Let f(x) = \(\left\{\begin{array}{c} \frac{1}{|x|} \text { for }|x| \geq 1 \\ a x^2+b \text { for }|x|<1 \end{array}\right.\). If f(x) is continuous and differentiable everywhere, then
1
\(a=\frac{1}{2}, b=-\frac{3}{2}\)
2
\(a=-\frac{1}{2}, b=\frac{3}{2}\)
3
a = 1, b = –1
4
a = b = 1