OCR Further Pure Core 1 2023 June — Question 9 14 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2023
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyChallenging +1.8 This is a substantial Further Maths question requiring multiple advanced techniques: deriving a trigonometric identity using de Moivre's theorem, differentiating a composite function involving inverse trig functions, and evaluating an improper integral. However, each part follows standard FM procedures without requiring novel insights—part (a) is routine binomial expansion with complex numbers, part (b) uses the chain rule systematically with the result from (a), and part (c) recognizes the antiderivative from part (b). The multi-step nature and FM content place it above average difficulty but below the most challenging proof-based questions.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions4.02q De Moivre's theorem: multiple angle formulae4.08c Improper integrals: infinite limits or discontinuous integrands

9 In this question you must show detailed reasoning.
  1. Use de Moivre's theorem to determine constants \(A\), \(B\) and \(C\) such that $$\sin ^ { 4 } \theta \equiv A \cos 4 \theta + B \cos 2 \theta + C .$$ The function f is defined by \(\mathrm { f } ( x ) = \sin \left( 4 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) - 8 \sin \left( 2 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) + 12 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) , \quad x \in \mathbb { R } , 0 \leqslant x < 1\).
  2. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 32 } { 5 \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }\). \includegraphics[max width=\textwidth, alt={}, center]{478c66d2-16a0-41ef-9444-25cfcd47d11d-7_894_842_1000_260} The diagram shows the curve with equation \(\mathrm { y } = \frac { 1 } { \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }\) for \(0 \leqslant x < 1\) and the asymptote \(x = 1\). The region \(R\) is the unbounded region between the curve, the \(x\)-axis, the line \(x = 0\) and the line \(x = 1\). You are given that the area of \(R\) is finite.
  3. Determine the exact area of \(R\).

Question 9:
AnswerMarks Guidance
9(a) DR
e i e -i 2 i s i n    − =
( e i e -i ) 4 1 6 s i n 4     − =
( e 4i 4 e 2i 6 4 e 2i e 4i )      − + − − + −
( 4i e 4 e 2i 6 4 e 2i e 4i ) ( e 4i e 4i ) 4 ( e 2i e 2i ) 6          − + − − + − = + − − + − +
2 c o s 4 8 c o s 2 6 1 6 s i n 4     − + =
1 1 3
s 4 i n c o s 4 c o s 2     = − +
8 2 8
1 1 3
i .e . A , B , C = = − =
AnswerMarks
8 2 8B1
M1
M1
M1
AnswerMarks
A1Or eiθ+ e–iθ= 2cosθ May use z without definition
4
oe, eg. (2isin𝜃)4 = 16sin4𝜃 = (𝑒i𝜃 −𝑒−i𝜃) . Award this
mark for sinθto the power of four, and for (2i)4 = 16. Note
that 16 may appear later.
Expanding (eiθ – e–iθ)4 with correct coefficients.
Grouping terms and using eiθ+ e–iθ = 2 cosθ.
cao, from fully correct reasoning.
Allow A, B, C seen in the expression only.
[5]
AnswerMarks Guidance
9(b) DR
1 d u 1 4 −
L e t u = x  = 5 x 5
d x 5
d v 1 1
L e t v = s i − 1  n u = =
d u 1 − u 2 25
− 1 x
d f
 f = s i n − 4 v 8 s i n 2 v + 1 2 v  = 4 c o s 4 v − 1 6 c o s 2 v + 1 2
d v
d f d f d v d u 1 1 − 45
= . = . ( 4 c o s 4 v − 1 6 c o + s 2 v 1 2 )  x
d x d v d u d x 25 5
1 − x
d f  1 1 3  45
= 3 2 − c o s 4 v c o s 2 v + =  ( 3 2 s i n v ) 4 = 3 2 u 4 = 3 2 x
d v 8 2 8
d f d f d v d u 4 1 1 − 45 3 2
T h e n = . . = 3 2 x  5  x =
d x d v d u d x 2 5 2
AnswerMarks
1 − x 5 5 − 1 x 5B1
M1
A1
M1
M1
AnswerMarks
A1d ( s i n − 1 u ) 1
Sight of =
d u 1 − u 2
Uses chain rule
Correct derivative, f’
Uses result from (a)
  1   45
Uses s i n 4 s i n − 1 x 5 = x
AG Clearly shown
[6]
AnswerMarks Guidance
9(c) k 1 5 5
R=lim dx= limf(x)k = lim(f(k)−f(0))
k→1 2 32 k→1 0 32 k→1
0 1−x5
f(0)=0
  1   1  1
f(k)=sin4sin−1k5−8sin2sin−1k5+12sin−1k5
   
