A) Continuous at \[x=1,\] but not differentiable at \[x=1\]
B) Both continuous and differentiable at \[x=1\]
C) Not continuous at \[x=1\]
D) Not differentiable at \[x=1\]
Correct Answer: A
Solution :
We have, \[f(x)=|x|+|x-1|\] \[=\left\{ \begin{matrix} -2x+1, & x<0 & {} \\ x-x+1, & 0\le x<1 & = \\ x+x-1, & x\ge 1 & {} \\ \end{matrix} \right.\left\{ \begin{matrix} -2x+1, & x<0 \\ 1 & 0\le x<1 \\ 2x-1, & x\ge 1 \\ \end{matrix} \right.\] Clearly,\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=1,\,\,\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=1,\,\,\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=1\] and \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=1\]. So, \[f(x)\] is continuous at \[x=0,\,\,1.\] Now \[f'(x)=\left\{ \begin{array}{*{35}{l}} -2,\,\,\,\,x<0 \\ \,\,0,\,\,\,\,\,0\le x<1 \\ \,\,2,\,\,\,\,\,x\ge 1 \\ \end{array} \right.\] Here x = 0, \[f'({{0}^{+}})=0\] while \[f'({{0}^{-}})=-2\] and at x = 1, \[f'({{1}^{+}})=2\] while \[f'({{1}^{-}})=0\] Thus, \[f(x)\] is not differentiable at x = 0 and 1.
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