Phương pháp giải:

Sử dụng công thức tính đạo hàm: \({\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right)^\prime } = \dfrac{{f'\left( x \right).g\left( x \right) - g'\left( x \right).f\left( x \right)}}{{{g^2}\left( x \right)}}.\)

Lời giải chi tiết:

Trên khoảng \(\left( {0; + \infty } \right)\), ta có:

\(\begin{array}{l}f\left( x \right) = x\left[ {\sin x + f'\left( x \right)} \right] + {\rm{cos}}\,x\\ \Leftrightarrow f\left( x \right) = x\sin x + xf'\left( x \right) + \cos x\\ \Leftrightarrow f\left( x \right) - xf'\left( x \right) = x\sin x + {\rm{cos}}\,x\\ \Leftrightarrow \dfrac{{f\left( x \right) - xf'\left( x \right)}}{{{x^2}}} = \dfrac{{x\sin x + {\rm{cos}}\,x}}{{{x^2}}}\\ \Leftrightarrow \dfrac{{xf'\left( x \right) - f\left( x \right)}}{{{x^2}}} = \dfrac{{ - x\sin x - {\rm{cos}}\,x}}{{{x^2}}}\\ \Leftrightarrow {\left[ {\dfrac{{f\left( x \right)}}{x}} \right]^\prime } = {\left( {\dfrac{{{\rm{cos}}\,x}}{x}} \right)^\prime }\\ \Leftrightarrow \dfrac{{f\left( x \right)}}{x} = \dfrac{{{\rm{cos}}\,x}}{x} + C \Leftrightarrow f\left( x \right) = {\rm{cos}}\,x + Cx\end{array}\)

Mà \(f\left( {\dfrac{\pi }{2}} \right) = \dfrac{\pi }{2}\)\( \Leftrightarrow \dfrac{\pi }{2} = \cos \dfrac{\pi }{2} + C.\dfrac{\pi }{2} \Leftrightarrow C = 1.\)

\( \Rightarrow f\left( x \right) = {\rm{cos}}\,x + x \Rightarrow f\left( \pi  \right) = {\rm{cos}}\,\pi  + \pi  =  - 1 + \pi .\)

Chọn B.