Cho hàm số \(f\left( x \right) = \left\{ \begin{array}{l}\sin x\,\,\,\,\,\,\,khi\,\,\left| x \right| \le \dfrac{\pi }{2}\\ax + b\,\,\,\,khi\,\,\left| x \right| > \dfrac{\pi }{2}\end{array} \right.\) liên tục trên $R.$ Khi đó giá trị của $a$ và $b$ là:

\(f\left( x \right) = \left\{ \begin{array}{l}\sin x\,\,\,\,\,\,\,khi\,\,\left| x \right| \le \dfrac{\pi }{2}\\ax + b\,\,\,\,khi\,\,\left| x \right| > \dfrac{\pi }{2}\end{array} \right. \) \(\Leftrightarrow f\left( x \right) = \left\{ \begin{array}{l}\sin x\,\,\,\,\,\,\,khi\,\, - \dfrac{\pi }{2} \le x \le \dfrac{\pi }{2}\\ax + b\,\,\,\,khi\,\,\left[ \begin{array}{l}x > \dfrac{\pi }{2}\\x <  - \dfrac{\pi }{2}\end{array} \right.\end{array} \right.\)

Ta có hàm số liên tục trên các khoảng \(\left( { - \infty ; - \dfrac{\pi }{2}} \right) \cup \left( { - \dfrac{\pi }{2};\dfrac{\pi }{2}} \right) \cup \left( {\dfrac{\pi }{2}; + \infty } \right)\)

Để hàm số liên tục trên $R$ thì hàm số phải liên tục tại các điểm \(x =  \pm \dfrac{\pi }{2} \)

\(\Rightarrow \left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to \frac{\pi }{2}} f\left( x \right) = f\left( {\dfrac{\pi }{2}} \right)\\\mathop {\lim }\limits_{x \to  - \frac{\pi }{2}} f\left( x \right) = f\left( { - \dfrac{\pi }{2}} \right)\end{array} \right.\)

Ta có

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

Từ (1) và (2) ta có hệ phương trình \(\left\{ \begin{array}{l}a.\dfrac{\pi }{2} + b = 1\\ - a.\dfrac{\pi }{2} + b =  - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = \dfrac{2}{\pi }\\b = 0\end{array} \right.\)