Question

若函数f (x)=(x^2-2x+k)/(x-1)在区间（2,+无穷大）上递增,则实数K的取值范围是（用定义法解答）

Answer 1

x1>x2>0

f(x1)-f(x2)

=(x1²-2x1+k)/(x1-1)-(x2²-2x2+k)/(x2-1)

通分

分母>0

分之=x1²x2-x1²-2x1x2+2x1+kx2-k-x1x2²+x2²+2x1x2-2x2-kx1+k

=x1x2(x1-x2)-(x1+x2)(x1-x2)+2(x1-x2)-k(x1-x2)

=(x1-x2)(x1x2-x1-x2+2-k)

x1-x2>0

所以要x1x2-x1-x2+2-k>0

(x1-1)(x2-1)>k-1

因为x1-1>1,x2-1>1

所以1>k-1

k