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 $x=0,$
$RHL\, f(0+h)=\displaystyle \lim _{h \rightarrow 0}|0+h-2|+(0+h)$
$=|-2|=2$
$LHL\, f(0-h)=\displaystyle \lim _{h \rightarrow 0}|0-h-2|+(0-h)$
$=|-2|=2$
At $x=2,$
$RHL\, f(2+h) =\displaystyle \lim _{h \rightarrow 0}|2+h-2|+(2+h)$
$=0+2+0=2$
$LHL\, f(2-h) =\displaystyle \lim _{h \rightarrow 0}|2-h-2|+(2+h)$
$=0+2-0=2$
and $f(0)=2, f(2)=2$
Hence, $f(x)$, is continuous at $x=0,2$
Now, we check differentiability
$f(x) = \begin{cases} (-x+2-x) & x < 0 \\ (-x + 2+2+x) & 0 \le x \le 2 \\ (x-2+2) & 2 \le x \end{cases} $
$f(x) = \begin{cases} 2-2x, x < 0 \\ 2,0 \le x \le 2 \\ 2x-2, 2 \le x \end{cases} $
Now, $f'(x) = \begin{cases} -2, x < 0 \\ 0,0 \le x \le 2 \\ 2, 2 \le x \end{cases} $
Hence, $f(x)$ is not differentiable at $x=0,\,2$.