Cho hàm số y = f(x) có đạo hàm liên tục trên đoạn [0;pi/4] và f(pi/4) = 0

* Hướng dẫn giải

Đáp án B.

Ta có $\underset{0}{\overset{\frac{\pi }{4}}{\int }}f\text{'}\left(x\right)\sin 2\text{x}d\text{x}$

Đặt $$\begin{gathered} \left\{\begin{array}{l}u=\sin 2\text{x}\\ dv=f\text{'}\left(x\right)\end{array}\right. \\ \Rightarrow \left\{\begin{array}{l}du=2\cos 2\text{xdx}\\ v=f\left(x\right)\end{array}\right. \end{gathered}$$ suy ra 

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\begin{array}{l}f\text{'}\left(x\right)\sin 2xdx\\ =\sin 2xf\left(x\right)\left|\begin{array}{l}{}^{\frac{\pi }{4}}\\ {}_0\end{array}\right.-\underset{0}{\overset{\frac{\pi }{4}}{\int }}2\cos 2x.f\left(x\right)\end{array}dx$

$$\begin{gathered} \Rightarrow -2\underset{0}{\overset{\frac{\pi }{4}}{\int }}cos2x.f\left(x\right)dx=-\frac{\pi }{4}\Rightarrow \\ \underset{0}{\overset{\frac{\pi }{4}}{\int }}cos2x.f\left(x\right)dx=\frac{\pi }{8} \end{gathered}$$

Lại có: $\underset{0}{\overset{\frac{\pi }{4}}{\int }}co{s}^{2}2x.f\left(x\right)dx=\frac{\pi }{8}\Rightarrow \underset{0}{\overset{\frac{\pi }{4}}{\int }}{\left[f\left(x\right)-cos2x\right]}^{2}dx=0\Rightarrow f\left(x\right)=cos2x$ 

Do đó $\underset{0}{\overset{\frac{\pi }{8}}{\int }}f\left(2x\right)dx=\underset{0}{\overset{\frac{\pi }{8}}{\int }}cos4xdx=\frac{\sin 4x}{4}\left|\begin{array}{l}{}^{\frac{\pi }{8}}\\ {}_0\end{array}\right.=\frac{1}{4}.$