Mathematics - Continuity - Differentiability Question with Solution | TestHub

$g\left[f\left(x\right)\right]=\left[\begin{array}{cc}f\left(x\right)+1& f\left(x\right)<0\\ \left(f\right(x)-1{)}^2+b& f\left(x\right)\ge 0\end{array}\right.$

$g\left[f\left(x\right)\right]=\left[\begin{array}{cc}x+a+1& x+a<0; x<0\\ |x-1|+1& |x-1|<0; x\ge 0\\ (x+a-1{)}^2+b& x+a\ge 0; x<0\\ \left(\right|x-1|-1{)}^2+b& |x-1|\ge 0; x\ge 0\end{array}\right.$

$g\left[f\left(x\right)\right]=\left[\begin{array}{cc}x+a+1& x\in (-\infty ,-a); x\in (-\infty ,0)\\ |x-1|+1& x\in \varphi \\ (x+a-1{)}^2+b& x\in [-a,\infty ); x\in (-\infty ,0)\\ \left(\right|x-1|-1{)}^2+b& x\in R; x\in [0,\infty )\end{array}\right.$

$g\left[f\left(x\right)\right]=\left[\begin{array}{cc}x+a+1& x\in (-\infty ,-a)\\ (x+a-1{)}^2+b& x\in [-a,0)\\ \left(\right|x-1|-1{)}^2+b& x\in [0,\infty )\end{array}\right.$

$g\left(f\left(x\right)\right)$ is continuous

at $x=-a \&$ at $x=0$

$1=b+1 \& (a-1{)}^2+b=b$

$b=0 \& a=1$

$\Rightarrow a+b=1$