Cho hàm số \(y=f\left( x \right)\) có đạo hàm liên...

D

- Hướng dẫn giải

Phương pháp giải:

Áp dụng kĩ thuật tổng bình phương để tìm hàm số \(y=f\left( x \right)\)

Giải chi tiết:

Đặt  \(\left\{ \begin{array}{l}
\sin 2x = u\\
f'\left( x \right){\rm{d}}x = {\rm{d}}v
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
2\cos 2x{\rm{d}}x = {\rm{d}}u\\
f\left( x \right) = v
\end{array} \right.\)

khi đó \(\int\limits_{0}^{\frac{\pi }{4}}{{f}'\left( x \right)\text{sin 2}x\text{d}x}=\left. \text{sin 2}x.f\left( x \right) \right|_{0}^{\frac{\pi }{4}}-2\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{cos2}x\text{d}x}\)

\(=\text{sin }\frac{\pi }{2}.f\left( \frac{\pi }{4} \right)-\sin 0.f\left( 0 \right)-2\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{cos2}x\text{d}x}\)\(=-\,2\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{cos2}x\text{d}x}\).

Theo đề bài ta có \(\int\limits_{0}^{\frac{\pi }{4}}{{f}'\left( x \right)\text{sin 2}x\text{d}x}=-\frac{\pi }{4}\)\(\Rightarrow \)\(\int\limits_{0}^{\frac{\pi }{4}}{f\left( x \right)\text{cos2}x\text{d}x}=\frac{\pi }{8}\). Mặt khác \(\int\limits_{0}^{\frac{\pi }{4}}{{{\cos }^{2}}2x\text{d}x}=\frac{\pi }{8}\).

Do \(\int\limits_{0}^{\frac{\pi }{4}}{{{\left( f\left( x \right)-\text{cos2}x \right)}^{2}}\text{d}x}=\int\limits_{0}^{\frac{\pi }{4}}{\left( {{f}^{2}}\left( x \right)-\text{2}f\left( x \right)\text{.cos2}x+{{\cos }^{2}}2x \right)\text{d}x}\) \(=\left( \frac{\pi }{8}-2\frac{\pi }{8}+\frac{\pi }{8} \right)=0\) nên \(f\left( x \right)=\cos 2x\).

Ta có \(I=\int\limits_{0}^{\frac{\pi }{8}}{\cos 4x\text{d}x}=\left. \frac{1}{4}\sin 4x \right|_{0}^{\frac{\pi }{8}}=\frac{1}{4}\).

Chọn D