The number of solution of the equastion `|x-1|+|x-2|=[{x}]+{[x]}+1` , where `[.]` and `{.}` denotes greatest integer function and fraction part funstion respectively is :

Options

A. `0`
B. `1`
C. `2`
D. Infinite

Explanation

To solve the equation \( |x-1| + |x-2| = [x] + \{x\} + 1 \), where \([.]\) denotes the greatest integer function and \(\{.\}\) denotes the fractional part function, we will follow these steps:

### Step 1: Understand the Right-Hand Side
The right-hand side of the equation is \([x] + \{x\} + 1\). We know that:
- \([x]\) is the greatest integer less than or equal to \(x\).
- \(\{x\} = x - [x]\), which is the fractional part of \(x\).

Thus, we can simplify the right-hand side:
\[ [x] + \{x\} = [x] + (x - [x]) = x. \]
So, the equation simplifies to:
\[ |x-1| + |x-2| = x + 1. \]

### Step 2: Analyze the Left-Hand Side
The left-hand side \( |x-1| + |x-2| \) changes based on the value of \(x\). We will consider three cases based on the critical points \(x = 1\) and \(x = 2\).

### Case 1: \(x < 1\)
In this case, both expressions inside the absolute values are negative:
\[ |x-1| = -(x-1) = -x + 1, \]
\[ |x-2| = -(x-2) = -x + 2. \]
Thus, the equation becomes:
\[ (-x + 1) + (-x + 2) = x + 1. \]
This simplifies to:
\[ -2x + 3 = x + 1. \]
Rearranging gives:
\[ -3x = -2 \implies x = \frac{2}{3}. \]
Since \(\frac{2}{3} < 1\), this solution is valid.

### Case 2: \(1 \leq x < 2\)
In this case, we have:
\[ |x-1| = x - 1, \]
\[ |x-2| = -(x-2) = -x + 2. \]
Thus, the equation becomes:
\[ (x - 1) + (-x + 2) = x + 1. \]
This simplifies to:
\[ 1 = x + 1. \]
Rearranging gives:
\[ 0 = x. \]
However, \(0\) is not in the interval \([1, 2)\), so there are no valid solutions in this case.

### Case 3: \(x \geq 2\)
In this case, both expressions inside the absolute values are positive:
\[ |x-1| = x - 1, \]
\[ |x-2| = x - 2. \]
Thus, the equation becomes:
\[ (x - 1) + (x - 2) = x + 1. \]
This simplifies to:
\[ 2x - 3 = x + 1. \]
Rearranging gives:
\[ x = 4. \]
Since \(4 \geq 2\), this solution is valid.

### Step 3: Summary of Solutions
From the analysis, we found:
- A valid solution \(x = \frac{2}{3}\) from Case 1.
- No valid solutions from Case 2.
- A valid solution \(x = 4\) from Case 3.

Additionally, for \(1 \leq x < 2\), the left-hand side \( |x-1| + |x-2| \) equals \(1\) for all \(x\) in that interval, meaning every \(x\) in \([1, 2)\) satisfies the equation.

### Final Count of Solutions
Thus, the solutions are:
1. \(x = \frac{2}{3}\)
2. All \(x\) in the interval \([1, 2)\) (which is infinite).
3. \(x = 4\)

Therefore, the total number of solutions is infinite due to the interval \([1, 2)\).

### Conclusion
The number of solutions of the equation is infinite.