MathematicsMedium179×since 2002Q4212Let f(x) = \left\{ {\matrix{ {{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr 0 & {,\,x = 0} \cr } } \right. Then at x=0x=0x=0Afff is continuous but f′f'f′ is not continuousBfff and f′f'f′ both are continuousCfff is continuous but not differentiableDf′f'f′ is continuous but not differentiableCheck answerSkip