Cho hàm số $\Large f(x)$ xác định trên $\Large \left[ 0;\dfrac{\pi }{2

Cho hàm số $\Large f(x)$ xác định trên $\Large \left[ 0;\dfrac{\pi }{2} \right]$ thỏa mãn $\Large \int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}(x)-2\sqrt{2}f(x)\sin \left( x-\dfrac{\pi }{4} \right) \right]dx=\dfrac{2-\pi }{2}}$ . Tích phân $\Large \int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}$ bằng

• A.

  $\Large \dfrac{\pi }{4}$

• B.

  $\Large 0$

• C.

  $\Large 1$

• D.

  $\Large \dfrac{\pi }{2}$

Ta có $\Large \int\limits_{0}^{\frac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx=\int\limits_{0}^{\frac{\pi }{2}}{\left[ 1-\cos \left( 2x-\dfrac{\pi }{2} \right) \right]dx=\int\limits_{0}^{\frac{\pi }{2}}{\left( 1-\sin 2x \right)dx}}}$ $\Large =\left( x+\dfrac{1}{2}\cos 2x \right)\left| \begin{align}  & \frac{\pi }{2} \\  & 0 \\ \end{align} \right.=\dfrac{\pi -2}{2}$ 

Do đó: 

$\Large \int\limits_{0}^{\frac{\pi }{2}}{\left[ {{f}^{2}}(x)-2\sqrt{2}f(x)\sin \left( x-\dfrac{\pi }{4} \right) \right]dx+\int\limits_{0}^{\frac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx\\\\\Large =\dfrac{2-\pi }{2}+\dfrac{\pi -2}{2}=0}}$

$\Large \Leftrightarrow \int\limits_{0}^{\frac{\pi }{2}}{\left[ {{f}^{2}}(x)-2\sqrt{2}f(x)\sin \left( x-\dfrac{\pi }{4} \right)+2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right) \right]dx=0}$

$\Large \Leftrightarrow \int\limits_{0}^{\frac{\pi }{2}}{{{\left[ f(x)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx=0}$

Suy ra $\Large f(x)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)=0$ hay $\Large f(x)=\sqrt{2}\sin x\left( x-\dfrac{\pi }{4} \right)$

Bởi vậy :

$\Large \int\limits_{0}^{\frac{\pi }{2}}{f(x)dx=\int\limits_{0}^{\frac{\pi }{2}}{\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)dx=-\sqrt{2}\cos \left( x-\dfrac{\pi }{4} \right)\left| \begin{align}  & \frac{\pi }{2} \\  & 0 \\ \end{align} \right.=0}}$