| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | March |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.8 Part (a) requires product rule differentiation of x³ln(x) and solving for the minimum, which is straightforward. Part (b) requires integration by parts of x³ln(x), a multi-step technique that students often find challenging. The combination of differentiation, finding stationary points, and integration by parts in one question elevates this above average difficulty, though it remains a standard Pure 3 exercise without requiring novel insight. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the product rule correctly | *M1 | \(x^3\frac{d}{dx}(\ln x) + \frac{d}{dx}(x^3)\ln x\) |
| Obtain the correct derivative in any form | A1 | e.g. \(\frac{x^3}{x} + 3x^2\ln x\) |
| Equate derivative to zero and solve exactly for \(x\) | DM1 | Reaching \(x = e^a\) |
| Obtain answer \(\left(\frac{1}{\sqrt[3]{e}}, -\frac{1}{3e}\right)\) or exact equivalent | A1 | ISW |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate by parts and reach \(ax^4\ln x + b\int(x^4/x)\,dx\) | *M1 | |
| Obtain \(\frac{x^4}{4}\ln x - \frac{1}{4}\int(x^4/x)\,dx\) | A1 | OE |
| Complete integration and obtain \(\frac{x^4}{4}\ln x - \frac{x^4}{16}\) | A1 | OE |
| Use limits \(x=\frac{1}{2}\) and \(x=1\) in correct order, having integrated twice | DM1 | Correct substitution \([(1/4)\ln 1\) or \(0 - 1/16] - [(1/64)\ln(1/2) - (1/16)^2]\) or minus this value CWO. Allow omission of \((1/4)\ln 1\) or \(0\) |
| Obtain \(\frac{15}{256} - \frac{1}{64}\ln 2\) or exact equivalent | A1 | |
| Total: 5 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the product rule correctly | *M1 | $x^3\frac{d}{dx}(\ln x) + \frac{d}{dx}(x^3)\ln x$ |
| Obtain the correct derivative in any form | A1 | e.g. $\frac{x^3}{x} + 3x^2\ln x$ |
| Equate derivative to zero and solve exactly for $x$ | DM1 | Reaching $x = e^a$ |
| Obtain answer $\left(\frac{1}{\sqrt[3]{e}}, -\frac{1}{3e}\right)$ or exact equivalent | A1 | ISW |
| **Total: 4** | | |
## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate by parts and reach $ax^4\ln x + b\int(x^4/x)\,dx$ | *M1 | |
| Obtain $\frac{x^4}{4}\ln x - \frac{1}{4}\int(x^4/x)\,dx$ | A1 | OE |
| Complete integration and obtain $\frac{x^4}{4}\ln x - \frac{x^4}{16}$ | A1 | OE |
| Use limits $x=\frac{1}{2}$ and $x=1$ in correct order, having integrated twice | DM1 | Correct substitution $[(1/4)\ln 1$ or $0 - 1/16] - [(1/64)\ln(1/2) - (1/16)^2]$ or minus this value CWO. Allow omission of $(1/4)\ln 1$ or $0$ |
| Obtain $\frac{15}{256} - \frac{1}{64}\ln 2$ or exact equivalent | A1 | |
| **Total: 5** | | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{8c26235b-c78c-40d8-9e8e-213dc1311186-12_437_686_274_719}
The diagram shows the curve $y = x ^ { 3 } \ln x$, for $x > 0$, and its minimum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Find the exact area of the shaded region bounded by the curve, the $x$-axis and the line $x = \frac { 1 } { 2 }$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q8 [9]}}