4. Find all values of parameters $a$ and $b$, for which the inequality

$$
\left|x^{2}+a x+b\right| \leqslant \frac{1}{8}
$$

holds for all values $x \in [0 ; 1]$.

(P. Alishev)
Answer: $a=-1, b=\frac{1}{8}$.

Solution. The maximum value of the function $f(x)=\left|x^{2}+a x+b\right|$ on the interval $[0 ; 1]$ is achieved at critical points within the interval or at the endpoints. $f^{\prime}(x)=\frac{x^{2}+a x+b}{\left|x^{2}+a x+b\right|} \cdot(2 x+a)$, the critical points are the roots of the equation $x^{2}+a x+b=0$ and $x=-0.5 a$. At zero and one, the inequality must be satisfied. Therefore, $f(0)=|b| \leqslant \frac{1}{8}, f(1)=|a+b+1| \leqslant \frac{1}{8}$. From this, it is not difficult to obtain the following constraints on $a$ and $b: -\frac{1}{8} \leqslant b \leqslant \frac{1}{8}$, $\frac{3}{4} \leqslant \frac{7}{8}+b \leqslant-a \leqslant \frac{9}{8}+b \leqslant \frac{5}{4}$, i.e., $-0.5 a \in[0 ; 1]$. For the roots of the equation $x^{2}+a x+b=0$, the original inequality will obviously be satisfied. All permissible values of the parameters are found from the solution of the system:

$$
\left\{\begin{array}{l}
-\frac{1}{8} \leqslant b \leqslant \frac{1}{8} \\
|a+b+1| \leqslant \frac{1}{8} \\
\left|-\frac{a^{2}}{4}+b\right| \leqslant \frac{1}{8}
\end{array}\right.
$$

We will plot all points on the plane that satisfy each of the inequalities. On the x-axis, we will plot the values of the parameter $a$, and on the y-axis, the values of the parameter $b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8c989a3fb7b602838580g-4.jpg?height=1196&width=1568&top_left_y=133&top_left_x=264)

Only the point $(-1 ; 0.125)$ falls within the solution of each inequality. We will prove more rigorously that there are no other solutions to the system

$$
\left\{\begin{array}{l}
-\frac{1}{8} \leqslant b \leqslant \frac{1}{8} \\
-a-\frac{9}{8} \leqslant b \leqslant-a-\frac{7}{8} \\
\frac{a^{2}}{4}-\frac{1}{8} \leqslant b \leqslant \frac{a^{2}}{4}+\frac{1}{8}
\end{array}\right.
$$

If $a^{2}>1$, then from the first and third inequalities we get: $b \leqslant \frac{1}{8}<\frac{a^{2}}{4}-\frac{1}{8} \leqslant b$, which is impossible. If $a^{2}<1$, then $b \leqslant-a-\frac{7}{8}<\frac{a^{2}}{4}-\frac{1}{8} \leqslant b$, which also has no solution. When $a^{2}=1$, from the third inequality we immediately get $b=\frac{1}{8}$, and then from the second $a=-1$.