Let f: R → R and g: R → R be defined as f(x) = \ cases x + a, & x <...

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

g\left[f\left(x\right)\right] = \left[\begin{cases}x + a + 1 & x \in ( - \infty , - a ) ; x \in ( - \infty , 0 ) \\ | x - 1 | + 1 & x \in ϕ \\ ( 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{cases}\right)</p><p>g\left[f\left(x\right)\right] = \left[$\begin{cases}x + a + 1 & x \in ( - \infty , - a ) \\ ( x + a - 1{)}^{2} + b & x \in [ - a, 0 ) \\ ( | x - 1 | - 1{)}^{2} + b & x \in [ 0, \infty )\end{cases}$\right)

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

at $x = - a \text{ and }$ at $x = 0$

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

$b = 0 \text{ and } a = 1$

$⇒ a + b = 1$