CAIE Further Paper 2 2023 November — Question 8 15 marks

Exam BoardCAIE
ModuleFurther Paper 2 (Further Paper 2)
Year2023
SessionNovember
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex numbers 2
TypeSum geometric series with complex terms
DifficultyChallenging +1.8 This is a sophisticated Further Maths question requiring multiple advanced techniques: geometric series with complex numbers, De Moivre's theorem, separating real/imaginary parts, and connecting to Riemann sums for integration bounds. Part (b) requires extended algebraic manipulation (7 marks), and parts (c)-(d) demand insight into upper/lower Riemann sums. While the individual components are standard Further Maths topics, the multi-stage synthesis and the non-trivial algebraic work place this well above average difficulty.
Spec1.08g Integration as limit of sum: Riemann sums4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02o Loci in Argand diagram: circles, half-lines4.06b Method of differences: telescoping series
  1. State the sum of the series \(1 + z + z^2 + \ldots + z^{n-1}\), for \(z \neq 1\). [1]
  2. By letting \(z = \cos\theta + i\sin\theta\), where \(\cos\theta \neq 1\), show that $$1 + \cos\theta + \cos 2\theta + \ldots + \cos(n-1)\theta = \frac{1}{2}\left(1 - \cos n\theta + \frac{\sin n\theta \sin\theta}{1 - \cos\theta}\right).$$ [7]
\includegraphics{figure_8} The diagram shows the curve with equation \(y = \cos x\) for \(0 \leq x \leq 1\), together with a set of \(n\) rectangles of width \(\frac{1}{n}\).
  1. By considering the sum of the areas of these rectangles, show that $$\int_0^1 \cos x\,dx < \frac{1}{2n}\left(1 - \cos 1 + \frac{\sin 1\sin\frac{1}{n}}{1 - \cos\frac{1}{n}}\right).$$ [4]
  2. Use a similar method to find, in terms of \(n\), a lower bound for \(\int_0^1 \cos x\,dx\). [3]
Question 8:
AnswerMarks
8(a)zn −1
z−1B1
1
AnswerMarks
8(b)zn −1 cosn−1+isinn
=
AnswerMarks
z−1 cos−1+isinB1
(cosn−1+isinn)(cos−1−isin)
AnswerMarks Guidance
(cos−1+isin)(cos−1−isin)M1 Multiplies numerator and denominator by complex
conjugate.
zn −1 cosncos+sinnsin−cosn−cos+1
Re =
AnswerMarks Guidance
 z−1  (cos−1)2 +sin2M1 Takes real part.
cos(n−1)=cosncos+sinnsin
cosncos+sinnsin−cosn 1
= +
AnswerMarks
2(1−cos) 2A1
cosn(cos−1)+sinnsin
1
= +
AnswerMarks Guidance
2(1−cos) 2M1 Factorises.
QuestionAnswer Marks
8(b)1 sinnsin
= 1−cosn+ 
AnswerMarks Guidance
2 1−cos M1 A1 Divides through by denominator. AG.
Alternative method for question 8(b)
zn −1 ein−1
=
AnswerMarks
z−1 ei−1B1
e i(n−1 2 ) −e −i1 2  cos (n−1)+isin (n−1)−cos1+isin (1)
= 2 2 2 2
i1 −e −i1  2isin1
e 2 2
AnswerMarks
2M1
zn −1 sin(n−1)+sin1
Re = 2 2
z−1 2sin1
 
