Câu hỏi 25

\(f\left( x \right) = \left\{ \matrix{  \cos {{\pi x} \over 2}\,\,\,\,khi\,\,\left| x \right| \le 1 \hfill \cr   \left| {x - 1} \right|\,\,\,\,\,\,\,khi\,\,\left| x \right| > 1 \hfill \cr}  \right. \Leftrightarrow f\left( x \right) = \left\{ \matrix{  \cos {{\pi x} \over 2}\,\,\,\,khi\,\, - 1 \le x \le 1 \hfill \cr   \left| {x - 1} \right|\,\,\,\,\,\,\,khi\,\,\left[ \matrix{  x > 1 \hfill \cr   x <  - 1 \hfill \cr}  \right. \hfill \cr}  \right.\)

Ta có:

\(\left. \matrix{  \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left| {x - 1} \right| = 0 \hfill \cr   \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \cos {{\pi x} \over 2} = \cos {\pi  \over 2} = 0 \hfill \cr   f\left( 1 \right) = \cos {\pi  \over 2} = 0 \hfill \cr}  \right\} \Rightarrow \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right) = 0 \Rightarrow \) Hàm số liên tục tại x = 1.

\(\left. \matrix{  \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \cos {{\pi x} \over 2} = \cos {{ - \pi } \over 2} = 0 \hfill \cr   \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \left| {x - 1} \right| = 2 \hfill \cr}  \right\} \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} f\left( x \right) \Rightarrow \) Hàm số không liên tục tại \(x =  - 1\).

Chọn B.