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