Q.Let $f(x) = \begin{cases} \dfrac{\sin(\pi x)}{\pi x} & x \ne 0 \ 1 & x = 0 \end{cases}$ . Then $f$ is differentiable at $x=0$ and $f'(0)=$

0

\pi

1

does not exist

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

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Q.Let f(x)={sin⁡(πx)πxx≠0 1x=0f(x) = \begin{cases} \dfrac{\sin(\pi x)}{\pi x} & x \ne 0 \ 1 & x = 0 \end{cases}f(x)={πxsin(πx)​​x=0 1​x=0​. Then fff is differentiable at x=0x=0x=0 and f′(0)=f'(0)=f′(0)=