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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục và có đạo hàm trên $\left( 0;\dfrac{\pi }{2} \right)$, thỏa mãn hệ thức $f\left( x \right)+\tan x{f}'\left( x \right)=\dfrac{x}{{{\cos }^{3}}x}$. Biết rằng $\sqrt{3}f\left( \dfrac{\pi }{3} \right)-f\left( \dfrac{\pi }{6} \right)=a\pi \sqrt{3}+b\ln 3$ trong đó $a,b\in \mathbb{Q}$. Tính giá trị của biểu thức $P=a+b.$
A. $P=-\dfrac{4}{9}.$
B. $P=-\dfrac{2}{9}.$
C. $P=\dfrac{7}{9}.$
D. $P=\dfrac{14}{9}.$
A. $P=-\dfrac{4}{9}.$
B. $P=-\dfrac{2}{9}.$
C. $P=\dfrac{7}{9}.$
D. $P=\dfrac{14}{9}.$
Từ giả thiết $\Rightarrow \cot x.f\left( x \right)+{f}'\left( x \right)=\dfrac{x}{{{\cos }^{3}}x}.\cot x$
Nhân thêm 2 vế với ${{e}^{\int{\cot xdx}}}=\sin x$ ta có
$\cos xf\left( x \right)+\sin x{f}'\left( x \right)=\dfrac{x}{{{\cos }^{2}}x}\Leftrightarrow {{\left[ \sin xf\left( x \right) \right]}^{\prime }}=\dfrac{x}{{{\cos }^{2}}x}.$
Suy ra $\sin xf\left( x \right)=\int{\dfrac{x}{{{\cos }^{2}}x}dx=x\tan x+\ln \left| \cos x \right|+C.}$
+ Với $x=\dfrac{\pi }{3}\xrightarrow{{}}\dfrac{\sqrt{3}}{2}f\left( \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}.\sqrt{3}-\ln 2\xrightarrow{{}}\sqrt{3}f\left( \dfrac{\pi }{3} \right)=\dfrac{2}{3}.\pi \sqrt{3}-2\ln 2+2C.$
+ Với $x=\dfrac{\pi }{6}\xrightarrow{{}}\dfrac{1}{2}f\left( \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}.\dfrac{\sqrt{3}}{3}+\dfrac{1}{2}\ln 3-\ln 2+C$
$\xrightarrow{{}}f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{9}..\pi \sqrt{3}+\ln 3-2\ln 2+2C.$
Suy ra$\sqrt{3}f\left( \dfrac{\pi }{3} \right)-f\left( \dfrac{\pi }{6} \right)=\dfrac{5}{9}\pi \sqrt{3}-\ln 3\xrightarrow{{}}\left\{ \begin{aligned}
& a=\dfrac{5}{9} \\
& b=-1 \\
\end{aligned} \right.\xrightarrow{{}}P=a+b=-\dfrac{4}{9}.$
Nhân thêm 2 vế với ${{e}^{\int{\cot xdx}}}=\sin x$ ta có
$\cos xf\left( x \right)+\sin x{f}'\left( x \right)=\dfrac{x}{{{\cos }^{2}}x}\Leftrightarrow {{\left[ \sin xf\left( x \right) \right]}^{\prime }}=\dfrac{x}{{{\cos }^{2}}x}.$
Suy ra $\sin xf\left( x \right)=\int{\dfrac{x}{{{\cos }^{2}}x}dx=x\tan x+\ln \left| \cos x \right|+C.}$
+ Với $x=\dfrac{\pi }{3}\xrightarrow{{}}\dfrac{\sqrt{3}}{2}f\left( \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}.\sqrt{3}-\ln 2\xrightarrow{{}}\sqrt{3}f\left( \dfrac{\pi }{3} \right)=\dfrac{2}{3}.\pi \sqrt{3}-2\ln 2+2C.$
+ Với $x=\dfrac{\pi }{6}\xrightarrow{{}}\dfrac{1}{2}f\left( \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}.\dfrac{\sqrt{3}}{3}+\dfrac{1}{2}\ln 3-\ln 2+C$
$\xrightarrow{{}}f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{9}..\pi \sqrt{3}+\ln 3-2\ln 2+2C.$
Suy ra$\sqrt{3}f\left( \dfrac{\pi }{3} \right)-f\left( \dfrac{\pi }{6} \right)=\dfrac{5}{9}\pi \sqrt{3}-\ln 3\xrightarrow{{}}\left\{ \begin{aligned}
& a=\dfrac{5}{9} \\
& b=-1 \\
\end{aligned} \right.\xrightarrow{{}}P=a+b=-\dfrac{4}{9}.$
Đáp án A.