Q.Let $f(x) = \begin{cases} x^2 \sin(1/x^2) & x \ne 0 \ 0 & x = 0 \end{cases}$ . Then at $x=0$ , $f$ is

twice differentiable but

f''(0) \ne 0

twice differentiable with

f''(0) = 0

differentiable but not twice differentiable

not differentiable

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.Let f(x)={x2sin⁡(1/x2)x≠0 0x=0f(x) = \begin{cases} x^2 \sin(1/x^2) & x \ne 0 \ 0 & x = 0 \end{cases}f(x)={x2sin(1/x2)​x=0 0​x=0​. Then at x=0x=0x=0, fff is