Đáp án:

Xét \(\int_{0}^{\frac{\pi}{4}} f'(x) \sin{2x} dx\).  

Đặt: \(\begin{cases} u = \sin{2x} \\ dv = f'(x) dx \end{cases} \Rightarrow \begin{cases} du = 2 \cos{2x} dx \\ v = f(x) \end{cases}\)  

\(-\frac{\pi}{4} = \int_{0}^{\frac{\pi}{4}} f'(x) \sin{2x} dx =  f(x) \sin{2x} \big|_{0}^{\frac{\pi}{4}} - 2\int_{0}^{\frac{\pi}{4}} f(x) \cdot \cos{2x} dx\)  

\(\Rightarrow \int_{0}^{\frac{\pi}{4}} f(x) \cos{2x} dx = \frac{\pi}{8}.\)  

Tìm \(k\) sao cho: \(\int_{0}^{\frac{\pi}{4}} \left[f(x) - k \cos{2x}\right]^2 dx = 0\)  

\(\Leftrightarrow \int_{0}^{\frac{\pi}{4}} (f(x))^2 dx - 2k \int_{0}^{\frac{\pi}{4}} f(x) \cos{2x} dx + k^2 \int_{0}^{\frac{\pi}{4}} \cos^2{2x} dx = 0\)  

\(\Leftrightarrow \frac{\pi}{8} -  \frac{\pi}{4}k + \frac{\pi}{8} k^2 = 0 \iff k^2 - 2k + 1 = 0 \iff k = 1.\)  

Suy ra: \(f(x) = \cos{2x}\).  

\(I = \int_{0}^{\frac{\pi}{8}} f(2x) dx = \int_{0}^{\frac{\pi}{8}} \cos{4x} dx = \frac{1}{4} \Rightarrow \boxed{B}\)