Question

The function $f(x)=|x-2|+x$ is ................

• A. continuous at $x=2$ but not at $x=0$
• B. continuous at both $x=2$ and $x=0$
• C. differentiable at both $x=2$ and $x=0$
• D. differentiable at $x=2$ but not at $x=0$

Answer 1

Answer: (B)

$f(x)=|x-2|+x$
First we check the continuity
At $x=0,$
$RHLf(0+h)=\underset{h\to 0}{\lim }|0+h-2|+(0+h)$
$=|-2|=2$
$LHLf(0-h)=\underset{h\to 0}{\lim }|0-h-2|+(0-h)$
$=|-2|=2$
At $x=2,$
$RHLf(2+h)=\underset{h\to 0}{\lim }|2+h-2|+(2+h)$
$=0+2+0=2$
$LHLf(2-h)=\underset{h\to 0}{\lim }|2-h-2|+(2+h)$
$=0+2-0=2$
and $f(0)=2,f(2)=2$
Hence, $f(x)$, is continuous at $x=0,2$
Now, we check differentiability
$f(x)=\{\begin{array}{ll}(-x+2-x)& x<0\\ (-x+2+2+x)& 0\le x\le 2\\ (x-2+2)& 2\le x\end{array}$
$f(x)=\{\begin{array}{l}2-2x,x<0\\ 2,0\le x\le 2\\ 2x-2,2\le x\end{array}$
Now, $f^{\prime }(x)=\{\begin{array}{l}-2,x<0\\ 0,0\le x\le 2\\ 2,2\le x\end{array}$
Hence, $f(x)$ is not differentiable at $x=0,2$.