
Problem 8'1. Given an inequality $|x-1|<a x$, where $a$ is a real parameter:

a) Solve the inequality.

b) Find all values of $a$ such that the inequality has exactly two integer solutions.

Chavdar Lozanov, Kiril Bankov, Teodosi Vitanov


Solution: a) I. Let $x \geq 1$. Then the inequality is equivalent to $x-10, a1 \Longleftrightarrow a>0$. Therefore $00 \Longleftrightarrow a>-1, \frac{1}{a+1}0$. Therefore $a>0, \frac{1}{a+1}<x<1$, when $-1<a \leq 0$ no solution exists.
2. $a+1=0 \Longleftrightarrow a=-1 \Longrightarrow 1<0 \cdot x$, so no solution exists.
3. $a+1<0 \Leftrightarrow a<-1 \Longrightarrow x<\frac{1}{a+1}<0$. Therefore $a<-1, x<\frac{1}{a+1}$.

So when $a<-1$, then $x<\frac{1}{a+1}$; for $-1 \leq a \leq 0$ no solution exists; when $0<a<1$, then $\frac{1}{a+1}<x<\frac{1}{1-a}$; when $1 \leq a$, then $1 \leq x$.

b) It follows from a) that the inequality could have two integer solutions only if $0<a<1$. Since in this case $0<\frac{1}{a+1}<1<\frac{1}{1-a}$, we find that there are exactly two integer solutions if and only if

$$
2<\frac{1}{1-a} \leq 3
$$

Therefore the answer is $\frac{1}{2}<a \leq \frac{2}{3}$.
