A) not continuous at \[x=2\]
B) differentiable at\[x=2\]
C) continuous but not differentiable at\[x=2\]
D) None of the above
Correct Answer: C
Solution :
\[LHL=\underset{x\to {{2}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,1+(2-h)=3\] \[RHL=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,+f(x)=\underset{h\to 0}{\mathop{\lim }}\,5-(2+h)=3,f(2)=3\] Hence,\[f\]is continuous at\[x=2\]. Now, \[Rf(2)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(2+h)-f(2)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{5-(2+h)-3}{h}=-1\] \[Lf(2)=\underset{h\to 0}{\mathop{\lim }}\,=\frac{f(2-h)-f(2)}{-h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,=\frac{1+(2-h)-3}{-h}=1\] \[\therefore \] \[Rf(2)\ne Lf(2)\] \[\therefore \]\[f\]is not differentiable at\[x=2\].
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