F'(x)=2x+a/x, domain (0, infinity) 1a =-2, f' (x) = 2x-2/x = 2 (x+1)/x makes F.

0 & ltx & lt= 1;

Let f (x) gt; =0,

x & gt= 1;

Let f(x)=0,

X= 1∴f(x) decreases at (0, 1) and monotonically increases at [1, infinity], and the extreme value is f (1) =12g (x) = X.

g'(x)=2x+a/x-2/x^2>； =0 is constant on [1, infinity], and immediately a >；; = 2/x-2x 2 hold,

Let h(x) = 2/x-2x 2, that is, find the maximum value of h(x) on [1, infinity] H' (x) =-2/x 2-2x.

The maximum value is obtained when x= 1, h (1) = 2-2 = 0 ∴ a >; =02)①f(x)= 1， g(x)=c(n)(0)x^0( 1-x)^n+c(n)( 1)x^ 1( 1-x)^(n- 1)+......+c(n)(n)x^n( 1-x)^0=[x+( 1-x)]^n= 1 ②f(x)=xg(x)=( 1/n)c(n)( 1)x^ 1( 1-x)^(n- 1)+......+(n/n)c(n)(n)x^n( 1-x)^0(k/n)c(n)(k)=(k/n)*n！ /[k！ (n-k)！ ]=(n- 1)！ /[(k- 1)！ (n-k)！ ]=c(n- 1)(k- 1)∴g(x)=c(n- 1)(0)x^ 1( 1-x)^(n- 1)+......+c(n- 1)(n- 1)x^n( 1-x) ^0=x[c(n- 1)(0)x 0( 1-x)(n- 1)+...+c(n- 1)(n- 1)x(n- 1)( 1)