Let $f:R\to R$ and $g:R\to R$ be defined as  $f\left(x\right)=\left\{\begin{array}{l}x+a,x<0\\ \left|x-1\right|,x\ge 0\end{array}\right.$ and ,$g\left(x\right)=\left\{\begin{array}{l}x+1,x<0\\ {\left(x-1\right)}^2+b,x\ge 0\end{array}\right.$

         $Where a, b are non negative real numbers. If gof\left(x\right) is continuous for all x\in R then the value of a+b is$_

$\begin{array}{l}f\left(x\right)=\left\{\begin{array}{l}x+a,x<0\\ \left|x-1\right|,x\ge 0\end{array}\right.\text{\hspace{0.17em} and }g\left(x\right)=\left\{\begin{array}{l}x+1,x<0\\ {\left(x-1\right)}^2+b,x\ge 0\end{array}\right.\\ g\left(f\left(x\right)\right)=\left\{\begin{array}{l}f\left(x\right)+1,iff\left(x\right)<0\\ {\left(f\left(x\right)-1\right)}^2+b,iff\left(x\right)\ge 0\end{array}\right.\\ =\left\{\begin{array}{l}x+a+1,ifx<-a\\ {\left(x+a-1\right)}^2+b,if-a\le x<0\\ {\left(\left|x-1\right|-1\right)}^2+b,ifx\ge 0\end{array}\right.\\ As g\left(f\left(x\right)\right)iscontinuousatx=-a\\ g\left(f\left(-a\right)\right)=g\left(f(-a^{+})\right)=g\left(f(-a^{-})\right)\\ g\left(f\left(0\right)\right)=g\left(f\left(0^{+}\right)\right)=g\left(f\left(0^{-}\right)\right),continuousatx=0\\ b={\left(a-1\right)}^2+b\Rightarrow a=1\Rightarrow a+b=1+0=1\end{array}$