| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Area under curve using integration |
| Difficulty | Standard +0.3 This is a straightforward two-part question requiring standard calculus techniques: (i) finding a stationary point by differentiation and solving e^(3x) = 8, and (ii) integrating standard exponential functions. Both parts are routine applications of core P2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | Differentiate to obtain \(e^x - 8e^{-2x}\) | B1 |
| Use correct process to solve equation of form \(ae^x + be^{-2x} = 0\) | M1 | |
| Confirm given answer \(\ln 2\) correctly | A1 | [3] |
| (ii) | Integrate to obtain expression of form \(pe^x + qe^{-2x}\) | M1 |
| Obtain correct \(e^x - 2e^{-2x}\) | A1 | |
| Apply both limits correctly | M1 depM | |
| Confirm given answer \(\frac{5}{2}\) | A1 | [4] |
(i) | Differentiate to obtain $e^x - 8e^{-2x}$ | B1 |
| Use correct process to solve equation of form $ae^x + be^{-2x} = 0$ | M1 |
| Confirm given answer $\ln 2$ correctly | A1 | [3]
(ii) | Integrate to obtain expression of form $pe^x + qe^{-2x}$ | M1 |
| Obtain correct $e^x - 2e^{-2x}$ | A1 |
| Apply both limits correctly | M1 depM |
| Confirm given answer $\frac{5}{2}$ | A1 | [4]
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The diagram shows the curve $y = \mathrm { e } ^ { x } + 4 \mathrm { e } ^ { - 2 x }$ and its minimum point $M$.\\
(i) Show that the $x$-coordinate of $M$ is $\ln 2$.\\
(ii) 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 $\frac { 5 } { 2 }$.
\hfill \mbox{\textit{CAIE P2 2015 Q4 [7]}}