Đặt \(\left\{ \begin{array}{l}u = \cos \dfrac{{\pi x}}{2}\\dv = f'\left( x \right)dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du =  - \dfrac{\pi }{2}\sin \dfrac{{\pi x}}{2}dx\\v = f\left( x \right)\end{array} \right.\)

\(\begin{array}{l} \Rightarrow \int\limits_0^1 {f'\left( x \right){\rm{cos}}\dfrac{{\pi x}}{2}{\rm{d}}x}  = \left. {\cos \dfrac{{\pi x}}{2}f\left( x \right)} \right|_0^1 + \dfrac{\pi }{2}\int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx} \\ = f\left( 1 \right).\cos \dfrac{\pi }{2} - f\left( 0 \right)\cos 0 + \dfrac{\pi }{2}\int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx} \\ = \dfrac{\pi }{2}\int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx}  = \dfrac{{3\pi }}{4} \Rightarrow \int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx}  = \dfrac{3}{2}\end{array}\)

Xét tích phân \(\int\limits_0^1 {{{\left[ {f\left( x \right) + k\sin \dfrac{{\pi x}}{2}} \right]}^2}dx}  = 0\)

\(\begin{array}{l} \Leftrightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2kf\left( x \right)\sin \dfrac{{\pi x}}{2} + {k^2}{{\sin }^2}\dfrac{{\pi x}}{2}} \right]dx}  = 0\\ \Leftrightarrow \int\limits_0^1 {{f^2}\left( x \right)dx}  + 2k\int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx}  + {k^2}\int\limits_0^1 {{{\sin }^2}\dfrac{{\pi x}}{2}dx}  = 0\\ \Leftrightarrow \dfrac{9}{2} + 2k\dfrac{3}{2} + \dfrac{1}{2}{k^2} = 0 \Leftrightarrow k =  - 3\end{array}\)

Khi đó ta có \(\int\limits_0^1 {{{\left[ {f\left( x \right) - 3\sin \dfrac{{\pi x}}{2}} \right]}^2}dx}  = 0 \Leftrightarrow f\left( x \right) - 3\sin \dfrac{{\pi x}}{2} = 0 \Leftrightarrow f\left( x \right) = 3\sin \dfrac{{\pi x}}{2}\)

Vậy \(\int\limits_0^1 {f\left( x \right){\rm{d}}x}  = 3\int\limits_0^1 {{\rm{sin}}\dfrac{{\pi x}}{2}{\rm{d}}x}  = \left. { - 3\dfrac{{\cos \dfrac{{\pi x}}{2}}}{{\dfrac{\pi }{2}}}} \right|_0^1 = \left. {\dfrac{{ - 6}}{\pi }\cos \dfrac{{\pi x}}{2}} \right|_0^1 =  - \dfrac{6}{\pi }\left( {\cos \dfrac{\pi }{2} - \cos 0} \right) = \dfrac{6}{\pi }\)

Chọn A.