October 2020 Paper 2 Q14
14 In this question you must show detailed reasoning.
Fig. 14 shows the graphs of \(y = \sin x\cos 2x\) and \(y = \frac{1}{2} - \sin 2x\cos x\).
Use integration to find the area between the two curves, giving your answer in an exact form. [8]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{2} - \sin 2x\cos x = \sin x\cos 2x\) | M1 | 3.1a |
| \(\sin 3x = \dfrac{1}{2}\) oe | M1 | 2.1 |
| \(x = \dfrac{\pi}{18}\) and \(x = \dfrac{5\pi}{18}\) | A1 A1 | 3.2a 1.1 |
| \(\pm\displaystyle\int \left(\sin x\cos 2x - \left(\frac{1}{2} - \sin 2x\cos x\right)\right)\mathrm{d}x\) oe | M1 | 1.1 |
| \(\mathrm{F}[x] = -\dfrac{x}{2} - \dfrac{\cos 3x}{3}\) | A1 | 1.1 |
| \(\mathrm{F}\left[\frac{5\pi}{18}\right] - \mathrm{F}\left[\frac{\pi}{18}\right]\) | M1 | 1.1 |
| \(\dfrac{\sqrt{3}}{3} - \dfrac{\pi}{9}\) or \(\dfrac{3\sqrt{3} - \pi}{9}\) cao | A1 | 3.2a |
| [8] |
Notes
M1: from compound angle formula
allow sign errors only
or \(4\sin^3 x - 3\sin x + \frac{1}{2} = 0\) oe
A1 A1: A1 for each
0.17453… A1
0.87266… A1
to 2 or more sf
M1: ignore limits
A1: allow the positive of this
\(\pm\left(-\frac{4}{3}\cos^3 x + \cos x - \frac{x}{2}\right)\) oe or \(\pm\left(-\frac{1}{3}\cos 2x\cos x + \frac{1}{3}\sin x\sin 2x - \frac{1}{2}x\right)\) oe for A1
M1: \(\mathrm{F}[x]\) must be one of the correct forms
\(\mathrm{F}[0.87266] - \mathrm{F}[0.17453]\) for M1