2. Find all real numbers $a$ and $b$ such that for every $x \in[-1,1]$, the inequality

$$
\left|2 x^{2}+a x+b\right| \leq 1
$$

holds.
2. Let $a$ and $b$ be such that for every $x \in [-1,1]$, the inequality $-1 \leq 2 x^{2} + a x + b \leq 1$ holds. By substituting $x=0$, $x=1$, and $x=-1$ respectively, we get

$$
\begin{aligned}
& -1 \leq b \leq 1 \\
& -3 \leq b + a \leq -1 \\
& -3 \leq b - a \leq -1
\end{aligned}
$$

10 points

By adding the last two inequalities, we get $-3 \leq b \leq -1$.

From this and the first inequality, it follows that $b = -1$.

5 points

Then $a$ satisfies the inequalities $-2 \leq a \leq 0$ and $-2 \leq -a \leq 0$, i.e., $a = 0$. \quad 5 points

Thus, if there exist numbers $a$ and $b$ for which the given inequality holds for every $x \in [-1,1]$, then it must be $a = 0$ and $b = -1$.

Let's verify that these numbers satisfy all the conditions in the problem, i.e., that for every $x \in [-1,1]$, the inequality $-1 \leq 2 x^{2} - 1 \leq 1$ holds. This inequality is equivalent to $0 \leq 2 x^{2} \leq 2$, which is satisfied for every $x \in [-1,1]$. 5 points