The Problems of Ukrainian Authors

Let's build the graph of the equation, namely the function $y=x^2-3x[x]+2x=a$. It is easy to understand that we are interested only when $x>0$. $$\begin{array}{rl}x\in [0,1)& \Rightarrow [x]=0\Rightarrow y=x^2+2x\\ x\in [1,2)& \Rightarrow [x]=1\Rightarrow y=x^2-x\\ x\in [2,3)& \Rightarrow [x]=2\Rightarrow y=x^2-4x\\ x\in [3,4)& \Rightarrow [x]=3\Rightarrow y=x^2-7x\end{array}$$ While $x\ge 4$, the apex of the parabola $y=x^2+x(2-3[x])$ is situated at the point $x_b=\frac{3|x|-2}{2}\ge [x]+1$, which is equivalent to $[x]\ge 4$. From this, it follows that the function on every interval like $[n,n+1)$ under the condition $n\ge 4$ is downward. In addition, it is downward when $x\ge 4$, which is shown by the following transformations: $$\begin{gathered} x\in [n-1,n)\Rightarrow [x]=n-1\Rightarrow y_{n-1}=x^2+x(5-3n) \\ \text{and }{y}_{n-1}^{\min }>y_{n-1}(n)=n^2+n(5-3n); \end{gathered}$$ $$x\in [n,n+1)\Rightarrow [x]=n\Rightarrow y_n=x^2+x(2-3n)\text{ and }{y}_n^{\max }=y_n(n)=n^2+n(2-3n)<y_{n-1}^{\min }.$$ Thus, exactly two positive solutions of the function can exist only when $x\le 4$, and here it is easy to portray the study graph function. For easy perception, the scale is changed. So we can see exactly two positive solutions under the condition $0<a<2$, also on the interval $-\frac{49}{4}<a<-12$.