Question:medium

A function \( f(x) = |1 - x + |x| | \) is:

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When analyzing the continuity of piecewise functions, always check the behavior in each interval and at the boundaries.
Updated On: Jan 13, 2026
  • discontinuous at \( x = 1 \) only
  • discontinuous at \( x = 0 \) only
  • discontinuous at \( x = 0, 1 \)
  • continuous everywhere
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The Correct Option is D

Solution and Explanation

Step 1: Analyze the function \( f(x) = |1 - x + |x|| \) by considering two cases. Case 1: If \( x \geq 0 \), then \( |x| = x \). The function simplifies to \( f(x) = |1 - x + x| = |1| = 1 \). Case 2: If \( x<0 \), then \( |x| = -x \). The function becomes \( f(x) = |1 - x - x| = |1 - 2x| \). Step 2: Check the continuity of the function. For \( x \geq 0 \), \( f(x) = 1 \), which is continuous. For \( x<0 \), \( f(x) = |1 - 2x| \), which is continuous as it's a piecewise linear function. At \( x = 0 \), the left-hand limit (LHL) and right-hand limit (RHL) are calculated as follows: LHL = \( f(0^-) = |1 - 2(0)| = 1 \) RHL = \( f(0^+) = 1 \) Since LHL = RHL, the function is continuous at \( x = 0 \). Step 3: Conclude the result. The function \( f(x) \) is continuous for all \( x \), therefore it is continuous everywhere.
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