Question

Let $f:R\to R$ be defined as

$$\begin{gathered} f\left(x\right)=\left\{x+a,x<0 \\ \left|x-1\right|,x\le 0\right\} \\ g\left(x\right)=\left\{x+1,x<0 \\ {\left(x-1\right)}^2+b,x\ge 0\right\} \end{gathered}$$

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

Step-1 Generating $\left(gof\right)\left(x\right)$:

Given that, $\left(gof\right)\left(x\right)$ is continuous for all $x\in R$

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

It is given that: $\left(gof\right)\left(x\right)$ is continuous.

Step-2 : Calculating the value of $a+b$:

Calculating $b$:

At

$$\begin{gathered} x=-a, \\ \underset{x\to a^{-}}{\lim }g\left(f\right(x\left)\right)=\underset{x\to a^{+}}{\lim }g\left(f\right(x\left)\right) \\ \Rightarrow -a+a+1={\left(-a+a-1\right)}^2+b \\ \Rightarrow 1=1+b \\ \Rightarrow b=0 \end{gathered}$$

Calculating $\mathrm{a}$:

At

$$\begin{gathered} x=0, \\ \underset{x\to 0^{-}}{\lim }g\left(f\right(x\left)\right)=g\left(f\right(0\left)\right) \\ \Rightarrow {\left(0+a-1\right)}^2+b={\left(\left|0-1\right|-1\right)}^2+b \\ \Rightarrow {\left(a-1\right)}^2=0 \\ \Rightarrow a=1 \end{gathered}$$

Adding $a\&b$we get,

$$\begin{gathered} \Rightarrow a+b=1+0 \\ =1 \end{gathered}$$

Hence, the value of $a+b$ is $1.$