cho f(x)=4x^2-4mx+m^2-2m+x.tìm m để min f(x) trên [0;2] = 5

Đáp án:

\(\,m = 1;\,\,m = 1 - \sqrt 5 \)

Giải thích các bước giải:

\(\eqalign{ & f\left( x \right) = 4{x^2} - 4mx + {m^2} - 2m + 1 \cr & Hoanh\,\,do\,\,dinh\,\,{x_I} = {m \over 2} \cr & Do\,\,a = 4 > 0 \Rightarrow Ham\,\,so\,DB/\left( {{m \over 2}; + \infty } \right);\,\,NB/\left( { - \infty ;{m \over 2}} \right) \cr & TH1:\,\,{m \over 2} \in \left[ {0;2} \right] \Rightarrow 0 \le {m \over 2} \le 2 \Leftrightarrow 0 \le m \le 4 \cr & \Rightarrow \mathop {\min }\limits_{\left[ {0;2} \right]} f\left( x \right) = f\left( {{m \over 2}} \right) = - 2{m^2} - 2m + 1 \cr & \Rightarrow - 2{m^2} - 2m + 1 = 5 \Leftrightarrow 2{m^2} + 2m - 4 = 0 \Leftrightarrow \left[ \matrix{ m = 1\,\,\left( {tm} \right) \hfill \cr m = - 2\,\,\left( {ktm} \right) \hfill \cr} \right. \cr & TH2:\,\,0 < 2 < {m \over 2} \Leftrightarrow m > 4 \cr & \Rightarrow \mathop {\min }\limits_{\left[ {0;2} \right]} f\left( x \right) = f\left( 2 \right) = {m^2} - 10m + 17 \cr & \Rightarrow {m^2} - 10m + 17 = 5 \Leftrightarrow {m^2} - 10m + 12 = 0 \cr & \Leftrightarrow \left[ \matrix{ m = 5 + \sqrt {13} \,\,\left( {ktm} \right) \hfill \cr m = 5 - \sqrt {13} \,\,\left( {ktm} \right) \hfill \cr} \right. \cr & TH3:\,\,{m \over 2} < 0 < 2 \Rightarrow m < 0 \cr & \Rightarrow \mathop {\min }\limits_{\left[ {0;2} \right]} f\left( x \right) = f\left( 0 \right) = {m^2} - 2m + 1 \cr & \Rightarrow {m^2} - 2m + 1 = 5 \Leftrightarrow {m^2} - 2m - 4 = 0 \Leftrightarrow \left[ \matrix{ m = 1 + \sqrt 5 \,\,\left( {ktm} \right) \hfill \cr x = 1 - \sqrt 5 \,\,\left( {tm} \right) \hfill \cr} \right. \cr & Vay\,\,m = 1;\,\,m = 1 - \sqrt 5 \cr} \)