Câu hỏi 39

Ta có: \(f\left( x \right) = \left\{ \begin{array}{l}\sin x\,\,\,\,khi\,\,\,\left| x \right| \le \frac{\pi }{2}\\ax + b\,\,\,\,khi\,\,\,\,\left| x \right| > \frac{\pi }{2}\end{array} \right. = \left\{ \begin{array}{l}\sin x\,\,\,\,khi\,\,\, - \frac{\pi }{2} \le x \le \frac{\pi }{2}\\ax + b\,\,\,\,khi\,\,\,\,x \in \left( { - \infty ; - \frac{\pi }{2}} \right) \cup \left( {\frac{\pi }{2}; + \infty } \right)\end{array} \right..\)

Hàm số liên tục trên các khoảng \(\left( { - \infty ; - \frac{\pi }{2}} \right);\,\left( { - \frac{\pi }{2};\frac{\pi }{2}} \right);\,\left( {\frac{\pi }{2}; + \infty } \right)\) .

Ta có:  \(f\left( {\frac{\pi }{2}} \right) = 1;\,\,f\left( { - \frac{\pi }{2}} \right) =  - 1.\)

\(\begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - \frac{\pi }{2}} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - \frac{\pi }{2}} \right)}^ + }} \sin x =  - 1;\\\mathop {\lim }\limits_{x \to {{\left( { - \frac{\pi }{2}} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( { - \frac{\pi }{2}} \right)}^ - }} \left( {ax + b} \right) =  - \frac{\pi }{2}a + b.\\\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ - }} \sin x = 1\\\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {ax + b} \right) = \frac{\pi }{2}a + b.\end{array}\)

Hàm số liên tục trên \(\mathbb{R} \Leftrightarrow \) hàm số liên tục tại \(x =  \pm \frac{\pi }{2}\)

 \( \Leftrightarrow \left\{ \begin{array}{l}\frac{\pi }{2}a + b = 1\\ - \frac{\pi }{2}a + b =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \frac{2}{\pi }\\b = 0\end{array} \right. \Rightarrow \frac{4}{{{a^2}}} + {b^2} = \frac{4}{{{{\left( {\frac{2}{\pi }} \right)}^2}}} + 0 = {\pi ^2}.\)

Chọn C.