Đáp án đúng: D
Giải chi tiết:Đặt \(t={{e}^{x}},\) với \(x\in \left[ 0;\ln 4 \right]\Rightarrow t\in \left[ 1;4 \right].\)
Khi đó, hàm số trở thành: \(g\left( t \right)=\left| {{t}^{2}}-4t+m \right|.\)
Xét hàm số \(u\left( t \right)={{t}^{2}}-4t+m\) trên \(\left[ 1;4 \right],\) có \({u}'\left( t \right)=2t-4=0\Leftrightarrow t=2.\)
Tính \(u\left( 1 \right)=m-3;\,\,u\left( 2 \right)=m-4;\,\,u\left( 4 \right)=m\) suy ra \(g\left( 1 \right)=\left| m-3 \right|;\,\,g\left( 2 \right)=\left| m-4 \right|;\,\,g\left( 4 \right)=\left| m \right|.\)
TH1. \(\left\{ \begin{array}{l}\left| {m - 4} \right| \le \left\{ {\left| {m - 3} \right|;\,\,\left| m \right|} \right\}\\\mathop {\min }\limits_{\left[ {1;4} \right]} g\left( t \right) = \left| {m - 4} \right| = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m - 4} \right| \le \left\{ {\left| {m - 3} \right|;\,\,\left| m \right|} \right\}\\\left[ \begin{array}{l}m = 10\\m = - \,2\end{array} \right.\end{array} \right.\,\, \Leftrightarrow \,\,m = 10.\)
TH2. \(\left\{ \begin{array}{l}\left| {m - 3} \right| \le \left\{ {\left| {m - 4} \right|;\,\,\left| m \right|} \right\}\\\mathop {\min }\limits_{\left[ {1;4} \right]} g\left( t \right) = \left| {m - 3} \right| = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| {m - 3} \right| \le \left\{ {\left| {m - 4} \right|;\,\,\left| m \right|} \right\}\\\left[ \begin{array}{l}m = 9\\m = - \,3\end{array} \right.\end{array} \right.\,\, \Rightarrow \) Vô nghiệm.
TH3. \(\left\{ \begin{array}{l}\left| m \right| \le \left\{ {\left| {m - 4} \right|;\,\,\left| {m - 3} \right|} \right\}\\\mathop {\min }\limits_{\left[ {1;4} \right]} g\left( t \right) = \left| m \right| = 6\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left| m \right| \le \left\{ {\left| {m - 4} \right|;\,\,\left| {m - 3} \right|} \right\}\\\left[ \begin{array}{l}m = 6\\m = - \,6\end{array} \right.\end{array} \right.\,\, \Leftrightarrow \,\,m = - \,6.\)
Vậy \(m=\left\{ 10;-\,6 \right\}\) là hai giá trị cần tìm.
Chọn D