| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume using cone or cylinder formula |
| Difficulty | Standard +0.8 This question requires finding a tangent equation to an exponential curve (involving differentiation of 3^x and using ln 3), then computing a volume of revolution by integrating the difference between the tangent line and exponential curve. The integration of 3^(2x) requires substitution and careful algebraic manipulation. While the techniques are C4 standard, the multi-step nature, exact form requirements, and need to subtract a cone volume from the rotated exponential make this moderately challenging. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| \(\{y = 3^x \Rightarrow\} \frac{dy}{dx} = 3^x \ln 3\) | B1 |
| \(\frac{dy}{dx} = 3^x \ln 3\) or \(\ln 3(e^{x\ln 3})\) or \(y \ln 3\). | |
| Either T: \(y - 9 = 3^x \ln 3(x-2)\) | M1 |
| or T: \(y = (3^x \ln 3)x + 9 - 18\ln 3\), where \(9 = (3^x \ln 3)(2) + c\) | |
| [Cuts x-axis \(\Rightarrow y = 0 \Rightarrow\)] Sets \(y = 0\) in their tangent equation and progresses to \(x = \ldots\). | M1 |
| \(-9 = 9\ln 3(x-2)\) or \(0 = (3^x \ln 3)x + 9 - 18\ln 3\). | |
| \(x = 2 - \frac{1}{\ln 3}\) or \(2 - \frac{\ln 3-1}{\ln 3}\) or \(\frac{2\ln 3-1}{\ln 3}\) o.e. | A1 cso |
| [4] |
| Answer | Marks |
|---|---|
| \(V = \pi\int_0^2(3^x)^2\{dx\}\) or \(\pi\int 3^{2x}\{dx\}\) or \(\pi\int 9^x\{dx\}\) | B1 o.e. |
| A correct expression for the volume with or without \(dx\), which can be implied | |
| Eg: either \(3^{2x} \rightarrow \frac{3^{2x}}{\pm\alpha(\ln 3)}\) or \(\pm\alpha(\ln 3)3^{2x}\) | M1 |
| or \(9^x \rightarrow \frac{9^x}{\pm\alpha(\ln 9)}\), \(\alpha \in \mathbb{ℝ}\) | |
| \(3^{2x} \rightarrow \frac{3^{2x}}{2\ln 3}\) or \(9^x \rightarrow \frac{9^x}{\ln 9}\) or \(e^{2x\ln 3} \rightarrow \frac{1}{2\ln 3}(e^{2x\ln 3})\) | A1 o.e. |
## Part (a)
$\{y = 3^x \Rightarrow\} \frac{dy}{dx} = 3^x \ln 3$ | B1 |
$\frac{dy}{dx} = 3^x \ln 3$ or $\ln 3(e^{x\ln 3})$ or $y \ln 3$. | |
Either T: $y - 9 = 3^x \ln 3(x-2)$ | M1 |
or T: $y = (3^x \ln 3)x + 9 - 18\ln 3$, where $9 = (3^x \ln 3)(2) + c$ | |
[Cuts x-axis $\Rightarrow y = 0 \Rightarrow$] Sets $y = 0$ in their tangent equation and progresses to $x = \ldots$. | M1 |
$-9 = 9\ln 3(x-2)$ or $0 = (3^x \ln 3)x + 9 - 18\ln 3$. | |
$x = 2 - \frac{1}{\ln 3}$ or $2 - \frac{\ln 3-1}{\ln 3}$ or $\frac{2\ln 3-1}{\ln 3}$ o.e. | A1 cso |
| [4] |
## Part (b)
$V = \pi\int_0^2(3^x)^2\{dx\}$ or $\pi\int 3^{2x}\{dx\}$ or $\pi\int 9^x\{dx\}$ | B1 o.e. |
A correct expression for the volume with or without $dx$, which can be implied | |
Eg: either $3^{2x} \rightarrow \frac{3^{2x}}{\pm\alpha(\ln 3)}$ or $\pm\alpha(\ln 3)3^{2x}$ | M1 |
or $9^x \rightarrow \frac{9^x}{\pm\alpha(\ln 9)}$, $\alpha \in \mathbb{ℝ}$ | |
$3^{2x} \rightarrow \frac{3^{2x}}{2\ln 3}$ or $9^x \rightarrow \frac{9^x}{\ln 9}$ or $e^{2x\ln 3} \rightarrow \frac{1}{2\ln 3}(e^{2x\ln 3})$ | A1 o.e. |
Substitutes limits of 9\includegraphics{figure_3}
Figure 3 shows a sketch of part of the curve $C$ with equation
$$y = 3^x$$
The point $P$ lies on $C$ and has coordinates $(2, 9)$.
The line $l$ is a tangent to $C$ at $P$. The line $l$ cuts the $x$-axis at the point $Q$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the $x$ coordinate of $Q$.
[4]
\end{enumerate}
The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $l$. This region $R$ is rotated through $360°$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Use integration to find the exact value of the volume of the solid generated.
Give your answer in the form $\frac{p}{q}$ where $p$ and $q$ are exact constants.
[You may assume the formula $V = \frac{1}{3}\pi r^2 h$ for the volume of a cone.]
[6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2015 Q8 [10]}}