| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area with trigonometric functions |
| Difficulty | Standard +0.3 This is a straightforward application of finding intersection points and computing area between curves using integration. Part (a) requires solving a trigonometric equation (likely equating two functions), and part (b) involves standard definite integration. While it requires multiple steps, the techniques are routine for P3 level with no novel problem-solving insight needed. |
| Spec | 1.05o Trigonometric equations: solve in given intervals1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct use of product rule to differentiate | *M1 | \(2\cos 2x(1+\sin 2x)+\sin 2x\times 2\cos 2x\), or \(2\cos^2 x - 2\sin^2 x + 8\sin x\cos^3 x - 8\cos x\sin^3 x\); all terms needed but could have errors in coefficients |
| Obtain \(2\cos 2x + 4\cos 2x\sin 2x\) | A1 | OE |
| Equate derivative to zero and solve for \(2x\) or \(x\) | DM1 | \(2\cos 2x(1+2\sin 2x)=0 \Rightarrow x = \frac{1}{2}\sin^{-1}\left(-\frac{1}{2}\right)\); condone if they only consider \(\cos 2x = 0\) |
| Obtain \(x=\frac{7}{12}\pi,\ y=-\frac{1}{4}\) | A1 | Mark degrees as misread; the question asks for an exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Use correct double angle formula to integrate, or use integration by parts and correct double angle formula | M1 | \(\int\sin 2x + \frac{1-\cos 4x}{2}\,dx\); or \(-\frac{1}{2}\cos 2x(1+\sin 2x)+\int\frac{1+\cos 4x}{2}\,dx\) |
| Obtain \(-\frac{1}{2}\cos 2x + \frac{x}{2} - \frac{1}{8}\sin 4x\ (+C)\) | A1 | Or \(-\frac{1}{2}\cos 2x - \frac{1}{2}\cos 2x\sin 2x + \frac{x}{2} + \frac{1}{8}\sin 4x\ (+C)\) |
| Use limits \(0\) and \(\frac{1}{2}\pi\) correctly in a solution containing \(p\cos 2x\) and \(q\sin 4x\) | M1 | \(\frac{1}{2}+\frac{1}{4}\pi - 0 + \frac{1}{2} - 0 + 0\) |
| Obtain \(\frac{1}{4}\pi + 1\) | A1 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct use of product rule to differentiate | *M1 | $2\cos 2x(1+\sin 2x)+\sin 2x\times 2\cos 2x$, or $2\cos^2 x - 2\sin^2 x + 8\sin x\cos^3 x - 8\cos x\sin^3 x$; all terms needed but could have errors in coefficients |
| Obtain $2\cos 2x + 4\cos 2x\sin 2x$ | A1 | OE |
| Equate derivative to zero and solve for $2x$ or $x$ | DM1 | $2\cos 2x(1+2\sin 2x)=0 \Rightarrow x = \frac{1}{2}\sin^{-1}\left(-\frac{1}{2}\right)$; condone if they only consider $\cos 2x = 0$ |
| Obtain $x=\frac{7}{12}\pi,\ y=-\frac{1}{4}$ | A1 | Mark degrees as misread; the question asks for an exact answer |
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## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct double angle formula to integrate, or use integration by parts and correct double angle formula | M1 | $\int\sin 2x + \frac{1-\cos 4x}{2}\,dx$; or $-\frac{1}{2}\cos 2x(1+\sin 2x)+\int\frac{1+\cos 4x}{2}\,dx$ |
| Obtain $-\frac{1}{2}\cos 2x + \frac{x}{2} - \frac{1}{8}\sin 4x\ (+C)$ | A1 | Or $-\frac{1}{2}\cos 2x - \frac{1}{2}\cos 2x\sin 2x + \frac{x}{2} + \frac{1}{8}\sin 4x\ (+C)$ |
| Use limits $0$ and $\frac{1}{2}\pi$ correctly in a solution containing $p\cos 2x$ and $q\sin 4x$ | M1 | $\frac{1}{2}+\frac{1}{4}\pi - 0 + \frac{1}{2} - 0 + 0$ |
| Obtain $\frac{1}{4}\pi + 1$ | A1 | |\begin{enumerate}[label=(\alph*)]
\item Given that the $x$-coordinate of $M$ lies in the interval $\frac { 1 } { 2 } \pi < x < \frac { 3 } { 4 } \pi$, find the exact coordinates of $M$.\\
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-10_2718_35_107_2012}\\
\includegraphics[max width=\textwidth, alt={}, center]{656df2a8-fc4d-49f3-a649-746103b4576e-11_2725_35_99_20}
\item Find the exact area of the region $R$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2024 Q6 [8]}}