| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with exponential functions |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring standard integration of exponential functions. Students need to find where the curve crosses the x-axis, set up the volume integral π∫y² dx, expand (e^(0.5x) - 2)², and integrate term-by-term. While it involves exponentials and requires algebraic manipulation to reach the ln 2 form, it follows a completely standard method with no novel insights needed. Slightly easier than average due to the routine nature of the technique. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Upper limit \(= 2\ln 2\) or \(\ln 4\) | B1 | Do not accept \(1.386\), \(\frac{\ln 2}{0.5}\) or anything else. Recovery from \(\frac{\ln 2}{0.5}\) allowed if candidate progresses to \(8\pi\ln 2 - 5\pi\) |
| \(\int\left(e^{0.5x}-2\right)^2 dx = \int\left(e^x - 4e^{0.5x}+4\right)dx = e^x - 8e^{0.5x}+4x\) | M1 A1 A1 | M1: Attempts to square \((e^{0.5x}-2)\) with \(\int e^{kx}dx \to e^{kx}\) seen at least once. Ignore \(\pi\). A1: Two terms correct. A1: All three terms correct |
| Volume \(= \pi\left[e^x - 8e^{0.5x}+4x\right]_0^{2\ln 2} = \ldots\) | dM1 | Full and complete method for volume. Dependent on previous M. Both limits used with evidence for 0 |
| Volume \(= 8\pi\ln 2 - 5\pi\) | A1 | cao. Also accept \(-5\pi+8\pi\ln 2\) or \(\pi(8\ln 2-5)\) |
# Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Upper limit $= 2\ln 2$ or $\ln 4$ | B1 | Do not accept $1.386$, $\frac{\ln 2}{0.5}$ or anything else. Recovery from $\frac{\ln 2}{0.5}$ allowed if candidate progresses to $8\pi\ln 2 - 5\pi$ |
| $\int\left(e^{0.5x}-2\right)^2 dx = \int\left(e^x - 4e^{0.5x}+4\right)dx = e^x - 8e^{0.5x}+4x$ | M1 A1 A1 | M1: Attempts to square $(e^{0.5x}-2)$ with $\int e^{kx}dx \to e^{kx}$ seen at least once. Ignore $\pi$. A1: Two terms correct. A1: All three terms correct |
| Volume $= \pi\left[e^x - 8e^{0.5x}+4x\right]_0^{2\ln 2} = \ldots$ | dM1 | Full and complete method for volume. Dependent on previous M. Both limits used with evidence for 0 |
| Volume $= 8\pi\ln 2 - 5\pi$ | A1 | cao. Also accept $-5\pi+8\pi\ln 2$ or $\pi(8\ln 2-5)$ |
---3.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-08_801_679_125_635}
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\caption{Figure 1}
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\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { e } ^ { 0.5 x } - 2$\\
The region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the $y$-axis.
The region $R$ is rotated $360 ^ { \circ }$ about the $x$-axis to form a solid of revolution.\\
Show that the volume of this solid can be written in the form $a \ln 2 + b$, where $a$ and $b$ are constants to be found.\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel P4 2020 Q3 [6]}}