Question

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

• A. discontinuous at \( x = 1 \) only
• B. discontinuous at \( x = 0 \) only
• C. discontinuous at \( x = 0, 1 \)
• D. continuous everywhere

Hint:

When analyzing the continuity of piecewise functions, always check the behavior in each interval and at the boundaries.

Answer 1

The Correct Option is D

Step 1: {Analyze the function for different cases}
The function is \( f(x) = |1 - x + |x|| \). Consider two cases:
Case 1: \( x \geq 0 \), then \( |x| = x \), so: \[ f(x) = |1 - x + x| = |1| = 1. \] Case 2: \( x<0 \), then \( |x| = -x \), so: \[ f(x) = |1 - x - x| = |1 - 2x|. \] Step 2: {Check continuity}
For \( x \geq 0 \), \( f(x) = 1 \), which is continuous.
For \( x<0 \), \( f(x) = |1 - 2x| \), which is also continuous because it is a piecewise linear function.
At \( x = 0 \), the left-hand limit (LHL) and right-hand limit (RHL) are: \[ {LHL} = f(0^-) = |1 - 2(0)| = 1, \quad {RHL} = f(0^+) = 1. \] Thus, \( f(x) \) is continuous at \( x = 0 \).
Step 3: {Conclude the result}
The function \( f(x) \) is continuous for all \( x \). Hence, it is continuous everywhere.