Question

已知f(x)=-2asin(2x+π/6)+2a+b,x∈[π/4,3π/4],是否存在常数a,b∈Z,使得f(x)的值

Answer 1

π/4≤X≤3π/4

π/2≤2X≤3π/2

2π/3=π/2+π/6≤2x+π/6≤3π/2+π/6=5π/3

-2≤2sin(2x+π/6)≤√3

当a>0时,2a≥-2asin(2x+π/6)≥-√3a

2a+2a+b≥-2asin(2x+π/6)+2a+b≥-√3a+2a+b

所以4a+b=√3-1,-√3a+2a+b=-3

因为a,b∈Z,所以方程组无解

当a