If `f(x)=|x|+|x-1|`, then which one of the following is correct?

Options

A. A) f(x) is continuous at x = 0 and x = 1
B. B) f(x) is continuous at x = 0 but not at x = 1
C. C) f(x) is continuous at x = 1 but not at x = 0
D. D) f(x) is neither continuous at x = 0 nor at x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) = |x| + |x - 1| \) and check its continuity at the points \( x = 0 \) and \( x = 1 \).

### Step-by-Step Solution:

1. **Identify the points of interest**:
We need to check the continuity of the function at \( x = 0 \) and \( x = 1 \).

2. **Evaluate \( f(0) \)**:
\[ f(0) = |0| + |0 - 1| = 0 + |-1| = 0 + 1 = 1 \]

3. **Evaluate \( f(1) \)**:
\[ f(1) = |1| + |1 - 1| = 1 + |0| = 1 + 0 = 1 \]

4. **Check the left-hand limit as \( x \) approaches 0**:
For \( x < 0 \):
\[ f(x) = -x + |x - 1| = -x + (1 - x) = -2x + 1 \]
As \( x \) approaches 0 from the left:
\[ \lim_{x \to 0^-} f(x) = -2(0) + 1 = 1 \]

5. **Check the right-hand limit as \( x \) approaches 0**:
For \( x \geq 0 \):
\[ f(x) = x + |x - 1| = x + (1 - x) = 1 \]
As \( x \) approaches 0 from the right:
\[ \lim_{x \to 0^+} f(x) = 1 \]

6. **Conclusion for continuity at \( x = 0 \)**:
Since \( f(0) = 1 \), \( \lim_{x \to 0^-} f(x) = 1 \), and \( \lim_{x \to 0^+} f(x) = 1 \), the function is continuous at \( x = 0 \).

7. **Check the left-hand limit as \( x \) approaches 1**:
For \( x < 1 \):
\[ f(x) = -x + (1 - x) = -2x + 1 \]
As \( x \) approaches 1 from the left:
\[ \lim_{x \to 1^-} f(x) = -2(1) + 1 = -2 + 1 = -1 \]

8. **Check the right-hand limit as \( x \) approaches 1**:
For \( x \geq 1 \):
\[ f(x) = x + (x - 1) = 2x - 1 \]
As \( x \) approaches 1 from the right:
\[ \lim_{x \to 1^+} f(x) = 2(1) - 1 = 2 - 1 = 1 \]

9. **Conclusion for continuity at \( x = 1 \)**:
Since \( f(1) = 1 \), \( \lim_{x \to 1^-} f(x) = -1 \), and \( \lim_{x \to 1^+} f(x) = 1 \), the function is not continuous at \( x = 1 \).

### Final Answer:
The function \( f(x) = |x| + |x - 1| \) is continuous at \( x = 0 \) but not continuous at \( x = 1 \).