Question

Let $f : R \rightarrow R$ and $g : R \rightarrow R$ be defined as
$f ( x )=\begin{cases} x + a , & x <0 \\ \mid x -1 l , & x \geq 0\end{cases}.$ and
$g(x)=\begin{cases} x+1, & x < 0 \\ (x-1)^{2}+b, & x \geq 0\end{cases}.$
where a, b are non-negative real numbers. If (gof) $( x )$ is continuous for all $x \in R$, then $a + b$ is equal to ________.

Answer 1

$g[ f ( x )]=\begin{bmatrix} f ( x )+1 & f ( x )<0 \\ ( f ( x )-1)^{2}+ b & f ( x ) \geq 0\end{bmatrix}$
$g[f(x)]=\begin{bmatrix} x+a+1 & x+a<0 \& x<0 \\ |x-1|+1 & |x-1|<0 \& x \geq 0 \\ (x+a-1)^{2}+b & x+a \geq 0 \& x<0 \\ (|x-1|-1)^{2}+b & |x-1| \geq 0 \& x \geq 0\end{bmatrix}$
$g[f(x)]=\begin{bmatrix} x+a+1 & x \in(-\infty,-a) \& x \in(-\infty, 0) \\ |x-1|+1 & x \in \phi \\ (x+a-1)^{2}+b & x \in[-a, \infty) \& x \in(-\infty, 0) \\ (|x-1|-1)^{2}+b & x \in R \& x \in[0, \infty)\end{bmatrix}$
$g[f(x)]=\begin{bmatrix} x+a+1 & x \in(-\infty,-a) \\ (x+a-1)^{2}+b & x \in[-a, 0) \\ (|x-1|-1)^{2}+b & x \in[0, \infty)\end{bmatrix}$
$g(f(x))$ is continuous
at $x=-a\,\,\&\,\,$at $x=0$
$1= b +1\,\,\&\,\,( a -1)^{2}+ b = b$
$b =0\,\,\&\,\,a=1$
$\Rightarrow a+b=1$