| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area with trigonometric functions |
| Difficulty | Challenging +1.2 This question requires finding intersection points of two trigonometric functions, setting up the integral of their difference, and applying trigonometric identities to simplify before integrating. While it involves multiple steps and trigonometric manipulation, it follows a standard 'area between curves' template with straightforward integration once simplified. The main challenge is algebraic manipulation of trig expressions rather than novel problem-solving, placing it moderately above average difficulty. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} - \sin 2x \cos x = \sin x \cos 2x\) | M1 (3.1a) | |
| \(\sin 3x = \frac{1}{2}\) oe | M1 (2.1) | from compound angle formula, allow sign errors only; *or* \(4\sin^3 x - 3\sin x + \frac{1}{2} = 0\) oe |
| \(x = \frac{\pi}{18}\) and \(x = \frac{5\pi}{18}\) | A1, A1 (3.2a, 1.1) | A1 for each; \(0.17453\ldots\) A1, \(0.87266\ldots\) A1, to 2 or more sf |
| \(\pm \int \left(\sin x \cos 2x - \left(\frac{1}{2} - \sin 2x \cos x\right)\right) dx\) oe | M1 (1.1) | ignore limits |
| \(F[x] = -\frac{x}{2} - \frac{\cos 3x}{3}\) | A1 (1.1) | allow the positive of this; \(\pm\left(-\frac{4}{3}\cos^3 x + \cos x - \frac{x}{2}\right)\) oe or \(\pm(-\frac{1}{3}\cos 2x \cos x + \frac{1}{3}\sin x \sin 2x - \frac{1}{2}x)\) oe for A1 |
| \(F\left[\frac{5\pi}{18}\right] - F\left[\frac{\pi}{18}\right]\) | M1 (1.1) | \(F[x]\) must be one of the correct forms; \(F[0.87266] - F[0.17453]\) for M1 |
| \(\frac{\sqrt{3}}{3} - \frac{\pi}{9}\) or \(\frac{3\sqrt{3}-\pi}{9}\) cao | A1 (3.2a) | |
| [8] |
# Question 14:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} - \sin 2x \cos x = \sin x \cos 2x$ | M1 (3.1a) | |
| $\sin 3x = \frac{1}{2}$ oe | M1 (2.1) | from compound angle formula, allow sign errors only; *or* $4\sin^3 x - 3\sin x + \frac{1}{2} = 0$ oe |
| $x = \frac{\pi}{18}$ and $x = \frac{5\pi}{18}$ | A1, A1 (3.2a, 1.1) | A1 for each; $0.17453\ldots$ A1, $0.87266\ldots$ A1, to 2 or more sf |
| $\pm \int \left(\sin x \cos 2x - \left(\frac{1}{2} - \sin 2x \cos x\right)\right) dx$ oe | M1 (1.1) | ignore limits |
| $F[x] = -\frac{x}{2} - \frac{\cos 3x}{3}$ | A1 (1.1) | allow the positive of this; $\pm\left(-\frac{4}{3}\cos^3 x + \cos x - \frac{x}{2}\right)$ oe or $\pm(-\frac{1}{3}\cos 2x \cos x + \frac{1}{3}\sin x \sin 2x - \frac{1}{2}x)$ oe for A1 |
| $F\left[\frac{5\pi}{18}\right] - F\left[\frac{\pi}{18}\right]$ | M1 (1.1) | $F[x]$ must be one of the correct forms; $F[0.87266] - F[0.17453]$ for M1 |
| $\frac{\sqrt{3}}{3} - \frac{\pi}{9}$ or $\frac{3\sqrt{3}-\pi}{9}$ cao | A1 (3.2a) | |
| **[8]** | | |
---14 In this question you must show detailed reasoning.
Fig. 14 shows the graphs of $y = \sin x \cos 2 x$ and $y = \frac { 1 } { 2 } - \sin 2 x \cos x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cea67565-8074-4703-8e1a-09b98e380baf-16_647_898_404_233}
\captionsetup{labelformat=empty}
\caption{Fig. 14}
\end{center}
\end{figure}
Use integration to find the area between the two curves, giving your answer in an exact form.
\hfill \mbox{\textit{OCR MEI Paper 2 2020 Q14 [8]}}