AnswerMarks Guidance
2M1 Takes real part
(n−1)
sin 1
= 2 +
2sin1 2
AnswerMarks
2A1
sinncos1−cosnsin1 1
= 2 2 +
2sin1 2
AnswerMarks Guidance
2M1 Uses compound angle identity
sinncos1 1 1
= 2 − cosn+
2sin1 2 2
AnswerMarks Guidance
2M1 Divides through by denominator.
sinnsin 1 1 sinnsin 1 1
− cosn+ = − cosn+
4sin2 1 2 2 2(1−cos) 2 2
AnswerMarks Guidance
2A1 AG. sin=2sin1cos1 and
2 2
2sin2 1=1−cos.
2
7
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
8(c)1 1 1 1 1 2 1 n−1
 cosxdx + cos + cos + cos
AnswerMarks Guidance
0 n n n n n n nM1 A1 Forms sum of areas of rectangles given in the diagram.
A0 if comparison with integral missing or unclear.
 n 1
sin sin
1 1 2 n−1 1  n n n 
= 1+cos +cos ++cos = 1−cos + 
n n n n  2n  n 1−cos 1 
AnswerMarks Guidance
 n M1 A1 1
Applies result from part (b) with = . AG.
n
4
AnswerMarks
8(d)1 1 1 1 2 1 n
 cosxdx cos + cos + cos
AnswerMarks Guidance
0 n n n n n nM1 A1 Forms sum of areas of rectangles. A0 if comparison
with integral missing or unclear.
 1  1
sin1sin sin1sin
1  n  1 1 1  n 
= 1−cos1+ + cos1− = cos1−1+ 
2n 1 n n 2n 1
 1−cos   1−cos 
AnswerMarks
 n   n A1
3
Question 8:
--- 8(a) ---
8(a) | zn −1
z−1 | B1
1
--- 8(b) ---
8(b) | zn −1 cosn−1+isinn
=
z−1 cos−1+isin | B1
(cosn−1+isinn)(cos−1−isin)
(cos−1+isin)(cos−1−isin) | M1 | Multiplies numerator and denominator by complex
conjugate.
zn −1 cosncos+sinnsin−cosn−cos+1
Re =
 z−1  (cos−1)2 +sin2 | M1 | Takes real part.
cos(n−1)=cosncos+sinnsin
cosncos+sinnsin−cosn 1
= +
2(1−cos) 2 | A1
cosn(cos−1)+sinnsin
1
= +
2(1−cos) 2 | M1 | Factorises.
Question | Answer | Marks | Guidance
8(b) | 1 sinnsin
= 1−cosn+ 
2 1−cos  | M1 A1 | Divides through by denominator. AG.
Alternative method for question 8(b)
zn −1 ein−1
=
z−1 ei−1 | B1
e i(n−1 2 ) −e −i1 2  cos (n−1)+isin (n−1)−cos1+isin (1)
= 2 2 2 2
i1 −e −i1  2isin1
e 2 2
2 | M1
zn −1 sin(n−1)+sin1
Re = 2 2
z−1 2sin1
 
2 | M1 | Takes real part
(n−1)
sin 1
= 2 +
2sin1 2
2 | A1
sinncos1−cosnsin1 1
= 2 2 +
2sin1 2
2 | M1 | Uses compound angle identity
sinncos1 1 1
= 2 − cosn+
2sin1 2 2
2 | M1 | Divides through by denominator.
sinnsin 1 1 sinnsin 1 1
− cosn+ = − cosn+
4sin2 1 2 2 2(1−cos) 2 2
2 | A1 | AG. sin=2sin1cos1 and
2 2
2sin2 1=1−cos.
2
7
Question | Answer | Marks | Guidance
--- 8(c) ---
8(c) | 1 1 1 1 1 2 1 n−1
 cosxdx + cos + cos + cos
0 n n n n n n n | M1 A1 | Forms sum of areas of rectangles given in the diagram.
A0 if comparison with integral missing or unclear.
 n 1
sin sin
1 1 2 n−1 1  n n n 
= 1+cos +cos ++cos = 1−cos + 
n n n n  2n  n 1−cos 1 
 n  | M1 A1 | 1
Applies result from part (b) with = . AG.
n
4
--- 8(d) ---
8(d) | 1 1 1 1 2 1 n
 cosxdx cos + cos + cos
0 n n n n n n | M1 A1 | Forms sum of areas of rectangles. A0 if comparison
with integral missing or unclear.
 1  1
sin1sin sin1sin
1  n  1 1 1  n 
= 1−cos1+ + cos1− = cos1−1+ 
2n 1 n n 2n 1
 1−cos   1−cos 
 n   n  | A1
3
\begin{enumerate}[label=(\alph*)]
\item State the sum of the series $1 + z + z^2 + \ldots + z^{n-1}$, for $z \neq 1$. [1]

\item By letting $z = \cos\theta + i\sin\theta$, where $\cos\theta \neq 1$, show that
$$1 + \cos\theta + \cos 2\theta + \ldots + \cos(n-1)\theta = \frac{1}{2}\left(1 - \cos n\theta + \frac{\sin n\theta \sin\theta}{1 - \cos\theta}\right).$$ [7]
\end{enumerate}

\includegraphics{figure_8}

The diagram shows the curve with equation $y = \cos x$ for $0 \leq x \leq 1$, together with a set of $n$ rectangles of width $\frac{1}{n}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item By considering the sum of the areas of these rectangles, show that
$$\int_0^1 \cos x\,dx < \frac{1}{2n}\left(1 - \cos 1 + \frac{\sin 1\sin\frac{1}{n}}{1 - \cos\frac{1}{n}}\right).$$ [4]

\item Use a similar method to find, in terms of $n$, a lower bound for $\int_0^1 \cos x\,dx$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q8 [15]}}
This paper (8 questions)
View full paper
Q1 4 Q2 5 Q3 6 Q4 10 Q5 10 Q6 14 Q7 11 Q8 15