f(x)=xe^ax- 1/x,x≠0

f(x)=x(e^ax- 1/x？ )x-& gt； 0+，f(x)-& gt； -∞; x-& gt； 0-，f(x)-& gt； +∞, x=0 is a breakpoint, but not a zero point.

F(x) has two zeros, which is equivalent to g (x) = e ax- 1/x? There are two zeros.

For example, y = (e a) x and y= 1/x? Intersection point, as shown in the figure below:

A = 0，e 0 = 1，y = (e A) x = 1 x = 1，y= 1/x？ There is an intersection on each side;

e^a>； 1=e^0,a>； 0 and one

As shown above, there must be an intersection point on the right side of the Y axis;

On the left side of the y axis, x

When tangent, the slopes of two curves are equal at the tangent point. Tangent points are also intersections.

Let the intersection (x0, y0), x0.

y 1=e^ax,y 1'=ae^x,>； 0y0=e^ax0

y2= 1/x？ ，y2'=-2/x？ & gt0,

ae^x0=-2/x0？ ;

y0=e^ax0

y0= 1/x0？

ay0=-2/x0？ ;

a=(-2/x0？ )/( 1/x0？ )=-2x0，x0=-a/2

y0= 1/(a？ /4)=4/a？

ay0=-2/(-a？ /8)= 16/a？ ,y0= 16/a^4

4/a？ = 16/a^4

1=4/a？

Answer? =4，a=2

∴-2≤a≤2