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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục và có đạo hàm trên khoảng $\left( 0;+\infty \right)$ thỏa mãn $\sin x+f\left( x \right)=x\left[ \cos x+{f}'\left( x \right) \right]$ và $f\left( \dfrac{\pi }{2} \right)=\pi -1.$ Giá trị của $f\left( \dfrac{\pi }{6} \right)$ bằng
A. $\dfrac{2\pi -3\sqrt{3}}{6}.$
B. $\dfrac{4\pi -3\sqrt{3}}{6}.$
C. $\dfrac{2\pi -3}{6}.$
D. $\dfrac{4\pi -3}{6}.$
A. $\dfrac{2\pi -3\sqrt{3}}{6}.$
B. $\dfrac{4\pi -3\sqrt{3}}{6}.$
C. $\dfrac{2\pi -3}{6}.$
D. $\dfrac{4\pi -3}{6}.$
Ta có $\dfrac{x.{f}'\left( x \right)-f\left( x \right)}{{{x}^{2}}}=-\dfrac{\cos x}{x}+\dfrac{\sin x}{{{x}^{2}}}\Rightarrow {{\left[ \dfrac{f\left( x \right)}{x} \right]}^{\prime }}=-\dfrac{\cos x}{x}+\dfrac{\sin x}{{{x}^{2}}}$
$\Rightarrow \dfrac{f\left( x \right)}{x}=-\int{\dfrac{\cos x}{x}dx}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}=-\int{\dfrac{1}{x}d\left( \sin x \right)}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}$
$=-\dfrac{\sin x}{x}+C+\int{\sin xd\left( \dfrac{1}{x} \right)}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}=-\dfrac{\sin x}{x}+C+\int{\sin x.\dfrac{-1}{{{x}^{2}}}dx+\int{\dfrac{\sin x}{{{x}^{2}}}dx}}=-\dfrac{\sin x}{x}+C$
$\Rightarrow f\left( x \right)=-\sin x+Cx.$
Mà $f\left( \dfrac{\pi }{2} \right)=\pi -1\Rightarrow -1+C.\dfrac{\pi }{2}=\pi -1\Rightarrow C=2\Rightarrow f\left( x \right)=-\sin x+2x$
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}+\dfrac{\pi }{3}.$
$\Rightarrow \dfrac{f\left( x \right)}{x}=-\int{\dfrac{\cos x}{x}dx}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}=-\int{\dfrac{1}{x}d\left( \sin x \right)}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}$
$=-\dfrac{\sin x}{x}+C+\int{\sin xd\left( \dfrac{1}{x} \right)}+\int{\dfrac{\sin x}{{{x}^{2}}}dx}=-\dfrac{\sin x}{x}+C+\int{\sin x.\dfrac{-1}{{{x}^{2}}}dx+\int{\dfrac{\sin x}{{{x}^{2}}}dx}}=-\dfrac{\sin x}{x}+C$
$\Rightarrow f\left( x \right)=-\sin x+Cx.$
Mà $f\left( \dfrac{\pi }{2} \right)=\pi -1\Rightarrow -1+C.\dfrac{\pi }{2}=\pi -1\Rightarrow C=2\Rightarrow f\left( x \right)=-\sin x+2x$
$\Rightarrow f\left( \dfrac{\pi }{6} \right)=-\dfrac{1}{2}+\dfrac{\pi }{3}.$
Đáp án C.