| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.3 Part (a) requires a standard product rule differentiation and solving f'(x)=0, yielding coordinates (1/a, 1/(ae)). Part (b) is integration by parts with given limits—both are routine A-level techniques with straightforward execution. The question is slightly easier than average due to its predictable structure and standard methods. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use correct product rule | *M1 | Or equivalent. Condone incorrect chain rule. M0 if a value is used for \(a\) (not equivalent work). |
| Obtain correct derivative | A1 | E.g. \(\frac{\mathrm{d}y}{\mathrm{d}x} = -axe^{-ax} + e^{-ax}\) |
| Equate derivative to zero and solve for \(x\) | DM1 | |
| Obtain \(x = \frac{1}{a}\), \(y = \frac{1}{ae}\) | A1 | ISW. Or exact equivalent. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use integration by parts to obtain \(pxe^{-ax} + q\int e^{-ax}\mathrm{d}x\) | *M1 | Condone sign error in parts formula and omission of \(\mathrm{d}x\). M0 if a value is used for \(a\) (not equivalent work). |
| Obtain \(-\frac{1}{a}xe^{-ax} + \frac{1}{a}\int e^{-ax}\mathrm{d}x\) | A1 | OE |
| Complete integration to obtain \(-\frac{1}{a}xe^{-ax} - \frac{1}{a^2}e^{-ax}\) | A1 | OE |
| Correct use of limits \(0\) and \(\frac{2}{a}\) in an expression of the form \(rxe^{-ax} + se^{-ax}\) | DM1 | \(\left(\frac{2}{a^2}e^{-2} - \frac{1}{a^2}e^{-2} + 0 + \frac{1}{a^2}\right)\) |
| Obtain \(\frac{1}{a^2}\left(1 - 3e^{-2}\right)\) | A1 | ISW. Or simplified 2-term equivalent, e.g. \(\frac{e^2-3}{a^2e^2}\). |
| 5 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use correct product rule | *M1 | Or equivalent. Condone incorrect chain rule. M0 if a value is used for $a$ (not equivalent work). |
| Obtain correct derivative | A1 | E.g. $\frac{\mathrm{d}y}{\mathrm{d}x} = -axe^{-ax} + e^{-ax}$ |
| Equate derivative to zero and solve for $x$ | DM1 | |
| Obtain $x = \frac{1}{a}$, $y = \frac{1}{ae}$ | A1 | ISW. Or exact equivalent. |
| | **4** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use integration by parts to obtain $pxe^{-ax} + q\int e^{-ax}\mathrm{d}x$ | *M1 | Condone sign error in parts formula and omission of $\mathrm{d}x$. M0 if a value is used for $a$ (not equivalent work). |
| Obtain $-\frac{1}{a}xe^{-ax} + \frac{1}{a}\int e^{-ax}\mathrm{d}x$ | A1 | OE |
| Complete integration to obtain $-\frac{1}{a}xe^{-ax} - \frac{1}{a^2}e^{-ax}$ | A1 | OE |
| Correct use of limits $0$ and $\frac{2}{a}$ in an expression of the form $rxe^{-ax} + se^{-ax}$ | DM1 | $\left(\frac{2}{a^2}e^{-2} - \frac{1}{a^2}e^{-2} + 0 + \frac{1}{a^2}\right)$ |
| Obtain $\frac{1}{a^2}\left(1 - 3e^{-2}\right)$ | A1 | ISW. Or simplified 2-term equivalent, e.g. $\frac{e^2-3}{a^2e^2}$. |
| | **5** | |6\\
\includegraphics[max width=\textwidth, alt={}, center]{5eb2657c-ed74-4ed2-b8c4-08e9e0f657b5-08_351_1031_264_516}
The diagram shows the curve $\mathrm { y } = \mathrm { xe } ^ { - \mathrm { ax } }$, where $a$ is a positive constant, and its maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Find the exact value of $\int _ { 0 } ^ { \frac { 2 } { a } } x e ^ { - a x } d x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q6 [9]}}