If `f(x) = {x} + { x + 1 } + {x + 2 }.........{x + 99),` then the value of `[f(sqrt2)]` is, where (.) denotes fractional part function & [.] denotes the greatest integer function

Options

A. 5050
B. 4950
C. 41
D. 14

Answer

To solve the problem, we need to evaluate the function \( f(x) = \{x\} + \{x + 1\} + \{x + 2\} + \ldots + \{x + 99\} \) at \( x = \sqrt{2} \), where \( \{.\} \) denotes the fractional part function and \( [.] \) denotes the greatest integer function.

### Step-by-step Solution:

1. **Understanding the fractional part function**:
The fractional part function \( \{x\} \) is defined as \( x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). The output of this function is always between 0 and 1, inclusive of 0 but exclusive of 1.

2. **Calculating \( \sqrt{2} \)**:
We know that \( \sqrt{2} \approx 1.414 \). Therefore, the integer part \( \lfloor \sqrt{2} \rfloor = 1 \).

3. **Finding the fractional part of \( \sqrt{2} \)**:
\[ \{\sqrt{2}\} = \sqrt{2} - \lfloor \sqrt{2} \rfloor = \sqrt{2} - 1 \approx 1.414 - 1 = 0.414 \]

4. **Calculating the fractional parts for \( x + k \)**:
For \( k = 0, 1, 2, \ldots, 99 \):
- \( \{ \sqrt{2} + k \} = \{ 1.414 + k \} = \{ k + 0.414 \} \)
- Since \( k \) is an integer, \( \lfloor k + 0.414 \rfloor = k \) when \( k + 0.414 < k + 1 \).
- Thus, \( \{ k + 0.414 \} = 0.414 \) for each \( k \).

5. **Summing the fractional parts**:
Since there are 100 terms (from \( k = 0 \) to \( k = 99 \)):
\[ f(\sqrt{2}) = \sum_{k=0}^{99} \{\sqrt{2} + k\} = 100 \times 0.414 = 41.4 \]

6. **Applying the greatest integer function**:
We need to find \( [f(\sqrt{2})] \):
\[ [f(\sqrt{2})] = [41.4] = 41 \]

### Final Answer:
The value of \( [f(\sqrt{2})] \) is \( 41 \).