, then
A) f(x) is continuous at \[x=0\]
B) discontinuous at; \[x=2\]
C) f(x) is discontinuous at \[x=1\]
D) None of these
Correct Answer: B
Solution :
\[\underset{x\to 0-}{\mathop{\lim }}\,\,f(x)=\,0,\,\,\,\,\,f(0)=0,\,\,\underset{x\to {{0}^{+}}}{\mathop{lim}}\,\,f(x)=-\,4\] \[\operatorname{f}(x) discontinuous at\,\,x =0.\] and \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=1\,\,and\,\,\underset{x\to {{1}^{+}}}{\mathop{lim}}\,\,f(x)=1,\,\,f(1)=1\] Hence f(x) is continuous at \[\operatorname{x} = 1\]. Also \[\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\,f(x)=4{{(2)}^{2}}-\,\,3.2\,\,=\,\,10\] \[\operatorname{f}(2)=10\,\,and\,\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,\,f(x)=\,\,3(2)+4=10\] Hence f(x) is continuous at\[\operatorname{x} = 2\].
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