A) discontinuous at\[x=0\]
B) discontinuous at \[x=1\]
C) continuous at all points
D) discontinuous at all points
Correct Answer: C
Solution :
Given, \[f(x)=|x|=\left\{ \begin{matrix} x, & x>0 \\ 0, & x=0 \\ -x, & x<0 \\ \end{matrix} \right.\] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,|-h|=0,\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,|h|=0\] \[\therefore \] \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\] \[\therefore \]At\[x=0,f(x)\]is continuous. Hence, function is continuous at every point. Alternative Method It is clear from the graph that\[f(x)\]is continuous everywhere.You need to login to perform this action.
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