Let $f\left(x\right)=\left\{\begin{array}{l}\frac{1}{\left|x\right|}for\left|x\right|\ge 1\\ a{x}^2+bfor\left|x\right|\text{<}1\end{array}\right.$. If is continuous and differentiable everywhere, then

$=\underset{x\to 1^{-}}{\lim }\left(a{x}^2+b\right)=\underset{x\to 1^{+}}{\lim }\frac{1}{\left|x\right|}\Rightarrow a+b=1.$.... (i) [ f(x) is continuous]