| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Integration with differentiation context |
| Difficulty | Standard +0.8 This question requires finding a maximum using the quotient rule (multi-step differentiation), then applying integration by parts with careful algebraic manipulation to prove an inequality. The proof in part (b) requires recognizing that the integral approaches a limit and showing it's bounded, which demands insight beyond routine application of techniques. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use quotient or product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for \(x\) | M1 | |
| Obtain \(x = \sqrt[4]{e}\) and \(y = \frac{1}{4e}\), or exact equivalents | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Commence integration and reach \(ax^{-3}\ln x + b\int x^{-3} \cdot \frac{1}{x}\,dx\) | *M1 | |
| Obtain \(-\frac{1}{3}x^{-3}\ln x + \frac{1}{3}\int x^{-3} \cdot \frac{1}{x}\,dx\) | A1 | OE |
| Complete integration and obtain \(-\frac{1}{3}x^{-3}\ln x - \frac{1}{9}x^{-3}\) | A1 | |
| Substitute limits correctly, having integrated twice | DM1 | |
| Obtain answer \(\frac{1}{9} - \frac{1}{3}a^{-3}\ln a - \frac{1}{9}a^{-3}\) | A1 | OE |
| Justify the given statement | A1 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use quotient or product rule | M1 | |
| Obtain correct derivative in any form | A1 | |
| Equate derivative to zero and solve for $x$ | M1 | |
| Obtain $x = \sqrt[4]{e}$ and $y = \frac{1}{4e}$, or exact equivalents | A1 | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Commence integration and reach $ax^{-3}\ln x + b\int x^{-3} \cdot \frac{1}{x}\,dx$ | *M1 | |
| Obtain $-\frac{1}{3}x^{-3}\ln x + \frac{1}{3}\int x^{-3} \cdot \frac{1}{x}\,dx$ | A1 | OE |
| Complete integration and obtain $-\frac{1}{3}x^{-3}\ln x - \frac{1}{9}x^{-3}$ | A1 | |
| Substitute limits correctly, having integrated twice | DM1 | |
| Obtain answer $\frac{1}{9} - \frac{1}{3}a^{-3}\ln a - \frac{1}{9}a^{-3}$ | A1 | OE |
| Justify the given statement | A1 | |8\\
\includegraphics[max width=\textwidth, alt={}, center]{1990cbac-d96f-4484-be4b-67dab35b3147-12_458_725_262_708}
The diagram shows the curve $y = \frac { \ln x } { x ^ { 4 } }$ and its maximum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item By using integration by parts, show that for all $a > 1 , \int _ { 1 } ^ { a } \frac { \ln x } { x ^ { 4 } } \mathrm {~d} x < \frac { 1 } { 9 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2021 Q8 [10]}}