If `f(x)={{:(,x([(1)/(x)]+[(2)/(x)]+.....+[(n)/(x)]),x ne 0),(,k,x=0):}` and `n in N`. Then the value of k for which f(x) is continuous at x=0 is

Options

A. n
B. n+1
C. `n(n+1)`
D. `(n(n+1))/(2)

Explanation

To find the value of \( k \) for which the function

\[ f(x) = \begin{cases} x \left( \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \ldots + \left\lfloor \frac{n}{x} \right\rfloor \right), & x \neq 0 \\ k, & x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches \( 0 \) and set it equal to \( k \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) \) for \( x \neq 0 \) involves the floor function \( \left\lfloor \frac{j}{x} \right\rfloor \) for \( j = 1, 2, \ldots, n \). As \( x \) approaches \( 0 \), \( \frac{j}{x} \) becomes very large, and thus \( \left\lfloor \frac{j}{x} \right\rfloor \) will also be large.

2. **Finding the Limit**:
We need to find the limit:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x \left( \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \ldots + \left\lfloor \frac{n}{x} \right\rfloor \right) \]

3. **Estimating Each Floor Function**:
For each \( j \):

\[ \left\lfloor \frac{j}{x} \right\rfloor \approx \frac{j}{x} - 1 \quad \text{(as } x \to 0\text{)} \]

Therefore,

\[ \left\lfloor \frac{j}{x} \right\rfloor \text{ lies between } \frac{j}{x} - 1 \text{ and } \frac{j}{x} \]

4. **Summing the Floor Functions**:
We can sum these estimates:

\[ \sum_{j=1}^{n} \left\lfloor \frac{j}{x} \right\rfloor \approx \sum_{j=1}^{n} \left( \frac{j}{x} - 1 \right) = \frac{1 + 2 + \ldots + n}{x} - n = \frac{\frac{n(n+1)}{2}}{x} - n \]

5. **Combining with \( x \)**:
Now, substituting back into the limit:

\[ f(x) \approx x \left( \frac{n(n+1)}{2x} - n \right) = \frac{n(n+1)}{2} - nx \]

6. **Taking the Limit**:
As \( x \to 0 \):

\[ \lim_{x \to 0} f(x) = \frac{n(n+1)}{2} \]

7. **Setting the Limit Equal to \( k \)**:
For continuity at \( x = 0 \):

\[ k = \frac{n(n+1)}{2} \]

### Final Answer:
The value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is

\[ \boxed{\frac{n(n+1)}{2}} \]