Explanation: We have, for \(-1Also, becomes negative and numerically less than 1 when is slightly greater than 1 , and so by definition of when \(1Thus, is constant and equal to 0 in the closed interval [-1,1] and is continuous and differentiable in the open interval (-1,1) At is discontinuous. since and is not differentiable at Hence, options (1),(2) and (4) are correct answers.