C3 June 2013 (R) Q3
3. \[\mathrm{f}(x)=7\cos x+\sin x\]Given that \(\mathrm{f}(x)=R\cos(x-\alpha)\), where \(R>0\) and \(0<\alpha<90^\circ\),
| Scheme | Marks |
|---|---|
| \(7\cos x+\sin x=R\cos(x-\alpha)\) | |
| \(R=\sqrt{(7^2+1^2)}=\sqrt{50}=(5\sqrt{2})\) | B1 |
| \(\alpha=\arctan\left(\dfrac{1}{7}\right)=8.13\ldots=\text{awrt }8.1^\circ\) | M1A1 |
| (3) |
Notes
B1 \(R=\sqrt{50}\). Accept \(5\sqrt{2}\) Accept \(R=\pm\sqrt{50}\)
Do not accept \(R=\sqrt{(7^2+1^2)}\) or the decimal equivalent 7.07…unless you see \(\sqrt{50}\) or \(5\sqrt{2}\) as well
M1 For \(\tan\alpha=\pm\dfrac{1}{7}\) or \(\tan\alpha=\pm\dfrac{7}{1}\). Condone if this comes from \(\cos\alpha=7,\ \sin\alpha=1\)
If \(R\) is used then only accept \(\sin\alpha=\pm\dfrac{1}{R}\) or \(\cos\alpha=\pm\dfrac{7}{R}\)
A1 \(\alpha=\text{awrt }8.1\).
Be aware that \(\tan\alpha=7\Rightarrow\alpha=81.9\) can easily be mistaken for the correct answer
Note that the radian answer awrt 0.1418… is A0
| Scheme | Marks |
|---|---|
| \(\sqrt{50}\cos(x-8.1)=5\Rightarrow\cos(x-8.1)=\dfrac{5}{\sqrt{50}}\) | M1 |
| \(x-8.1=45\Rightarrow x=53.1^\circ\) | M1,A1 |
| AND \(\ x-8.1=315\Rightarrow x=323.1^\circ\) | M1A1 |
| (5) |
Notes
M1 For using their answers to part (a) and moving from \(R\cos(x\pm\alpha)=5\Rightarrow\cos(x\pm\alpha)=\dfrac{5}{R}\) using their numerical values of \(R\) and \(\alpha\)
This may be implied for sight of 53.1 if \(R\) and \(\alpha\) were correct
M1 For achieving \(x\pm\alpha=\text{awrt }45^\circ\) or 315, leading to one value of \(x\) in the range
Note that for this to be scored \(R\) has to be correct (to 2sf) as awrt 45, 315 must be achieved
This may be implied for achieving an answer of either \(45+\textit{their}\ \alpha\) or \(315+\textit{their}\ \alpha\)
A1 One correct answer, either awrt 53.1° or 323.1°
M1 For an attempt at finding a secondary value of \(x\) in the range.
Usually this is an attempt at solving \(x-\textit{their}\ 8.1^\circ=360^\circ-\textit{their}\ 45^\circ\Rightarrow x=..\)
A1 Both values correct awrt 53.1° and 323.1°.
Withhold this mark if there are extra values in the range.
Ignore extra values outside the range
| Scheme | Marks |
|---|---|
| One solution if \(\dfrac{k}{\sqrt{50}}=\pm 1,\Rightarrow k=\pm\sqrt{50}\) ft on \(R\) | M1A1ft |
| (2) | |
| (10 marks) |
Notes
M1 For stating that \(\dfrac{k}{\textit{their}\ R}=1\) OR \(\dfrac{k}{\textit{their}\ R}=-1\)
This may be implied by seeing \(k=(\pm)\textit{their}\ R\)
A1ft Both values \(k=\pm\sqrt{50}\) oe. Follow through on their numerical R
Answers all in radians. Lose the first time that it appears but demand an accuracy of 2dp.
Part (a) \(\quad R=\sqrt{50}\quad\alpha=\textit{awrt}\ 0.14\)
Part (b) \(\quad x=\textit{awrt}\ 0.927,\ 5.64\). Accuracy must be to 3 sf.
With correct working this would score (a) B1M1A0 (b) M1A1A1M1A1
Mixed degrees and radians refer to the main scheme