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Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm liên tục trên $\mathbb{R}$ thỏa mãn $f(x)+f\left( \dfrac{\pi }{2}-x \right)=\sin x.\cos x$ với mọi $x\in \mathbb{R}$ và $f(0)=0$. Giá trị của tích phân $\int\limits_{0}^{\dfrac{\pi }{2}}{x.{f}'(x)d\text{x}}$ bằng
A. $-\dfrac{\pi }{4}$
B. $\dfrac{1}{4}$
C. $\dfrac{\pi }{4}$
D. $-\dfrac{1}{4}$
A. $-\dfrac{\pi }{4}$
B. $\dfrac{1}{4}$
C. $\dfrac{\pi }{4}$
D. $-\dfrac{1}{4}$
Theo giả thiết, $f(0)=0$ và $f(x)+f\left( \dfrac{\pi }{2}-x \right)=\sin x.\cos x$ nên $f(0)+f\left( \dfrac{\pi }{2} \right)=0\Leftrightarrow f\left( \dfrac{\pi }{2} \right)=0$.
Ta có: $\int\limits_{0}^{\dfrac{\pi }{2}}{x.{f}'(x)d\text{x}}=\int\limits_{0}^{\dfrac{\pi }{2}}{x\text{d}\left[ f(x) \right]}=\left. \left[ xf(x) \right] \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}$. Suy ra: $I=-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}$. Mặt khác, ta có:
$f(x)+f\left( \dfrac{\pi }{2}-x \right)=\sin x.\cos x\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\cos xdx}=\dfrac{1}{2}$
Suy ra: $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}=\dfrac{1}{2}\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}=\dfrac{1}{4}$. Vậy $I=-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}=-\dfrac{1}{4}$.
Ta có: $\int\limits_{0}^{\dfrac{\pi }{2}}{x.{f}'(x)d\text{x}}=\int\limits_{0}^{\dfrac{\pi }{2}}{x\text{d}\left[ f(x) \right]}=\left. \left[ xf(x) \right] \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}$. Suy ra: $I=-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}$. Mặt khác, ta có:
$f(x)+f\left( \dfrac{\pi }{2}-x \right)=\sin x.\cos x\Rightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}=\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x.\cos xdx}=\dfrac{1}{2}$
Suy ra: $\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( \dfrac{\pi }{2}-x \right)dx}=\dfrac{1}{2}\Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{f(x)dx}=\dfrac{1}{4}$. Vậy $I=-\int\limits_{0}^{\dfrac{\pi }{2}}{f(x)d\text{x}}=-\dfrac{1}{4}$.
Đáp án D.