Cho hàm số f ( x ) = tan (x − 2pi/ 3) .

a) Sai.

Dùng công thức: \(\left( {\tan u} \right)' = \frac{{u'}}{{{{\cos }^2}u}}\).

Ta có: \(f'\left( x \right) = \frac{1}{{{{\cos }^2}\left( {x - \frac{{2\pi }}{3}} \right)}}\).

Hoặc bấm máy:

\({\left. {\frac{d}{{dx}}\left( {\tan \left( {x - \frac{{2\pi }}{3}} \right)} \right)} \right|_{x = x}} - \frac{3}{{{{\cos }^2}x{{\left( {1 - \sqrt 3 \tan x} \right)}^2}}}\)            \(\begin{array}{*{20}{c}}{Calc:x = \frac{\pi }{3}}&{}&{ \Rightarrow kq: = 1 \ne 0}\end{array}\).

b) Đúng.

Bấm máy

\(\frac{{\sqrt 3 }}{4}.\frac{1}{{{{\cos }^2}\left( {x - \frac{{2\pi }}{3}} \right)}} - \tan \left( {x - \frac{{2\pi }}{3}} \right)\)            \(\begin{array}{*{20}{c}}{Calc:x = 0}&{}&{ \Rightarrow kq: = 0}\end{array}\).

Hoặc bấm máy

\({\left. {\frac{{\sqrt 3 }}{4}.\frac{d}{{dx}}\left( {\tan \left( {x - \frac{{2\pi }}{3}} \right)} \right)} \right|_{x = x}} - \tan \left( {x - \frac{{2\pi }}{3}} \right)\)            \(\begin{array}{*{20}{c}}{Calc:x = 0}&{}&{ \Rightarrow kq: = 0}\end{array}\).

c) Đúng.

\(\begin{array}{*{20}{c}}{f\left( x \right) = \tan \left( {x - \frac{{2\pi }}{3}} \right) = 0}&{ \Leftrightarrow x - \frac{{2\pi }}{3} = 0}&{ \Leftrightarrow x = \frac{{2\pi }}{3} + k\pi }&{\left( {k \in \mathbb{Z}} \right)}\end{array}\).

Với \(\begin{array}{*{20}{c}}{k =  - 1}&{}&{ \Rightarrow x = \frac{{2\pi }}{3} - \pi  =  - \frac{\pi }{3}}\end{array}\).

d) Sai.

Ta có:\(\begin{array}{*{20}{c}}{\frac{1}{{f'\left( x \right)}} = {{\cos }^2}\left( {x - \frac{{2\pi }}{3}} \right)}&{}&{}&{ \Rightarrow M = 1 \notin \left( {0;1} \right)}\end{array}\).