A. \( - 15 < S < - 10.\)
B. \( - 20 < S < - 15.\)
C. \( - 5 < S < 0.\)
D. \( - 10 < S < - 5.\)
C
\(\begin{array}{l}
y = f\left( x \right) = {x^3} - 2m{x^2} - 4{m^2}x + 100 \Rightarrow y' = 3{x^2} - 4mx - 4{m^2}\\
y' = 0 \Leftrightarrow 3{x^2} - 4mx - 4{m^2} = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 2m\\
x = - \frac{2}{3}m
\end{array} \right.
\end{array}\)
Do \(m<0\) nên \(2m < 0 < - \frac{2}{3}m\)
TH1: \( - \frac{2}{3}m < 1 < 2 \Leftrightarrow m > - \frac{3}{2}\)
\(\mathop {\min }\limits_{\left[ {1;2} \right]} f\left( x \right) = f\left( 1 \right) = 101 - 2m - 4{m^2} = 12 \Rightarrow 4{m^2} + 2m - 89 = 0 \Leftrightarrow m = \frac{{ - 1 \pm \sqrt {357} }}{4}\,(ktm)\)
TH2: \(1 \le - \frac{2}{3}m \le 2 \Leftrightarrow - 3 \le m \le - \frac{3}{2}\)
\(\mathop {\min }\limits_{\left[ {1;2} \right]} f\left( x \right) = f\left( { - \frac{2}{3}m} \right) = \frac{{40}}{{27}}{m^3} + 100 = 12 \Rightarrow m = - \sqrt[3]{{\frac{{297}}{5}}}(ktm)\)
TH3: \(1 < 2 < - \frac{2}{3}m \Leftrightarrow m < - 3\)
\(\mathop {\min }\limits_{\left[ {1;2} \right]} f\left( x \right) = f\left( 2 \right) = 8 - 8m - 8{m^2} + 100 = 12 \Rightarrow 8{m^2} + 8m - 96 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = 3\,(ktm)\\
m = - 4\,(tm)
\end{array} \right.\)
Vậy \(m = - 4 \Rightarrow S = - 4 \in \left( { - 5;0} \right) \Rightarrow - 5 < S < 0\)
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247