Đường cong trong hình có thể là đồ thị của hàm số nào dưới đây?

\(R\backslash \left\{ {k\pi ,k \in Z} \right\}\)

\(R\backslash \left\{ {\dfrac{{k\pi }}{2},k \in Z} \right\}\)

\(R\backslash \left\{ {\dfrac{\pi }{2} + k\pi ,k \in Z} \right\}\)

\(R\)

\(\max y =  - 2,\min y = 4\)

\(\max y = 2,\min y = 4\)

\(\max y =  - 2,\min y = 3\)

\(\max y = 4,\min y =  - 2\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{k2\pi }}{3},k \in \mathbb{Z}} \right\}\)      

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{6} + \dfrac{{k\pi }}{3},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{k\pi }}{3},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\)

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x = \pi  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)    

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x = \pi  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x =  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)      

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x =  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x = \dfrac{{5\pi }}{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)          

\(\left[ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\x =  - \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)          

\(\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x =  - \dfrac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

$k2\pi \left( {k \in Z} \right)$

$k2\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

$\dfrac{{k\pi }}{2}\left( {k \in Z} \right)$

$k\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$