| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| 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 an exponential equation, and (ii) computing a definite integral using reverse chain rule. Both parts are routine applications of core P2 methods with no novel problem-solving required, making it 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 |
|---|---|---|
| (i) Differentiate to obtain expression of form \(ke^{3x} + m\) | M1 | |
| Obtain correct \(2e^{3x} - 6\) | A1 | |
| Equate attempt at first derivative to zero and attempt solution | DM1 | |
| Obtain \(\frac{1}{3} x = \ln 3\) or equivalent | A1 | |
| Conclude \(x = \ln 9\) or \(a = 9\) | A1 | [5] |
| (ii) Integrate to obtain expression of form \(ae^{3x} + bx^2 + cx\) | M1 | |
| Obtain correct \(8e^{3x} - 3x^2 + 3x\) | A1 | |
| Substitute correct limits and attempt simplification | DM1 | |
| Obtain \(8e - 14\) | A1 | [4] |
**(i)** Differentiate to obtain expression of form $ke^{3x} + m$ | M1 | |
| Obtain correct $2e^{3x} - 6$ | A1 | |
| Equate attempt at first derivative to zero and attempt solution | DM1 | |
| Obtain $\frac{1}{3} x = \ln 3$ or equivalent | A1 | |
| Conclude $x = \ln 9$ or $a = 9$ | A1 | [5] |
**(ii)** Integrate to obtain expression of form $ae^{3x} + bx^2 + cx$ | M1 | |
| Obtain correct $8e^{3x} - 3x^2 + 3x$ | A1 | |
| Substitute correct limits and attempt simplification | DM1 | |
| Obtain $8e - 14$ | A1 | [4] |
5\\
\includegraphics[max width=\textwidth, alt={}, center]{0a45a806-007f-4840-85e7-16d4c1a2c599-3_528_757_251_694}
The diagram shows the curve $y = 4 e ^ { \frac { 1 } { 2 } x } - 6 x + 3$ and its minimum point $M$.\\
(i) Show that the $x$-coordinate of $M$ can be written in the form $\ln a$, where the value of $a$ is to be stated.\\
(ii) Find the exact value of the area of the region enclosed by the curve and the lines $x = 0 , x = 2$ and $y = 0$.
\hfill \mbox{\textit{CAIE P2 2012 Q5 [9]}}