C3 June 2007 Q6
6.
| Scheme | Marks |
|---|---|
| Complete method for \(R\): e.g. \(R\cos\alpha = 3\), \(R\sin\alpha = 2\), \(R = \sqrt{(3^2 + 2^2)}\) | M1 |
| \(R = \sqrt{13}\) or 3.61 (or more accurate) | A1 |
| Complete method for \(\tan\alpha = \dfrac{2}{3}\) [Allow \(\tan\alpha = \dfrac{3}{2}\)] | M1 |
| \(\alpha = 0.588\) (Allow \(33.7^\circ\)) | A1 |
| (4) |
Notes
1st M1 on Epen for correct method for R, even if found second
2nd M1 for correct method for \(\tan\alpha\)
No working at all: M1A1 for \(\sqrt{13}\), M1A1 for 0.588 or \(33.7^\circ\).
N.B. \(R\cos\alpha = 2\), \(R\sin\alpha = 3\) used, can still score M1A1 for R, but loses the A mark for \(\alpha\).
\(\cos\alpha = 3\), \(\sin\alpha = 2\): apply the same marking.
| Scheme | Marks |
|---|---|
| Greatest value \(= \left(\sqrt{13}\right)^4 = 169\) | M1, A1 |
| (2) |
Notes
M1 for realising \(\sin(x + \alpha) = \pm 1\), so finding \(R^4\).
| Scheme | Marks |
|---|---|
| \(\sin(x + 0.588) = \dfrac{1}{\sqrt{13}}\) \((= 0.27735\ldots)\) \(\sin(x + \text{their }\alpha) = \frac{1}{\text{their }R}\) | M1 |
| \((x + 0.588) = 0.281(03\ldots\) or \(16.1^\circ\) | A1 |
| \((x + 0.588) = \pi - 0.28103\ldots\) Must be \(\pi -\) their 0.281 or \(180^\circ -\) their \(16.1^\circ\) | M1 |
| or \((x + 0.588) = 2\pi + 0.28103\ldots\) Must be \(2\pi +\) their 0.281 or \(360^\circ +\) their \(16.1^\circ\) | M1 |
| \(x = 2.273\) or \(x = 5.976\) (awrt) Both (radians only) | A1 |
| If 0.281 or \(16.1^\circ\) not seen, correct answers imply this A mark | |
| (5) | |
| (11 marks) |
Notes
Working in mixed degrees/rads: first two marks available
Working consistently in degrees: Possible to score first 4 marks
[Degree answers, just for reference, Only are \(130.2^\circ\) and \(342.4^\circ\)]
Third M1 can be gained for candidate’s 0.281 – candidate’s \(0.588 + 2\pi\) or equiv. in degrees
One of the answers correct in radians or degrees implies the corresponding M mark.Alternative (c) (i)
| Squaring to form quadratic in \(\sin x\) or \(\cos x\) \([13\cos^2 x - 4\cos x - 8 = 0, \quad 13\sin^2 x - 6\sin x - 3 = 0]\) | M1 |
| Correct values for \(\cos x = 0.953\ldots,\ -0.646\); or \(\sin x = 0.767,\ 2.27\) awrt | A1 |
| For any one value of \(\cos x\) or \(\sin x\), correct method for two values of \(x\) | M1 |
| \(x = 2.273\) or \(x = 5.976\) (awrt) Both seen anywhere | A1 |
| Checking other values (0.307, 4.011 or 0.869, 3.449) and discarding | M1 |
Alternative (c) (ii)
| Squaring and forming equation of form \(a\cos 2x + b\sin 2x = c\) \(9\sin^2 x + 4\cos^2 x + 12\sin 2x = 1 \Rightarrow 12\sin 2x + 5\cos 2x = 11\) | |
| Setting up to solve using R formula e.g. \(13\cos(2x - 1.176) = 11\) | M1 |
| \((2x - 1.176) = \cos^{-1}\left(\dfrac{11}{13}\right) = 0.562(0\ldots \quad (\alpha)\) | A1 |
| \((2x - 1.176) = 2\pi - \alpha,\ 2\pi + \alpha, \ldots\ldots\) | M1 |
| \(x = 2.273\) or \(x = 5.976\) (awrt) Both seen anywhere | A1 |
| Checking other values and discarding | M1 |
Notes
(corrected from the printed mark scheme: Alt (c)(ii) is printed with \(\sqrt{13}\cos(2x - 1.176) = 11\) and \(\cos^{-1}\left(\dfrac{11}{\sqrt{13}}\right)\); here \(12\sin 2x + 5\cos 2x = 13\cos(2x - 1.176)\), so \(R = 13\), and \(\cos^{-1}\left(\dfrac{11}{13}\right) = 0.562\ldots\))