| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Trigonometric power reduction |
| Difficulty | Challenging +1.8 This is a Further Maths F3 reduction formula question requiring integration by parts to derive the recurrence relation, then recursive application to find I_5. The derivation involves standard IBP technique with careful boundary evaluation at π/4, but the multi-step recursive calculation and exact arithmetic with surds requires sustained accuracy. Harder than typical A-level integration but standard for FM reduction formulae. |
| Spec | 4.08a Maclaurin series: find series for function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_n = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{\pi}{4}} - (-)\int_0^{\frac{\pi}{4}}(n-1)\cos^{n-2}\theta\sin^2\theta\,d\theta\) | M1A1 | M1: Attempt parts the correct way round. A1: Correct expression |
| \(I_n = \left(\frac{1}{\sqrt{2}}\right)^n + \ldots\) | B1 | Uses limits to obtain \(\left(\frac{1}{\sqrt{2}}\right)^n\) |
| \(I_n = \ldots + \int_0^{\frac{\pi}{4}}(n-1)\cos^{n-2}\theta(1-\cos^2\theta)\,d\theta\) | dM1 | Replaces \(\sin^2\theta\) by \(1-\cos^2\theta\). Dependent on previous method mark |
| \(nI_n = \left(\frac{1}{\sqrt{2}}\right)^n + (n-1)I_{n-2}\) * | ddM1A1cso | M1: Replaces expressions for \(I_n\) and \(I_{n-1}\), dependent on both previous method marks. A1: Achieves printed answer with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_1 = \int_0^{\frac{\pi}{4}}\cos\theta\,d\theta = [\sin\theta]_0^{\frac{\pi}{4}} = \frac{1}{\sqrt{2}}\) | M1A1 | M1: Attempt \(I_1\). A1: \(\frac{1}{\sqrt{2}}\) |
| \(I_3 = \frac{1}{3}\left(\frac{1}{2\sqrt{2}}+2I_1\right),\quad I_5 = \frac{1}{5}\left(\frac{1}{4\sqrt{2}}+4I_3\right)\) | M1M1 | M1: Uses reduction formula first time (allow slips). M1: Uses reduction formula second time (allow slips) |
| \(I_5 = \frac{43\sqrt{2}}{120}\) or \(\frac{43}{60\sqrt{2}}\) | A1 | Correct answer |
# Question 5:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_n = \left[\cos^{n-1}\theta\sin\theta\right]_0^{\frac{\pi}{4}} - (-)\int_0^{\frac{\pi}{4}}(n-1)\cos^{n-2}\theta\sin^2\theta\,d\theta$ | M1A1 | M1: Attempt parts the correct way round. A1: Correct expression |
| $I_n = \left(\frac{1}{\sqrt{2}}\right)^n + \ldots$ | B1 | Uses limits to obtain $\left(\frac{1}{\sqrt{2}}\right)^n$ |
| $I_n = \ldots + \int_0^{\frac{\pi}{4}}(n-1)\cos^{n-2}\theta(1-\cos^2\theta)\,d\theta$ | dM1 | Replaces $\sin^2\theta$ by $1-\cos^2\theta$. Dependent on previous method mark |
| $nI_n = \left(\frac{1}{\sqrt{2}}\right)^n + (n-1)I_{n-2}$ * | ddM1A1cso | M1: Replaces expressions for $I_n$ and $I_{n-1}$, dependent on both previous method marks. A1: Achieves printed answer with no errors seen |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_1 = \int_0^{\frac{\pi}{4}}\cos\theta\,d\theta = [\sin\theta]_0^{\frac{\pi}{4}} = \frac{1}{\sqrt{2}}$ | M1A1 | M1: Attempt $I_1$. A1: $\frac{1}{\sqrt{2}}$ |
| $I_3 = \frac{1}{3}\left(\frac{1}{2\sqrt{2}}+2I_1\right),\quad I_5 = \frac{1}{5}\left(\frac{1}{4\sqrt{2}}+4I_3\right)$ | M1M1 | M1: Uses reduction formula first time (allow slips). M1: Uses reduction formula second time (allow slips) |
| $I_5 = \frac{43\sqrt{2}}{120}$ or $\frac{43}{60\sqrt{2}}$ | A1 | Correct answer |\begin{enumerate}
\item Given that
\end{enumerate}
$$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 4 } } \cos ^ { n } \theta \mathrm {~d} \theta , \quad n \geqslant 0$$
(a) prove that, for $n \geqslant 2$,
$$n I _ { n } = \left( \frac { 1 } { \sqrt { 2 } } \right) ^ { n } + ( n - 1 ) I _ { n - 2 }$$
(b) Hence find the exact value of $I _ { 5 }$, showing each step of your working.\\
\hfill \mbox{\textit{Edexcel F3 2014 Q5 [11]}}