1. Given the quadratic trinomial $f(x)=a x^{2}-a x+1$. It is known that $|f(x)| \leqslant 1$ for all $x \in[0,1]$. What is the greatest value that $a$ can take?
Answer: 8

Solution. It is not difficult to check that $a=8$ works. Indeed, $|2 x-1| \leqslant 1$ for $x \in[0,1]$, so $f(x)=8 x^{2}-8 x+1=2(2 x-1)^{2}-1 \leqslant 1$, and the inequality $f(x) \geqslant-1$ holds for all $x$.

Suppose that $a>8$. Then

$$
f\left(\frac{1}{2}\right)=\frac{a}{4}-\frac{a}{2}+1=1-\frac{a}{4}=\frac{4-a}{4}<-1
$$

which is impossible by the condition.