Let f(x) = sin(π/6 sin(π/2 sin x)) for all x ∈ R and g(x) = π/2 sin x for all x ∈ R. Let (f◦g)(x) denote f(g(x)) and (g◦f)(x) denote g(f(x)). Then which of the following is (are) true?

f(x) = sin(π/6 sin(π/2 sin x)) π sin x /2 varies from -π/2 to π/2. Let u = π/2 sin x. Then -π/2 ≤ u ≤ π/2. sin u varies from -1 to 1. Therefore, π/6 sin u varies from -π/6 to π/6. Finally, sin(π/6 sin u) varies from sin(-π/6) to sin(π/6), which is from -1/2 to 1/2. Therefore the range of f(x) is [-1/2, 1/2]. However, if we consider the case when sin(π/2 sin x) is near 0, we have sin(π/6 sin(π/2 sin x)) ≈ π/6 sin(π/2 sin x) ≈ π/6 (π/2 sin x) = π²/12 sin x. The range of this approximation is [-π²/12, π²/12] which is approximately [-0.822, 0.822]. Let's examine the range of f(x) more precisely. Since -1 ≤ sin x ≤ 1, we have -π/2 ≤ π/2 sin x ≤ π/2. Then -1 ≤ sin(π/2 sin x) ≤ 1. Then -π/6 ≤ π/6 sin(π/2 sin x) ≤ π/6. Therefore -sin(π/6) ≤ sin(π/6 sin(π/2 sin x)) ≤ sin(π/6). -1/2 ≤ f(x) ≤ 1/2. For g(x) = π/2 sin x, we have -π/2 ≤ g(x) ≤ π/2. Then f(g(x)) = sin(π/6 sin(g(x)) = sin(π/6 sin(π/2 sin x)) which has range [-1/2, 1/2]. However, option C states that the range of f◦g is [-1/4, 1/2]. This is incorrect. Let's look at (g◦f)(x) = π/2 sin(f(x)). Since the range of f(x) is [-1/2, 1/2], the range of sin(f(x)) is approximately [-1/2, 1/2]. Therefore, the range of (g◦f)(x) is [-π/4, π/4]. Thus, there exists an x such that (g◦f)(x) = 1 is false.