| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2013 |
| 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 Part (a) is a straightforward integration of an exponential function using reverse chain rule, requiring only substitution of limits. Part (b) requires differentiation using quotient rule, setting equal to zero, and solving a simple trigonometric equation. Both parts are standard techniques with no novel insight required, making this slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Obtain one term of form \(ke^{2x-1}\) with any non-zero \(k\) | M1 | |
| Obtain correct integral \(x + \frac{1}{2}e^{2x-1}\) | A1 | |
| Substitute limits, giving exact values | M1 | |
| Correct answer \(\frac{1}{2}e^3 + 1\) | A1 | [4] |
| (b) Use product or quotient rule | M1✱ | |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1✱ | |
| dep | ||
| Obtain \(2x = 1\) | A1 | |
| Obtain \(x = \frac{\pi}{8}\) | A1 | [5] |
**(a)** Obtain one term of form $ke^{2x-1}$ with any non-zero $k$ | M1 |
Obtain correct integral $x + \frac{1}{2}e^{2x-1}$ | A1 |
Substitute limits, giving exact values | M1 |
Correct answer $\frac{1}{2}e^3 + 1$ | A1 | [4]
**(b)** Use product or quotient rule | M1✱ |
Obtain correct derivative in any form | A1 |
Equate derivative to zero and solve for $x$ | M1✱ |
dep |
Obtain $2x = 1$ | A1 |
Obtain $x = \frac{\pi}{8}$ | A1 | [5]7
\begin{enumerate}[label=(\alph*)]
\item Find the exact area of the region bounded by the curve $y = 1 + \mathrm { e } ^ { 2 x - 1 }$, the $x$-axis and the lines $x = \frac { 1 } { 2 }$ and $x = 2$.
\item \\
\includegraphics[max width=\textwidth, alt={}, center]{e3ee4932-8219-4332-9cd2-e7f835522469-3_469_719_397_753}
The diagram shows the curve $y = \frac { \mathrm { e } ^ { 2 x } } { \sin 2 x }$ for $0 < x < \frac { 1 } { 2 } \pi$, and its minimum point $M$. Find the exact $x$-coordinate of $M$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2013 Q7 [9]}}