0 k k + 1 1 4
1 4 ( k 1 ) 5 6 k = 9 k ( k + 1 )
AnswerMarks
 9 k 2 − 6 1 k − 1 4 = 0M1
A1
M1
A1
AnswerMarks
M13.1a
1.1
3.1a
2.2b
AnswerMarks
1.1Correct form for partial fractions
or A = 1, B = 3, C = –4
Writing out sufficient terms so
that cancellation pattern becomes
evident. Could see N rather than
98.
5 k + 1 1 2 4
Could see −
k ( k + 1 ) 2 4 7 5
Equating their expression for
LHS to 20539/34650 and
AnswerMarks
rearranging to 3 term quadraticIf use r = a to N then M1 awarded
when N = 98 and N = k – 1 both
considered.
2
( 9 k + 2 ) ( k − 7 ) = 0  k = − o r 7
9
2
But k is an integer so k  − so k = 7
AnswerMarks
9A1
A1
AnswerMarks
[7]1.1
3.2a− − 6 1  ( − 6 2 1 ) − 4  9  ( − 1 4 )
 2 9
6 1  4 2 2 5 6 1  6 5
= =
1 8 1 8
Explicitly rejecting non-integer
value, with reason, and deducing
correct value...
...or stating that the original
function is undefined if r is
AnswerMarks
allowed to take the value 0. 2 2 
 6 1   6 1 
or 9 k − − − 1 4 = 0
1 8 1 8
2
 6 1  4 2 2 5
9 k − − = 0
1 8 3 6
NB k ≥ 0 (or k > 0) alone is not a
valid reason to reject –2/9
Question 8:
8 | DR
5r+2 A B C
= + +
r(r+1)(r+2) r r+1 r+2
1 3 4
= + −
r r + 1 r + 2
98 5r+2 98 1 3 4 
 =  + − 
r(r+1)(r+2) r r+1 r+2
r=k r=k
1 3 4
= + −
k k+1 k+2
1 3 4
+ + −
k+1 k+2 k+3
1 3 4
+ + −
k+2 k+3 k+4
+...
1 3 4
+ + −
96 97 98
1 3 4
+ + −
97 98 99
1 3 4
+ + −
98 99 100
1 4 1 2 4 1 4 1 4
= + − or + − − oe
k k + 1 2 4 7 5 k k + 1 9 9 1 0 0
1 4 1 2 4 2 0 5 3 9 1 4 9
+ − =  + =
k k + 1+ 2+ 4 7 5 3 4 6 5 0 k k + 1 1 4
1 4 ( k 1 ) 5 6 k = 9 k ( k + 1 )
 9 k 2 − 6 1 k − 1 4 = 0 | M1
A1
M1
A1
M1 | 3.1a
1.1
3.1a
2.2b
1.1 | Correct form for partial fractions
or A = 1, B = 3, C = –4
Writing out sufficient terms so
that cancellation pattern becomes
evident. Could see N rather than
98.
5 k + 1 1 2 4
Could see −
k ( k + 1 ) 2 4 7 5
Equating their expression for
LHS to 20539/34650 and
rearranging to 3 term quadratic | If use r = a to N then M1 awarded
when N = 98 and N = k – 1 both
considered.
2
( 9 k + 2 ) ( k − 7 ) = 0  k = − o r 7
9
2
But k is an integer so k  − so k = 7
9 | A1
A1
[7] | 1.1
3.2a | − − 6 1  ( − 6 2 1 ) − 4  9  ( − 1 4 )
 2 9
6 1  4 2 2 5 6 1  6 5
= =
1 8 1 8
Explicitly rejecting non-integer
value, with reason, and deducing
correct value...
...or stating that the original
function is undefined if r is
allowed to take the value 0. |  2 2 
 6 1   6 1 
or 9 k − − − 1 4 = 0
1 8 1 8
2
 6 1  4 2 2 5
9 k − − = 0
1 8 3 6
NB k ≥ 0 (or k > 0) alone is not a
valid reason to reject –2/9
\begin{enumerate}[label=(\alph*)]
\item Show that $\operatorname { Re } \left( \mathrm { e } ^ { \mathrm { Ai } \theta } \left( \mathrm { e } ^ { \mathrm { i } \theta } + \mathrm { e } ^ { - \mathrm { i } \theta } \right) ^ { 4 } \right) = a \cos 4 \theta \cos ^ { 4 } \theta$, where $a$ is an integer to be determined.
\item Hence show that $\cos \frac { 1 } { 12 } \pi = \frac { 1 } { 2 } \sqrt [ 4 ] { \mathrm { b } + \mathrm { c } \sqrt { 3 } }$, where $b$ and $c$ are integers to be determined.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2022 Q8 [7]}}
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