| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| 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 x cos 2x (routine but with chain rule). Part (b) requires integration by parts of x cos 2x, which is a standard technique but requires careful execution and finding correct limits where y=0. The combination of differentiation and integration by parts in one question, plus exact answers, elevates this slightly above average difficulty. |
| Spec | 1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use the product rule correctly on \(y=x\cos 2x\) | M1 | \(dx/dx\cos 2x + x\,d/dx(\cos 2x)\) attempted |
| Obtain the correct derivative in any form | A1 | e.g. \(\cos 2x - 2x\sin 2x\). If \(\cos 2x + x - 2\sin 2x\), not recovered, max M1A0A1FTA0 but can recover for full marks by seeing correct substitution. |
| Obtain \(y=-\frac{\pi}{2}\) and \(\frac{dy}{dx}=-1\) when \(x=\frac{\pi}{2}\) | A1 FT | FT their \(\frac{dy}{dx}\) with \(x=\frac{\pi}{2}\) substituted |
| Obtain answer \(x+y=0\) | A1 | OE CWO. Need to see \(y\) and \(dy/dx\) at \(x=\frac{\pi}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Integrate by parts and reach \(ax\sin 2x + b\int\sin 2x\,dx\) | *M1 | |
| Obtain \(\frac{1}{2}x\sin 2x - \frac{1}{2}\int\sin 2x\,dx\) | A1 | OE |
| Complete integration and obtain \(\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x\) | A1 | OE |
| Use limits of \(x=0\) and \(x=\frac{\pi}{4}\) in correct order, having integrated twice to obtain \(ax\sin 2x + c\cos 2x\) | DM1 | If correct, \(\frac{1}{2}\left(\frac{\pi}{4}\right)\sin\frac{2\pi}{4}+\frac{1}{4}\cos\frac{2\pi}{4}-\frac{1}{4}\cos 0\). Max one substitution error. |
| Obtain answer \(\frac{\pi}{8}-\frac{1}{4}\) or exact simplified two-term equivalent | A1 | ISW. Accept \(\frac{\pi-2}{8}\). Accept \(\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x\) then final answer. |
## Question 10(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use the product rule correctly on $y=x\cos 2x$ | M1 | $dx/dx\cos 2x + x\,d/dx(\cos 2x)$ attempted |
| Obtain the correct derivative in any form | A1 | e.g. $\cos 2x - 2x\sin 2x$. If $\cos 2x + x - 2\sin 2x$, not recovered, max M1A0A1FTA0 but can recover for full marks by seeing correct substitution. |
| Obtain $y=-\frac{\pi}{2}$ and $\frac{dy}{dx}=-1$ when $x=\frac{\pi}{2}$ | A1 FT | FT their $\frac{dy}{dx}$ with $x=\frac{\pi}{2}$ substituted |
| Obtain answer $x+y=0$ | A1 | OE CWO. Need to see $y$ and $dy/dx$ at $x=\frac{\pi}{2}$ |
**Total: 4 marks**
---
## Question 10(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Integrate by parts and reach $ax\sin 2x + b\int\sin 2x\,dx$ | *M1 | |
| Obtain $\frac{1}{2}x\sin 2x - \frac{1}{2}\int\sin 2x\,dx$ | A1 | OE |
| Complete integration and obtain $\frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x$ | A1 | OE |
| Use limits of $x=0$ and $x=\frac{\pi}{4}$ in correct order, having integrated twice to obtain $ax\sin 2x + c\cos 2x$ | DM1 | If correct, $\frac{1}{2}\left(\frac{\pi}{4}\right)\sin\frac{2\pi}{4}+\frac{1}{4}\cos\frac{2\pi}{4}-\frac{1}{4}\cos 0$. Max one substitution error. |
| Obtain answer $\frac{\pi}{8}-\frac{1}{4}$ or exact simplified two-term equivalent | A1 | ISW. Accept $\frac{\pi-2}{8}$. Accept $\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x$ then final answer. |
**Total: 5 marks**
---10\\
\includegraphics[max width=\textwidth, alt={}, center]{a49b720b-f8d2-42ff-b147-5d676993aa4c-16_611_689_274_721}
The diagram shows the curve $y = x \cos 2 x$, for $x \geqslant 0$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the tangent to the curve at the point where $x = \frac { 1 } { 2 } \pi$.
\item Find the exact area of the shaded region shown in the diagram, bounded by the curve and the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q10 [9]}}