Question

一道数学题 着急要

Answer 1

1) g'(x)=2x-2+2a/x=2[(x-1/2)^2+(a-1/4)]/x,
Since a>1/4, when x<0, g'(x)<0, so g(x) is decreasing.
For the same reason, g(x) is increasing for x>0.

2) When a=0, g=(x-1)^2-1, a parabolic curve.
#1 Let's pick two points on the curve: A(Xa,Ya) and B(Xb,Yb),
then we get P(X0,Y0), with X0=(Xa+Xb)/2, Y0=(Ya+Yb)/2, and
Q(X0,g(X0)), with g(X0)=X0^2-2X0;
#2 Because Ya=Xa^2-2Xa, Yb=Xb^2-2Xb, we have
Y0=(Xa^2-2Xa+Xb^2-2Xb)/2=[(Xa+Xb)^2-2(Xa+Xb)-2XaXb]/2;
#3 Line AB: (y-Ya)/(x-Xa)=(Ya-Yb)/(Xa-Xb)=(Xa+Xb-2);
y=Ya+(Xa+Xb-2)(x-Xa) =(Xa+Xb-2)x+Xa^2-2Xa-(Xa+Xb-2)Xa
=(Xa+Xb-2)x-XaXb
#4 The tangent line of g(x) at Q(X0,g(x0)) is: y-g(X0)=2(X0-1)(x-X0), (X0, Y0) being P.
Since g(x0)=X0^2-2X0, we have the tangent line:
y=2(X0-1)(x-X0)+X0^2-2X0=2(X0-1)x-X0^2=(Xa+Xb-2)x-(Xa+Xb)^2/4
#5 from #3 and #4 we know the distance between the two lines are:
yAB-yTan=-XaXb+(Xa+Xb)^2/4=(Xa-Xb)^2/4