Cho hàm số f(x) liên tục và có đạo hàm

* Hướng dẫn giải

Đáp án A

Ta có $\cos xf\left(x\right)+\sin x{f}^{\text{'}}\left(x\right)=\frac{x}{{\cos }^{2}x}\iff {\left[\sin xf\left(x\right)\right]}^{\text{'}}=\frac{x}{{\cos }^{2}x}.$

Suy ra $\sin xf\left(x\right)=\int \frac{x}{{\cos }^{2}x}dx=x\tan x+\ln \left|\cos x\right|+C.$

Với $x=\frac{\pi }{3}\Rightarrow \frac{\sqrt{3}}{2}f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}.\sqrt{3}-\ln 2+C\Rightarrow \sqrt{3}f\left(\frac{\pi }{3}\right)=\frac{2}{3}.\pi \sqrt{3}-2\ln 2+2C.$

Với $x=\frac{\pi }{6}\Rightarrow \frac{1}{2}f\left(\frac{\pi }{6}\right)=\frac{\pi }{6}.\frac{\sqrt{3}}{3}+\frac{1}{2}\ln 3-\ln 2+C\Rightarrow f\left(\frac{\pi }{6}\right)=\frac{1}{9}.\pi \sqrt{3}+\ln 3-2\ln 2+2C.$

Vậy $\sqrt{3}f\left(\frac{\pi }{3}\right)-f\left(\frac{\pi }{6}\right)=\frac{5}{9}\pi \sqrt{3}-\ln 3\Rightarrow \left\{\begin{array}{l}a=\frac{5}{9}\\ b=-1\end{array}\right.\Rightarrow P=a+b=-\frac{4}{9}.$