a = 1，b = 0，f(x)=[ln( 1-x)]/(x- 1)
Domain x < 1
f '(x)= {[ln( 1-x)]'(x- 1)-[ln( 1-x)](x- 1)' }/(x- 1)？ =[ 1-ln( 1-x)]/(x- 1)？ f'(x) = 0， 1 - ln( 1 - x) = 0， 1 - x = e，x = 1 - e
x & lt 1 - e，f '(x)& lt； 0, negative function
1-e & lt； X< 1, f'(x)>0, increasing function.
The minimum value f (1-e) = ln (1-kloc-0/+e)/(1-e-1) =-1/e.
(2)?
B = 1, f (x) = [ln (a-x)]/(x-a)-x, and the domain x < a.
f '(x)= {[ln(a-x)]'(x-a)-[ln(a-x)](x-a)' }/(x-a)& amp； # 178; ? - 1 =[ 1-ln(a-x)]/(x-a)？ - 1
f'(x) = 0， 1 - ln(a - x) = (a - x)？
For convenience, let t = a-x > 0, 1-lnt = t?
On the left is a monotonic decreasing function, and on the right is a monotonic increasing function with only one intersection. It is easy to see that t = 1 is its only solution: a-x = 1.
x = a - 1
x & lta - 1，f '(x)& lt； 0, negative function. Suppose that X is a negative number with great absolute value, ln(a-x) is a positive number,1-LN (a-x) <)
a- 1 & lt； X & lta:f'(x)>0, increasing function.
The red line in the figure is f(x) when a = 2, and the black line is x = 1 (a- 1).
(3)
Don't ask, continue later.
In this case, f (x) = [ln (3-x)]/(x-3)+X.
(x - 3)f(x) = c，ln(3 - x) + x(x - 3) = c
G(x) = ln(3-x)+x(x-3)-c, and the domain X.
g '(x)= 1/(x-x)+2x-3 =(x-2)(2x-5)/(x-3)= 0
X<3, the molecule is always less than 0.
X<2, the molecule is positive, g' (x)
2<x & lt5/2, the molecule is negative, g' (x) >; 0
5/2 & lt； x & lt3:? The molecule is positive, g' (x)
G(2) = -2-c is the minimum value.
G(5/2) = -ln2-5/4-c is the maximum value.
To make it have three unequal real roots in [1, 2 1/8], the following conditions must be met at the same time:
(i) g( 1) = ln2 - 2 - c？ ≥ 0，c？ ≤ ln2 - 2
(ii)g(2)=-2-c & lt； 0，c & gt-2
(3) g (5/2) =-LN2-5/4-C > 0，c & lt-ln2 - 5/4
(iv)g(2 1/8)= ln(3/8)-63/64-c？ ≤ 0，c？ ≥ ln(3/8) - 63/64？
Compared with the picture below, imagine translating the blue line up and down, which is easy to understand.
Looking at the above lines, the result is ln(3/8)-63/64? ≤c & lt； -ln2 - 5/4