已知函数f(x)=asinx+bcosx,且f(派/3)=1,则对任意实数a,b,函数f(x)的最大值的取值范围是

f(派/3)=1
1/2(√3a+b)=1
b=2-√3a
f(x)=asinx+bcosx
=√(a^2+b^2)sin(x+t),tant=b/a
=√(a^2+3a^2-4√3a+4)sin(x+t),
=2(√[(a-√3/2)^2+1/4]sin(x+t),
>=sin(x+t),a=√3/2等号成立
>=1(取得最大sin(x+t)=1）
f(x)>=1

f(π/3) =1, √3/2 a + 1/2 b =1, b = 2- √3a

f(x)=asinx+bcosx
= √(a^2+b^2) sin(x+θ) (其中cosθ = a/√(a^2+b^2))
f(x) ≤ √(a^2+b^2)
a^2+b^2
= a^2 + (2- √3a)^2
= 4a^2 - 4√3a + 4
= 4 [ (a- √3/2)^2 + 1/4]
≥ 1
f(x) ≥ 1