Question

Function f(x)=$\left|x-2\right|-2\left|x-4\right|is$ discontinous at:

• A. $x=2,4$
• B. $x=2$
• C. No where
• D. Except $x=2$

Answer 1

The correct option is C No where

$f\left(x\right)=\mid x-2\mid -2\mid x-4\mid$

$\therefore f\left(x\right)=\left\{\begin{array}{c}-(x-2)+2(x-4),x<2\\ (x-2)+2(x-4),2\le x\le 4\\ (x-2)-2(x-4)x>4\end{array}\right\}$

At $x=2$

L.H.L(Left Hand Limit)

$\underset{x\to 2^{-}}{lim}=-(2-2)+2(2-4)=-4$

R.H.LRight Hand Limit/0

$\underset{x\to 2^{+}}{lim}=(2-2)+2(2-4)=-4$

$\therefore \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(x\right)$

At $x=4$

L.H.L(Left Hand Limit)

$\underset{x\to 4^{-}}{lim}=(4-2)+2(4-4)=2$

R.H.L(Right Hand Limit)

$\underset{x\to 4^{+}}{lim}=(4-2)-2(4-4)=2$

$\therefore \underset{x\to 4^{-}}{lim}f\left(x\right)=\underset{x\to 4^{+}}{lim}f\left(x\right)$

$\therefore f\left(x\right)$ is continuous at everywhere.