| 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 multi-part question requiring standard techniques: (i) differentiate and solve dy/dx=0 to find the minimum (routine calculus), (ii) integrate to find area under curve. Both parts use standard integrals (exponential and polynomial) with no complex manipulation. The 'show that' format and 'exact value' requirement add minimal difficulty. Slightly above average due to the multi-step nature and need for careful algebraic manipulation, but well within typical A-level expectations. |
| 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 \(8c - 14\) | 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 $8c - 14$ | A1 | [4]5\\
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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]}}