OCR MEI C4 2005 June — Question 5 7 marks

Exam BoardOCR MEI
ModuleC4 (Core Mathematics 4)
Year2005
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeSolve equation with sin2x/cos2x by substitution
DifficultyModerate -0.3 This is a straightforward double angle equation requiring the standard substitution cos 2x = 2cos²x - 1, leading to a quadratic in cos x, then solving for x in the given range. It's slightly easier than average as it follows a well-practiced procedure with no conceptual surprises, though it does require multiple steps and careful attention to finding all solutions in the domain.
Spec1.05l Double angle formulae: and compound angle formulae1.05o Trigonometric equations: solve in given intervals
5 Solve the equation \(2 \cos 2 x = 1 + \cos x\), for \(0 ^ { \circ } \leqslant x < 360 ^ { \circ }\).
AnswerMarks Guidance
\(\frac{1}{\phi} = \frac{1}{1+\sqrt{5}} \cdot \frac{1}{1+\sqrt{5}} = \frac{2}{1+\sqrt{5}} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}\)M1 Must discount ±
\(= \frac{2(\sqrt{5}-1)}{(\sqrt{5})^2-1} = \frac{2(\sqrt{5}-1)}{4} = \frac{\sqrt{5}-1}{2}\)M1 M1 Must discount ±; Substituting for φ and simplifying
OR \(\frac{1}{\phi} = \phi - 1\)M1 Must discount ±
\(= \frac{\sqrt{5}+1}{2} - 1 = \frac{\sqrt{5}-1}{2}\)M1 E1 [3] 
$\frac{1}{\phi} = \frac{1}{1+\sqrt{5}} \cdot \frac{1}{1+\sqrt{5}} = \frac{2}{1+\sqrt{5}} \cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}$ | M1 | Must discount ±
$= \frac{2(\sqrt{5}-1)}{(\sqrt{5})^2-1} = \frac{2(\sqrt{5}-1)}{4} = \frac{\sqrt{5}-1}{2}$ | M1 M1 | Must discount ±; Substituting for φ and simplifying
**OR** $\frac{1}{\phi} = \phi - 1$ | M1 | Must discount ±
$= \frac{\sqrt{5}+1}{2} - 1 = \frac{\sqrt{5}-1}{2}$ | M1 E1 [3] |
5 Solve the equation $2 \cos 2 x = 1 + \cos x$, for $0 ^ { \circ } \leqslant x < 360 ^ { \circ }$.

\hfill \mbox{\textit{OCR MEI C4 2005 Q5 [7]}}
This paper (6 questions)
View full paper
Q2 6 Q3 4 Q4 5 Q5 7 Q6 8 Q7 18