Suppose x> when f(x) has only one zero at 0, then ax? ≤ e x is a constant, that is, a ≤ e x/x? Let f (x) = a = e x/x? So a' = f' (x) = (e x x? -e^x 2x)/x^4=e^x(x-2)/x？ =0, F(x)min=2, that is, a≤e? /4 when f(x) has only one zero point on (0, +∞), then when a >, f(x) has two zero points on (0, +∞); e？ /4

(2) We might as well set 0; 2-x 1 >0

By f(x)=ax? -ax when there are two zeros (0, +∞) on e x? -e^x>； 0 has ax? & gtE x is 4/e? & gtx？ /e^x>； 0

Let F(x)=x? /e x so f' (x) = (2x e x-x? e^x)/(e^x)？ = x (2-x)/e x is obviously 0.