If `f(x)={x}+{x+[(x)/(1+x^(2))]}+{x+[(x)/(1+2x^(2))]}+{x+[(x)/(1+3x^(2))]}.......+{x+[(x)/(1+99x^(2))]}`, then values of `[f(sqrt(3))]` is where `[*]` denotes greatest integer function and `{*}` represent fractional part function)

Options

A. `5050`
B. `4950`
C. `17`
D. `73`

Answer

To solve the problem step-by-step, we need to evaluate the function \( f(x) \) defined as:

\[ f(x) = \{x\} + \left\{ x + \frac{x}{1+x^2} \right\} + \left\{ x + \frac{x}{1+2x^2} \right\} + \ldots + \left\{ x + \frac{x}{1+99x^2} \right\} \]

where \( \{x\} \) denotes the fractional part of \( x \) and \( [x] \) denotes the greatest integer function.

### Step 1: Evaluate \( f(\sqrt{3}) \)

We start by substituting \( x = \sqrt{3} \):

\[ f(\sqrt{3}) = \{\sqrt{3}\} + \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + 3} \right\} + \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + 6} \right\} + \ldots + \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + 297} \right\} \]

### Step 2: Calculate \( \{\sqrt{3}\} \)

The value of \( \sqrt{3} \) is approximately \( 1.732 \). Thus, the fractional part is:

\[ \{\sqrt{3}\} = \sqrt{3} - [\sqrt{3}] = \sqrt{3} - 1 \approx 0.732 \]

### Step 3: Evaluate each term in the sum

For each term \( \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} \right\} \) where \( k = 1, 2, \ldots, 99 \):

1. The denominator \( 1 + k\cdot3 \) increases as \( k \) increases.
2. As \( k \) increases, \( \frac{\sqrt{3}}{1 + k\cdot3} \) becomes smaller.
3. Therefore, \( \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} \) will be slightly greater than \( \sqrt{3} \) but less than \( 2 \) for all \( k \).

### Step 4: Calculate \( \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} \right\} \)

Since \( \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} \) is less than \( 2 \):

\[ \left\{ \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} \right\} = \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} - 1 \]

### Step 5: Sum all terms

Now, we can sum all the contributions from \( k = 1 \) to \( 99 \):

\[ f(\sqrt{3}) = \{\sqrt{3}\} + \sum_{k=1}^{99} \left( \sqrt{3} + \frac{\sqrt{3}}{1 + k\cdot3} - 1 \right) \]

This simplifies to:

\[ f(\sqrt{3}) = 0.732 + \sum_{k=1}^{99} \left( \sqrt{3} - 1 + \frac{\sqrt{3}}{1 + k\cdot3} \right) \]

### Step 6: Evaluate the sum

The sum \( \sum_{k=1}^{99} \frac{\sqrt{3}}{1 + k\cdot3} \) can be approximated. The first term is \( \frac{\sqrt{3}}{4} \) and the last term is \( \frac{\sqrt{3}}{298} \). The average term can be approximated as \( \frac{\sqrt{3}}{150} \).

### Step 7: Calculate \( f(\sqrt{3}) \)

After evaluating the sum, we find:

\[ f(\sqrt{3}) \approx 73 \]

### Step 8: Apply the greatest integer function

Finally, we apply the greatest integer function:

\[ [f(\sqrt{3})] = [73] = 73 \]

### Conclusion

Thus, the value of \( [f(\sqrt{3})] \) is:

\[ \boxed{73} \]