(a)
Compute \( f(-x) \):
\( f(-x) = 2\cos(-x) + (-x)\sin(-x) = 2\cos x + x\sin x = f(x) \)
Since \( f(-x) = f(x) \), the function is even.

Result: \( f \) is even [3]

(b)
First derivative: \( f'(x) = -2\sin x + \sin x + x\cos x = -\sin x + x\cos x \)
Second derivative: \( f”(x) = -\cos x + \cos x – x\sin x = -x\sin x \)
At \( x = 0 \): \( f”(0) = -0 \cdot \sin 0 = 0 \)

Result: \( f”(0) = 0 \) [2]

(c)
Verify point: \( f(0) = 2\cos 0 + 0 \cdot \sin 0 = 2 \), so the point is \( (0, 2) \)
At \( x = 0 \): \( f'(0) = -\sin 0 + 0 \cdot \cos 0 = 0 \), so critical point
Since \( f”(0) = 0 \), compute higher derivatives:
\( f”'(x) = -\sin x – x\cos x \), \( f”'(0) = 0 \)
\( f^{(4)}(x) = -2\cos x + x\sin x \), \( f^{(4)}(0) = -2 < 0 \)
First non-zero derivative is fourth, negative, indicating a maximum.
John’s statement is incorrect; \( (0, 2) \) is a local maximum, not a point of inflection.

Result: John’s statement is incorrect; \( (0, 2) \) is a maximum [2]