Lời giải
Ta xét hàm số \(u\left( x \right) = {x^4} - 2{x^3} + {x^2} + a\) trên đoạn \(\left[ { - 1;2} \right]\).
Ta có \(u'\left( x \right) = 0 \Leftrightarrow 4{x^3} - 6{x^2} + 2x = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\\x = \frac{1}{2}\end{array} \right.\).
\(M = \mathop {max}\limits_{\left[ { - 1;2} \right]} {\mkern 1mu} u\left( x \right) = max\left\{ {u\left( { - 1} \right);u\left( 0 \right);u\left( 1 \right);u\left( {\frac{1}{2}} \right);u\left( 2 \right)} \right\}\).
\( = max\left\{ {a + 4;a + 4;a;a;a + \frac{1}{{16}}} \right\} = a + 4\)
Và \(m = \mathop {\min }\limits_{\left[ { - 1;2} \right]} {\mkern 1mu} u\left( x \right) = a\)
\( \Rightarrow \mathop {max}\limits_{\left[ { - 1;2} \right]} {\mkern 1mu} y = max\left\{ {\left| {a + 4} \right|;\left| a \right|} \right\} \le 2020\)
TH1: \(\left| {a + 4} \right| \le \left| a \right| \le 2020 \Leftrightarrow \left\{ \begin{array}{l}{\left( {a + 4} \right)^2} \le {a^2}\\ - 2020 \le a \le 2020\end{array} \right. \Leftrightarrow - 2020 \le a \le - 2\)
TH2: \(\left| a \right| \le \left| {a + 4} \right| \le 2020 \Leftrightarrow \left\{ \begin{array}{l}{a^2} \le {\left( {a + 4} \right)^2}\\ - 2020 \le a + 4 \le 2020\end{array} \right. \Leftrightarrow - 2 \le a \le 2016\)
Vậy \(a \in \left\{ { - 2020;...;2016} \right\}\) Þ có \(2020 + 2017 = 4037\) số.
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