已知f(x)=x^3+ax+b定义在区间[-1,1]上,且f(0)=f(1),又P(x1,y1),Q(x2,y2)是其图象上任意两点(x1≠x2).
1.设直线PQ的斜率为k,求证:|k|<2.
2.若O≤x1<x2≤1,求证:|y1-y2|<1.

f(0) = b,
f(1) = 1+a+b,
f(0) = f(1);
b = 1+ a +b, a = -1;
f(x) = x^3 -x +b;

1)
y1 = x1^3 - x1 + b;
y2 = x2^3 - x2 + b;
y2 - y1 = x2^3 - x1^3 - (x2 -x1)
=(x2-x1)*(x2^2 + x1^2 + x1*x2 -1);
k = (y2-y1)/(x2-x1)
= (x2^2 + x1^2+ x1*x2-1)
<=x2^2 + x1^2 + (x1^2+x2^2)/2 -1
|x1|^2 <=1,
|x2|^2 <=1,

|k| < 3(x1^2+x2^2)/2 -1
= 3-1 = 2 , 由于 x1 <>x2, 故不能取得等号,

|k| <2;


2)
0<=x1 < x2<=1,
g(x) = x^3-x;
g'(x) = 3x^2 -1;
x = sqrt(3)/3, g(x) = 0,
0<=x<=sqrt(3)/3, g'(x)<0, g(x) 单调下降,
sqrt(3)/3 < x <=1, g'(x) >0, g(x)单调上升,
g(0)=g(1) = 0,
0<x<1,
g(x) <0;
g(x) >= g(sqrt(3)/3);
g(sqrt(3)/3) = sqrt(3)/9 - sqrt(3)/3
= -2*sqrt(3)/9

y2 - y1 = g(x2) - g(x1);
0<=x1<x2<=1;
|y1-y2| =|g(x1)-g(x2)|
<|g(sqrt(3))/3| -|g(0)|
= 2*sqrt(3)/9 <1;

从而，O≤x1<x2≤1,
|y1-y2|<1.