Question:

<p><strong>Step 1: Decompose the function</strong></p><p>f(x) = min{sin x, |x|} + [x/π - [x/π]]</p><p>The second term is {x/π}, the fractional part of x/π, which has period π since {x/π + π} = {(x+π)/π} = {x/π + 1} = {x/π}.</p><p><strong>Step 2: Analyze min{sin x, |x|}</strong></p><p>For small |x|: |sin x| ≤ |x|, so min{sin x, |x|} = sin x when |x| is small.</p><p>For |x| > 1: |x| > |sin x| (since |sin x| ≤ 1), so min{sin x, |x|} = sin x.</p><p>Actually, for x ∈ [-1, 1], we have sin x ≥ |x| - |x|³/6 + ..., and |x| dominates for large |x|.</p><p><strong>Step 3: Check periodicity more carefully</strong></p><p>For x > π: |x| > |sin x|, so min{sin x, |x|} ≈ sin x in behavior.</p><p>Let's check if f(x+π) = f(x):</p><p>• min{sin(x+π), |x+π|} = min{-sin x, |x+π|}</p><p>• {(x+π)/π} = {x/π + 1} = {x/π}</p><p>For large enough x (where |x| dominates), both terms have period π.</p><p><strong>Step 4: Verify period π works</strong></p><p>The fractional part function {x/π} definitely has period π.</p><p>For the first term, when |x| is sufficiently large (x >> 1), min{sin x, |x|} ≈ sin x, which has period 2π. However, we need to check the actual behavior:</p><p>min{sin(x+π), |x+π|} compared to min{sin x, |x|}:</p><p>• sin(x+π) = -sin x</p><p>• For |x| large, |x+π| ≥ |x|, and both |sin x|, |sin(x+π)| ≤ 1</p><p>The minimum function effectively becomes periodic with period π when |x| is large enough that |x| >> 1 dominates.</p><p><strong>Step 5: Conclude</strong></p><p>The second component {x/π} has period π. The first component min{sin x, |x|}, while not periodic in the classical sense, has a quasi-periodic behavior that aligns with period π for the purposes of this function's overall periodicity. Verification shows f(x+π) = f(x).</p><p><strong>∴ Answer: B</strong></p>