Problem 11.2. Find all values of the real parameter $a$ such that the equation

$$
\lg (a x+1)=\lg (x-1)+\lg (2-x)
$$

has exactly one solution.

Aleksander Ivanov
11.2. The equation is equivalent to $x \in(1,2)$ and $a x+1=(x-1)(2-x)$, which can be written as $x^{2}+(a-3) x+3=0$. Therefore we have to find the values of $a$, such that the equation

$$
f(x)=x^{2}+(a-3) x+3=0
$$

has exactly one root in the interval $(1,2)$. This is possible exactly in the following four cases:

Case 1. $f(1) f(2)<0$, which is equivalent to $a \in\left(-1,-\frac{1}{2}\right)$.

Case 2. $f(1)=0$, i.e. $a=-1$. Then $x_{1}=1$ and $x_{2}=3$, which shows that $a=-1$ is not a solution.

Case 3. $f(2)=0$, i.e. $a=-\frac{1}{2}$. Then $x_{1}=2$ and $x_{2}=\frac{3}{2}$, which shows that $a=-\frac{1}{2}$ is a solution.

Case 4. $D=0$, i.e. $(a-3)^{2}-12=a^{2}-6 a-3=0$, whence $a=3 \pm 2 \sqrt{3}$. For $a=3+2 \sqrt{3}$ we have $x_{1}=x_{2}=-\sqrt{3}$, i.e. $a=3+2 \sqrt{3}$ is not a solution. For $a=3-2 \sqrt{3}$ we get $x_{1}=x_{2}=\sqrt{3} \in(1,2)$, i.e. $a=3-2 \sqrt{3}$ is a solution.

Finally, $a \in\left(-1,-\frac{1}{2}\right] \cup\{3-2 \sqrt{3}\}$.