How many real values of x satisfy the equation `|x|+|x-1|=1`?

Options

A. 1
B. 2
C. Infinite
D. No value of x

Explanation

To solve the equation \( |x| + |x-1| = 1 \), we will analyze it by considering different cases based on the values of \( x \).

### Step 1: Identify the cases based on the absolute values
The absolute values \( |x| \) and \( |x-1| \) change based on the value of \( x \). We will consider three cases:
1. Case 1: \( x < 0 \)
2. Case 2: \( 0 \leq x < 1 \)
3. Case 3: \( x \geq 1 \)

### Step 2: Solve Case 1: \( x < 0 \)
In this case:
- \( |x| = -x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the equation gives:
\[ -x + (-x + 1) = 1 \]
Simplifying this:
\[ -2x + 1 = 1 \]
\[ -2x = 0 \implies x = 0 \]
Since \( x = 0 \) does not satisfy \( x < 0 \), there are no solutions in this case.

### Step 3: Solve Case 2: \( 0 \leq x < 1 \)
In this case:
- \( |x| = x \)
- \( |x-1| = -(x-1) = -x + 1 \)

Substituting these into the equation gives:
\[ x + (-x + 1) = 1 \]
Simplifying this:
\[ 1 = 1 \]
This is true for all \( x \) in the interval \( [0, 1) \). Thus, every value of \( x \) in this range is a solution.

### Step 4: Solve Case 3: \( x \geq 1 \)
In this case:
- \( |x| = x \)
- \( |x-1| = x - 1 \)

Substituting these into the equation gives:
\[ x + (x - 1) = 1 \]
Simplifying this:
\[ 2x - 1 = 1 \]
\[ 2x = 2 \implies x = 1 \]
Since \( x = 1 \) satisfies \( x \geq 1 \), it is a valid solution.

### Step 5: Summary of solutions
From the analysis:
- In Case 1, there are no solutions.
- In Case 2, all values \( x \) in the interval \( [0, 1) \) are solutions.
- In Case 3, \( x = 1 \) is a solution.

Thus, the total number of real values of \( x \) that satisfy the equation is infinite (all values in the interval \( [0, 1) \)) plus one specific value \( x = 1 \).

### Final Answer
The equation \( |x| + |x-1| = 1 \) has infinitely many real solutions in the interval \( [0, 1) \) and one solution at \( x = 1 \).