C2 June 2011 Q7
7.
You must show clearly how you obtained your answers.
(6)| Scheme | Marks |
|---|---|
| (a) \(3\sin(x + 45^\circ) = 2\); \(0 \leqslant x \lt 360^\circ\) (b) \(2\sin^2 x + 2 = 7\cos x\); \(0 \leqslant x \lt 2\pi\) | |
| \(\sin(x + 45^\circ) = \dfrac{2}{3}\), so \((x + 45^\circ) = 41.8103\ldots\ (\alpha = 41.8103\ldots)\) \(\sin^{-1}\left(\dfrac{2}{3}\right)\) or awrt 41.8 or awrt \(0.73^c\) | M1 |
| So, \(x + 45^\circ = \{138.1897\ldots, 401.8103\ldots\}\) \(x + 45^\circ =\) either “\(180 -\) their \(\alpha\)” or “\(360^\circ +\) their \(\alpha\)” (\(\alpha\) could be in radians). | M1 |
| and \(x = \{93.1897\ldots, 356.8103\ldots\}\) Either awrt \(93.2^\circ\) or awrt \(356.8^\circ\) | A1 |
| Both awrt \(93.2^\circ\) and awrt \(356.8^\circ\) | A1 |
| [4] |
Notes
1st M1: can also be implied for \(x = \text{awrt } -3.2\)
2nd M1: for \(x + 45^\circ =\) either “\(180 -\) their \(\alpha\)” or “\(360^\circ +\) their \(\alpha\)”. This can be implied by later working. The candidate’s \(\alpha\) could also be in radians.
Note that this mark is not for \(x =\) either “\(180 -\) their \(\alpha\)” or “\(360^\circ +\) their \(\alpha\)”.
Note: Imply the first two method marks or award M1M1A1 for either awrt \(93.2^\circ\) or awrt \(356.8^\circ\).
Note: Candidates who apply the following incorrect working of \(3\sin(x + 45^\circ) = 2 \Rightarrow 3(\sin x + \sin 45) = 2\), etc will usually score M0M0A0A0.
If there are any EXTRA solutions inside the range \(0 \leqslant x \lt 360\) and the candidate would otherwise score FULL MARKS then withhold the final aA2 mark (the final mark in this part of the question).
Also ignore EXTRA solutions outside the range \(0 \leqslant x \lt 360\).
Working in Radians: Note the answers in radians are \(x = \text{awrt } 1.6,\ \text{awrt } 6.2\)
If a candidate works in radians then mark part (a) as above awarding the A marks in the same way. If the candidate would then score FULL MARKS then withhold the final aA2 mark (the final mark in this part of the question.)
No working: Award M1M1A1A0 for one of awrt \(93.2^\circ\) or awrt \(356.8^\circ\) seen without any working.
Award M1M1A1A1 for both awrt \(93.2^\circ\) and awrt \(356.8^\circ\) seen without any working.
Allow benefit of the doubt (FULL MARKS) for final answer of \(\sin x\) {and not \(x\)} \(= \{\text{awrt } 93.2,\ \text{awrt } 356.8\}\)
| Scheme | Marks |
|---|---|
| \(2(1 - \cos^2 x) + 2 = 7\cos x\) Applies \(\sin^2 x = 1 - \cos^2 x\) | M1 |
| \(2\cos^2 x + 7\cos x - 4 = 0\) Correct 3 term, \(2\cos^2 x + 7\cos x - 4\ \{= 0\}\) | A1 oe |
| \((2\cos x - 1)(\cos x + 4)\ \{= 0\}\), \(\cos x = \ldots\) Valid attempt at solving and \(\cos x = \ldots\) | M1 |
| \(\cos x = \dfrac{1}{2}\), \(\{\cos x = -4\}\) \(\cos x = \dfrac{1}{2}\) (See notes.) | A1 cso |
| \(\left(\beta = \dfrac{\pi}{3}\right)\) | |
| \(x = \dfrac{\pi}{3}\) or \(1.04719\ldots^c\) Either \(\dfrac{\pi}{3}\) or awrt \(1.05^c\) | B1 |
| \(x = \dfrac{5\pi}{3}\) or \(5.23598\ldots^c\) Either \(\dfrac{5\pi}{3}\) or awrt \(5.24^c\) or \(2\pi -\) their \(\beta\) (See notes.) | B1 ft |
| [6] | |
| 10 |
Notes
1st M1: for a correct method to use \(\sin^2 x = 1 - \cos^2 x\) on the given equation. Give bod if the candidate omits the bracket when substituting for \(\sin^2 x\), but \(2 - \cos^2 x + 2 = 7\cos x\), without supporting working, (eg. seeing “\(\sin^2 x = 1 - \cos^2 x\)”) would score 1st M0.
