Q.Let $f(x) = \begin{cases} x^2 & x \le 0 \ \sqrt{x} & x > 0 \end{cases}$ . Then at $x=0$ , $f$ is

continuous and differentiable

continuous but not differentiable

not continuous

differentiable from right only

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

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Q.Let f(x)={x2x≤0 xx>0f(x) = \begin{cases} x^2 & x \le 0 \ \sqrt{x} & x > 0 \end{cases}f(x)={x2​x≤0 x​​x>0​. Then at x=0x=0x=0, fff is