| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Hard +2.3 This is a sophisticated Further Maths question requiring multiple advanced techniques: binomial expansion with complex numbers, de Moivre's theorem, deriving a reduction formula via integration by parts, and combining these results to evaluate a definite integral. The multi-part structure demands sustained reasoning across different areas, and part (c) requires strategic application of both previous results—significantly beyond standard A-level integration. |
| Spec | 1.05l Double angle formulae: and compound angle formulae4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z + z^{-1} = 2\cos\theta\) | B1 | For LHS |
| \(\left(z+z^{-1}\right)^7 = \left(z^7+z^{-7}\right)+7\left(z^5+z^{-5}\right)+21\left(z^3+z^{-3}\right)+35\left(z+z^{-1}\right)\) | M1A1 | Expands and groups. A0 if not grouped clearly |
| \(2^7\cos^7\theta = 2\cos7\theta + 7(2\cos5\theta) + 21(2\cos3\theta) + 35(2\cos\theta)\) | M1 | Substitutes \(z^n + z^{-n} = 2\cos n\theta\) |
| \(\cos^7\theta = \frac{1}{64}\cos7\theta + \frac{7}{64}\cos5\theta + \frac{21}{64}\cos3\theta + \frac{35}{64}\cos\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \int_0^{\frac{1}{4}\pi}\cos^{n-1}\theta\cos\theta\,d\theta = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{1}{4}\pi} + (n-1)\int_0^{\frac{1}{4}\pi}\cos^{n-2}\theta\sin^2\theta\,d\theta\) | M1A1 | Applies integration by parts |
| \(I_n = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{1}{4}\pi} + (n-1)\int_0^{\frac{1}{4}\pi}\cos^{n-2}\theta\left(1-\cos^2\theta\right)d\theta\) | M1 | Applies \(\sin^2\theta = 1-\cos^2\theta\) |
| \(I_n = 2^{-\frac{n}{2}} + (n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n = 2^{-\frac{n}{2}} + (n-1)I_{n-2}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(9I_9 = 2^{-\frac{9}{2}} + 8I_7 = \frac{1}{16}\left(2^{-\frac{1}{2}}\right) + 8I_7\) | M1A1 | Applies reduction formula from (b) |
| \(I_7 = \int_0^{\frac{1}{4}\pi}\frac{1}{64}\cos7\theta + \frac{7}{64}\cos5\theta + \frac{21}{64}\cos3\theta + \frac{35}{64}\cos\theta\,d\theta\) | M1 | Applies identity from (a). Allow missing/incorrect limits |
| \(I_7 = \frac{1}{64}\left[\frac{1}{7}\sin7\theta + \frac{7}{5}\sin5\theta + \frac{21}{3}\sin3\theta + 35\sin\theta\right]_0^{\frac{1}{4}\pi}\) | A1 | |
| \(9I_9 = \frac{1}{16}\left(2^{-\frac{1}{2}}\right) + \frac{1}{8}\left(\frac{1}{7}\left(-2^{-\frac{1}{2}}\right) + \frac{7}{5}\left(-2^{-\frac{1}{2}}\right) + \frac{21}{3}\left(2^{-\frac{1}{2}}\right) + 35\left(2^{-\frac{1}{2}}\right)\right)\) \(I_9 = \frac{2867}{5040}\left(2^{-\frac{1}{2}}\right) = \frac{2867}{10080}\sqrt{2}\) | A1 | Check exact answer when \(I_9\) is subject, like terms collected. ISW. 0.402237 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z + z^{-1} = 2\cos\theta$ | B1 | For LHS |
| $\left(z+z^{-1}\right)^7 = \left(z^7+z^{-7}\right)+7\left(z^5+z^{-5}\right)+21\left(z^3+z^{-3}\right)+35\left(z+z^{-1}\right)$ | M1A1 | Expands and groups. A0 if not grouped clearly |
| $2^7\cos^7\theta = 2\cos7\theta + 7(2\cos5\theta) + 21(2\cos3\theta) + 35(2\cos\theta)$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$ |
| $\cos^7\theta = \frac{1}{64}\cos7\theta + \frac{7}{64}\cos5\theta + \frac{21}{64}\cos3\theta + \frac{35}{64}\cos\theta$ | A1 | |
**Total: 5 marks**
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \int_0^{\frac{1}{4}\pi}\cos^{n-1}\theta\cos\theta\,d\theta = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{1}{4}\pi} + (n-1)\int_0^{\frac{1}{4}\pi}\cos^{n-2}\theta\sin^2\theta\,d\theta$ | M1A1 | Applies integration by parts |
| $I_n = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{1}{4}\pi} + (n-1)\int_0^{\frac{1}{4}\pi}\cos^{n-2}\theta\left(1-\cos^2\theta\right)d\theta$ | M1 | Applies $\sin^2\theta = 1-\cos^2\theta$ |
| $I_n = 2^{-\frac{n}{2}} + (n-1)I_{n-2} - (n-1)I_n \Rightarrow nI_n = 2^{-\frac{n}{2}} + (n-1)I_{n-2}$ | A1 | AG |
**Total: 4 marks**
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $9I_9 = 2^{-\frac{9}{2}} + 8I_7 = \frac{1}{16}\left(2^{-\frac{1}{2}}\right) + 8I_7$ | M1A1 | Applies reduction formula from (b) |
| $I_7 = \int_0^{\frac{1}{4}\pi}\frac{1}{64}\cos7\theta + \frac{7}{64}\cos5\theta + \frac{21}{64}\cos3\theta + \frac{35}{64}\cos\theta\,d\theta$ | M1 | Applies identity from (a). Allow missing/incorrect limits |
| $I_7 = \frac{1}{64}\left[\frac{1}{7}\sin7\theta + \frac{7}{5}\sin5\theta + \frac{21}{3}\sin3\theta + 35\sin\theta\right]_0^{\frac{1}{4}\pi}$ | A1 | |
| $9I_9 = \frac{1}{16}\left(2^{-\frac{1}{2}}\right) + \frac{1}{8}\left(\frac{1}{7}\left(-2^{-\frac{1}{2}}\right) + \frac{7}{5}\left(-2^{-\frac{1}{2}}\right) + \frac{21}{3}\left(2^{-\frac{1}{2}}\right) + 35\left(2^{-\frac{1}{2}}\right)\right)$ $I_9 = \frac{2867}{5040}\left(2^{-\frac{1}{2}}\right) = \frac{2867}{10080}\sqrt{2}$ | A1 | Check exact answer when $I_9$ is subject, like terms collected. ISW. 0.402237 |
**Total: 5 marks**8
\begin{enumerate}[label=(\alph*)]
\item By considering the binomial expansion of $\left( z + \frac { 1 } { z } \right) ^ { 7 }$, where $z = \cos \theta + \mathrm { i } \sin \theta$, use de Moivre's theorem to show that
$$\cos ^ { 7 } \theta = a \cos 7 \theta + b \cos 5 \theta + c \cos 3 \theta + d \cos \theta$$
where $a , b , c$ and $d$ are constants to be determined.\\
Let $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { n } \theta \mathrm {~d} \theta$.
\item Show that
$$n I _ { n } = 2 ^ { - \frac { 1 } { 2 } n } + ( n - 1 ) I _ { n - 2 }$$
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-18_2716_40_109_2009}
\item Using the results given in parts (a) and (b), find the exact value of $I _ { 9 }$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q8 [14]}}