C3 June 2017 Q4

Notes

B1: \(R = \sqrt{29}\)
Condone \(R = \pm\sqrt{29}\)   (Do not allow decimals for this mark Eg 5.39 but remember to isw after \(\sqrt{29}\))

M1: \(\tan\alpha = \pm\dfrac{2}{5}, \tan\alpha = \pm\dfrac{5}{2} \Rightarrow \alpha = \ldots\)
If \(R\) is used to find \(\alpha\) accept \(\sin\alpha = \pm\dfrac{2}{R}\) or \(\cos\alpha = \pm\dfrac{5}{R} \Rightarrow \alpha = \ldots\)

A1: \(\alpha = \text{awrt } 0.381\)    Note that the degree equivalent \(\alpha = \text{awrt } 21.8^\circ\) is A0

Notes

M1: Replaces \(\cot 2x\) by \(\dfrac{\cos 2x}{\sin 2x}\) and \(\mathrm{cosec}\,2x\) by \(\dfrac{1}{\sin 2x}\) in the lhs
Do not be concerned by the coefficients 5 and -3.
Replacing \(\cot 2x\) by \(\dfrac{1}{\tan 2x}\) does not score marks until the \(\tan 2x\) has been replaced by \(\dfrac{\sin 2x}{\cos 2x}\)
They may state \(\times\sin 2x \Rightarrow 5\cos 2x - 3 = 2\sin 2x\) which implies this mark

A1: cso \(5\cos 2x - 2\sin 2x = 3\)    There is no need to state the value of '\(c\)'
The notation must be correct. They cannot mix variables within their equation
Do not accept for the final A1 \(\tan 2x = \dfrac{\sin}{\cos}2x\) within their equations

Alt I (c)

Alt II (c)

Notes

M1: Attempts to use part (a) and (b). They must be using their \(R\) and \(\alpha\) from part (a) and their \(c\) from part (b)
Accept \(\cos(2x \pm \text{'}\alpha\text{'}) = \dfrac{\text{'}c\text{'}}{\text{'}R\text{'}}\) Condone \(\cos(\theta \pm \text{'}\alpha\text{'}) = \dfrac{\text{'}c\text{'}}{\text{'}R\text{'}}\) or even \(\cos(x \pm \text{'}\alpha\text{'}) = \dfrac{\text{'}c\text{'}}{\text{'}R\text{'}}\) for the first M

dM1: Score for dealing with the cos, the \(\alpha\) and the 2 correctly and in that order to reach \(x = ..\)
Don't be concerned if they change the variable in the question and solve for \(\theta =\) (as long as all operations have been undone).You may not see any working. It is implied by one correct answer.
You may need to check with a calculator.
Eg for an incorrect \(\alpha\) \(\cos(2x + 1.19) = \dfrac{3}{\sqrt{29}} \Rightarrow x = -0.105\) would score M1 dM1 A0 A0

A1: One solution correct, usually \(x = 0.3/0.30\) or \(x = 2.46\) or in degrees \(17.2^\circ\) or \(141.(0)^\circ\)

A1: Both solutions correct awrt \(x = \text{awrt } 0.30, 2.46\) and no extra values in the range.
Condone candidates who write 0.3 and 2.46 without any (more accurate) answers
In degrees accept awrt 1 dp \(17.2^\circ, 141.(0)^\circ\) and no extra values in the range.

Special case: For candidates who are misreading the question and using their part (a) with 2 on the rhs.
They will be allowed to score a maximum of SC M1 dM1 A0 A0

M1: Attempts to use part (a) with 2. They must be using their \(R\) and \(\alpha\) from part (a)
Accept \(\cos(2x \pm \text{'}\alpha\text{'}) = \dfrac{2}{\text{'}R\text{'}}\) Condone \(\cos(\theta \pm \text{'}\alpha\text{'}) = \dfrac{2}{\text{'}R\text{'}}\) or even \(\cos(x \pm \text{'}\alpha\text{'}) = \dfrac{2}{\text{'}R\text{'}}\) for the first M

dM1: Score for dealing with the cos, the \(\alpha\) and the 2 correctly and in that order to reach \(x = ..\)
You may not see any working. It is implied by one correct answer. You may need to check with a calculator.
Eg for an correct \(\alpha\) and \(R\) \(\cos(2x + 0.381) = \dfrac{2}{\sqrt{29}} \Rightarrow x = 0.405\)

Alt to part (c)

M1: Attempts both double angle formulae condoning sign slips on \(\cos 2x\), divides by \(\cos^2 x\) and forms a quadratic in tan by using the identity \(\pm 1 \pm \tan^2 x = \sec^2 x\)

dM1: Attempts to solve their quadratic in \(\tan x\) leading to a solution for \(x\).

A1 A1: As above