Question

Let $f(x)=\left\{\begin{array}{cc}x+a, & \text { if } x<0 \\ |x-1|, \text { if } x \geq 0\end{array}\right.$ and
$g(x)=\left\{\begin{array}{cc}x+1, & \text { if } x<0 \\ (x-1)^{2}+b, & \text { if } x \geq 0\end{array}\right.$
where $a$ and $b$ are non-negative real numbers. If the composite function $g o f \left(\right. x \left.\right)$ is continuous for all real $x,$ then the values of $a+b$ is _____.

Answer 1

$\begin{aligned} g o f(x) &=\left\{\begin{array}{cc}f(x)+1, & \text { if } f^{(x)}<0 \\ \left\{f^{(x)}-1\right\}^{2}+b, & \text { if } f(x) \geq 0\end{array}\right.\\ &=\left\{\begin{array}{cc}x+a+1, & \text { if } x<-a \\ (x+a-1)^{2}+b, & \text { if }-a \leq x<0 \\ (|x-1|-1)^{2}+b, & \text { if } x \geq 0\end{array}\right.\end{aligned}$
As $g o f \left(\right. x \left.\right)$ is continuous at $x = - a$ ,
$L.H.L=1$ and $R.H.L=1+b$
$\therefore $ $b = 0$
$g o f \left(- a\right) = g o f \left(- a^{+}\right) = g o f \left(- a^{-}\right)$
$\Rightarrow 1 + b = 1 + b = 1 \Rightarrow b = 0$
Also, $g o f \left(\right. x \left.\right)$ is continuous at $x = 0$ .
$\Rightarrow g o f \left(0\right) = g o f \left(0^{+}\right) = g o f \left(0^{-}\right)$
$\Rightarrow b = \left(a - 1\right)^{2} + b$
$\Rightarrow a = 1$
$\Rightarrow a+b=1$