2)f(x)+2b & lt； =0 has a solution, which is equivalent to B.

That is, the maximum value of b < = f(x)/2 (because if b is greater than the maximum value of f(x)/2, the inequality has no solution).

Just find the minimum value of f(x).

f(x)= x-lnx/x f '(x)= 1-( 1-lnx)/x =(x？ - 1+lnx)/x

Let f'(x)=0 x= 1, then the minimum value of f(x) is f( 1)= 1.

The maximum value of -f(x)/2 is-1/2.

So b < =- 1/2.

3)y=f(x) monotonically increases in the definition domain, and f'(x)=(x? -a+alnx)/x & gt； =0 holds.

Let g(x)=x? -a+alnx domain (0, +∞), then g(x)>=0 holds.

g'(x)=2x+a/x=(2x？ +a)/x

From a∈[m, 0], we can know that a < 0, so that g' (x) has zero x=√(-a/2).

So g(X) has a minimum value g (√ (-a/2)) =-a/2-a+a ln (√ (-a/2)) > = 0.

That is -3/2+ln (√ (-a/2)) < =0.

ln(-a/2)& lt； = 3

Get a>= -2 e^3

So the minimum value of m is -2 e 3.