A) Continuous but non-differentiable
B) Discontinuous and differentiable
C) Discontinuous and non-differentiable
D) Continuous and differentiable
Correct Answer: A
Solution :
Since this function is continuous at \[x=0\] Now for differentiability \[f(x)=\,|\,\,x\,\,|\,\,=\,\,|0|\,\,=0\] and \[f(0+h)=f(h)=\,\,|h|\] \[\therefore \,\,\underset{h\to 0-}{\mathop{\lim }}\,\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0-}{\mathop{\lim }}\,\,\frac{|h|}{h}=-1\] and \[\underset{h\to 0+}{\mathop{\lim }}\,\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0+}{\mathop{\lim }}\,\,\frac{|h|}{h}=1\]. Therefore it is continuous and non-differentiable.
You need to login to perform this action.
You will be redirected in
3 sec
