| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Difficulty | Standard +0.8 This P4 question requires finding a volume of revolution using integration with a non-trivial integrand that simplifies nicely when squared. The key insight is that y² = 3x/(3x²+5) leads to an integral solvable by substitution (u = 3x²+5), requiring recognition of the derivative relationship. The multi-step process (squaring, setting up volume integral, substitution, evaluation at awkward limits √5 and 5, simplifying logarithms) and the requirement for exact form with irrational coefficient makes this moderately challenging but still within standard P4 technique application. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume \(= \int_{\sqrt{5}}^{5} \pi\left(\sqrt{\frac{3x}{3x^2+5}}\right)^2 dx\) | B1 | States or implies correct volume formula with limits; \(\pi\) may be implied by later work; limits may be implied; \(dx\) may be missing |
| \(\int\left(\sqrt{\frac{3x}{3x^2+5}}\right)^2 dx = \int\frac{3x}{(3x^2+5)}dx = \frac{1}{2}\ln(3x^2+5)\) | M1A1 | M1: Attempts to achieve \(k\ln(3x^2+5)\); may use substitution \(u=3x^2\) or \(u=3x^2+5\); A1: Correct result of integration, may be unsimplified |
| Volume \(= \{\pi\}\left(\frac{1}{2}\ln(3\times25+5)-\frac{1}{2}\ln(3\times5+5)\right)\) | M1 | Substitutes correct limits for their variable and subtracts; \(\pi\) may be missing |
| \(= \pi\ln 2\) | A1 | cao; note \(\frac{\pi}{2}\ln 4\) is A0 as form not as specified |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume $= \int_{\sqrt{5}}^{5} \pi\left(\sqrt{\frac{3x}{3x^2+5}}\right)^2 dx$ | **B1** | States or implies correct volume formula with limits; $\pi$ may be implied by later work; limits may be implied; $dx$ may be missing |
| $\int\left(\sqrt{\frac{3x}{3x^2+5}}\right)^2 dx = \int\frac{3x}{(3x^2+5)}dx = \frac{1}{2}\ln(3x^2+5)$ | **M1A1** | M1: Attempts to achieve $k\ln(3x^2+5)$; may use substitution $u=3x^2$ or $u=3x^2+5$; A1: Correct result of integration, may be unsimplified |
| Volume $= \{\pi\}\left(\frac{1}{2}\ln(3\times25+5)-\frac{1}{2}\ln(3\times5+5)\right)$ | **M1** | Substitutes correct limits for their variable and subtracts; $\pi$ may be missing |
| $= \pi\ln 2$ | **A1** | cao; note $\frac{\pi}{2}\ln 4$ is A0 as form not as specified |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c46ca445-cf59-4664-931e-add9f2f81851-08_419_665_255_708}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\section*{In this question you must show all stages of your working.}
\section*{Solutions relying entirely on calculator technology are not acceptable.}
Figure 1 shows a sketch of the curve with equation
$$y = \sqrt { \frac { 3 x } { 3 x ^ { 2 } + 5 } } \quad x \geqslant 0$$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the lines with equations $x = \sqrt { 5 }$ and $x = 5$
The region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.\\
Use integration to find the exact volume of the solid generated. Give your answer in the form $a \ln b$, where $a$ is an irrational number and $b$ is a prime number.
\hfill \mbox{\textit{Edexcel P4 2023 Q3 [5]}}