The maximum value of g(t) is g(2)= 1.
F(s)>= 1 is established in Shanghang.
a/x+xlnx & gt; = 1
a & gt=x-x^2lnx
Let h (x) = x-x 2 lnx.
h`(x)= 1-2xlnx-x
Let h ` (x) = 0 and x = 1.
H(x) is declining.
The maximum value of h(x) is h( 1)= 1.
∴a>; = 1
(1)f'(x)= 1/x-a, which is the decreasing function in the interval (1, +∞), that is, when x >: 1, f'(x).
So 1/x-a
1/x & lt; a
Get a> 1.
g(x)'=e^x-a
According to the meaning of the question, (1, +∞) should have a minimum value, that is, when x >; 1, g'(x)>0, used to increase the function, so:
e^x-a>; 0
e^x>; a
Namely: e>a.
So the range of a is: (1, e).
(2) g (x)' = e x-a, and the interval (-1, +∞) is monotone increasing function, that is, when x >;; -1, g' (x) >: 0 is an increasing function, so:
e^x-a>; 0
e^x>; a
e^x>; e^(- 1)>; a
Then: a <1/e.
At this time f'(x)= 1/x-a,
When 0
When e
When x>1/a > e,f' (x)
So there is only one zero point.