| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Stationary points then area/volume |
| Difficulty | Standard +0.8 Part (a) requires finding stationary points by differentiating a product (sin x cos 2x), applying product and chain rules, solving a trigonometric equation, and selecting the correct root. Part (b) requires integration by parts on sin x cos 2x, which involves careful manipulation and potentially multiple applications. The combination of non-trivial differentiation, equation solving, and integration by parts with trigonometric functions makes this moderately challenging, though each technique is standard for Further Maths Pure. |
| Spec | 1.08f Area between two curves: using integration1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct product rule | \*M1 | As far as \(p\cos x\cos 2x + q\sin x\sin 2x\) or full working (\(u\), \(v\), \(du/dx\), \(dv/dx\)) shown. |
| Obtain \(\frac{dy}{dx} = \cos x\cos 2x - 2\sin x\sin 2x\) | A1 | OE |
| Equate derivative to zero and use correct double angle formulae | DM1 | Allow if only have one double angle in their derivative. |
| Obtain \(\cos x(1 - 6\sin^2 x) = 0\) or equivalent | A1 | e.g. \(\cos x(6\cos^2 x - 5) = 0\), \(5\tan^2 x = 1\). Simplified but not necessarily factorised - like terms must be collected. |
| Obtain \(a = 0.42\) | A1 | Only. Accept \(x = 0.42\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct double angle formula | \*M1 | |
| Obtain \(\sin x - 2\sin^3 x\) or equivalent | A1 | |
| Use correct chain rule or product rule to differentiate and equate the derivative to zero | DM1 | |
| Obtain \(\cos x(1 - 6\sin^2 x) = 0\) | A1 | OE |
| Obtain \(a = 0.42\) | A1 | Only. Accept \(x = 0.42\). |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use double angle formula and obtain \(p\cos^3 x + q\cos x\) correctly | \*M1 | e.g. from \(\int 2\cos^2 x\sin x - \sin x\ dx\) |
| Obtain \(\pm\left(-\frac{2}{3}\cos^3 x + \cos x\right)\) | A1 | Correct for *their* integral. |
| Correct use of limits \(\frac{1}{4}\pi\) and \(\frac{3}{4}\pi\) (or use double the integral from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\)) | DM1 | OE: \(\pm\left(-\frac{2}{3}\left[\left(\frac{-1}{\sqrt{2}}\right)^3 - \left(\frac{1}{\sqrt{2}}\right)^3\right] - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\) |
| Obtain \(\frac{2\sqrt{2}}{3}\) | A1 | Or simplified exact equivalent. Final answer must be positive. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use integration by parts twice and obtain \(r\cos x\cos 2x + s\sin x\sin 2x\) | \*M1 | Seen, not just implied. |
| Obtain \(\frac{1}{3}\cos x\cos 2x + \frac{2}{3}\sin x\sin 2x\) | A1 | Accept \(\pm\) (correct for *their* integral). |
| Correct use of limits \(\frac{1}{4}\pi\) and \(\frac{3}{4}\pi\) (or use double the integral from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\)) | DM1 | OE: \(\pm\frac{1}{3}\left(0 + 2\times\frac{1}{\sqrt{2}}\times -1 - 0 - 2\times\frac{1}{\sqrt{2}}\times 1\right)\) |
| Obtain \(\frac{2\sqrt{2}}{3}\) | A1 | Or simplified exact equivalent. Final answer must be positive. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use factor formula and integrate to obtain \(g\cos 3x + h\cos x\) | \*M1 | \(\int\frac{1}{2}(\sin 3x - \sin x)dx\) |
| Obtain \(\pm\left(-\frac{1}{6}\cos 3x + \frac{1}{2}\cos x\right)\) | A1 | Correct for *their* integral. |
| Correct use of limits \(\frac{1}{4}\pi\) and \(\frac{3}{4}\pi\) (or use double the integral from \(\frac{1}{4}\pi\) to \(\frac{1}{2}\pi\)) | DM1 | OE: \(\mp\frac{1}{\sqrt{2}}\left(\frac{1}{6} + \frac{1}{2} + \frac{1}{6} + \frac{1}{2}\right)\) |
| Obtain \(\frac{2\sqrt{2}}{3}\) | A1 | Or exact equivalent. Final answer must be positive. |
