C3 June 2014 (R) Q3
3.
| Scheme | Marks |
|---|---|
| \(2\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\sin x}=\dfrac{5}{\sin x}\) | B1 |
| Uses common denominator to give \(2\sin^2x-\cos^2x=5\cos x\) | M1 |
| Replaces \(\sin^2x\) by \((1-\cos^2x)\) to give \(2(1-\cos^2x)-\cos^2x=5\cos x\) | M1 |
| Obtains \(3\cos^2x+5\cos x-2=0\quad(a=3,\ b=5,\ c=-2)\) | A1 |
| (4) |
Alternative to main scheme 3. (i) (a)
| Scheme | Marks |
|---|---|
| \(2\tan x-\dfrac{1}{\tan x}=\dfrac{5}{\sin x}\) does not score any marks until | |
| \(\times\tan x\Rightarrow 2\tan^2x+1=5\sec x\) | B1, M1 |
| Replaces \(\tan^2x\) by \((\sec^2x-1)\) to give \(2(\sec^2x-1)+1=5\sec x\) (corrected from the printed mark scheme, which has \(2(\sec x^2-1)+1=5\sec x\)) | M1 |
| Obtains \(3\cos^2x+5\cos x-2=0\quad(a=3,\ b=5,\ c=-2)\) | A1 |
| (4) |
Notes
B1 Uses definitions \(\tan x=\dfrac{\sin x}{\cos x}\), \(\cot x=\dfrac{\cos x}{\sin x}\) and \(\operatorname{cosec}x=\dfrac{1}{\sin x}\) to write the equation in terms of \(\cos x\) and \(\sin x\). Condone \(5\operatorname{cosec}x=\dfrac{1}{5\sin x}\) as the intention is clear.
Alternatively uses \(\cot x=\dfrac{1}{\tan x}\) and \(\operatorname{cosec}x=\dfrac{1}{\sin x}\) to write the equation in terms of \(\tan x\) and \(\sin x\)
This may be implied by later work that achieves \(A\tan^2x\pm B=C\sec x\)
M1 Either uses common denominator and cross multiples, or multiplies each term by \(\sin x\cos x\) to achieve an equation of the form equivalent to \(A\sin^2x\pm B\cos^2x=C\cos x\). It may be seen on the numerator of a fraction
Alternatively multiplies by \(\tan x\) to achieve \(A\tan^2x\pm B=C\sec x\)
M1 Uses a correct Pythagorean relationship, usually \(\sin^2x=1-\cos^2x\) to form a quadratic equation in terms of \(\cos x\). In the alternative uses \(\tan^2x=\sec^2x-1\) to form a quadratic in sec \(x\), followed by \(\sec x=\dfrac{1}{\cos x}\) to form a quadratic equation in terms of \(\cos x\)
A1 Obtains \(\pm K\left(3\cos^2x+5\cos x-2\right)=0\quad(a=3,\ b=5,\ c=-2)\)
| Scheme | Marks |
|---|---|
| Solves \(3\cos^2x+5\cos x-2=0\) to give \(\cos x=\) | M1 |
| \(\cos x=\tfrac{1}{3}\) only (rejects \(\cos x=-2\)) | A1 |
| So \(x=1.23\) or \(5.05\) | dM1A1 |
| (4) |
Alternative to main scheme 3. (i) (b)
| Scheme | Marks |
|---|---|
| Solves \(3\cos^2x+5\cos x-2=0\) to give \(\cos x=\) or \(2\sec^2x-5\sec x-3=0\Rightarrow\sec x=..\) | M1 |
| \(\cos x=\tfrac{1}{3}\) only (rejects \(\cos x=-2\)) | A1 |
| So \(x=1.23\) or \(5.05\) | dM1A1 |
| (4) |
Notes
M1 Uses a standard method to solve their quadratic equation in \(\cos x\) from (i)(a) OR sec\(x\) from an earlier line in (a)
See General Principles for Core Mathematics on how to solve quadratics
A1 \(\cos x=\tfrac{1}{3}\) only Do not need to see \(-2\) rejected
dM1 Uses arcos on their value to obtain at least one answer. It is dependent upon the previous M.
It may be implied by one correct answer
A1 Both values correct awrt 3sf \(x=1.23\) and \(5.05\).
Ignore any solutions outside the range. Any extra solutions in the range will score A0.
Answers in degrees will score A0.
| Scheme | Marks |
|---|---|
| Either \(\tan\theta+\cot\theta\equiv\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}\) Or \(\tan\theta+\cot\theta\equiv\tan\theta+\dfrac{1}{\tan\theta}\) | B1 |
| Either \(\equiv\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\) Or \(\equiv\dfrac{\tan^2\theta+1}{\tan\theta}\) | M1 |
| Either \(\equiv\dfrac{2}{\sin 2\theta}\) Or \(\equiv\dfrac{1}{\cos^2\theta\times\frac{\sin\theta}{\cos\theta}}\equiv\dfrac{2}{\sin 2\theta}\) | M1 |
| \(\equiv 2\operatorname{cosec}2\theta\quad(\text{so }\lambda=2)\) | A1 |
| (4) | |
| (12 marks) |
Alternative to main scheme 3. (ii)
| Scheme | Marks |
|---|---|
| \(\tan\theta+\cot\theta=\lambda\operatorname{cosec}2\theta\Rightarrow\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\lambda}{\sin 2\theta}=\left(\xcancel{\dfrac{\lambda}{2\sin\theta\cos\theta}}\right)\) | B1 |
| \(\times 2\sin\theta\cos\theta\Rightarrow 2\sin^2\theta+2\cos^2\theta=\lambda\) | M1 |
| Factorises \(2(\sin^2\theta+\cos^2\theta)=\lambda\Rightarrow 2=\lambda\) | M1 |
| All above correct + a statement like ‘hence true’, ‘QED’ | A1 |
| (4) |
Notes
B1 Uses a definition of cot with matching expression for tan. Acceptable answers are
\(\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta},\ \dfrac{\sin\theta}{\cos\theta}+\dfrac{1}{\frac{\sin\theta}{\cos\theta}},\ \tan\theta+\dfrac{1}{\tan\theta}\). Condone a miscopy on the sign. Eg Allow \(\tan\theta-\dfrac{1}{\tan\theta}\)
M1 Uses common denominator, writing their expression as a single fraction. In the examples given above, example 2 would need to be inverted. The denominator has to be correct and one of the terms must be adapted.
M1 Uses identities \(\sin^2\theta+\cos^2\theta=1\) and \(\sin 2\theta=2\sin\theta\cos\theta\) specifically to achieve an expression of the form \(\dfrac{\lambda}{\sin 2\theta}\)
Alternatively uses \(1+\tan^2\theta=\sec^2\theta=\dfrac{1}{\cos^2\theta}\), \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) and \(\sin 2\theta=2\sin\theta\cos\theta\) specifically to achieve an expression of the form \(\dfrac{\lambda}{\sin 2\theta}\). A line of \(\dfrac{1}{\sin\theta\cos\theta}\) achieved on the lhs followed by \(\lambda=\dfrac{1}{2}\) or 2 would imply this mark
A1 Achieves printed answer with no errors.
Allow for a different variable as long as it is used consistently.