Đáp án B

+ \[\begin{array}{l}{9.6^{f\left( x \right)}} + \left( {4 - {f^2}\left( x \right)} \right){.9^{f\left( x \right)}} \le \left( { - {m^2} + 5m} \right){.4^{f\left( x \right)}}\\ \Leftrightarrow - {m^2} + 5m \ge 9.{\left( {\frac{3}{2}} \right)^{f\left( x \right)}} + \left( {4 - {f^2}\left( x \right)} \right){\left( {\frac{3}{2}} \right)^{2f\left( x \right)}}\;\;\;\left( 1 \right).\end{array}\]

+ Từ đồ thị suy ra \[f\left( x \right) \le - 2,\forall x \Rightarrow 9.{\left( {\frac{3}{2}} \right)^{f\left( x \right)}} \le 4,\forall x\] và \[\left( {4 - {f^2}\left( x \right)} \right){\left( {\frac{3}{2}} \right)^{2f\left( x \right)}} \le 0,\forall x\].

+ Suy ra \[g\left( x \right) = 9.{\left( {\frac{3}{2}} \right)^{f\left( x \right)}} + \left( {4 - {f^2}\left( x \right)} \right){\left( {\frac{3}{2}} \right)^{2f\left( x \right)}} \le 4,\forall x \Rightarrow \mathop {\max }\limits_\mathbb{R} g\left( x \right) = 4\].

+ Bất phương trình (1) nghiệm đúng \[\forall x \in \mathbb{R} \Leftrightarrow - {m^2} + 5m \ge 4 \Leftrightarrow 1 \le m \le 4\].

Vậy \[m \in \left\{ {1;2;3;4} \right\}\].