Question 8

**Analysis** For x ≠ 0, f(x) = (6x + sin x)/(2x + sin x) = 1 + 4/(2 + (sin x)/x). Since lim_{x→0} (sin x)/x = 1, f is continuous at 0 with f(0) = 7/3. Let h(x) = (sin x)/x. Since h(x) < 1 for all x ≠ 0, we get f(x) = 1 + 4/(2 + h(x)) > 1 + 4/3 = 7/3 = f(0) for x ≠ 0. So x = 0 is a local minimum → **(B) true, (A) false**. f is a decreasing function of h, so local maxima of f ↔ local minima of h, and local minima of f ↔ local maxima of h. h'(x) = (x cos x − sin x)/x² = 0 ⇔ tan x = x, with exactly one root in each interval (nπ, nπ + π/2), n ≥ 1: - (π, 3π/2): h local min → f local max - (2π, 5π/2): h local max → f local min - (3π, 7π/2): h local min → f local max - (4π, 9π/2): h local max → f local min - (5π, 11π/2): h local min → f local max In [π, 6π]: local maxima of f at 3 points → **(C) true**. In [2π, 4π]: local minimum of f at 1 point (in (2π, 5π/2)) → **(D) true**.

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