4. All values of the quadratic trinomial $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed 1 in absolute value. What is the greatest value that the quantity $|a|+|b|+|c|$ can have under these conditions? For which function $f(x)$ is this value achieved?
Answer: the maximum value is 7; for example, it is achieved for $f(x)=2 x^{2}-4 x+1$.

Solution. By the condition, the values of $f(x)=a x^{2}+b x+c$ on the interval $[0 ; 2]$ do not exceed one in absolute value. In particular, $|f(0)| \leqslant 1,|f(1)| \leqslant 1,|f(2)| \leqslant 1$, which is equivalent to the system

$$
\left\{\begin{array}{r}
|c| \leqslant 1 \\
|a+b+c| \leqslant 1 \\
|4 a+2 b+c| \leqslant 1
\end{array}\right.
$$

Since the absolute value of the sum of numbers does not exceed the sum of their absolute values, we obtain the inequalities

$$
\begin{aligned}
& |2 a|=|(4 a+2 b+c)-2(a+b+c)+c| \leqslant|4 a+2 b+c|+2|(a+b+c)|+|c| \leqslant 4 \\
& |2 b|=|4(a+b+c)-(4 a+2 b+c)-3 c| \leqslant 4|a+b+c|+|4 a+2 b+c|+3|c| \leqslant 8
\end{aligned}
$$

from which $|a| \leqslant 2,|b| \leqslant 4$, and thus, $|a|+|b|+|c| \leqslant 7$.

For the function $f(x)=2 x^{2}-4 x+1$, the quantity $|a|+|b|+|c|$ is exactly 7. Moreover, the quadratic trinomial $f(x)=2 x^{2}-4 x+1=2(x-1)^{2}-1$ satisfies the condition of the problem, since for $x \in[0 ; 2]$

$$
-1=f(1) \leqslant f(x) \leqslant f(0)=f(2)=1
$$

Thus, the maximum value of the quantity $|a|+|b|+|c|$ under the given conditions is 7.

Criteria. Correct answer -0 points. The inequality $|c| \leqslant 1$ is proven -1 point. Only one of the inequalities $|a| \leqslant 2$ or $|b| \leqslant 4$ is proven - another 2 points. It is proven that the sum of the absolute values does not exceed $7,-5$ points (not cumulative with previous points). Example of a function without justification of the estimate 2 points. Full solution -7 points.