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

Hướng dẫn giải:

Đáp án B.

Ta có:  \( \int\limits_{0}^{\frac{\pi }{2}}{\left[ {{f}^{2}}(x)-2\sqrt{2}f(x)\sin \left( x-\frac{\pi }{4} \right) \right]dx}=\frac{2-\pi }{2} \)

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

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

Xét  \( 2\int\limits_{0}^{\frac{\pi }{2}}{{{\sin }^{2}}\left( x-\frac{\pi }{4} \right)dx}=\int\limits_{0}^{\frac{\pi }{2}}{\left[ 1-\cos \left( 2x-\frac{\pi }{2} \right) \right]dx}=\int\limits_{0}^{\frac{\pi }{2}}{(1-\sin 2x)dx} \)

 \( =\left. \left( x+\frac{1}{2}\cos 2x \right) \right|_{0}^{\frac{\pi }{2}}=\frac{\pi -2}{2} \).

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

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

Suy ra:  \( f(x)-\sqrt{2}\sin \left( x-\frac{\pi }{4} \right)=0\Leftrightarrow f(x)=\sqrt{2}\sin \left( x-\frac{\pi }{4} \right) \) .

Vậy:  \( \int\limits_{0}^{\frac{\pi }{2}}{f(x)dx}=\int\limits_{0}^{\frac{\pi }{2}}{\sqrt{2}\sin \left( x-\frac{\pi }{4} \right)dx}=\left. -\sqrt{2}\cos \left( x-\frac{\pi }{4} \right) \right|_{0}^{\frac{\pi }{2}}=0 \).