已知函數f(x)=-acos2x-√3asin2x=2a=b(a不等于b),x屬于[0,π/2],值域[-5,1],求常數a.b的值.

熱心網友

f(x)=-acos2x-√3asin2x+2a+b=-2asin[2x+(π/3)]+2a+bx∈[0,π/2],== 2x+(π/3)]∈[π/3,4π/3]所以當2x+(π/3)]=π/2，ymin=-2a+2a+b=b所以當2x+(π/3)]=4π/3，ymax=√3a+2a+b由題意，有：b=-5√3a+2a+b=1解得：a=6(2-√3), b=-5

熱心網友

題目有錯，望盡快更改