| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Standard integral of 1/√(x²+a²) |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring completion of the square to recognize standard inverse hyperbolic/trig forms, then applying the formula for volume of revolution. While it involves FP3 content (inherently harder), the steps are methodical: complete the square x²+2x-3=(x+1)²-4, recognize the standard form 1/√(u²-a²), integrate to get arsinh or ln form, then compute a straightforward volume integral. The techniques are standard for FP3 students, making this moderately above average difficulty overall but not requiring novel insight. |
| Spec | 4.07f Inverse hyperbolic: logarithmic forms4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(x^2+2x-3 = (x+1)^2-4\) | M1 | \(x^2+2x-3=(x\pm1)^2\pm\alpha\pm3,\ \alpha\neq 0\) |
| \(\int\frac{1}{\sqrt{(x+1)^2-4}}\,dx = \text{arcosh}\frac{(x+1)}{2}\ (+c)\) or \(\ln\left\{(x+1)+\sqrt{(x+1)^2-4}\right\}\) | M1 A1 | M1: use of arcosh (allow arccosh, \(\cosh^{-1}\)) or uses \(\ln\{x+\sqrt{x^2-a^2}\}\); A1: \(\text{arcosh}\frac{(x+1)}{2}\) (\(+c\) not required) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| \(S = \pi\int y^2\,dx = \pi\int\left(\frac{1}{\sqrt{x^2+2x-3}}\right)^2dx\) | M1 | Use of \(\pi\int y^2\,dx\) |
| \(= \int\frac{1}{(x+1)^2-4}\,dx = \left[\frac{1}{4}\ln\left(\frac{x-1}{x+3}\right)\right]\) | M1 A1 | M1: use of \(\ln\left(\frac{x\pm p}{x\pm q}\right)\); A1: \(\frac{1}{4}\ln\left(\frac{x-1}{x+3}\right)\) |
| \(= \frac{\pi}{4}\left(\ln\frac{1}{3}-\ln\frac{1}{5}\right) = \frac{\pi}{4}\ln\frac{5}{3}\) | A1 | \(\frac{\pi}{4}\ln\frac{5}{3}\) |
## Question 4(a):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $x^2+2x-3 = (x+1)^2-4$ | M1 | $x^2+2x-3=(x\pm1)^2\pm\alpha\pm3,\ \alpha\neq 0$ |
| $\int\frac{1}{\sqrt{(x+1)^2-4}}\,dx = \text{arcosh}\frac{(x+1)}{2}\ (+c)$ or $\ln\left\{(x+1)+\sqrt{(x+1)^2-4}\right\}$ | M1 A1 | M1: use of arcosh (allow arccosh, $\cosh^{-1}$) or uses $\ln\{x+\sqrt{x^2-a^2}\}$; A1: $\text{arcosh}\frac{(x+1)}{2}$ ($+c$ not required) |
## Question 4(b):
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $S = \pi\int y^2\,dx = \pi\int\left(\frac{1}{\sqrt{x^2+2x-3}}\right)^2dx$ | M1 | Use of $\pi\int y^2\,dx$ |
| $= \int\frac{1}{(x+1)^2-4}\,dx = \left[\frac{1}{4}\ln\left(\frac{x-1}{x+3}\right)\right]$ | M1 A1 | M1: use of $\ln\left(\frac{x\pm p}{x\pm q}\right)$; A1: $\frac{1}{4}\ln\left(\frac{x-1}{x+3}\right)$ |
| $= \frac{\pi}{4}\left(\ln\frac{1}{3}-\ln\frac{1}{5}\right) = \frac{\pi}{4}\ln\frac{5}{3}$ | A1 | $\frac{\pi}{4}\ln\frac{5}{3}$ |
*Special case:* Uses $S = k\int y^2\,dx$ scores maximum M0M1A1A0\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 1 } { \sqrt { x ^ { 2 } + 2 x - 3 } } , \quad x > 1$$
(a) Find $\int y \mathrm {~d} x$
The region $R$ is bounded by the curve $C$, the $x$-axis and the lines with equations $x = 2$ and $x = 3$. The region $R$ is rotated through $2 \pi$ radians about the $x$-axis.\\
(b) Find the volume of the solid generated. Give your answer in the form $p \pi \ln q$, where $p$ and $q$ are rational numbers to be found.\\
\hfill \mbox{\textit{Edexcel FP3 2015 Q4 [7]}}