A) \[x\]
B) \[x\]except at\[x=0\]
C) \[x\]except at\[x=1\]
D) \[x\]except at\[x=0\]and\[x=1\]
Correct Answer: C
Solution :
We have, \[f(x)=\left\{ \begin{matrix} \frac{{{x}^{2}}-x}{{{x}^{2}}-x}=1, & if\,\,x<0\,\,or\,\,x>1 \\ -\frac{({{x}^{2}}-x)}{({{x}^{2}}-x)}=-1, & if\,\,0<x<1 \\ 1, & if\,\,x=0 \\ -1, & if\,\,x=1 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} 1,\,\,\,\,if & x\le 0\,\,or\,\,x>1 \\ -1,\,\,\,\text{if} & 0<x\le 1 \\ \end{matrix} \right.\] Now\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\]\[1=1\]and\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,-1=1\] Clearly,\[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)\] So,\[f(x)\]not continuous at\[x=0\]. It can be easily seen that it is not continuous at\[x=1\]also.
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