Cho hàm số f(x) có đạo hàm liên tục trên đoạn [0; 1] và f( 0) + f(1) = 0. Biết

* Hướng dẫn giải

Đáp án B.

Ta có$\underset{0}{\overset{1}{\int }}f\text{'}\left(x\right).cos\pi xdx$

$=\underset{0}{\overset{1}{\int }}cos\pi xd\left(f\left(x\right)\right)=f\left(x\right).cos\pi x\left|{}_0^1\right.-\underset{0}{\overset{1}{\int }}f\left(x\right).{\left(cos\pi x\right)}^{\text{'}}dx$ 

$$\begin{gathered} =-\left[f\left(1\right)+f\left(0\right)\right]+\pi \underset{0}{\overset{1}{\int }}f\left(x\right).\sin \pi xdx=\frac{\pi }{2} \\ \Rightarrow \underset{0}{\overset{1}{\int }}f\left(x\right).\sin \pi xdx=\frac{1}{2}. \end{gathered}$$ 

Xét $\underset{0}{\overset{1}{\int }}{\left[f\left(x\right)+k.\sin \pi x\right]}^{2}dx=0$

$\iff \underset{0}{\overset{1}{\int }}{f}^2\left(x\right)dx+2k.\underset{0}{\overset{1}{\int }}f\left(x\right).\sin \pi xdx+k^2.\underset{0}{\overset{1}{\int }}{\sin }^2\left(\pi x\right)dx=0$ 

$\iff \frac{1}{2}{k}^2+2k.\frac{1}{2}+\frac{1}{2}=0\iff {\left(k+1\right)}^2=0\iff k=-1.$

Suy ra $\underset{0}{\overset{1}{\int }}{\left[f\left(x\right)-\sin \pi x\right]}^{2}dx=0.$ 

Vậy$f\left(x\right)=\sin \pi x\Rightarrow \underset{0}{\overset{1}{\int }}f\left(x\right)dx=\underset{0}{\overset{1}{\int }}\sin \pi xdx=\frac{2}{\pi }.$