A) differentiable at both \[x=2\] and \[x=0\]
B) differentiable at \[x=2\] but not at \[x=0\]
C) continuous at \[x=2\]but not at \[x=0\]
D) continuous at both \[x=2\] and \[x=0\]
Correct Answer: D
Solution :
\[f(x)=|x-2|+x\] First we check the continuity At \[x=0,\] \[RHL\,\,f(0+h)=\underset{h\to 0}{\mathop{lim}}\,|0+h-2|+(0+h)\] \[=|-2|=2\] \[LHL\,\,f(0-h)=\underset{h\to 0}{\mathop{\lim }}\,|0-h-2|+(0-h)\] \[=|-2|=2\] At \[x=2,\] \[RHL\,f(2+h)=\underset{h\to 0}{\mathop{\lim }}\,|0-h-2|+(2+h)\] \[=0+2+0=2\] \[LHL,\,f(2-h)=\underset{h\to 0}{\mathop{\lim }}\,|2-h-2|+(2+h)\] \[=0+2-0=2\] and \[f(0)=2,f(2)=2\] Hence, \[f(x),\] is continuous at \[x=0,2\] Now, we check differentiability \[f(x)=\left\{ \begin{matrix} (-x+2-x),\,x<0 \\ (-x+2+x),0\le x\le 2 \\ (x-2+2),2\le x \\ \end{matrix} \right.\] \[f(x)=\left\{ \begin{matrix} 2-2x,\,x<0 \\ 2,0\le x\le 2 \\ 2x-2,2\le x \\ \end{matrix} \right.\] Now, \[f'(x)=\left\{ \begin{matrix} -2,x<0 \\ 0,0\le x\le 2 \\ 2,2\le x \\ \end{matrix} \right.\] Hence, \[f(x)\] is not differentiable at \[x=0,2.\]
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