Lời giải của Tự Học 365

Ta có: $\int {\tan {\mkern 1mu} xdx} {\rm{\;}} = \int {\dfrac{{\sin {\mkern 1mu} xdx}}{{\cos x}}} {\rm{\;}} = {\rm{\;}} - \int {\dfrac{{d(\cos x)}}{{\cos x}}} {\rm{\;}} = {\rm{\;}} - \ln \left| {\cos x} \right| + C$

$f'(x) = \tan {\mkern 1mu} x,{\mkern 1mu} {\mkern 1mu} \forall x \in \left( { - \dfrac{\pi }{4};\dfrac{{3\pi }}{4}} \right)\backslash \left\{ {\dfrac{\pi }{2}} \right\}$ $ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = {\rm{\;}} - \ln \left( {\cos x} \right) + {C_1},{\mkern 1mu} {\mkern 1mu} x \in \left( { - \dfrac{\pi }{4};\dfrac{\pi }{2}} \right)}\\{f(x) = {\rm{\;}} - \ln \left( { - \cos x} \right) + {C_2},{\mkern 1mu} {\mkern 1mu} x \in \left( {\dfrac{\pi }{2};\dfrac{{3\pi }}{4}} \right)}\end{array}} \right.$

Mà $f(0) = 0,{\mkern 1mu} {\mkern 1mu} f(\pi ) = 1 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{ - \ln 1 + {C_1} = 0}\\{ - \ln 1 + {C_2} = 1}\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{{C_1} = 0}\\{{C_2} = 1}\end{array}} \right.$

$ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f(x) = {\rm{\;}} - \ln \left( {\cos x} \right),{\mkern 1mu} {\mkern 1mu} x \in \left( { - \dfrac{\pi }{4};\dfrac{\pi }{2}} \right)}\\{f(x) = {\rm{\;}} - \ln \left( { - \cos x} \right) + 1,{\mkern 1mu} {\mkern 1mu} x \in \left( {\dfrac{\pi }{2};\dfrac{{3\pi }}{4}} \right)}\end{array}} \right.$

$ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{f\left( {\dfrac{{2\pi }}{3}} \right) = {\rm{\;}} - \ln \left( { - \cos \dfrac{{2\pi }}{3}} \right) + 1 = \ln 2 + 1}\\{f\left( {\dfrac{\pi }{4}} \right) = {\rm{\;}} - \ln \left( {\cos \dfrac{\pi }{4}} \right) = \dfrac{1}{2}\ln 2}\end{array}} \right.$$ \Rightarrow \dfrac{{f\left( {\dfrac{{2\pi }}{3}} \right)}}{{f\left( {\dfrac{\pi }{4}} \right)}} = \dfrac{{2\left( {\ln 2 + 1} \right)}}{{\ln 2}} = 2\left( {1 + {{\log }_2}e} \right)$

Đáp án cần chọn là: a