已知函数f(x）=√3sinxcosx-cosx平方+1/2（x∈R)（1）求函数f(x）的最小正周期（2）求函数f(x

f(x）=√3sinxcosx-cos^2x+1/2
=√3/2*sin2x-(cos2x+1)/2+1/2
=√3/2*sin2x-1/2*cos2x
=sin(2x-π/6)
（1）求函数f(x）的最小正周期T=2π/2=π
（2）因0≤x≤π/4
所以-π/6≤2x-π/6≤π/3
-1/2≤sin(2x-π/6)≤√3/2
函数f(x）[0,π/4]上的值域
[-1/2,√3/2]
(3）函数f(x)的单调递增区间2kπ-π/2≤2x-π/6≤2kπ+π/2（k∈Z）即
kπ-π/6≤x≤kπ+π/3（k∈Z）
函数f(x)的单调递减区间2kπ+π/2≤2x-π/6≤2kπ+3π/2（k∈Z）即
kπ+π/3≤x≤kπ+5π/6（k∈Z）