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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left( 0;\pi \right)$ thỏa mãn ${f}'\left( x \right)=f\left( x \right).\cot x+2x.\sin x$. Biết $f\left( \dfrac{\pi }{2} \right)=\dfrac{{{\pi }^{2}}}{4}$. Tính $f\left( \dfrac{\pi }{6} \right)$.
A. $\dfrac{{{\pi }^{2}}}{36}$.
B. $\dfrac{{{\pi }^{2}}}{72}$.
C. $\dfrac{{{\pi }^{2}}}{54}$.
D. $\dfrac{{{\pi }^{2}}}{80}$.
A. $\dfrac{{{\pi }^{2}}}{36}$.
B. $\dfrac{{{\pi }^{2}}}{72}$.
C. $\dfrac{{{\pi }^{2}}}{54}$.
D. $\dfrac{{{\pi }^{2}}}{80}$.
$\begin{aligned}
& {f}'\left( x \right)=f\left( x \right).\cot x+2x.\sin x\Leftrightarrow \sin x.{f}'\left( x \right)-f\left( x \right).\cos x+2x.{{\sin }^{2}}x \\
& \Leftrightarrow \sin x.{f}'\left( x \right)-f\left( x \right).\cos x=2x.{{\sin }^{2}}x\Leftrightarrow \dfrac{sinx.{f}'\left( x \right)-f\left( x \right).\cos x}{{{\sin }^{2}}x}=2x \\
& \Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{sinx.{f}'\left( x \right)-f\left( x \right).\cos x}{{{\sin }^{2}}x}dx}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{2x.dx}\Leftrightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{{{\left( \dfrac{f\left( x \right)}{\sin x} \right)}^{'}}dx}=\left. {{x}^{2}} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} \\
& \Leftrightarrow \left. \dfrac{f\left( x \right)}{\sin x} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{36}\Leftrightarrow \dfrac{f\left( \dfrac{\pi }{2} \right)}{1}-\dfrac{f\left( \dfrac{\pi }{6} \right)}{\dfrac{1}{2}}=\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{36}\Leftrightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{{{\pi }^{2}}}{72} \\
\end{aligned}$
& {f}'\left( x \right)=f\left( x \right).\cot x+2x.\sin x\Leftrightarrow \sin x.{f}'\left( x \right)-f\left( x \right).\cos x+2x.{{\sin }^{2}}x \\
& \Leftrightarrow \sin x.{f}'\left( x \right)-f\left( x \right).\cos x=2x.{{\sin }^{2}}x\Leftrightarrow \dfrac{sinx.{f}'\left( x \right)-f\left( x \right).\cos x}{{{\sin }^{2}}x}=2x \\
& \Rightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{sinx.{f}'\left( x \right)-f\left( x \right).\cos x}{{{\sin }^{2}}x}dx}=\int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{2x.dx}\Leftrightarrow \int\limits_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{{{\left( \dfrac{f\left( x \right)}{\sin x} \right)}^{'}}dx}=\left. {{x}^{2}} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}} \\
& \Leftrightarrow \left. \dfrac{f\left( x \right)}{\sin x} \right|_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}=\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{36}\Leftrightarrow \dfrac{f\left( \dfrac{\pi }{2} \right)}{1}-\dfrac{f\left( \dfrac{\pi }{6} \right)}{\dfrac{1}{2}}=\dfrac{{{\pi }^{2}}}{4}-\dfrac{{{\pi }^{2}}}{36}\Leftrightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{{{\pi }^{2}}}{72} \\
\end{aligned}$
Đáp án B.