Question

函数y=x^2+ax-2/x^2-x+1的值域y≤2,求a的值.

Answer 1

函数y=x^2+ax-2/x^2-x+1的值域y≤2,求a的值.

y=(x^2+ax-2)/(x^2-x+1)

=[(x^2-x+1)+(a+1)x-3]/(x^2-x+1)

=1+[(a+1)x-3]/(x^2-x+1)

值域为(-∝,2],则

1+[(a+1)x-3]/(x^2-x+1)=1

f(x2)-f(x1)=f(x2)-f(x2+(x1-x2))=f(x2)-f(x2)*f(x1-x2)=f(x2)[1-f(x1-x2)]