Cho hàm số y=f(x) xác định trên đoạn

* Hướng dẫn giải

Đáp án B.

Đặt $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[f^2\left(x\right)-2\sqrt{2}.f\left(x\right).\sin \left(x-\frac{\pi }{4}\right)\right]dx$.

Ta có $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)-2\sqrt{2}.f\left(x\right).\sin \left(x-\frac{\pi }{4}\right)+2{\sin }^2\left(x-\frac{\pi }{4}\right)dx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}2{\sin }^2\left(x-\frac{\pi }{4}\right)dx$

$\iff I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\left[f\left(x\right)-\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)\right]}^{2}dx-\underset{0}{\overset{\frac{\pi }{2}}{\int }}2{\sin }^2\left(x-\frac{\pi }{4}\right)dx$

Có $$\begin{gathered} \underset{0}{\overset{\frac{\pi }{2}}{\int }}2{\sin }^2\left(x-\frac{\pi }{4}\right)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[1-\cos \left(2x-\frac{\pi }{2}\right)\right]dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[1-\sin 2x\right]dx \\ ={\left(x+\frac{1}{2}\cos 2x\right)}_0^{\frac{\pi }{2}}=\frac{\pi -2}{2} \end{gathered}$$

Mà $I=\frac{\pi -2}{2}\Rightarrow \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\left[f\left(x\right)-\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)\right]}^{2}dx=0\left(1\right)$

Vì $y={\left[f\left(x\right)-\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)\right]}^2$ liên tục và không âm nên $\Rightarrow \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\left[f\left(x\right)-\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)\right]}^{2}dx\ge 0$

Dấu ‘=’ xảy ra $\iff f\left(x\right)-\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)=0$.

$\iff f\left(x\right)=\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)$

$\iff \underset{0}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{2}.\sin \left(x-\frac{\pi }{4}\right)dx=0$.