C3 June 2015 Q3
3.\[\mathrm{g}(\theta) = 4\cos 2\theta + 2\sin 2\theta\]Given that \(\mathrm{g}(\theta) = R\cos(2\theta - \alpha)\), where \(R > 0\) and \(0 < \alpha < 90^\circ\),
Given that \(k\) is a constant and the equation \(\mathrm{g}(\theta) = k\) has no solutions,
| Scheme | Marks |
|---|---|
| \(4\cos 2\theta + 2\sin 2\theta = R\cos(2\theta - \alpha)\) | |
| \(R = \sqrt{4^2 + 2^2} = \sqrt{20} = \left(2\sqrt{5}\right)\) | B1 |
| \(\alpha = \arctan\left(\dfrac{1}{2}\right) = 26.565^\circ\ldots = \text{awrt } 26.57^\circ\) | M1A1 |
| (3) |
Notes
You can marks parts (a) and (b) together as one.
B1: For \(R = \sqrt{20} = 2\sqrt{5}\). Condone \(R = \pm\sqrt{20}\)
M1: For \(\alpha = \arctan\left(\pm\dfrac{1}{2}\right)\) or \(\alpha = \arctan(\pm 2)\) leading to a solution of \(\alpha\)
Condone any solutions coming from \(\cos\alpha = 4, \sin\alpha = 2\)
Condone for this mark \(2\alpha = \arctan\left(\pm\dfrac{1}{2}\right) \Rightarrow \alpha = ..\)
If \(R\) has been used to find \(\alpha\) award for only \(\alpha = \arccos\left(\pm\dfrac{4}{\text{'}R\text{'}}\right)\) \(\alpha = \arcsin\left(\pm\dfrac{2}{\text{'}R\text{'}}\right)\)
A1: \(\alpha = \text{awrt } 26.57^\circ\)
| Scheme | Marks |
|---|---|
| \(\sqrt{20}\cos(2\theta - 26.6) = 1 \Rightarrow \cos(2\theta - 26.57) = \dfrac{1}{\sqrt{20}}\) | M1 |
| \(\Rightarrow (2\theta - 26.57) = +77.1.. \Rightarrow \theta = \ldots\) | dM1 |
| \(\theta = \text{awrt } 51.8^\circ\) | A1 |
| \(2\theta - 26.57 = \text{'}{-}77.1...\text{'} \Rightarrow \theta = -\text{awrt } 25.3^\circ\) | ddM1A1 |
| (5) |
Notes
M1: Using part (a) and proceeding as far as \(\cos(2\theta \pm \text{their } 26.57) = \dfrac{1}{\text{their } R}\).
This may be implied by \((2\theta \pm \text{their } 26.57) = \arccos\left(\dfrac{1}{\text{their} R}\right)\)
Allow this mark for \(\cos(\theta \pm \text{their } 26.57) = \dfrac{1}{\text{their } R}\)
dM1: Dependent upon the first M1- it is for a correct method to find \(\theta\) from their principal value
Look for the correct order of operations, that is dealing with the "26.57" before the "2".
Condone subtracting 26.57 instead of adding.
\(\cos(2\theta \pm \text{their } 26.57) = \ldots \Rightarrow 2\theta \pm \text{their } 26.57 = \beta \Rightarrow \theta = \dfrac{\beta \pm \text{their } 26.57}{2}\)
A1: awrt \(\theta = 51.8^\circ\)
ddM1: For a correct method to find a secondary value of \(\theta\) in the range
Either \(2\theta \pm 26.57 = \text{'}{-}\beta\text{'} \Rightarrow \theta =\) OR \(2\theta \pm 26.57 = 360 - \text{'}\beta\text{'} \Rightarrow \theta =\) THEN MINUS 180
A1: awrt \(\theta = -25.3^\circ\)
Withhold this mark if there are extra solutions in the range.
Radian solution: Only lose the first time it occurs.
FYI. In radians desired accuracy is awrt 2 dp (a) \(\alpha = 0.46\) and (b) \(\theta_1 = 0.90, \theta_2 = -0.44\)
Mixing degrees and radians only scores the first M
| Scheme | Marks |
|---|---|
| \(k < -\sqrt{20},\ k > \sqrt{20}\) | B1ft either B1ft both |
| (2) | |
| (10 marks) |
Notes
B1ft: Follow through on their \(R\). Accept decimals here including \(\sqrt{20} \approx\) awrt 4.5.
Score for one of the ends \(k > \sqrt{20}\), \(k < -\sqrt{20}\)
Condone versions such as \(\mathrm{g}(\theta) > \sqrt{20}\), \(y > \sqrt{20}\)
or both ends including the boundaries \(k \geqslant \sqrt{20}\), \(k \leqslant -\sqrt{20}\)
B1ft: For both intervals in terms of \(k\).
Accept \(k > \sqrt{20}\) or \(k < -\sqrt{20}\). Accept \(|k| > \sqrt{20}\) Accept \(k \in \left(\sqrt{20}, \infty\right) \cup \left(-\infty, -\sqrt{20}\right)\)
Condone \(k > \sqrt{20}, k < -\sqrt{20}\) \(k > \sqrt{20}\) and \(k < -\sqrt{20}\) for both marks
but \(-\sqrt{20} > k > \sqrt{20}\) is B1 B0