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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left[ 0;\dfrac{\pi }{2} \right]$, biết $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}\left( x \right)-2\sqrt{2}f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right) \right]d\text{x}}=\dfrac{2-\pi }{2}$. Tính tích phân $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)d\text{x}}$.
A. $I=0$
B. $I=\dfrac{\pi }{4}$
C. $I=1$
D. $I=\dfrac{\pi }{2}$
A. $I=0$
B. $I=\dfrac{\pi }{4}$
C. $I=1$
D. $I=\dfrac{\pi }{2}$
Ta có $\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)d\text{x}}=-\dfrac{2-\pi }{2}$.
Do đó giả thiết tương đương với
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}\left( x \right)-2\sqrt{2}f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right)+2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right) \right]d\text{x}}=0$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}d\text{x}}=0\Leftrightarrow f\left( x \right)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)=0,\forall x\in \left[ 0;\dfrac{\pi }{2} \right]$.
Suy ra $f\left( x \right)=\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)$.
Vậy $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)d\text{x}}=\sqrt{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( x-\dfrac{\pi }{4} \right)d\text{x}}=0$.
Do đó giả thiết tương đương với
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}\left( x \right)-2\sqrt{2}f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right)+2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right) \right]d\text{x}}=0$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}d\text{x}}=0\Leftrightarrow f\left( x \right)-\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)=0,\forall x\in \left[ 0;\dfrac{\pi }{2} \right]$.
Suy ra $f\left( x \right)=\sqrt{2}\sin \left( x-\dfrac{\pi }{4} \right)$.
Vậy $I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)d\text{x}}=\sqrt{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin \left( x-\dfrac{\pi }{4} \right)d\text{x}}=0$.
Đáp án A.