4. (20 points) For what values of the parameter $a$ does the equation

$$
(a+1)(|x-2.3|-1)^{2}-2(a-3)(|x-2.3|-1)+a-1=0
$$

have exactly two distinct solutions?
Solution. Let $|x-2.3|-1=t$ (1), then the original equation will take the form: $(a+1) t^{2}-2(a-3) t+a-1=0$ (2). Let's analyze equation (1): when $t>-1$, it corresponds to two different values of x. Thus, the original equation can have from zero to four solutions. It has two distinct roots in the following three cases for equation (2):

1) the linear case, if the only root is greater than $(-1)$;
2) $D=0 ; t_{b}>-1$;
3) equation (2) has two distinct roots, one of which is greater than (-1), and the other is less than (-1).

The case where one root is greater than (-1) and the other is equal to (-1) does not suit us, as in this case there will be three solutions. Let's investigate the above cases. 1) $a=-1 ; 8 t-2=0 ; t=\frac{1}{4}>-1$ therefore, this value of the parameter is included in the answer.

2) $\frac{D}{4}=(a-3)^{2}-(a+1)(a-1)=10-6 a=0 ; \quad a=\frac{5}{3}$. The only root $-t_{b}=\frac{a-3}{a+1}$; $t_{6}\left(\frac{5}{3}\right)=\frac{\frac{5}{3}-3}{\frac{5}{3}+1}=-\frac{1}{2}>-1, a=\frac{5}{3}$ is included in the answer.
3) The general case - the roots are on different sides of (-1) - is described by the inequality $(a+1) f(-1) the correct answer is insufficiently justified. | 15 |
| The substitution of the variable is made and the correct restrictions for the new variable are defined. Attempts are made to write down some restrictions for the coefficients in connection with the conditions for the new variable, but they are only partially correct. | 10 |
| The substitution of the variable is made, the problem is reduced to the study of a quadratic trinomial with a parameter, but the reasoning is limited to considering the discriminant. | 5 |
| The solution does not correspond to any of the criteria listed above. | 0 |
| Maximum score | 20 |