| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Area under curve using integration |
| Difficulty | Moderate -0.3 This is a straightforward integration question requiring standard exponential differentiation and integration. Part (a) involves routine differentiation of exponentials at a given point. Part (b) requires solving a simple exponential equation and computing a definite integral using standard results. While it has multiple parts, each step uses direct application of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to obtain form \(k_1e^{-x} + k_2e^{2x}\) | M1 | Where \(k_1k_2 \neq 0\), \(k_1 \neq 8\) and \(k_2 \neq -1\) |
| Obtain \(-8e^{-x} - 2e^{2x}\) | A1 | |
| Substitute \(x = 0\) to obtain \(-10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to find \(x\)-coordinate of \(B\) | M1 | \(8e^{-x} - e^{2x} = 0\) |
| Obtain \(e^{3x} = 8\) and hence \(x = \ln 2\) | A1 | AG so necessary detail needed. A0 if decimals used. |
| Integrate to obtain \(-8e^{-x} - \frac{1}{2}e^{2x}\) | B1 | |
| Use limits \(0\) and \(\ln 2\) correctly to find area | M1 | For integral of form \(k_3e^{-x} + k_4e^{2x}\) where \(k_3k_4 \neq 0\), \(k_1 \neq 8\) and \(k_2 \neq -1\) |
| Obtain \(\frac{5}{2}\) | A1 | OE |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain form $k_1e^{-x} + k_2e^{2x}$ | M1 | Where $k_1k_2 \neq 0$, $k_1 \neq 8$ and $k_2 \neq -1$ |
| Obtain $-8e^{-x} - 2e^{2x}$ | A1 | |
| Substitute $x = 0$ to obtain $-10$ | A1 | |
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to find $x$-coordinate of $B$ | M1 | $8e^{-x} - e^{2x} = 0$ |
| Obtain $e^{3x} = 8$ and hence $x = \ln 2$ | A1 | AG so necessary detail needed. A0 if decimals used. |
| Integrate to obtain $-8e^{-x} - \frac{1}{2}e^{2x}$ | B1 | |
| Use limits $0$ and $\ln 2$ correctly to find area | M1 | For integral of form $k_3e^{-x} + k_4e^{2x}$ where $k_3k_4 \neq 0$, $k_1 \neq 8$ and $k_2 \neq -1$ |
| Obtain $\frac{5}{2}$ | A1 | OE |3\\
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_776_483_310_769}
The diagram shows the curve with equation $y = 8 \mathrm { e } ^ { - x } - \mathrm { e } ^ { 2 x }$. The curve crosses the $y$-axis at the point $A$ and the $x$-axis at the point $B$. The shaded region is bounded by the curve and the two axes.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{76df3465-9617-4f2b-a8b7-f474b2817504-04_2715_35_141_2011}
\begin{center}
\end{center}
\item Show that the $x$-coordinate of $B$ is $\ln 2$ and hence find the area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2024 Q3 [8]}}