JEE Main 2003Annual, Paper 1MathematicsMediumLimit, Continuity and Differentiability420× since 2002If f(x)={xe−(1∣x∣+1x),x≠00,x=0f(x) = \left\{ {\matrix{ {x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}} & {,x \ne 0} \cr 0 & {,x = 0} \cr } } \right.f(x)={xe−(∣x∣1+x1)0,x=0,x=0 then f(x)f(x)f(x) isAOption A. discontinuous everywhereBOption B. continuous as well as differentiable for all xCOption C. continuous for all x but not differentiable at x = 0DOption D. neither differentiable nor continuous at x = 0CheckSkip · show solution