CAIE FP1 2018 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeIntegration using De Moivre identities
DifficultyChallenging +1.3 This is a Further Maths question requiring binomial expansion with complex exponentials and De Moivre's theorem to derive a multiple angle formula, followed by straightforward integration. While it involves multiple techniques and careful algebraic manipulation, it follows a well-established method taught in FP1 with no novel insight required. The 10-mark allocation and multi-step nature place it above average difficulty, but it remains a standard Further Maths exercise.
Spec1.08d Evaluate definite integrals: between limits4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae
  1. By considering the binomial expansion of \(\left(z + \frac{1}{z}\right)^6\), where \(z = \cos \theta + \mathrm{i} \sin \theta\), express \(\cos^6 \theta\) in the form $$\frac{1}{32}(p + q \cos 2\theta + r \cos 4\theta + s \cos 6\theta),$$ where \(p, q, r\) and \(s\) are integers to be determined. [6]
  2. Hence find the exact value of $$\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos^6\left(\frac{1}{2}x\right) \mathrm{d}x.$$ [4]
Question 8:
AnswerMarks Guidance
8(i)z+z−1 =2cosθ B1
( )6 ( ) ( ) ( )
AnswerMarks Guidance
z+z−1 = z6 +z−6 +6 z4 +z−4 +15 z2 +z−2 +20M1A1 Expands and groups.
64cos6θ=2cos6θ+12cos4θ+30cos2θ+20M1A1 Substitutes zn +z−n =2cosnθ.
⇒ cos6θ= 1 ( 10+15cos2θ+6cos4θ+cos6θ )
AnswerMarks Guidance
32A1 (Allow p=10, q=15, r =6, s=1.)
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(ii)1 1
π π
2 x 1 2
∫ cos6 dx= ∫ 10+15cosx+6cos2x+cos3xdx
2 32
1 1
− π − π
AnswerMarks Guidance
2 2M1 Applies part (i)
M1Integrates correctly (3/4 terms correct).
1
π
1  1 2
10x+15sinx+3sin2x+ sin3x
 
32 3  1
− π
AnswerMarks Guidance
2M1 Inserts limits and evaluates.
1  1  1 1  44
=  5π+15+0−  −  −5π−15+0+  =  5π+ 
AnswerMarks
32 3  3 16 3 A1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(i) ---
8(i) | z+z−1 =2cosθ | B1 | Use of z+z−1 =2cosθ.
( )6 ( ) ( ) ( )
z+z−1 = z6 +z−6 +6 z4 +z−4 +15 z2 +z−2 +20 | M1A1 | Expands and groups.
64cos6θ=2cos6θ+12cos4θ+30cos2θ+20 | M1A1 | Substitutes zn +z−n =2cosnθ.
⇒ cos6θ= 1 ( 10+15cos2θ+6cos4θ+cos6θ )
32 | A1 | (Allow p=10, q=15, r =6, s=1.)
6
Question | Answer | Marks | Guidance
--- 8(ii) ---
8(ii) | 1 1
π π
2 x 1 2
∫ cos6 dx= ∫ 10+15cosx+6cos2x+cos3xdx
2 32
1 1
− π − π
2 2 | M1 | Applies part (i)
M1 | Integrates correctly (3/4 terms correct).
1
π
1  1 2
10x+15sinx+3sin2x+ sin3x
 
32 3  1
− π
2 | M1 | Inserts limits and evaluates.
1  1  1 1  44
=  5π+15+0−  −  −5π−15+0+  =  5π+ 
32 3  3 16 3  | A1
4
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\roman*)]
\item By considering the binomial expansion of $\left(z + \frac{1}{z}\right)^6$, where $z = \cos \theta + \mathrm{i} \sin \theta$, express $\cos^6 \theta$ in the form
$$\frac{1}{32}(p + q \cos 2\theta + r \cos 4\theta + s \cos 6\theta),$$
where $p, q, r$ and $s$ are integers to be determined. [6]

\item Hence find the exact value of
$$\int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos^6\left(\frac{1}{2}x\right) \mathrm{d}x.$$ [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q8 [10]}}
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