A) Not continuous at \[x=2\]
B) Differentiable at \[x=2\]
C) Continuous but not differentiable at \[x=2\]
D) None of these
Correct Answer: C
Solution :
\[\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,1+(2-h)=3\], \[\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,5-(2+h)=3\], \[f(2)=3\] Hence, f is continuous at \[x=2\] Now \[R{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{5-(2+h)-3}{h}=-1\] \[L{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{1+(2-h)-3}{-h}=1\] \[\because R{f}'(x)\ne L{f}'(x)\]; \ f is not differentiable at \[x=2\].
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