       
 1 
As k→1sin−1k5→sin−1(1)=
  2
lim(f(k))=sin(2)−8sin+6=6
k→1
5 15
R= 6=
AnswerMarks
32 16M1
M1
AnswerMarks
A1For use of part (b) and an upper limit of k < 1
Integral must be found in terms of k (which could be 1)
For correct use of limits
[3]
PMT
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Question 9:
9 | (a) | DR
e i e -i 2 i s i n    − =
( e i e -i ) 4 1 6 s i n 4     − =
( e 4i 4 e 2i 6 4 e 2i e 4i )      − + − − + −
( 4i e 4 e 2i 6 4 e 2i e 4i ) ( e 4i e 4i ) 4 ( e 2i e 2i ) 6          − + − − + − = + − − + − +
2 c o s 4 8 c o s 2 6 1 6 s i n 4     − + =
1 1 3
s 4 i n c o s 4 c o s 2     = − +
8 2 8
1 1 3
i .e . A , B , C = = − =
8 2 8 | B1
M1
M1
M1
A1 | Or eiθ+ e–iθ= 2cosθ May use z without definition
4
oe, eg. (2isin𝜃)4 = 16sin4𝜃 = (𝑒i𝜃 −𝑒−i𝜃) . Award this
mark for sinθto the power of four, and for (2i)4 = 16. Note
that 16 may appear later.
Expanding (eiθ – e–iθ)4 with correct coefficients.
Grouping terms and using eiθ+ e–iθ = 2 cosθ.
cao, from fully correct reasoning.
Allow A, B, C seen in the expression only.
[5]
9 | (b) | DR
1 d u 1 4 −
L e t u = x  = 5 x 5
d x 5
d v 1 1
L e t v = s i − 1  n u = =
d u 1 − u 2 25
− 1 x
d f
 f = s i n − 4 v 8 s i n 2 v + 1 2 v  = 4 c o s 4 v − 1 6 c o s 2 v + 1 2
d v
d f d f d v d u 1 1 − 45
= . = . ( 4 c o s 4 v − 1 6 c o + s 2 v 1 2 )  x
d x d v d u d x 25 5
1 − x
d f  1 1 3  45
= 3 2 − c o s 4 v c o s 2 v + =  ( 3 2 s i n v ) 4 = 3 2 u 4 = 3 2 x
d v 8 2 8
d f d f d v d u 4 1 1 − 45 3 2
T h e n = . . = 3 2 x  5  x =
d x d v d u d x 2 5 2
1 − x 5 5 − 1 x 5 | B1
M1
A1
M1
M1
A1 | d ( s i n − 1 u ) 1
Sight of =
d u 1 − u 2
Uses chain rule
Correct derivative, f’
Uses result from (a)
  1   45
Uses s i n 4 s i n − 1 x 5 = x
AG Clearly shown
[6]
9 | (c) | k 1 5 5
R=lim dx= limf(x)k = lim(f(k)−f(0))
k→1 2 32 k→1 0 32 k→1
0 1−x5
f(0)=0
  1   1  1
f(k)=sin4sin−1k5−8sin2sin−1k5+12sin−1k5
   
       
 1 
As k→1sin−1k5→sin−1(1)=
  2
lim(f(k))=sin(2)−8sin+6=6
k→1
5 15
R= 6=
32 16 | M1
M1
A1 | For use of part (b) and an upper limit of k < 1
Integral must be found in terms of k (which could be 1)
For correct use of limits
[3]
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
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/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.
9 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Use de Moivre's theorem to determine constants $A$, $B$ and $C$ such that

$$\sin ^ { 4 } \theta \equiv A \cos 4 \theta + B \cos 2 \theta + C .$$

The function f is defined by\\
$\mathrm { f } ( x ) = \sin \left( 4 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) - 8 \sin \left( 2 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) \right) + 12 \sin ^ { - 1 } \left( x ^ { \frac { 1 } { 5 } } \right) , \quad x \in \mathbb { R } , 0 \leqslant x < 1$.
\item Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 32 } { 5 \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{478c66d2-16a0-41ef-9444-25cfcd47d11d-7_894_842_1000_260}

The diagram shows the curve with equation $\mathrm { y } = \frac { 1 } { \sqrt { 1 - x ^ { \frac { 2 } { 5 } } } }$ for $0 \leqslant x < 1$ and the asymptote $x = 1$. The region $R$ is the unbounded region between the curve, the $x$-axis, the line $x = 0$ and the line $x = 1$.

You are given that the area of $R$ is finite.
\item Determine the exact area of $R$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q9 [14]}}