C3 June 2016 Q8
8.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
(6)| Scheme | Marks |
|---|---|
| \(2\cot 2x + \tan x \equiv \dfrac{2}{\tan 2x} + \tan x\) | B1 |
| \(\equiv \dfrac{(1 - \tan^2 x)}{\tan x} + \dfrac{\tan^2 x}{\tan x}\) | M1 |
| \(\equiv \dfrac{1}{\tan x}\) | M1 |
| \(\equiv \cot x\) | A1* |
| (4) |
8 (a) alt 1
| Scheme | Marks |
|---|---|
| \(2\cot 2x + \tan x \equiv \dfrac{2\cos 2x}{\sin 2x} + \tan x\) | B1 |
| \(\equiv 2\dfrac{\cos^2 x - \sin^2 x}{2\sin x\cos x} + \dfrac{\sin x}{\cos x}\) | M1 |
| \(\equiv \dfrac{\cos^2 x - \sin^2 x}{\sin x\cos x} + \dfrac{\sin^2 x}{\sin x\cos x} \equiv \dfrac{\cos^2 x}{\sin x\cos x}\) | M1 |
| \(\equiv \dfrac{\cos x}{\sin x}\) \(\equiv \cot x\) | A1* |
8 (a) alt 2
| Scheme | Marks |
|---|---|
| \(2\cot 2x + \tan x \equiv 2\dfrac{(1 - \tan^2 x)}{2\tan x} + \tan x\) | B1M1 |
| \(\equiv \dfrac{2}{2\tan x} - \dfrac{2\tan^2 x}{2\tan x} + \tan x \qquad \text{or} \quad \dfrac{(1 - \tan^2 x) + \tan^2 x}{\tan x}\) | |
| \(\equiv \dfrac{2}{2\tan x} = \cot x\) | M1A1* |
Notes
B1: States or uses the identity \(2\cot 2x = \dfrac{2}{\tan 2x}\) or alternatively \(2\cot 2x = \dfrac{2\cos 2x}{\sin 2x}\)
This may be implied by \(2\cot 2x = \dfrac{1 - \tan^2 x}{\tan x}\). Note \(2\cot 2x = \dfrac{1}{2\tan 2x}\) is B0
M1: Uses the correct double angle identity \(\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x}\)
Alternatively uses \(\sin 2x = 2\sin x\cos x\), \(\cos 2x = \cos^2 x - \sin^2 x\) oe and \(\tan x = \dfrac{\sin x}{\cos x}\)
M1: Writes their two terms with a single common denominator and simplifies to a form \(\dfrac{ab}{cd}\).
For this to be scored the expression must be in either \(\sin x\) and \(\cos x\) or just \(\tan x\).
In alternative 2 it is for splitting the complex fraction into parts and simplifying to a form \(\dfrac{ab}{cd}\).
You are awarding this for a correct method to proceed to terms like \(\dfrac{\cos^2 x}{\sin x\cos x}, \dfrac{2\cos^3 x}{2\sin x\cos^2 x}, \dfrac{2}{2\tan x}\)
A1*: cso. For proceeding to the correct answer. This is a given answer and all aspects must be correct including the consistent use of variables. If the candidate approaches from both sides there must be a conclusion for this mark to be awarded. Occasionally you may see a candidate attempting to prove \(\cot x - \tan x \equiv 2\cot 2x\). This is fine but again there needs to be a conclusion for the A1*
If you are unsure of how some items should be marked then please use review
| Scheme | Marks |
|---|---|
| \(6\cot 2x + 3\tan x = \mathrm{cosec}^2 x - 2 \Rightarrow 3\cot x = \mathrm{cosec}^2 x - 2\) | |
| \(\Rightarrow 3\cot x = 1 + \cot^2 x - 2\) | M1 |
| \(\Rightarrow 0 = \cot^2 x - 3\cot x - 1\) | A1 |
| \(\Rightarrow \cot x = \dfrac{3 \pm \sqrt{13}}{2}\) | M1 |
| \(\Rightarrow \tan x = \dfrac{2}{3 \pm \sqrt{13}} \Rightarrow x = ..\) | M1 |
| \(\Rightarrow x = 0.294, -2.848, -1.277, 1.865\) | A2,1,0 |
| (6) | |
| (10 marks) |
Alt (b)
| Scheme | Marks |
|---|---|
| \(6\cot 2x + 3\tan x = \mathrm{cosec}^2 x - 2 \Rightarrow \dfrac{3\cos x}{\sin x} = \dfrac{1}{\sin^2 x} - 2\) | |
| \(\left(\times\sin^2 x\right) \Rightarrow 3\sin x\cos x = 1 - 2\sin^2 x\) | M1 |
| \(\Rightarrow \dfrac{3}{2}\sin 2x = \cos 2x\) | M1A1 |
| \(\Rightarrow \tan 2x = \dfrac{2}{3} \Rightarrow x = ..\) | M1 |
| \(\Rightarrow x = 0.294, -2.848, -1.277, 1.865\) | A2,1,0 |
| (6) |
Notes
M1: For using part (a) and writing \(6\cot 2x + 3\tan x\) as \(k\cot x\), \(k \neq 0\) in their equation (or equivalent)
WITH an attempt at using \(\mathrm{cosec}^2 x = \pm 1 \pm \cot^2 x\) to produce a quadratic equation in just \(\cot x\) / \(\tan x\)
A1: \(\cot^2 x - 3\cot x - 1 = 0\) The = 0 may be implied by subsequent working
Alternatively accept \(\tan^2 x + 3\tan x - 1 = 0\)
M1: Solves a 3TQ=0 in \(\cot x\) (or tan) using the formula or any suitable method for their quadratic to find at least one solution. Accept answers written down from a calculator. You may have to check these from an incorrect quadratic. FYI answers are \(\cot x = \text{awrt } 3.30, \ -0.30\)
Be aware that \(\cot x = \dfrac{3 \pm \sqrt{13}}{2} \Rightarrow \tan x = \dfrac{-3 \pm \sqrt{13}}{2}\)
M1: For \(\tan x = \dfrac{1}{\cot x}\) and using arctan producing at least one answer for \(x\) in degrees or radians.
You may have to check these with your calculator.
A1: Two of \(x = 0.294, -2.848, -1.277, 1.865\) (awrt 3dp) in radians or degrees.
In degrees the answers you would accept are (awrt 2dp) \(x = 16.8^\circ, 106.8^\circ, -73.2^\circ, -163.2^\circ\)
A1: All four of \(x = 0.294, -2.848, -1.277, 1.865\) (awrt 3 dp) with no extra solutions in the range \(-\pi \leqslant x < \pi\)
See main scheme for Alt to (b) using Double Angle formulae still entered M A M M A A in epen
1st M1 For using part (a) and writing \(6\cot 2x + 3\tan x\) as \(k\cot x\), \(k \neq 0\) in their equation (or equivalent) then using \(\cot x = \dfrac{\cos x}{\sin x}\), \(\mathrm{cosec}^2 x = \dfrac{1}{\sin^2 x}\) and \(\times\sin^2 x\) to form an equation sin and cos
1st A1 For \(\dfrac{3}{2}\sin 2x = \cos 2x\) or equivalent. Attached to the next M
2nd M1 For using both correct double angle formula
3rd M1 For moving from \(\tan 2x = C\) to \(x\)=..using the correct order of operations.