Question

A function $f\left(x\right)$ is defined as
$f\left(x\right)=\{\begin{array}{c}x+a;x<0\\ x;0\le x<1\\ b-x;x\ge 1\end{array}$
If $f\left(x\right)$is continuous in its domain. Find $a+b$.

Answer 1

Domain of $f=R$

$\because f$ is continuous in its domain

$\therefore f$ is continuous at $x=0$ and at $x=1$.

Continuity at $x=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(x+a)=a$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}x=0$

$f\left(0\right)=0$

If $f\left(x\right)$ is continuous at $x=0$, then

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

$\Rightarrow a=0=0$

$\Rightarrow a=0$

Continuity at $x=1$

$f\left(1\right)=b-1$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}x=1$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(b-x)=b-1$

If $f\left(x\right)$ is continuous at $x=1$, then

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$

$\Rightarrow 1=b-1$

$\Rightarrow b-1=1$

$\Rightarrow b=2$

Hence, $a+b=0+2=2$.