a. [3 marks]

Sketch the graph of \( C(x) = \cos x + \frac{1}{2} \cos 2x \) for \( -2\pi \leq x \leq 2\pi \).
Key points:
– At \( x = 0 \): \( C(0) = \cos 0 + \frac{1}{2} \cos 0 = 1 + \frac{1}{2} = 1.5 \) (maximum).
– Period is \( 2\pi \) (from part b).
– Symmetry: \( C(x) = C(-x) \) (from part e(i)).
Graph shows maxima at \( x = -2\pi, 0, 2\pi \), and zeros around \( x \approx \pm 1.2, \pm 5.1 \) (M1)(A1)(A1).

b. [2 marks]

Prove periodicity: \( C(x + 2\pi) = \cos(x + 2\pi) + \frac{1}{2} \cos(2(x + 2\pi)) \).
Since \( \cos(x + 2\pi) = \cos x \), \( \cos(2x + 4\pi) = \cos 2x \) (M1),
\( C(x + 2\pi) = \cos x + \frac{1}{2} \cos 2x = C(x) \) (A1).
Period is \( 2\pi \) (smallest positive value satisfying periodicity).
Answer: \( C(x) \) is periodic with period \( 2\pi \).

c. [3 marks]

Find maximum of \( C(x) \): Differentiate \( C(x) = \cos x + \frac{1}{2} \cos 2x \).
\( C'(x) = -\sin x – \frac{1}{2} \cdot 2 \sin 2x = -\sin x – \sin 2x \) (M1).
Set \( C'(x) = 0 \): \( \sin x + \sin 2x = 0 \implies \sin x + 2 \sin x \cos x = 0 \implies \sin x (1 + 2 \cos x) = 0 \).
Solutions: \( \sin x = 0 \implies x = -2\pi, -\pi, 0, \pi, 2\pi \); or \( \cos x = -\frac{1}{2} \implies x = \pm \frac{2\pi}{3}, \pm \frac{4\pi}{3} \) (A1).
Test second derivative or values: At \( x = -2\pi, 0, 2\pi \), \( C(x) = 1.5 \); at \( x = \pm \frac{2\pi}{3}, \pm \frac{4\pi}{3} \), \( C(x) = -0.25 \) (A1).
Maxima occur at \( x = -2\pi, 0, 2\pi \).
Answer: \( x = -2\pi, 0, 2\pi \).

d. [2 marks]

Solve \( C(x) = \cos x + \frac{1}{2} \cos 2x = 0 \) for smallest positive \( x_0 \).
Use numerical methods or graphing: Approximate solution at \( x_0 \approx 1.231 \) (M1).
To two significant figures: \( x_0 \approx 1.2 \) (A1).
Answer: \( x_0 = 1.2 \).

e. [2 + 2 marks]

(i) Prove \( C(x) = C(-x) \):
\( C(-x) = \cos(-x) + \frac{1}{2} \cos(2(-x)) = \cos x + \frac{1}{2} \cos 2x = C(x) \) (M1)(A1).
Answer: \( C(x) = C(-x) \) for all \( x \).
(ii) Find \( x_1 \), \( \pi < x_1 < 2\pi \), where \( C(x_1) = 0 \).
Since \( C(x) = C(-x) \), if \( C(x_0) = 0 \), then \( C(-x_0) = 0 \).
Periodicity: \( C(x + 2\pi) = C(x) \), so \( C(2\pi – x_0) = C(-x_0) = 0 \) (M1).
Check: \( \pi < 2\pi – x_0 < 2\pi \implies 0 < x_0 < \pi \), satisfied since \( x_0 \approx 1.2 \).
Thus, \( x_1 = 2\pi – x_0 \) (A1).
Answer: \( x_1 = 2\pi – x_0 \).