Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án B.

Đặt $I = \int_{0}^{\frac{π}{2}} [f^{2} (x) - 2 \sqrt{2} . f (x) . \sin (x - \frac{π}{4})] d x$ .

Ta có $I = \int_{0}^{\frac{π}{2}} f (x) - 2 \sqrt{2} . f (x) . \sin (x - \frac{π}{4}) + 2 \sin^{2} (x - \frac{π}{4}) d x - \int_{0}^{\frac{π}{2}} 2 \sin^{2} (x - \frac{π}{4}) d x$

$\Leftrightarrow I = \int_{0}^{\frac{π}{2}} {[f (x) - \sqrt{2} . \sin (x - \frac{π}{4})]}^{2} d x - \int_{0}^{\frac{π}{2}} 2 \sin^{2} (x - \frac{π}{4}) d x$

Có $\int_{0}^{\frac{π}{2}} 2 \sin^{2} (x - \frac{π}{4}) d x = \int_{0}^{\frac{π}{2}} [1 - \cos (2 x - \frac{π}{2})] d x = \int_{0}^{\frac{π}{2}} [1 - \sin 2 x] d x = {(x + \frac{1}{2} \cos 2 x)}_{0}^{\frac{π}{2}} = \frac{π - 2}{2}$

Mà $I = \frac{π - 2}{2} \Rightarrow \int_{0}^{\frac{π}{2}} {[f (x) - \sqrt{2} . \sin (x - \frac{π}{4})]}^{2} d x = 0 (1)$

Vì $y = {[f (x) - \sqrt{2} . \sin (x - \frac{π}{4})]}^{2}$ liên tục và không âm nên $\Rightarrow \int_{0}^{\frac{π}{2}} {[f (x) - \sqrt{2} . \sin (x - \frac{π}{4})]}^{2} d x \geq 0$

Dấu ‘=’ xảy ra $\Leftrightarrow f (x) - \sqrt{2} . \sin (x - \frac{π}{4}) = 0$ .

$\Leftrightarrow f (x) = \sqrt{2} . \sin (x - \frac{π}{4})$

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x = \int_{0}^{\frac{π}{2}} \sqrt{2} . \sin (x - \frac{π}{4}) d x = 0$ .

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Answer 2