=(a/2)sin2x-cos2x

f(0)=- 1，f(-π/3)=-(√3)a/4+ 1/2

Then: -(√3)a/4+ 1/2=- 1

-(√3)a/4=-3/2

a=2√3

Therefore, f(x)=√3sin2x-cos2x=2sin(2x-π/6).

x∈

That is, the maximum value of f(x) on x∈[π/4, 1 1π/24] is 2, and the minimum value is √2.