| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse trig function |
| Difficulty | Standard +0.8 This is a Further Maths question requiring chain rule differentiation of an inverse trig function with a composite argument (part a), then recognizing how to reverse this to evaluate a non-standard integral (part b). The integration requires either spotting the connection to part (a) or making a substitution like u = √(3x), which is not immediately obvious. The multi-step reasoning and the need to connect differentiation to integration elevates this above standard A-level, though it's within reach for FP2 students who recognize the pattern. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\left(\frac{dy}{dx}\right) = \frac{1}{1+(\sqrt{3x})^2}\) | M1 | \(\frac{dy}{dx} = \frac{1}{1+3x}\) |
| \(\times \frac{1}{2} \times \sqrt{3} x^{\frac{1}{2}}\) OE | A1 | 2 marks |
| For guidance \(\frac{dy}{dx} = \frac{\sqrt{3}}{2(1+3x)\sqrt{x}}\) | ||
| (b) \(\left(\int\right) = k\tan^{-1}\sqrt{3x}\) | M1 | |
| \(\left(\int\right) = \frac{2}{\sqrt{3}}\tan^{-1}\sqrt{3x}\) | A1 | |
| \(k\left(\frac{\pi}{3} - \frac{\pi}{4}\right)\) | m1 | or \(k\frac{\pi}{12}\) |
| \(= \frac{\sqrt{3}\pi}{18}\) | A1 | 4 marks |
**(a)** $\left(\frac{dy}{dx}\right) = \frac{1}{1+(\sqrt{3x})^2}$ | M1 | $\frac{dy}{dx} = \frac{1}{1+3x}$
$\times \frac{1}{2} \times \sqrt{3} x^{\frac{1}{2}}$ OE | A1 | **2 marks** | may have $\frac{3}{\sqrt{3}}$ instead of $\sqrt{3}$
| | | For guidance $\frac{dy}{dx} = \frac{\sqrt{3}}{2(1+3x)\sqrt{x}}$
**(b)** $\left(\int\right) = k\tan^{-1}\sqrt{3x}$ | M1 |
$\left(\int\right) = \frac{2}{\sqrt{3}}\tan^{-1}\sqrt{3x}$ | A1 |
$k\left(\frac{\pi}{3} - \frac{\pi}{4}\right)$ | m1 | or $k\frac{\pi}{12}$ | PI by correct answer
$= \frac{\sqrt{3}\pi}{18}$ | A1 | **4 marks**
**Total: 6 marks**
**Notes:**
- **(a)** Alternative 1: $\sec^2 y \frac{dy}{dx} = kx^{-\frac{1}{2}}$ M1 leading to correct $\frac{dy}{dx}$ in terms of $x$ A1
- **(a)** Alternative 2: $x = A\tan^2 y \Rightarrow \frac{dx}{dy} = k\sec^2 y \tan y$ M1 leading to correct $\frac{dy}{dx}$ in terms of $x$ A1
- **(b)** If a substitution such as $u = \sqrt{x}$ is used giving $\int\frac{2}{1+3u^2}du$ then M1 is still only earned for $k\tan^{-1}\sqrt{3}u$ and A1 for $\frac{2}{\sqrt{3}}\tan^{-1}\sqrt{3}u$ and m1 A1 as above
---\begin{enumerate}[label=(\alph*)]
\item Given that $y = \tan^{-1} \sqrt{3x}$, find $\frac{dy}{dx}$, giving your answer in terms of $x$. [2 marks]
\item Hence, or otherwise, show that $\int_{\frac{1}{3}}^1 \frac{1}{(1+3x)\sqrt{x}} \, dx = \frac{\sqrt{3}\pi}{n}$, where $n$ is an integer. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2016 Q4 [6]}}