Question

$f\left(x\right)=\left\{\begin{array}{ccc}|x-2|+1& if& x<2\\ 2& if& x=2,\\ |x-2|-1& if& x>2\end{array}\right\}$
If $\underset{x\to a}{lim}f\left(x\right)$ exists then

• A. $a=2$
• B. $aϵR$
• C. $aϵR-\left\{2\right\}$
• D. $aϵR-\{1,2\}$

Answer 1

The correct option is C $aϵR-\left\{2\right\}$

At $x=2$

$\underset{x\to 2}{lim}f\left(x\right)=2$

$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(x-3)=-1[|x-2|>0forx\to 2^{-}]$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(3-x)=1[|x-2|<0forx\to 2^{+}]$

LHL$\ne$RHL

$\therefore$ At $x=2$ it is discontinuous, everywhere else it is continuous

$\Rightarrow aϵR-\left\{2\right\}$