lehoanghiep

Đây là T8/416

Ta có $P\left ( 0 \right )=q;P\left ( 1 \right )=1+p+q;P\left ( -1 \right )=1-p+q$. Khi đó
$\left | P\left ( 1 \right ) \right | + \left | P\left ( 0 \right ) \right |\geq$$\left | P\left ( 1 \right )-P\left ( 0 \right ) \right |\geq \left | 1+p \right |$
và
$\left | P\left ( -1 \right ) \right | + \left | P\left ( 0 \right ) \right |\geq$$\left | P\left ( -1 \right )-P\left ( 0 \right ) \right |\geq \left | 1-p \right |$
Nếu $p> 0 $
$\Rightarrow \left | 1+p \right |> 1\Rightarrow\left | P\left ( 1 \right ) \right |> \frac{1}{2}$ hoặc $\left | P\left ( 0 \right ) \right |> \frac{1}{2}$$\Rightarrow max\left | P\left ( x \right ) \right |> \frac{1}{2}$
Nếu $p<0$ $\Rightarrow \left | 1-p \right | >1\Rightarrow \left | P\left ( -1 \right ) \right | >\frac{1}{2}$ hoặc $\left | P\left ( 0 \right ) \right | >\frac{1}{2}\Rightarrow max\left | P\left ( x \right ) \right | >\frac{1}{2}$
Nếu $p=0$ thì $P\left ( x \right )=x^{2}+q;P\left ( 0 \right )=0,P\left ( 1 \right )=1+q$
nếu $q >-\frac{1}{2}\Rightarrow 1+q >\frac{1}{2}\Rightarrow \left | P\left ( 1 \right ) \right | >\frac{1}{2}\Rightarrow max\left | P\left ( x \right ) \right |>\frac{1}{2}$
nếu $q <-\frac{1}{2}\Rightarrow \left | q \right |>\frac{1}{2}\Rightarrow \left | P\left ( 0 \right ) \right | >\frac{1}{2}\Rightarrow max\left | P\left ( x \right ) \right |>\frac{1}{2}$
nếu $q=\frac{1}{2}$ thì $P\left ( x \right )=x^{2}-\frac{1}{2}\leq \frac{1}{2}$
Vậy $p=0$, $q=\frac{1}{2}$. Giá trị nhỏ nhất đó là $\frac{1}{2}$