C3 June 2013 (R) Q6
6.
| Scheme | Marks |
|---|---|
| \(\operatorname{cosec}2x=\dfrac{1}{\sin 2x}\) | M1 |
| \(=\dfrac{1}{2\sin x\cos x}\) | M1 |
| \(=\dfrac{1}{2}\operatorname{cosec}x\sec x\Rightarrow\quad\lambda=\dfrac{1}{2}\) | A1 |
| (3) |
Notes
M1 Uses the identity \(\operatorname{cosec}2x=\dfrac{1}{\sin 2x}\)
M1 Uses the correct identity for \(\sin 2x=2\sin x\cos x\) in their expression.
Accept \(\sin 2x=\sin x\cos x+\cos x\sin x\)
A1 \(\lambda=\dfrac{1}{2}\) following correct working
| Scheme | Marks |
|---|---|
| \(3\sec^2\theta+3\sec\theta=2\tan^2\theta\Rightarrow 3\sec^2\theta+3\sec\theta=2(\sec^2\theta-1)\) | M1 |
| \(\sec^2\theta+3\sec\theta+2=0\) | |
| \((\sec\theta+2)(\sec\theta+1)=0\) | M1 |
| \(\sec\theta=-2,-1\) | A1 |
| \(\cos\theta=-0.5,\ -1\) | M1 |
| \(\theta=\dfrac{2\pi}{3},\dfrac{4\pi}{3},\pi\) | A1A1 |
| (6) | |
| (9 marks) |
Notes
M1 Replaces \(\tan^2\theta\) by \(\pm\sec^2\theta\pm 1\) to produce an equation in just \(\sec\theta\)
M1 Award for a forming a 3TQ=0 in \(\sec\theta\) and applying a correct method for factorising, or using the formula, or completing the square to find two answers to \(\sec\theta\)
If they replace \(\sec\theta=\dfrac{1}{\cos\theta}\) it is for forming a 3TQ in \(\cos\theta\) and applying a correct method for finding two answers to \(\cos\theta\)
A1 Correct answers to \(\sec\theta=-2,-1\) or \(\cos\theta=-\dfrac{1}{2},-1\)
M1 Award for using the identity \(\sec\theta=\dfrac{1}{\cos\theta}\) and proceeding to find at least one value for \(\theta\).
If the 3TQ was in cosine then it is for finding at least one value of \(\theta\).
A1 Two correct values of \(\theta\). All method marks must have been scored.
Accept two of \(120^\circ,180^\circ,240^\circ\) or two of \(\dfrac{2\pi}{3},\dfrac{4\pi}{3},\pi\) or two of awrt 2dp 2.09, 3.14, 4.19
A1 All three answers correct. They must be given in terms of \(\pi\) as stated in the question.
Accept \(0.\dot{6}\pi,\ 1.\dot{3}\pi,\ \pi\)
Withhold this mark if further values in the range are given. All method marks must have been scored.
Ignore any answers outside the range.
ALT (ii)
| Scheme | Marks |
|---|---|
| \(3\sec^2\theta+3\sec\theta=2\tan^2\theta\Rightarrow 3\times\dfrac{1}{\cos^2\theta}+3\times\dfrac{1}{\cos\theta}=2\times\dfrac{\sin^2\theta}{\cos^2\theta}\) | |
| \(3+3\cos\theta=2\sin^2\theta\) | |
| \(3+3\cos\theta=2(1-\cos^2\theta)\) | M1 |
| \(2\cos^2\theta+3\cos\theta+1=0\) | |
| \((2\cos\theta+1)(\cos\theta+1)=0\Rightarrow\cos\theta=-0.5,\ -1\) | M1A1 |
| \(\theta=\dfrac{2\pi}{3},\dfrac{4\pi}{3},\pi\) | M1,A1,A1 |
| (6) | |
| (9 marks) |
M1 Award for replacing \(\sec^2\theta\) with \(\dfrac{1}{\cos^2\theta}\), \(\sec\theta\) with \(\dfrac{1}{\cos\theta}\), \(\tan^2\theta\) with \(\dfrac{\sin^2\theta}{\cos^2\theta}\) multiplying through by \(\cos^2\theta\) (seen in at least 2 terms) and replacing \(\sin^2\theta\) with \(\pm 1\pm\cos^2\theta\) to produce an equation in just \(\cos\theta\)
M1 Award for a forming a 3TQ=0 in \(\cos\theta\) and applying a correct method for factorising, or using the formula, or completing the square to find two answers to \(\cos\theta\)
A1 \(\cos\theta=-\dfrac{1}{2},-1\)
M1 Proceeding to finding at least one value of \(\theta\) from an equation in \(\cos\theta\).
A1 Two correct values of \(\theta\). All method marks must have been scored
Accept two of \(120^\circ,180^\circ,240^\circ\) or two of \(\dfrac{2\pi}{3},\dfrac{4\pi}{3},\pi\) or two of awrt 2dp 2.09, 3.14, 4.19
A1 All three answers correct. They must be given in terms of \(\pi\) as stated in the question.
Accept \(0.\dot{6}\pi,\ 1.\dot{3}\pi,\ \pi\)
All method marks must have been scored. Withhold this mark if further values in the range are given.
Ignore any answers outside the range