Question

已知关于x的不等式(kx-k^2-1)(x-1)<0

1.当k<=0时，解该不等式
2.当k>=1时，设该不等式的解集为(a,b),求(1/a)-(1/b)的取值范围
要过程

Answer 1

1. k<0, 根为x1=1,x2= k+1/k<=-2,
(x-x1)(x-x2)>0, x>1 or x<k+1/k
k=0, -(x-1)<0, x-1>0, x>1
2. k>=1, k+1/k>=2,
1<x<k+1/k
a=1
b=k+1/k>=2, 0<1/b<=1/2
1/2=<1/a-1/b<1,

Answer 2

1. k+1/k<x<1
2. 1< x<k+1/k 所以a=1 b=k+1/k>=2
以(1/a)-(1/b)<=1/2
取值范围小于等1/2