| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/(a²+x²) |
| Difficulty | Moderate -0.3 This is a straightforward application of standard inverse trig integral formulas from the Further Maths syllabus. Part (i) uses arctan with simple substitution recognition (a=4), and part (ii) uses arcsin with a=3/2. Both require careful arithmetic but no problem-solving insight beyond pattern matching to memorized formulas. Slightly easier than average A-level due to being direct formula application, though the Further Maths context and exact value manipulation keep it near average. |
| Spec | 4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \((8)\int \frac{1}{16+x^2}\,dx = (8)\left(\frac{1}{4}\arctan\left(\frac{x}{4}\right)\right)\) | M1 | Obtains \(\ldots\arctan(kx)\); allow \(k=1\) |
| \(2\left[\arctan\left(\frac{x}{4}\right)\right]_4^{4\sqrt{3}} = 2(\arctan\sqrt{3} - \arctan 1) = \ldots\) | dM1 | Substitutes the given limits, subtracts either way round and obtains a value (could be decimal). Substitution does not need to be seen explicitly and may be implied by their value. |
| \(\frac{\pi}{6}\) or \(p = \frac{1}{6}\) | A1 | Correct exact value (or value for \(p\)). Accept equivalent exact expressions e.g. \(\frac{2\pi}{12}\) or \(p = \frac{2}{12}\); isw if necessary. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(2\int \frac{1}{\sqrt{9-4x^2}}\,dx = 2\left(\frac{1}{2}\arcsin\frac{2x}{3}\right)\) | M1 A1 | M1: Obtains \(\ldots\arcsin(kx)\); allow \(k=1\) so allow just \(\arcsin x\). A1: Fully correct integration but allow unsimplified as above. |
| \(\left[\arcsin\left(\frac{2x}{3}\right)\right]_{\frac{1}{2}}^{k} = \arcsin\left(\frac{2k}{3}\right) - \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{12}\) | dM1 | Substitutes the given limits, subtracts either way round, sets \(= \frac{\pi}{12}\), uses \(\arcsin\!\left(\frac{1}{2}\right)=\frac{\pi}{6}\) and correct order of operations condoning sign errors only to reach a value for \(k\). Note \(k\) may be inexact (decimal) or in terms of "sin" but must have simplified argument e.g. \(k = \frac{3\sin\!\left(\frac{\pi}{4}\right)}{2}\). |
| \(k = \dfrac{3\sqrt{2}}{4}\) or exact equivalent e.g. \(\dfrac{3}{2\sqrt{2}}\) | A1 | Note a common incorrect answer is \(k = \frac{3}{2}\sin\!\left(\frac{5\pi}{24}\right)(= 0.913\ldots)\) from incorrect integral of \(2\arcsin\!\left(\frac{2x}{3}\right)\) (generally scoring 1010). Condone \(x = \frac{3\sqrt{2}}{4}\). |
## Question 1:
### Part (i):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $(8)\int \frac{1}{16+x^2}\,dx = (8)\left(\frac{1}{4}\arctan\left(\frac{x}{4}\right)\right)$ | M1 | Obtains $\ldots\arctan(kx)$; allow $k=1$ |
| $2\left[\arctan\left(\frac{x}{4}\right)\right]_4^{4\sqrt{3}} = 2(\arctan\sqrt{3} - \arctan 1) = \ldots$ | dM1 | Substitutes the given limits, subtracts either way round and obtains a value (could be decimal). Substitution does not need to be seen explicitly and may be implied by their value. |
| $\frac{\pi}{6}$ or $p = \frac{1}{6}$ | A1 | Correct exact value (or value for $p$). Accept equivalent exact expressions e.g. $\frac{2\pi}{12}$ or $p = \frac{2}{12}$; isw if necessary. |
**Total: 3 marks**
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### Part (ii):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $2\int \frac{1}{\sqrt{9-4x^2}}\,dx = 2\left(\frac{1}{2}\arcsin\frac{2x}{3}\right)$ | M1 A1 | M1: Obtains $\ldots\arcsin(kx)$; allow $k=1$ so allow just $\arcsin x$. A1: Fully correct integration but allow unsimplified as above. |
| $\left[\arcsin\left(\frac{2x}{3}\right)\right]_{\frac{1}{2}}^{k} = \arcsin\left(\frac{2k}{3}\right) - \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{12}$ | dM1 | Substitutes the given limits, subtracts either way round, sets $= \frac{\pi}{12}$, uses $\arcsin\!\left(\frac{1}{2}\right)=\frac{\pi}{6}$ and correct **order** of operations condoning sign errors only to reach a value for $k$. Note $k$ may be inexact (decimal) or in terms of "sin" but must have simplified argument e.g. $k = \frac{3\sin\!\left(\frac{\pi}{4}\right)}{2}$. |
| $k = \dfrac{3\sqrt{2}}{4}$ or exact equivalent e.g. $\dfrac{3}{2\sqrt{2}}$ | A1 | Note a common incorrect answer is $k = \frac{3}{2}\sin\!\left(\frac{5\pi}{24}\right)(= 0.913\ldots)$ from incorrect integral of $2\arcsin\!\left(\frac{2x}{3}\right)$ (generally scoring 1010). Condone $x = \frac{3\sqrt{2}}{4}$. |
**Total: 4 marks**
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**Question 1 Total: 7 marks**\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.\\
(i) Show that
$$\int _ { 4 } ^ { 4 \sqrt { 3 } } \frac { 8 } { 16 + x ^ { 2 } } d x = p \pi$$
where $p$ is a rational number to be determined.\\
(ii) Determine the exact value of $k$ for which
$$\int _ { \frac { 3 } { 4 } } ^ { k } \frac { 2 } { \sqrt { 9 - 4 x ^ { 2 } } } d x = \frac { \pi } { 12 }$$
\hfill \mbox{\textit{Edexcel F3 2024 Q1 [7]}}