Question

Let $f:R\to R$ and $g:R\to R$ be defined as
$f\left(x\right)=\{\begin{array}{cc}x+a,& x<0\\ |x-1|,& x\ge 0\end{array}$ and $g\left(x\right)=\{\begin{array}{cc}x+1,& x<0\\ (x-1{)}^2+b,& x\ge 0\end{array},$
where $a,b$ are non-negative real numbers. If $\left(gof\right)\left(x\right)$ is continuous for all $x\in R,$ then $a+b$ is equal to

Answer 1

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

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

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

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

Given $g\left(f\right(x\left)\right)$ is continuous for all $x\in R.$

So, at $x=-a$ and at $x=0,$

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

$\Rightarrow b=0$ and $a=1$

$\Rightarrow a+b=1$