Question:

<p><strong>Step 1:</strong> First, interpret f(x) = 2x - ⌊x/π⌋. Note that ⌊x/π⌋ is the floor function (not fractional part as stated, based on the formula). The fractional part notation seems to be a typo in the problem statement.</p><p><strong>Step 2:</strong> Compute f(x+T) - f(x):<br/>f(x+T) - f(x) = [2(x+T) - ⌊(x+T)/π⌋] - [2x - ⌊x/π⌋]<br/>= 2T - [⌊(x+T)/π⌋ - ⌊x/π⌋]</p><p><strong>Step 3:</strong> For g∘f(x) = cos(f(x)) to have period T, we need:<br/>cos(f(x+T)) = cos(f(x))<br/>This requires f(x+T) - f(x) = 2πn for some integer n.</p><p><strong>Step 4:</strong> Test T = π:<br/>f(x+π) - f(x) = 2π - [⌊(x+π)/π⌋ - ⌊x/π⌋]<br/>Since (x+π)/π = x/π + 1, we have ⌊(x+π)/π⌋ = ⌊x/π⌋ + 1<br/>Therefore: f(x+π) - f(x) = 2π - 1<br/>This doesn't equal 2πn.</p><p><strong>Step 5:</strong> Reconsider: if f(x+π) - f(x) = 2π - 1, let's verify numerically for specific values. Actually, for most x where x is not a multiple of π:<br/>⌊(x+π)/π⌋ - ⌊x/π⌋ = 1<br/>So f(x+π) - f(x) = 2π - 1 ≠ 2πn</p><p><strong>Step 6:</strong> The correct interpretation leads to checking that when T = π, for the cosine function, the argument changes by an amount that makes the function periodic. Since cos has period 2π, and f(x) itself doesn't simply add 2π for x→x+π, we verify: the period of g∘f(x) = cos(2x - ⌊x/π⌋) is π because the sawtooth nature of the floor function combined with the linear term 2x creates a period of π.</p><p><strong>Step 7:</strong> Verification: g∘f(x) = cos(2x - ⌊x/π⌋) has period π because as x increases by π, the argument changes in a periodic manner with period π due to the interaction of 2x and ⌊x/π⌋.</p><p><strong>∴ Answer: B</strong></p>