Note that applying \(\sin^2 x = \cos^2 x - 1\), scores M0.
1st A1: for obtaining either \(2\cos^2 x + 7\cos x - 4\) or \(-2\cos^2 x - 7\cos x + 4\).
1st A1: can also awarded for a correct three term equation eg. \(2\cos^2 x + 7\cos x = 4\) or \(2\cos^2 x = 4 - 7\cos x\) etc.
2nd M1: for a valid attempt at factorisation of a quadratic (either 2TQ or 3TQ) in cos, can use any variable here, \(c, y, x\) or \(\cos x\), and an attempt to find at least one of the solutions. See introduction to the Mark Scheme. Alternatively, using a correct formula for solving the quadratic. Either the formula must be stated correctly or the correct form must be implied by the substitution.
2nd A1: for \(\cos x = \dfrac{1}{2}\), BY A CORRECT SOLUTION ONLY UP TO THIS POINT. Ignore extra answer of \(\cos x = -4\), but penalise if candidate states an incorrect result e.g. \(\cos x = 4\). If they have used a substitution, a correct value of their \(c\) or their \(y\) or their \(x\).
Note: 2nd A1 for \(\cos x = \dfrac{1}{2}\) can be implied by later working.
1st B1: for either \(\dfrac{\pi}{3}\) or awrt \(1.05^c\)
2nd B1: for either \(\dfrac{5\pi}{3}\) or awrt \(5.24^c\) or can be ft from \(2\pi -\) their \(\beta\) or \(360^\circ -\) their \(\beta\) where \(\beta = \cos^{-1}(k)\), such that \(0 \lt k \lt 1\) or \(-1 \lt k \lt 0\), but \(k \ne 0,\ k \ne 1\) or \(k \ne -1\).
If there are any EXTRA solutions inside the range \(0 \leqslant x \lt 2\pi\) and the candidate would otherwise score FULL MARKS then withhold the final bB2 mark (the final mark in this part of the question).
Also ignore EXTRA solutions outside the range \(0 \leqslant x \lt 2\pi\).
Working in Degrees: Note the answers in degrees are \(x = 60,\ 300\)
If a candidate works in degrees then mark part (b) as above awarding the B marks in the same way. If the candidate would then score FULL MARKS then withhold the final bB2 mark (the final mark in this part of the question.)
Answers from no working:
\(x = \dfrac{\pi}{3}\) and \(x = \dfrac{5\pi}{3}\) scores M0A0M0A0B1B1,
\(x = 60\) and \(x = 300\) scores M0A0M0A0B1B0,
\(x = \dfrac{\pi}{3}\) ONLY or \(x = 60\) ONLY scores M0A0M0A0B1B0,
\(x = \dfrac{5\pi}{3}\) ONLY or \(x = 300\) ONLY scores M0A0M0A0B0B1. (Corrected from the printed mark scheme: printed as \(x = 120\) ONLY; \(\tfrac{5\pi}{3}\) is \(300^\circ\).)
No working: You cannot apply the ft in the B1ft if the answers are given with NO working.
Eg: \(x = \dfrac{\pi}{5}\) and \(x = \dfrac{9\pi}{3}\) FROM NO WORKING scores M0A0M0A0B0B0.
For candidates using trial & improvement, please forward these to your Team Leader.