Hướng dẫn giải:

Đáp án A.

Từ giả thiết:  \( f(x)+x\left( {f}'(x)-2\sin x \right)={{x}^{2}}\cos x,\text{ }\forall x\in \mathbb{R}\Leftrightarrow f(x)+x{f}'(x)={{x}^{2}}\cos x+2x\sin x \)

 \( \Leftrightarrow {{\left( xf(x) \right)}^{\prime }}={{\left( {{x}^{2}}\sin x \right)}^{\prime }}\Leftrightarrow \int{{{\left( xf(x) \right)}^{\prime }}dx=\int{{{\left( {{x}^{2}}\sin x \right)}^{\prime }}dx}}\Leftrightarrow xf(x)={{x}^{2}}\sin x+C \).

Mặt khác:  \( f\left( \frac{\pi }{2} \right)=\frac{\pi }{2}\Rightarrow \frac{\pi }{2}.\frac{\pi }{2}={{\left( \frac{\pi }{2} \right)}^{2}}.\sin \frac{\pi }{2}+C\Rightarrow C=0 \).

 \( \Rightarrow f(x)=x\sin x\Rightarrow {f}'(x)=\sin x+x\cos x \).

Xét  \( \int\limits_{0}^{\frac{\pi }{2}}{x{f}”(x)dx} \).

Đặt: \( \left\{ \begin{align}  & u=x\Rightarrow du=dx \\ & dv={f}”(x)dx\Rightarrow v={f}'(x) \\ \end{align} \right. \).

Khi đó:  \( \int\limits_{0}^{\frac{\pi }{2}}{x{f}”(x)dx}=\left. x{f}'(x) \right|_{0}^{\frac{\pi }{2}}-\int\limits_{0}^{\frac{\pi }{2}}{{f}'(x)dx}=\left. \left[ x.(\sin x+x\cos x)-f(x) \right] \right|_{0}^{\frac{\pi }{2}} \)

 \( =\left. \left[ x.(\sin x+x\cos x)-x\sin x \right] \right|_{0}^{\frac{\pi }{2}}=\left. {{x}^{2}}\cos x \right|_{0}^{\frac{\pi }{2}}=0 \).