f(x)=ab=√3msinxcosx-mcosxcosx

=(√3/2)msin2x-(m/2)[2(cosx)^2- 1+ 1]

=msin2xcosπ/6-mcos2xsinπ/6-m/2

=msin(2x-π/6)-m/2

X∈[0，π/2]，2x-π/6 ∈ [-π/6，5π/6]，sin (2x-π/6) ∈ [- 1/2， 1]。

M>0, f (x) =-m/2-m/2 =-m =-4m = the minimum value of 4.

Maximum value =m-m/2=m/2=2, 2x-π/6=π/2 x=π/3.

The minimum value of m<0 and f(x) =m-m/2=m/2=-4 m=-8.

Maximum value =-m/2-m/2 =-m = 8, 2x-π/6 =-π/6, and x = 0.

2.f(x)=[ 1/(2^x- 1)+ 1/2]x^3

2 x- 1 ≠ 0, that is, x≠0.

The domain of ∴f(x) is (-∞, 0)∞(0,+∞).

f(-x))=[ 1/(2^(-x)- 1)+ 1/2](-x^3)

=[2^x/( 1-2^x)+ 1/2](-x^3)

={[2*2^x+ 1-2^x]/2(2^x- 1)}(x^3)

=[(2^x+ 1)/2(2^x- 1)]x^3

=[(2^x- 1 +2)/2(2^x- 1)]x^3

=[ 1/(2^x- 1)+ 1/2]x^3

=f(x)

∴f(x) is an even function

f(x)=[ 1/(2^x- 1)+ 1/2]x^3

When x>0, x 3 > 0

2^x>； 1,2^x- 1>； 0, 1/(2^x- 1)+ 1/2>； 0

∴f(x)>； 0

F(x) is an even function. When x, the image is symmetric about y ∴.

∴x∈(-∞,0)∪(0,+∞),f(x)>； 0