| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indefinite & Definite Integrals |
| Type | Exponential and logarithmic integration |
| Difficulty | Standard +0.3 This is a standard C3 integration question involving exponential functions. Part (a) tests basic transformations, (b) is trivial substitution, (c) involves a routine quadratic substitution with e^(2x), and (d) requires integrating exponentials and finding area between curves. All techniques are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| Stretch parallel to \(x\)-axis, scale factor \(\frac{1}{2}\) | B1 B1 |
| Translation by \(\begin{pmatrix}0\\-1\end{pmatrix}\) | B1 B1 |
| Answer | Marks |
|---|---|
| \(A = (0, 6)\) since \(y = 4e^0 + 2 = 6\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{2x} - 1 = 4e^{-2x} + 2\); multiply by \(e^{2x}\): \((e^{2x})^2 - 3e^{2x} - 4 = 0\) | M1 A1 | Shown |
| Answer | Marks |
|---|---|
| \((e^{2x} - 4)(e^{2x} + 1) = 0\) | M1 |
| \(e^{2x} = 4\) (since \(e^{2x} > 0\)) | A1 |
| \(2x = \ln 4 \Rightarrow x = \frac{\ln 4}{2} = \ln 2\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(\int_0^{\ln 2}\left[(4e^{-2x}+2)-(e^{2x}-1)\right]dx\) | M1 |
| \(= \int_0^{\ln 2}(4e^{-2x} - e^{2x} + 3)\,dx\) | A1 |
| \(= \left[-2e^{-2x} - \frac{e^{2x}}{2} + 3x\right]_0^{\ln 2}\) | A1 |
| \(= \left(-\frac{1}{2} - 2 + 3\ln 2\right) - \left(-2 - \frac{1}{2}\right) = 3\ln 2\) | M1 A1 |
# Question 8:
## Part (a)
| Stretch parallel to $x$-axis, scale factor $\frac{1}{2}$ | B1 B1 | |
| Translation by $\begin{pmatrix}0\\-1\end{pmatrix}$ | B1 B1 | |
## Part (b)
| $A = (0, 6)$ since $y = 4e^0 + 2 = 6$ | B1 | |
## Part (c)(i)
| $e^{2x} - 1 = 4e^{-2x} + 2$; multiply by $e^{2x}$: $(e^{2x})^2 - 3e^{2x} - 4 = 0$ | M1 A1 | Shown |
## Part (c)(ii)
| $(e^{2x} - 4)(e^{2x} + 1) = 0$ | M1 | |
| $e^{2x} = 4$ (since $e^{2x} > 0$) | A1 | |
| $2x = \ln 4 \Rightarrow x = \frac{\ln 4}{2} = \ln 2$ | M1 A1 | |
## Part (d)
| $\int_0^{\ln 2}\left[(4e^{-2x}+2)-(e^{2x}-1)\right]dx$ | M1 | |
| $= \int_0^{\ln 2}(4e^{-2x} - e^{2x} + 3)\,dx$ | A1 | |
| $= \left[-2e^{-2x} - \frac{e^{2x}}{2} + 3x\right]_0^{\ln 2}$ | A1 | |
| $= \left(-\frac{1}{2} - 2 + 3\ln 2\right) - \left(-2 - \frac{1}{2}\right) = 3\ln 2$ | M1 A1 | |8 The diagram shows the curves $y = \mathrm { e } ^ { 2 x } - 1$ and $y = 4 \mathrm { e } ^ { - 2 x } + 2$.\\
\includegraphics[max width=\textwidth, alt={}, center]{33ca7e6d-b9eb-46be-b5b0-c5685212d7ff-6_958_1492_372_242}
The curve $y = 4 \mathrm { e } ^ { - 2 x } + 2$ crosses the $y$-axis at the point $A$ and the curves intersect at the point $B$.
\begin{enumerate}[label=(\alph*)]
\item Describe a sequence of two geometrical transformations that maps the graph of $y = \mathrm { e } ^ { x }$ onto the graph of $y = \mathrm { e } ^ { 2 x } - 1$.
\item Write down the coordinates of the point $A$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of the point $B$ satisfies the equation
$$\left( \mathrm { e } ^ { 2 x } \right) ^ { 2 } - 3 \mathrm { e } ^ { 2 x } - 4 = 0$$
\item Hence find the exact value of the $x$-coordinate of the point $B$.
\end{enumerate}\item Find the exact value of the area of the shaded region bounded by the curves $y = \mathrm { e } ^ { 2 x } - 1$ and $y = 4 \mathrm { e } ^ { - 2 x } + 2$ and the $y$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2010 Q8 [15]}}