A) Continuous and differentiable
B) Continuous and not differentiable
C) Discontinuous and differentiable
D) Discontinuous and not differentiable
Correct Answer: B
Solution :
We have \[Rf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(1+h)-f(1)}{h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{\left\{ {{(1+h)}^{3}}-1 \right\}-0}{h}=3\] \[Lf'(1)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{f(1-h)-f(1)}{-h}\]\[=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{\left\{ (1-h)-1 \right\}-0}{-h}=1\] \[\therefore \,\,\,Rf'(1)\ne Lf'(1)\]\[\Rightarrow \,\,f(x)\] is not differentiable at \[x=1.\] Now, \[f(1+0)=\underset{h\to 0}{\mathop{\lim }}\,\,f(1+h)=0\] and \[f(1-0)=\underset{h\to 0}{\mathop{\lim }}\,\,f(1-h)=0\] \[\therefore \,\,\,f(1+0)=f(1-0)=f(0)\]\[\Rightarrow \,\,f(x)\] is continuous at \[x=1.\]Hence at \[x=1,\,\,\,f(x)\]is continuous and not differentiable.
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