Question

y=Sin(π/4-wx)且（W>0) Y最小正周期为π,求单调递增区间?若X∈【0,π/2】求函数值域?

Answer 1

①最小正周期T=2π/w=π,所以w=2,即y=sin(π/4-2x)

②y=sin(π/4-2x)=-sin(2x-π/4)

令t=2x-π/4,则2kπ+π/2≤t≤2kπ+3π/2为增区间,从而求出kπ+3π/8≤x≤kπ+7π/8

③因为x∈[0,π/2],所以(π/4-2x)∈[-3π/4,π/4],所以函数的值域为[-1,√2/2]

当π/4-2x=-π/2时有最小值-1,当π/4-2x=π/4时有最大值√2/2