Q.The function $f(x) = \begin{cases} x & |x| \le 1 \ \operatorname{sgn}(x) & |x| > 1 \end{cases}$ is

continuous everywhere

discontinuous at

x=\pm1

not differentiable at

x=\pm1

both (a) and (c)

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

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Q.The function f(x)={x∣x∣≤1 sgn⁡(x)∣x∣>1f(x) = \begin{cases} x & |x| \le 1 \ \operatorname{sgn}(x) & |x| > 1 \end{cases}f(x)={x​∣x∣≤1 sgn(x)​∣x∣>1​ is