Let `f(x)={{:(,sum_(r=0)^(x^(2)[(1)/(|x|)])r,x ne 0),(,k,x=0):}` where [.] denotes the greatest integer function. The value of k for which is continuous at x=0, is

Options

A. 1
B. 2
C. 4
D. `(1)/(2)`

Explanation

To solve the problem, we need to determine the value of \( k \) for which the function \( f(x) \) is continuous at \( x = 0 \). The function is defined as follows:

\[ f(x) = \begin{cases} \sum_{r=0}^{\left\lfloor \frac{x^2}{|x|} \right\rfloor} r & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

### Step 1: Analyze the function for \( x \neq 0 \)

For \( x \neq 0 \), we need to evaluate \( \left\lfloor \frac{x^2}{|x|} \right\rfloor \).

Since \( |x| = x \) when \( x > 0 \) and \( |x| = -x \) when \( x < 0 \), we can analyze both cases:

- If \( x > 0 \):
\[ \frac{x^2}{|x|} = \frac{x^2}{x} = x \]
Hence, \( \left\lfloor \frac{x^2}{|x|} \right\rfloor = \lfloor x \rfloor \).

- If \( x < 0 \):
\[ \frac{x^2}{|x|} = \frac{x^2}{-x} = -x \]
Hence, \( \left\lfloor \frac{x^2}{|x|} \right\rfloor = \lfloor -x \rfloor \).

### Step 2: Evaluate the summation

The summation \( \sum_{r=0}^{n} r \) for a non-negative integer \( n \) is given by the formula:
\[ \sum_{r=0}^{n} r = \frac{n(n+1)}{2} \]

Thus, we can express \( f(x) \) for \( x > 0 \) and \( x < 0 \):

- For \( x > 0 \):
\[ f(x) = \frac{\lfloor x \rfloor (\lfloor x \rfloor + 1)}{2} \]

- For \( x < 0 \):
\[ f(x) = \frac{\lfloor -x \rfloor (\lfloor -x \rfloor + 1)}{2} \]

### Step 3: Find the limit as \( x \) approaches 0

To ensure continuity at \( x = 0 \), we need to find:
\[ \lim_{x \to 0} f(x) \]

As \( x \) approaches 0 from the right (\( x \to 0^+ \)):
\[ \lim_{x \to 0^+} f(x) = \frac{\lfloor 0 \rfloor (\lfloor 0 \rfloor + 1)}{2} = \frac{0(0 + 1)}{2} = 0 \]

As \( x \) approaches 0 from the left (\( x \to 0^- \)):
\[ \lim_{x \to 0^-} f(x) = \frac{\lfloor 0 \rfloor (\lfloor 0 \rfloor + 1)}{2} = \frac{0(0 + 1)}{2} = 0 \]

### