Lời giải chi tiết:

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

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

\(\begin{array}{l} \Leftrightarrow \int\limits_0^1 {\left[ {{f^2}\left( x \right) + 2f\left( x \right)k\sin \dfrac{{\pi x}}{2} + {k^2}{{\sin }^2}\dfrac{{\pi x}}{2}} \right]dx}  = 0\\ \Leftrightarrow \int\limits_0^1 {{f^2}\left( x \right)dx}  + 2k\int\limits_0^1 {f\left( x \right)\sin \dfrac{{\pi x}}{2}dx}  + {k^2}\int\limits_0^1 {{{\sin }^2}\dfrac{{\pi x}}{2}dx}  = 0\\ \Leftrightarrow \dfrac{9}{2} + 2k\dfrac{3}{2} + {k^2}.\dfrac{1}{2} = 0 \Leftrightarrow {k^2} + 6k + 9 = 0 \Leftrightarrow {\left( {k + 3} \right)^2} = 0 \Leftrightarrow k =  - 3\\ \Rightarrow \int\limits_0^1 {{{\left( {f\left( x \right) - 3\sin \dfrac{{\pi x}}{2}} \right)}^2}dx}  = 0 \Leftrightarrow f\left( x \right) - 3\sin \dfrac{{\pi x}}{2} = 0 \Leftrightarrow f\left( x \right) = 3\sin \dfrac{{\pi x}}{2}\\ \Rightarrow \int\limits_0^1 {f\left( x \right) = dx}  = \int\limits_0^1 {3\sin \dfrac{{\pi x}}{2}dx}  = \left. {\dfrac{{ - 3\cos \dfrac{{\pi x}}{2}}}{{\dfrac{\pi }{2}}}} \right|_0^1 =  - \dfrac{6}{\pi }\left( {\cos \dfrac{\pi }{2} - \cos 0} \right) = \dfrac{6}{\pi }\end{array}\)

Chọn C.