(Adapted from 12 questions of the 2018 Zhejiang Provincial Competition)
Let $a\in R$, and for any real number $b$ have $\max\limits_{x\in[0,1]}|x^2+ax+b |\ge\dfrac{1}{4}$ find the range of $a$.
Answer: According to the meaning of the question $\min\limits_{b\in R}{\max\limits_{x\in[0,1]}{|x^2+ax+b|}}\ge\dfrac{1} {4}$
remember $N=\max\limits_{x\in[0,1]}{|x^2+ax+b|}$
$$\because\min\limits_{b\in R}N=
\begin{cases}
\dfrac{1+a}{2},&a\ge0,\\
\dfrac{(a+2)^2}{8},&a\in[-1,0),\\
\dfrac{a^2}{8},&a\in[-2,-1),\\
-\dfrac{1+a}{2},&a<-2,
\end{cases}
\therefore a\in(-\infty,-\sqrt{2}]\cup [\sqrt{2}-2,+\infty)$$
Note: The right side of the original question of the provincial competition is 1, which can be solved by the absolute value inequality, but after changing to $\dfrac{1}{4}$, the absolute value is very risky and several absolute value inequalities must be considered. The importance of general law.