A) Is continuous but not differentiable
B) Is discontinuous
C) Is having continuous derivative
D) Is continuous and differentiable
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,f(x)={{x}^{2}}\sin \left( \frac{1}{x} \right)\] , but \[-1\le \sin \left( \frac{1}{x} \right)\le 1\] and \[x\to 0\] \ \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=0=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f(x)=f(0)\] Therefore \[f(x)\] is continuous at \[x=0\] . Also, the function \[f(x)={{x}^{2}}\sin \frac{1}{x}\] is differentiable because \[R{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}\sin \frac{1}{h}-0}{h}=0\] , \[L{f}'(x)=\underset{h\to 0}{\mathop{\lim }}\,\frac{{{h}^{2}}\sin (1/-h)}{-h}=0\] .
You need to login to perform this action.
You will be redirected in
3 sec
