5. Find all values of the parameter $a$, for each of which the common solutions of the inequalities $x^{2}-4 x+2-a \leqslant 0$ and $x^{2}-5 x+2 a+8 \leqslant 0$ form a segment on the number line with a length of one.
Answer: $a=-1 ; a=-1.75$

Solution. In the plane $O x a$ we construct the sets of solutions of the inequalities (the part of the plane bounded by parabolas) $a \geqslant x^{2}-4 x+2$ and $a \leqslant-0.5 x^{2}+2.5 x-4$. In the common part of the solutions, we find the segments $A B$ and $C D$ of length one, corresponding to the values $a=-1$ and $a=-1.75$ respectively. There are no other values, as within the strip between the lines $A D$ and $B C$, all segments of length one are present, for $a \in(-1.75 ;-1)$ the solution forms a segment longer than one, outside the specified interval - shorter than one.

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