Giả sử

B

- Hướng dẫn giải

Ta có:

$$\begin{gathered} {\int }_{-1}^2\frac{e^{x}dx}{2+e^x}=\ln \left(\frac{ae+e^3}{ae+b}\right) \\ =\ln {\left(2+e^x\right)}_{-1}^2=\ln \left(2+e^2\right)-\ln \left(2+e^{-1}\right) \\ \ln \left(\frac{2+e^x}{2+e^{-1}}\right)=\ln \left(\frac{2e+e^3}{2e+1}\right)=\ln \left(\frac{ae+e^3}{ae+b}\right) \end{gathered}$$

Suy ra a = 2; b = 1 hay 

$P=\sin \left(\frac{\mathrm{\pi b}}{a}+2017\mathrm{\pi }\right)+\cos \left(\frac{\mathrm{\pi b}}{a}-\sin \left(2018\mathrm{\pi }\right)\right)$ = -1

Đáp án B