Question

Đã trả lời bởi chuyên gia

Answer 1

Đáp án đúng là: D

+) Thay x = 0 vào phương trình $f (x) + f (\frac{π}{2} - x) = \sin x . \cos x$ ta có

$f (0) + f (\frac{π}{2}) = 0 \Rightarrow f (\frac{π}{2}) = 0$

$f (x) + f (\frac{π}{2} - x) = \sin x . \cos x$

$\Rightarrow \int_{0}^{\frac{π}{2}} [f (x) + f (\frac{π}{2} - x)] d x = \int_{0}^{\frac{π}{2}} \sin x . \cos x d x$

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x + \int_{0}^{\frac{π}{2}} f (\frac{π}{2} - x) d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} 2. \sin x . \cos x d x$

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x + \int_{0}^{\frac{π}{2}} f (\frac{π}{2} - x) d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 x d x$

Đặt: $u = \frac{π}{2} - x \Rightarrow d u = - d x$

Đổi cận

+) $x = 0 \Rightarrow u = \frac{π}{2}$

+) $x = \frac{π}{2} \Rightarrow u = 0$

Phương trình (1) trở thành

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x - \int_{\frac{π}{2}}^{0} f (u) d u = \frac{1}{2} \int_{0}^{\frac{π}{2}} \sin 2 x d x$

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x + \int_{0}^{\frac{π}{2}} f (u) d u = - \frac{1}{4} \int_{0}^{\frac{π}{2}} - 2 \sin 2 x d x$

$\Leftrightarrow 2 \int_{0}^{\frac{π}{2}} f (x) d x = {- \frac{1}{4} \cos 2 x|}_{0}^{\frac{π}{2}}$

$\Leftrightarrow \int_{0}^{\frac{π}{2}} f (x) d x = {- \frac{1}{8} \cos 2 x|}_{0}^{\frac{π}{2}} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$

Ta có:

$\int_{0}^{\frac{π}{2}} x . f' (x) d x$

Đặt $\{\begin{matrix} u = x \Rightarrow d u = d x \\ d v = f' (x) d x \Rightarrow v = f (x) \end{matrix}$

Vậy suy ra

$\int_{0}^{\frac{π}{2}} x . f' (x) d x = {x . f (x)|}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} f (x) d x$

$= \frac{π}{2} . f (\frac{π}{2}) - \frac{1}{4} = - \frac{1}{4} .$

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Answer 2