Xác định đa thức $f(x)=x^{2}+ax+b$ biết rằng $\left | f(x) \right |\leq \frac{1}{2}$ với mọi x thỏa mãn $-1\leq x\leq 1$
$\left | f(-1) \right |\leq \frac{1}{2}\Rightarrow \left | 1-a+b \right |\leq \frac{1}{2}$
$\left | f(1) \right |\leq \frac{1}{2}\Rightarrow \left | 1+a+b \right |\leq \frac{1}{2}$
$\left | f(0) \right |\leq \frac{1}{2}\Rightarrow -\frac{1}{2}\leq b\leq \frac{1}{2}$(1)
$1\geq \left | 1-a+b \right |+\left | 1+a+b \right |\geq \left | 2+2b \right |\Rightarrow 1\geq 2+2b\geqslant -1\Rightarrow -\frac{1}{2}\geq b\geq -\frac{3}{2}$(2)
Từ (1) và (2) suy ra $b=-\frac{1}{2}$$\Rightarrow f(x)=x^2+a-\frac{1}{2}$
Mặt khác $\left | 1+a-\frac{1}{2} \right |\leq \frac{1}{2}\Rightarrow \left | \frac{1}{2}+a \right |\leq \frac{1}{2}\Rightarrow -1\leq a\leq 0$
$\left | 1-a-\frac{1}{2} \right |\leq \frac{1}{2}\Rightarrow \left | \frac{1}{2}-a \right |\leq \frac{1}{2}\Rightarrow 0\geq -a\geq -1\Rightarrow a\geq 0$
Do đó a=0
Vậy $f(x)=x^2-\frac{1}{2}$