Question

已知函数fx=cos²(x-30°)-sin²x 若对于任意的x∈【0，π/2】，都有f（x)≤c，求实数c的范围

Answer 1

f(x)=cos²(x-30°)-sin²x
=1/2[1+cos(2x-60º)]-1/2(1-cos2x)
=1/2(cos2xcos60º+sin2xsin60º)+1/2cos2x
=√3/4sin2x+3/4cos2x
=√3/2(1/2sin2x+√3/2cos2x)
=√3/2sin(2x+π/3)
∵x∈【0，π/2】
∴2x∈[0,π]
∴2x+π/3∈[π/3,4π/3]
∴sin(2x+π/3)∈[-√3/2,1]
∴f(x)∈[-3/4,√3/2]
∵对于任意的x∈【0，π/2】，都有f（x)≤c
∴f(x)max=√3/2≤c
∴实数c的范围是c≥√3/2

Answer 2

f(x)=cos²(x-30°)-sin²x
=1/2[cos(2x-60°)+1+cos2x-1]
=1/2*2cos(2x-30°)cos(-30°)
=√3/2cos(2x-30°)
∵x∈[0,π/2]
2x-30°∈[-π/6,5π/6]
∴cos(2x-30°)∈[-1/2,1]
f(x)∈[-√3/4,√3/2]
∴C≥√3/2