If \(A\) and \(B\) are the domain and range of the real valued function, \[ f(x)=\dfrac{|x|}{\sqrt{1-|x|}} \] then \(A \cup B =\ ?\)
If \[ f(x)= \begin{cases} \dfrac{x-2}{|x-2|}+a, & x<2, \\[6pt] a+b, & x=2, \\[6pt] \dfrac{x-2}{|x-2|}+b, & x>2, \end{cases} \] is continuous at \(x=2\), then