f(x)=√3sin2x+cos2x=2 (√3/2sin2x+1/2cos2x)=2 (cosπ/6sin2x+sinπ/6cos2x)=2sin(2x+π/6)因为x∈ [0,π/2]所以2x+π/6∈ [π/6,7π/6]所以sin(2x+π/6)∈ [-1/2,1]所以2sin(2x+π/6)∈ [-1,2]所以函数f（x）在区...