| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a standard Further Maths question combining de Moivre's theorem with integration. Part (a) is a routine application of binomial expansion and de Moivre (well-practiced technique), while part (b) requires recognizing that (1-x²)^(3/2) becomes cos⁴θ after substitution, then applying the result from (a). The question is structured to guide students through each step, making it moderately challenging but not requiring novel insight—typical of Further Maths integration questions. |
| Spec | 1.04a Binomial expansion: (a+b)^n for positive integer n1.08h Integration by substitution4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| \(z + z^{-1} = 2\cos\theta\) | B1 | Use of \(z + z^{-1} = 2\cos\theta\) |
| \((z+z^{-1})^4 = (z^4+z^{-4}) + 4(z^2+z^{-2}) + 6\) | M1 A1 | Expands and groups. M1 A0 for no clear grouping. Correct substitution of \(z^n = \cos n\theta + i\sin n\theta\) for each term scores M1 A1 |
| \((2\cos\theta)^4 = 2\cos 4\theta + 4(2\cos 2\theta) + 6\) | M1 | Substitutes \(z^n + z^{-n} = 2\cos n\theta\). If \(z^n = \cos n\theta + i\sin n\theta\) used then must cancel sin |
| \(\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)\) | A1 | AG. SC B1 expands \((\cos\theta + i\sin\theta)^4\) and uses trigonometric identities |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{1}{2}}(1-x^2)^{\frac{3}{2}}\,dx = \int_0^{\frac{1}{6}\pi}\cos^4\theta\,d\theta = \frac{1}{8}\left[\frac{1}{4}\sin 4\theta + 2\sin 2\theta + 3\theta\right]_0^{\frac{1}{6}\pi}\) | M1 A1 | Applies substitution (M1), gets to \(\int\cos^4\theta\,d\theta\), changes limits, integration correct (A1) |
| \(\frac{1}{16}\left(\frac{9}{4}\sqrt{3} + \pi\right)\) | A1 |
## Question 3(a):
| $z + z^{-1} = 2\cos\theta$ | B1 | Use of $z + z^{-1} = 2\cos\theta$ |
|---|---|---|
| $(z+z^{-1})^4 = (z^4+z^{-4}) + 4(z^2+z^{-2}) + 6$ | M1 A1 | Expands and groups. M1 A0 for no clear grouping. Correct substitution of $z^n = \cos n\theta + i\sin n\theta$ for each term scores M1 A1 |
| $(2\cos\theta)^4 = 2\cos 4\theta + 4(2\cos 2\theta) + 6$ | M1 | Substitutes $z^n + z^{-n} = 2\cos n\theta$. If $z^n = \cos n\theta + i\sin n\theta$ used then must cancel sin |
| $\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)$ | A1 | AG. SC B1 expands $(\cos\theta + i\sin\theta)^4$ and uses trigonometric identities |
---
## Question 3(b):
| $\int_0^{\frac{1}{2}}(1-x^2)^{\frac{3}{2}}\,dx = \int_0^{\frac{1}{6}\pi}\cos^4\theta\,d\theta = \frac{1}{8}\left[\frac{1}{4}\sin 4\theta + 2\sin 2\theta + 3\theta\right]_0^{\frac{1}{6}\pi}$ | M1 A1 | Applies substitution (M1), gets to $\int\cos^4\theta\,d\theta$, changes limits, integration correct (A1) |
|---|---|---|
| $\frac{1}{16}\left(\frac{9}{4}\sqrt{3} + \pi\right)$ | A1 | |
---3
\begin{enumerate}[label=(\alph*)]
\item By considering the binomial expansion of $\left( z + z ^ { - 1 } \right) ^ { 4 }$, where $z = \cos \theta + \mathrm { i } \sin \theta$, use de Moivre's theorem to show that $\cos ^ { 4 } \theta = \frac { 1 } { 8 } ( \cos 4 \theta + 4 \cos 2 \theta + 3 )$.
\item Use the substitution $x = \sin \theta$ to find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } } \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q3 [8]}}