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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. $P=\dfrac{14}{9}.$
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 $x=\dfrac{\pi }{3}\Rightarrow \dfrac{\sqrt{3}}{2}f\left( \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}.\sqrt{3}-\ln 2+C\Rightarrow \sqrt{3}f\left( \dfrac{\pi }{3} \right)=\dfrac{2}{3}.\pi \sqrt{3}-2\ln 2+2C.$
Với $x=\dfrac{\pi }{6}\Rightarrow \dfrac{1}{2}f\left( \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}.\dfrac{\sqrt{3}}{3}+\dfrac{1}{2}\ln 3-\ln 2+C\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{9}.\pi \sqrt{3}+\ln 3-2\ln 2+2C.$
Vậy $\sqrt{3}f\left( \dfrac{\pi }{3} \right)-f\left( \dfrac{\pi }{6} \right)=\dfrac{5}{9}\pi \sqrt{3}-\ln 3\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{5}{9} \\
& b=-1 \\
\end{aligned} \right.\Rightarrow P=a+b=-\dfrac{4}{9}.$
Suy ra $\sin xf\left( x \right)=\int{\dfrac{x}{{{\cos }^{2}}x}dx}=x\tan x+\ln \left| \cos x \right|+C.$
Với $x=\dfrac{\pi }{3}\Rightarrow \dfrac{\sqrt{3}}{2}f\left( \dfrac{\pi }{3} \right)=\dfrac{\pi }{3}.\sqrt{3}-\ln 2+C\Rightarrow \sqrt{3}f\left( \dfrac{\pi }{3} \right)=\dfrac{2}{3}.\pi \sqrt{3}-2\ln 2+2C.$
Với $x=\dfrac{\pi }{6}\Rightarrow \dfrac{1}{2}f\left( \dfrac{\pi }{6} \right)=\dfrac{\pi }{6}.\dfrac{\sqrt{3}}{3}+\dfrac{1}{2}\ln 3-\ln 2+C\Rightarrow f\left( \dfrac{\pi }{6} \right)=\dfrac{1}{9}.\pi \sqrt{3}+\ln 3-2\ln 2+2C.$
Vậy $\sqrt{3}f\left( \dfrac{\pi }{3} \right)-f\left( \dfrac{\pi }{6} \right)=\dfrac{5}{9}\pi \sqrt{3}-\ln 3\Rightarrow \left\{ \begin{aligned}
& a=\dfrac{5}{9} \\
& b=-1 \\
\end{aligned} \right.\Rightarrow P=a+b=-\dfrac{4}{9}.$
Đáp án A.