| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find stationary points |
| Difficulty | Moderate -0.3 Part (i) requires finding the minimum by setting dy/dx = 0 and solving e^x = 2e^{-2x}, which is straightforward differentiation and algebra. Part (ii) is a standard definite integration of exponential functions with given limits. Both parts are routine A-level calculus exercises with clear methods and no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Differentiate to obtain \(e^x - 8e^{-2x}\); Use correct process to solve equation of form \(ae^x + be^{-2x} = 0\); Confirm given answer \(\ln 2\) correctly | B1, M1, A1 | [3] |
| Integrate to obtain expression of form \(pe^x + qe^{-2x}\); Obtain correct \(e^x - 2e^{-2x}\); Apply both limits correctly; Confirm given answer \(\frac{5}{2}\) | M1, A1, M1 depM, A1 | [4] |
| Answer | Marks | Guidance |
|--------|-------|----------|
| Differentiate to obtain $e^x - 8e^{-2x}$; Use correct process to solve equation of form $ae^x + be^{-2x} = 0$; Confirm given answer $\ln 2$ correctly | B1, M1, A1 | [3] |
| Integrate to obtain expression of form $pe^x + qe^{-2x}$; Obtain correct $e^x - 2e^{-2x}$; Apply both limits correctly; Confirm given answer $\frac{5}{2}$ | M1, A1, M1 depM, A1 | [4] |
---\includegraphics{figure_4}
The diagram shows the curve $y = e^x + 4e^{-2x}$ and its minimum point $M$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of $M$ is $\ln 2$. [3]
\item The region shaded in the diagram is enclosed by the curve and the lines $x = 0$, $x = \ln 2$ and $y = 0$. Use integration to show that the area of the shaded region is $\frac{5}{2}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2015 Q4 [7]}}