已知函数f(x)=2sin(2x+π/6)若存在常数a,b,使得函数F（x）=af(x)+2a+b,x∈[π/4,3π/4]的值域_数学解答

已知函数f(x)=2sin(2x+π/6)
若存在常数a,b,使得函数F（x）=af(x)+2a+b,x∈[π/4,3π/4]的值域为[-√3,(√3)-1],求ab的值

人气：452 ℃　时间：2020-03-26 20:59:43

解答

π/4≤x≤3π/4,
π/2≤2x≤3π/2,
2π/3≤2x+π/6≤5π/3,
-2≤2sin(2x+π/6)≤√3,
-2≤f(x)≤√3.
a>0
b≤F(x)≤√3a+2a+b
b=-√3,
√3a+2a+b=(√3)-1,
a=(2(√3)-1)/(√3+2)>0,
ab=(√3-6)/(√3+2).
a