Question:

Let \( f \) be the function defined by \[ f(x) = \begin{cases} x^2 - 1, & x \neq 1 \\ x^2 - 2|x-1|^{-1}, & x = 1 \end{cases} \] The function is continuous at:

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To check for continuity, ensure that the function is defined at the point and that the left-hand limit, right-hand limit, and function value all match.
Updated On: Jan 12, 2026
  • \( x = 1 \)
  • \( x = 2 \)
  • \( x = 0 \)
  • None of these
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The Correct Option is D

Solution and Explanation

By evaluating the function and considering the behavior at \( x = 1 \), the function is not continuous at any point.
Final Answer: \[ \boxed{\text{None of these}} \]
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