The absolute valued function f is defined as `f(x) = {{:(x,, x ge 0),(-x ,, x lt 0):}}` and fractional part function g(x) as g(x) = x-[x], graphically the number of real solution(s) of the equation f(x) = g(x) is obtained by finding the point(s) of interaction of the graph of y = f(x) and y = g(x).
The number of solutions (s) of `|x-1| = {x}, x in [-1, 1]` is

Options

A. 1
B. 2
C. 3
D. Infinitely many

Answer

To solve the equation \( |x - 1| = \{x\} \) where \( x \) is in the interval \([-1, 1]\), we will analyze the two functions involved: the absolute value function \( f(x) = |x - 1| \) and the fractional part function \( g(x) = \{x\} = x - \lfloor x \rfloor \).

### Step 1: Define the Functions
1. **Absolute Value Function**:
- For \( x \geq 1 \), \( f(x) = x - 1 \)
- For \( -1 \leq x < 1 \), \( f(x) = 1 - x \)

2. **Fractional Part Function**:
- For \( -1 \leq x < 0 \), \( g(x) = x + 1 \)
- For \( 0 \leq x < 1 \), \( g(x) = x \)

### Step 2: Analyze the Functions on the Interval \([-1, 1]\)
- We will analyze the functions piecewise over the interval \([-1, 1]\).

#### For \( x \in [-1, 0) \):
- \( f(x) = 1 - x \)
- \( g(x) = x + 1 \)

Set the equations equal:
\[ 1 - x = x + 1 \]
Solving this gives:
\[ 1 - x = x + 1 \implies 1 - 1 = 2x \implies 0 = 2x \implies x = 0 \]
However, \( x = 0 \) is not in the interval \([-1, 0)\).

#### For \( x \in [0, 1) \):
- \( f(x) = 1 - x \)
- \( g(x) = x \)

Set the equations equal:
\[ 1 - x = x \]
Solving this gives:
\[ 1 = 2x \implies x = \frac{1}{2} \]
Since \( \frac{1}{2} \) is in the interval \([0, 1)\), this is a valid solution.

### Step 3: Check the Endpoints
- **At \( x = -1 \)**:
\[ f(-1) = 1 - (-1) = 2, \quad g(-1) = -1 + 1 = 0 \quad \text{(not equal)} \]

- **At \( x = 1 \)**:
\[ f(1) = 1 - 1 = 0, \quad g(1) = 1 \quad \text{(not equal)} \]

### Conclusion
The only valid solution in the interval \([-1, 1]\) is \( x = \frac{1}{2} \).

Thus, the number of solutions \( s \) of the equation \( |x - 1| = \{x\} \) for \( x \in [-1, 1] \) is:
\[ \boxed{1} \]