Chọn phương án A.

\(\begin{aligned}\text{Ta có }f(1)=3\Leftrightarrow&\,M\sin\pi+N\cos\pi=3\\
\Leftrightarrow&\,N=-3\\
\Rightarrow&\,f(x)=M\sin\pi x-3\cos\pi x.\end{aligned}\)

\(\begin{aligned}
\text{Lại có }&\,\displaystyle\int\limits_0^{\tfrac{1}{2}}f(x)\mathrm{\,d}x=-\dfrac{1}{\pi}\\
\Leftrightarrow&\,\displaystyle\int\limits_0^{\tfrac{1}{2}}\left(M\sin\pi x-3\cos\pi x\right)\mathrm{\,d}x=-\dfrac{1}{\pi}\\
\Leftrightarrow&\,\left(-\dfrac{M}{\pi}\cos\pi x-\dfrac{3}{\pi}\sin\pi x\right)\bigg|_0^{\tfrac{1}{2}}=-\dfrac{1}{\pi}\\
\Leftrightarrow&\,-\dfrac{3}{\pi}+\dfrac{M}{\pi}=-\dfrac{1}{\pi}\\
\Leftrightarrow&\,M=2.\end{aligned}\)

Khi đó \(f'(x)=2\pi\cos\pi x+3\pi\sin\pi x\).
Suy ra \(f'\left(\dfrac{1}{4}\right)=\dfrac{5\pi\sqrt{2}}{2}\).