| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Polynomial times trigonometric |
| Difficulty | Challenging +1.8 This is a Further Maths F3 reduction formula question requiring substitution (u = x²), integration by parts, and recursive application. While the setup demands careful algebraic manipulation and the boundary term evaluation is non-trivial, the techniques are standard for this module and the question provides clear scaffolding through parts (a) and (b). |
| Spec | 1.08h Integration by substitution8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \int_0^{\sqrt{\frac{\pi}{2}}} x^{n-1} \cdot x\cos(x^2)\,dx = \left[x^{n-1}\cdot\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}(n-1)x^{n-2}\cdot\frac{1}{2}\sin(x^2)\,dx\) | M1A1 | Applies IBP in correct direction with correct split; fully correct expression |
| \(= \left[x^{n-1}\cdot\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \frac{1}{2}(n-1)\left(\left[x^{n-3}\cdot\left(-\frac{1}{2}\cos(x^2)\right)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}(n-3)x^{n-4}\cdot\left(-\frac{1}{2}\cos(x^2)\right)dx\right)\) | dM1A1 | Second application of IBP in correct direction; fully correct second application |
| \(= \left(\frac{1}{2}\left(\sqrt{\frac{\pi}{2}}\right)^{n-1}\sin\frac{\pi}{2} - 0\right) - \frac{1}{2}(n-1)\left[(0-0) + \frac{1}{2}(n-3)I_{n-4}\right]\) | dM1 | Applies limits completely and replaces final integral by \(I_{n-4}\); depends on both previous M marks |
| \(= \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n-1}{2}} - \frac{1}{4}(n-1)(n-3)I_{n-4}\) * | A1* | Achieves printed answer from completely correct work with evidence of limits applied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n = \left[\frac{x^{n+1}}{n+1}\cdot\cos(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}\frac{x^{n+1}}{n+1}\cdot(-2x\sin(x^2))\,dx\) | M1A1 | IBP in correct direction; fully correct expression |
| \(= \left[\frac{x^{n+1}}{n+1}\cdot\cos(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} + \frac{2}{n+1}\left(\left[\frac{x^{n+3}}{n+3}\cdot\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}\frac{x^{n+3}}{n+3}\cdot 2x\cos(x^2)\,dx\right)\) | dM1A1 | Second IBP application; fully correct |
| \(= (0-0) + \frac{2}{n+1}\left(\frac{1}{n+3}\left(\sqrt{\frac{\pi}{2}}\right)^{n+3}\sin\frac{\pi}{2} - 0 - \frac{2}{n+3}I_{n+4}\right)\) | dM1 | Applies limits and replaces integral by \(I_{n+4}\) |
| \(\Rightarrow I_{n+4} = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n+3}{2}} - \frac{1}{4}(n+1)(n+3)I_n\), replacing \(n\) by \(n-4\): \(I_n = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n-1}{2}} - \frac{1}{4}(n-1)(n-3)I_{n-4}\) * | A1* | Achieves printed answer with clear statement that \(n\) is replaced by \(n-4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_1 = \int_0^{\sqrt{\frac{\pi}{2}}} x\cos(x^2)\,dx = \left[\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} = \frac{1}{2}\) | B1 | Correct \(I_1\); may be seen after attempting the reduction |
| \(I_5 = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{5-1}{2}} - \frac{1}{4}(5-1)(5-3)\times\text{"}\frac{1}{2}\text{"}\) | M1 | Applies reduction formula with their \(I_1\) and \(n=5\); condone slips evaluating \(\frac{1}{4}(n-1)(n-3)\) |
| \(= \frac{\pi^2}{8} - 1\) oe e.g. \(\frac{\pi^2-8}{8}\), \(\frac{1}{2}\left(\frac{\pi}{2}\right)^2 - 1\) | A1 | Correct answer |
