A) continuous and differentiable at \[x=3\]
B) continuous at \[x=3,\] but not differentiable at \[x=3\]
C) continuous and differentiable everywhere
D) continuous at \[x=1,\]but not differentiable at \[x=1,\]
Correct Answer: B
Solution :
Clearly, \[f(x)\] is not differentiable at \[x=3.\] Now, \[\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,\,\,\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(3-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,|3-h-3|\] \[\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,\,\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(3+h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,|3+h-3|=0\] and \[f(3)=|3-3|=0\] \[\therefore \] \[f(x)\] is continuous at \[x=3.\]
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