Google
Câu hỏi: Giả sử $f\left( x \right)$ là hàm có đạo hàm liên tục trên khoảng $\left( 0;\pi \right)$ và $f'\left( x \right)\sin x=x+f\left( x \right)\cos x, \forall x\in \left( 0;\pi \right).$ Biết $f\left( \dfrac{\pi }{2} \right)=1, f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{12}\left( a+b\ln 2+c\pi \sqrt{3} \right)$, với $a, b, c$ là các số nguyên. Giá trị $a+b+c$ bằng
A. $-1$.
B. $1$.
C. $11$.
D. $-11$.
A. $-1$.
B. $1$.
C. $11$.
D. $-11$.
$\forall x\in \left( 0;\pi \right)$, ta có: $f'\left( x \right)\sin x=x+f\left( x \right)\cos x, \forall x\in \left( 0;\pi \right).$
$\begin{aligned}
& \Leftrightarrow \dfrac{f'\left( x \right)\sin x-f\left( x \right)\cos x}{{{\sin }^{2}}x}=\dfrac{x}{{{\sin }^{2}}x} \\
& \Leftrightarrow {{\left( \dfrac{f\left( x \right)}{\sin x} \right)}^{'}}=\dfrac{x}{{{\sin }^{2}}x} \\
& \Rightarrow \dfrac{f\left( x \right)}{\sin x}=\int{\dfrac{x}{{{\sin }^{2}}x}dx}=-x.\cot x+\ln \left| \sin x \right|+C \\
& \Rightarrow f\left( x \right)=-x.\cos x+\sin x.\ln \left| \sin x \right|+C.\sin x \\
\end{aligned}$
Ta lại có: $f\left( \dfrac{\pi }{2} \right)=1\Leftrightarrow C=1$
$\begin{aligned}
& \Rightarrow f\left( \dfrac{\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}.\dfrac{\pi }{6}+\dfrac{1}{2}.\ln \dfrac{1}{2}+\dfrac{1}{2}=\dfrac{1}{12}\left( 6-6\ln 2-\pi \sqrt{3} \right) \\
& \Rightarrow a=6, b=-6, c=-1. \\
\end{aligned}$
Vậy $a+b+c=-1$.
$\begin{aligned}
& \Leftrightarrow \dfrac{f'\left( x \right)\sin x-f\left( x \right)\cos x}{{{\sin }^{2}}x}=\dfrac{x}{{{\sin }^{2}}x} \\
& \Leftrightarrow {{\left( \dfrac{f\left( x \right)}{\sin x} \right)}^{'}}=\dfrac{x}{{{\sin }^{2}}x} \\
& \Rightarrow \dfrac{f\left( x \right)}{\sin x}=\int{\dfrac{x}{{{\sin }^{2}}x}dx}=-x.\cot x+\ln \left| \sin x \right|+C \\
& \Rightarrow f\left( x \right)=-x.\cos x+\sin x.\ln \left| \sin x \right|+C.\sin x \\
\end{aligned}$
Ta lại có: $f\left( \dfrac{\pi }{2} \right)=1\Leftrightarrow C=1$
$\begin{aligned}
& \Rightarrow f\left( \dfrac{\pi }{6} \right)=-\dfrac{\sqrt{3}}{2}.\dfrac{\pi }{6}+\dfrac{1}{2}.\ln \dfrac{1}{2}+\dfrac{1}{2}=\dfrac{1}{12}\left( 6-6\ln 2-\pi \sqrt{3} \right) \\
& \Rightarrow a=6, b=-6, c=-1. \\
\end{aligned}$
Vậy $a+b+c=-1$.
Đáp án A.