# Question 6:
## Part (a) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \int_0^{\sqrt{\frac{\pi}{2}}} x^{n-1} \cdot x\cos(x^2)\,dx = \left[x^{n-1}\cdot\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}(n-1)x^{n-2}\cdot\frac{1}{2}\sin(x^2)\,dx$ | M1A1 | Applies IBP in correct direction with correct split; fully correct expression |
| $= \left[x^{n-1}\cdot\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \frac{1}{2}(n-1)\left(\left[x^{n-3}\cdot\left(-\frac{1}{2}\cos(x^2)\right)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}(n-3)x^{n-4}\cdot\left(-\frac{1}{2}\cos(x^2)\right)dx\right)$ | dM1A1 | Second application of IBP in correct direction; fully correct second application |
| $= \left(\frac{1}{2}\left(\sqrt{\frac{\pi}{2}}\right)^{n-1}\sin\frac{\pi}{2} - 0\right) - \frac{1}{2}(n-1)\left[(0-0) + \frac{1}{2}(n-3)I_{n-4}\right]$ | dM1 | Applies limits completely and replaces final integral by $I_{n-4}$; depends on both previous M marks |
| $= \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n-1}{2}} - \frac{1}{4}(n-1)(n-3)I_{n-4}$ * | A1* | Achieves printed answer from completely correct work with evidence of limits applied |
## Part (a) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n = \left[\frac{x^{n+1}}{n+1}\cdot\cos(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}\frac{x^{n+1}}{n+1}\cdot(-2x\sin(x^2))\,dx$ | M1A1 | IBP in correct direction; fully correct expression |
| $= \left[\frac{x^{n+1}}{n+1}\cdot\cos(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} + \frac{2}{n+1}\left(\left[\frac{x^{n+3}}{n+3}\cdot\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} - \int_0^{\sqrt{\frac{\pi}{2}}}\frac{x^{n+3}}{n+3}\cdot 2x\cos(x^2)\,dx\right)$ | dM1A1 | Second IBP application; fully correct |
| $= (0-0) + \frac{2}{n+1}\left(\frac{1}{n+3}\left(\sqrt{\frac{\pi}{2}}\right)^{n+3}\sin\frac{\pi}{2} - 0 - \frac{2}{n+3}I_{n+4}\right)$ | dM1 | Applies limits and replaces integral by $I_{n+4}$ |
| $\Rightarrow I_{n+4} = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n+3}{2}} - \frac{1}{4}(n+1)(n+3)I_n$, replacing $n$ by $n-4$: $I_n = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{n-1}{2}} - \frac{1}{4}(n-1)(n-3)I_{n-4}$ * | A1* | Achieves printed answer with clear statement that $n$ is replaced by $n-4$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_1 = \int_0^{\sqrt{\frac{\pi}{2}}} x\cos(x^2)\,dx = \left[\frac{1}{2}\sin(x^2)\right]_0^{\sqrt{\frac{\pi}{2}}} = \frac{1}{2}$ | B1 | Correct $I_1$; may be seen after attempting the reduction |
| $I_5 = \frac{1}{2}\left(\frac{\pi}{2}\right)^{\frac{5-1}{2}} - \frac{1}{4}(5-1)(5-3)\times\text{"}\frac{1}{2}\text{"}$ | M1 | Applies reduction formula with their $I_1$ and $n=5$; condone slips evaluating $\frac{1}{4}(n-1)(n-3)$ |
| $= \frac{\pi^2}{8} - 1$ oe e.g. $\frac{\pi^2-8}{8}$, $\frac{1}{2}\left(\frac{\pi}{2}\right)^2 - 1$ | A1 | Correct answer |
---6.
$$I _ { n } = \int _ { 0 } ^ { \sqrt { \frac { \pi } { 2 } } } x ^ { n } \cos \left( x ^ { 2 } \right) \mathrm { d } x \quad n \geqslant 1$$
\begin{enumerate}[label=(\alph*)]
\item Prove that, for $n \geqslant 5$
$$I _ { n } = \frac { 1 } { 2 } \left( \frac { \pi } { 2 } \right) ^ { \frac { n - 1 } { 2 } } - \frac { 1 } { 4 } ( n - 1 ) ( n - 3 ) I _ { n - 4 }$$
\item Hence, determine the exact value of $I _ { 5 }$, giving your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q6 [9]}}