Question

函数f（x）=2cos平方x+2*根号3sinxcosx+1x∈【π/6,π/2】的值域

Answer 1

f(x)＝2cos²x＋2√3sinxcosx＋1＝(2cos²x－1)＋√3×(2sinxcosx)＋2

＝cos(2x)＋√3sin(2x)＋2＝2[1/2×cos(2x)＋√3/2×sin(2x)]＋2

＝2sin(2x＋π/6)＋2

∵x∈[π/6,π/2] ∴2x∈[π/3,π] ∴2x＋π/6∈[π/2,7π/6]

∴sin(2x＋π/6)∈[﹣1/2,1] ∴f(x)∈[1,4]