CAIE FP1 2019 November — Question 9 11 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeExpress roots in trigonometric form
DifficultyChallenging +1.8 This is a challenging Further Maths question requiring de Moivre's theorem to derive a complex trigonometric identity, then connecting it to a sextic equation through substitution. Part (i) demands algebraic manipulation of complex exponentials and careful handling of secant functions (6 marks indicates substantial work). Part (ii) requires recognizing the connection between the derived identity and the polynomial, then extracting roots in a specific form. While the techniques are standard for FP1, the multi-step reasoning, algebraic complexity, and non-routine connection between parts elevate this significantly above average A-level difficulty.
Spec4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae
  1. Use de Moivre's theorem to show that $$\sec 6\theta = \frac{\sec^6 \theta}{32 - 48 \sec^2 \theta + 18 \sec^4 \theta - \sec^6 \theta}.$$ [6]
  2. Hence obtain the roots of the equation $$3x^6 - 36x^4 + 96x^2 - 64 = 0$$ in the form \(\sec q\pi\), where \(q\) is rational. [5]
Question 9:
AnswerMarks
9(i)c=cosθ, s=sinθ.
Write
AnswerMarks Guidance
cos6θ+isin6θ=( c+is )6M1 Uses binomial theorem.
⇒cos6θ=c6 −15c4s2 +15c2s4 −s6A1
( ) ( ) ( )
AnswerMarks Guidance
c6 −15c4s2 +15c2s4 −s6 =c6 −15c4 1−c2 +15c2 1−c2 2 − 1−c2 3M1 Uses c2 =1−s2.
( ) ( ) ( )
AnswerMarks Guidance
=c6 −15c4 1−c2 +15c2 1−2c2 +c4 − 1−3c2 +3c4 −c6A1
=32c6 −48c4 +18c2 −1M1 Divides numerator and denominator by c6.
1 sec6θ
⇒sec6θ= =
AnswerMarks Guidance
32c6 −48c4 +18c2 −1 32−48sec2θ+18sec4θ−sec6θA1 AG
6
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
9(ii)( ) x6
x6 =2 32−48x2 +18x4 −x6 ⇒ =2
AnswerMarks Guidance
32−48x2 +18x4 −x6M1 A1 Relates with equation in part (i).
1
sec6θ=2⇒cos6θ=
AnswerMarks Guidance
2M1 1
cos6θ=
Solves .
2
π
x=sec
AnswerMarks Guidance
18A1 Gives one correct solution.
5 7 11 13 17
x=secqπ, q= , , , ,
AnswerMarks Guidance
18 18 18 18 18A1 Gives five other solutions. Allow different values of q as long
as all six solutions are found.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 9:
--- 9(i) ---
9(i) | c=cosθ, s=sinθ.
Write
cos6θ+isin6θ=( c+is )6 | M1 | Uses binomial theorem.
⇒cos6θ=c6 −15c4s2 +15c2s4 −s6 | A1
( ) ( ) ( )
c6 −15c4s2 +15c2s4 −s6 =c6 −15c4 1−c2 +15c2 1−c2 2 − 1−c2 3 | M1 | Uses c2 =1−s2.
( ) ( ) ( )
=c6 −15c4 1−c2 +15c2 1−2c2 +c4 − 1−3c2 +3c4 −c6 | A1
=32c6 −48c4 +18c2 −1 | M1 | Divides numerator and denominator by c6.
1 sec6θ
⇒sec6θ= =
32c6 −48c4 +18c2 −1 32−48sec2θ+18sec4θ−sec6θ | A1 | AG
6
Question | Answer | Marks | Guidance
--- 9(ii) ---
9(ii) | ( ) x6
x6 =2 32−48x2 +18x4 −x6 ⇒ =2
32−48x2 +18x4 −x6 | M1 A1 | Relates with equation in part (i).
1
sec6θ=2⇒cos6θ=
2 | M1 | 1
cos6θ=
Solves .
2
π
x=sec
18 | A1 | Gives one correct solution.
5 7 11 13 17
x=secqπ, q= , , , ,
18 18 18 18 18 | A1 | Gives five other solutions. Allow different values of q as long
as all six solutions are found.
5
Question | Answer | Marks | Guidance
\begin{enumerate}[label=(\roman*)]
\item Use de Moivre's theorem to show that
$$\sec 6\theta = \frac{\sec^6 \theta}{32 - 48 \sec^2 \theta + 18 \sec^4 \theta - \sec^6 \theta}.$$ [6]

\item Hence obtain the roots of the equation
$$3x^6 - 36x^4 + 96x^2 - 64 = 0$$
in the form $\sec q\pi$, where $q$ is rational. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q9 [11]}}
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