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Câu hỏi: Cho hàm số $y=f\left( x \right)$ xác định trên đoạn $\left[ 0;\dfrac{\pi }{2} \right]$ thỏa mãn $\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}\left( x \right)-2\sqrt{2}.f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right) \right]dx=\dfrac{\pi -2}{2}}$. Tích phân $\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}$ bằng
A. $\dfrac{\pi }{4}$.
B. 0.
C. $\dfrac{\pi }{2}$.
D. 1.
A. $\dfrac{\pi }{4}$.
B. 0.
C. $\dfrac{\pi }{2}$.
D. 1.
Đặt $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ {{f}^{2}}\left( x \right)-2\sqrt{2}.f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right) \right]dx}$.
Ta có
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)-2\sqrt{2}.f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right)+2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}$
$\Leftrightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}$
Có $\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 1-\cos \left( 2x-\dfrac{\pi }{2} \right) \right]dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 1-\sin 2x \right]dx}=\left( x+\dfrac{1}{2}\cos 2x \right)_{0}^{\dfrac{\pi }{2}}=\dfrac{\pi -2}{2}$
Mà $I=\dfrac{\pi -2}{2}\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}=0 \left( 1 \right)$
Vì $y={{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}$ liên tục và không âm nên
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}\ge 0$
Dấu '=' xảy ra $\Leftrightarrow f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)=0$.
$\Leftrightarrow f\left( x \right)=\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)dx}=0$.
Ta có
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)-2\sqrt{2}.f\left( x \right).\sin \left( x-\dfrac{\pi }{4} \right)+2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}$
$\Leftrightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}$
Có $\int\limits_{0}^{\dfrac{\pi }{2}}{2{{\sin }^{2}}\left( x-\dfrac{\pi }{4} \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 1-\cos \left( 2x-\dfrac{\pi }{2} \right) \right]dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ 1-\sin 2x \right]dx}=\left( x+\dfrac{1}{2}\cos 2x \right)_{0}^{\dfrac{\pi }{2}}=\dfrac{\pi -2}{2}$
Mà $I=\dfrac{\pi -2}{2}\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}=0 \left( 1 \right)$
Vì $y={{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}$ liên tục và không âm nên
$\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left[ f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right) \right]}^{2}}dx}\ge 0$
Dấu '=' xảy ra $\Leftrightarrow f\left( x \right)-\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)=0$.
$\Leftrightarrow f\left( x \right)=\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)$
$\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{2}.\sin \left( x-\dfrac{\pi }{4} \right)dx}=0$.
Đáp án B.