Phương pháp giải:

\(\begin{array}{l}y = f\left( x \right) = \,\left\{ {\begin{array}{*{20}{c}}{{f_1}\left( x \right)\,\,\,{\rm{ }}khi\,\,x \in {D_1}{\rm{ }}}\\{{f_2}\left( x \right){\rm{ }}\,\,{\rm{ }}khi\,\,x \in {D_2}}\\{{f_3}\left( x \right){\rm{  }}khi\,\,x \in {D_3}}\end{array}} \right.\\{D_f} = {D_1} \cup {D_2} \cup {D_3}\\f\left( {{x_1}} \right) = {f_1}\left( {{x_1}} \right){\rm{ }};{\rm{ }}{x_1} \in {D_1}\\f\left( {{x_2}} \right) = {f_2}\left( {{x_2}} \right){\rm{ ; }}{x_2} \in {D_2}\\f\left( {{x_3}} \right) = {f_3}\left( {{x_3}} \right){\rm{ ; }}{x_3} \in {D_3}\end{array}\)

\({x_4} \notin D \Rightarrow \) không tồn tại \(f\left( {{x_4}} \right).\)

Lời giải chi tiết:

\(\begin{array}{l}f\left( { - 1} \right) = \left| { - 5.\left( { - 1} \right)} \right| = \left| 5 \right| = 5\\f\left( 2 \right) = \left| { - 5.2} \right| = \left| { - 10} \right| = 10\\f\left( { - 2} \right) = \left| { - 5.\left( { - 2} \right)} \right| = \left| {10} \right| = 10\\f\left( {\frac{1}{5}} \right) = \left| { - 5.\frac{1}{5}} \right| = \left| { - 1} \right| = 1\end{array}\)

Vậy đáp án D sai.

Chọn  D.