Q.If $f(x) = \begin{cases} x^2 & x \le 1 \ 2-x & x > 1 \end{cases}$ , then at $x=1$

continuous and differentiable

continuous but not differentiable

not continuous

none

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

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Q.If f(x)={x2x≤1 2−xx>1f(x) = \begin{cases} x^2 & x \le 1 \ 2-x & x > 1 \end{cases}f(x)={x2​x≤1 2−x​x>1​, then at x=1x=1x=1