A. $\left\{\begin{array}{l}{a=\dfrac{2}{\pi}} \\ {b=1}\end{array}\right.$
B. $\left\{\begin{array}{l}{a=\dfrac{2}{\pi}} \\ {b=2}\end{array}\right.$
C. $\left\{\begin{array}{l}{a=\dfrac{1}{\pi}} \\ {b=0}\end{array}\right.$
D. $\left\{\begin{array}{l}{a=\dfrac{2}{\pi}} \\ {b=0}\end{array}\right.$
Hướng dẫn giải (chi tiết)
$f(x)=\left\{\begin{array}{ll}{\sin x} & {k h i|x| \leq \dfrac{\pi}{2}} \\ {a x+b} & {k h i|x|>\dfrac{\pi}{2}}\end{array}\right.$ $\begin{array}{l} \Leftrightarrow f\left( x \right) = \left\{ \begin{array}{l} \sin x\,\,\,\,\,\,\,\,;\, - \dfrac{\pi }{2} \le x \le \dfrac{\pi }{2}\\ ax + b\,\,\,\,;\,\left[ \begin{array}{l} x > \dfrac{\pi }{2}\\ x < - \dfrac{\pi }{2} \end{array} \right. \end{array} \right.\,\,\\ \end{array}$
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 $\mathbb R$ thì hàm số phải liên tục tại các điểm $x=\pm \dfrac{\pi}{2}$
$\Rightarrow\left\{\begin{array}{l}{\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\dfrac{\pi}{2}\right)} \\ {\lim _{x \rightarrow-\frac{\pi}{2}} f(x)=f\left(-\dfrac{\pi}{2}\right)}\end{array}\right.$
Ta có:
$\begin{array}{l}{\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f(x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}}(a x+b)=a \cdot \dfrac{\pi}{2}+b} \\ {\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f(x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(\sin x)=\sin \dfrac{\pi}{2}=1} \\ {f\left(\dfrac{\pi}{2}\right)=\sin \dfrac{\pi}{2}=1}\end{array}$
$\Rightarrow \lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}} f(x)=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} f(x)=f\left(\dfrac{\pi}{2}\right) \Leftrightarrow a \cdot \dfrac{\pi}{2}+b=1$ (1)
$\begin{array}{l}{\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}} f(x)=\lim _{x \rightarrow\left(-\frac{-\pi}{2}\right)^{+}}(\sin x)=-1} \\ {\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}} f(x)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}}(a x+b)=-a \cdot \dfrac{\pi}{2}+b} \\ {f\left(-\dfrac{\pi}{2}\right)=\sin \dfrac{-\pi}{2}=-1}\end{array}$
$\Rightarrow \lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{+}} f(x)=\lim _{x \rightarrow\left(-\frac{\pi}{2}\right)^{-}} f(x)=f\left(-\dfrac{\pi}{2}\right) \Leftrightarrow-a \cdot \dfrac{\pi}{2}+b=-1$ (2)
Từ (1) và (2) ta có hệ phương trình $\left\{\begin{array}{l}{a \cdot \dfrac{\pi}{2}+b=1} \\ {-a \cdot \dfrac{\pi}{2}+b=-1}\end{array} \Leftrightarrow\left\{\begin{array}{l}{a=\dfrac{2}{\pi}} \\ {b=0}\end{array}\right.\right.$
Đáp án: D