| Total | 4 |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct product rule | \*M1 | As far as $p\cos x\cos 2x + q\sin x\sin 2x$ or full working ($u$, $v$, $du/dx$, $dv/dx$) shown. |
| Obtain $\frac{dy}{dx} = \cos x\cos 2x - 2\sin x\sin 2x$ | A1 | OE |
| Equate derivative to zero and use correct double angle formulae | DM1 | Allow if only have one double angle in their derivative. |
| Obtain $\cos x(1 - 6\sin^2 x) = 0$ or equivalent | A1 | e.g. $\cos x(6\cos^2 x - 5) = 0$, $5\tan^2 x = 1$. Simplified but not necessarily factorised - like terms must be collected. |
| Obtain $a = 0.42$ | A1 | Only. Accept $x = 0.42$. |
**Alternative method for question 9(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct double angle formula | \*M1 | |
| Obtain $\sin x - 2\sin^3 x$ or equivalent | A1 | |
| Use correct chain rule or product rule to differentiate and equate the derivative to zero | DM1 | |
| Obtain $\cos x(1 - 6\sin^2 x) = 0$ | A1 | OE |
| Obtain $a = 0.42$ | A1 | Only. Accept $x = 0.42$. |
| **Total** | **5** | |
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use double angle formula and obtain $p\cos^3 x + q\cos x$ correctly | \*M1 | e.g. from $\int 2\cos^2 x\sin x - \sin x\ dx$ |
| Obtain $\pm\left(-\frac{2}{3}\cos^3 x + \cos x\right)$ | A1 | Correct for *their* integral. |
| Correct use of limits $\frac{1}{4}\pi$ and $\frac{3}{4}\pi$ (or use double the integral from $\frac{1}{4}\pi$ to $\frac{1}{2}\pi$) | DM1 | OE: $\pm\left(-\frac{2}{3}\left[\left(\frac{-1}{\sqrt{2}}\right)^3 - \left(\frac{1}{\sqrt{2}}\right)^3\right] - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ |
| Obtain $\frac{2\sqrt{2}}{3}$ | A1 | Or simplified exact equivalent. Final answer must be positive. |
**Alternative method 1 for question 9(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use integration by parts **twice** and obtain $r\cos x\cos 2x + s\sin x\sin 2x$ | \*M1 | Seen, not just implied. |
| Obtain $\frac{1}{3}\cos x\cos 2x + \frac{2}{3}\sin x\sin 2x$ | A1 | Accept $\pm$ (correct for *their* integral). |
| Correct use of limits $\frac{1}{4}\pi$ and $\frac{3}{4}\pi$ (or use double the integral from $\frac{1}{4}\pi$ to $\frac{1}{2}\pi$) | DM1 | OE: $\pm\frac{1}{3}\left(0 + 2\times\frac{1}{\sqrt{2}}\times -1 - 0 - 2\times\frac{1}{\sqrt{2}}\times 1\right)$ |
| Obtain $\frac{2\sqrt{2}}{3}$ | A1 | Or simplified exact equivalent. Final answer must be positive. |
**Alternative method 2 for question 9(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Use factor formula and integrate to obtain $g\cos 3x + h\cos x$ | \*M1 | $\int\frac{1}{2}(\sin 3x - \sin x)dx$ |
| Obtain $\pm\left(-\frac{1}{6}\cos 3x + \frac{1}{2}\cos x\right)$ | A1 | Correct for *their* integral. |
| Correct use of limits $\frac{1}{4}\pi$ and $\frac{3}{4}\pi$ (or use double the integral from $\frac{1}{4}\pi$ to $\frac{1}{2}\pi$) | DM1 | OE: $\mp\frac{1}{\sqrt{2}}\left(\frac{1}{6} + \frac{1}{2} + \frac{1}{6} + \frac{1}{2}\right)$ |
| Obtain $\frac{2\sqrt{2}}{3}$ | A1 | Or exact equivalent. Final answer must be positive. |
| **Total** | **4** | |
---9\\
\includegraphics[max width=\textwidth, alt={}, center]{39cf66af-095b-404b-a38c-0aa7684c4a27-14_428_787_274_671}
The diagram shows the curve $y = \sin x \cos 2 x$, for $0 \leqslant x \leqslant \pi$, and a maximum point $M$, where $x = a$. The shaded region between the curve and the $x$-axis is denoted by $R$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $a$ correct to 2 decimal places.
\item Find the exact area of the region $R$, giving your answer in simplified form.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q9 [9]}}