Chọn A
Hàm số \(f\left(x\right)\) có đạo hàm tại x=2 thì \(f\left(x\right)\) liên tục tại x=2
\({\mathop{\lim }\limits_{x\to 2^{-} }} \left(2a.\sin \pi x+b\right)=b ;{\mathop{\lim }\limits_{x\to 2^{+} }} \left(x+\cos \pi x\; \right)=3 ;f\left(2\right)=b \)
Hàm số liên tục tại x=2 khi
\({\mathop{\lim }\limits_{x\to 2^{-} }} f\left(x\right)={\mathop{\lim }\limits_{x\to 2^{+} }} f\left(x\right)=f\left(2\right)\Leftrightarrow b=3\)
Với b=3, ta có \(y=\left\{\begin{array}{l} {2a\sin \pi x+3\; \; \; \; \; \; khi\; x\le 2} \\ {x+\cos \pi x\; \; \; \; \; \; \; khi\; x>2} \end{array}\right. \)
Khi đó\({\mathop{\lim }\limits_{x\to 2^{-} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{-} }} \frac{\left(2a.\sin \pi x+3\right)-3}{x-2} ={\mathop{\lim }\limits_{x\to 2^{-} }} \frac{2a.\sin \pi x}{x-2} \)
\(={\mathop{\lim }\limits_{x\to 2^{-} }} \left(a\pi \, .\, \cos \frac{\pi \left(x+2\right)}{2} \, .\, \frac{\sin \frac{\pi \left(x-2\right)}{2} }{\frac{\pi \left(x-2\right)}{2} } \right)=a\pi . \)
\({\mathop{\lim }\limits_{x\to 2^{+} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{\left(x+\cos \pi x\right)-3}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{x-2+\cos \pi x-1}{x-2} \)
\(=\begin{array}{l} {{\mathop{\lim }\limits_{x\to 2^{+} }} \frac{x-2+\cos \pi x-1}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{\left(x-2\right)-2\sin \frac{\pi \left(x+2\right)}{2} .\sin \frac{\pi \left(x-2\right)}{2} }{x-2} =} \\ {} \end{array} \)
\(={\mathop{\lim }\limits_{x\to 2^{+} }} \left(1-\pi .\sin \frac{\pi \left(x+2\right)}{2} \, .\, \frac{\sin \frac{\pi \left(x-2\right)}{2} }{\frac{\pi \left(x-2\right)}{2} } \right)=1. \)
Hàm số có đạo hàm tại x=2 khi :
\({\mathop{\lim }\limits_{x\to 2^{-} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} ={\mathop{\lim }\limits_{x\to 2^{+} }} \frac{f\left(x\right)-f\left(2\right)}{x-2} \Leftrightarrow a\, .\, \pi =1\Leftrightarrow a=\frac{1}{\pi } \)
Vậy \(\frac{b}{a} =3\pi .\)