题目

已知函数f(x)=-acos2x-2√3asinxcosx+2a+b的定义域为【0,2/π],值域为【-5,1】.求常数a,b的值

提问时间：2020-06-13

答案

∵f(x)=-acos2x-√3asin2x+2a+b=-2a(cos2x/2+√3/2*sin2x)+2a+b=-2asin(2x+π/6)+2a+b∵0≤x≤π/2∴π/6≤2x+π/6≤7π/6,∴-1/2≤sin(2x+π/6)≤1,当a＞0,f(x)max=a+2a+b=3a+b=1,f(x)min=-2a+2a+b=b=-5,解得a=2,b=...