f(x)=ax^2+lnx,x>； 0

f'(x)=2ax+( 1/x)

Because f'( 1)=0, there is.

2a+ 1=0, and the solution is a=- 1/2.

That is f (x) = (- 1/2) x 2+lnx, x >；; 0

So f' (x) =-x+(1/x) = (-x2+1)/x.

Let f'(x)=0, x >；; 0

It is found that f(x) is increasing function at (0, 1); It is a decreasing function on [1,+infinity].

And f( 1) is the largest, so the minimum value of f(x) on [1/e, e] may be f(e) or f( 1/e).

f( 1/e)=(- 1/2e^2)- 1,f(e)=(-e^2/2)+ 1

F(e)-f( 1/e) difference comparison found that there are

f(e)-f( 1/e)=(4e^2-e^4+ 1)/2e^2

f(e)-f( 1/e)=[4-e^2+( 1/e^2)]/2<； 0

That is, f (e) < f( 1/e)

So f (x) min = f (e) = (-e 2/2)+ 1, choose a.