已知函数f(x)=2asin^2x-(2根号3)asinxcosx+a+b的定义域为[0,π/2],值域为[-5,1],

题目

已知函数f(x)=2asin^2x-(2根号3)asinxcosx+a+b的定义域为[0,π/2],值域为[-5,1],

答案

2asin^x-2根号3asinxcosx+a+b
=2asin²x-√3a(2sinxcosx)+a+b
=-(a-2asin²x)+a-√3a(2sinxcosx)+a+b
=-acos2x-√3asin2x+2a+b
=-2a[1/2cos2x+√3/2sin2x]+2a+b
=-2asin(2x+π/6)+2a+b
又x∈[π/2,π],
则2x+π/6∈[7π/6,13π/6]
sin(2x+π/6)∈[-1,1/2]
①若a>0
则值域是[a+b,4a+b]
有a+b=2 4a+b=5
得a=b=1
②若a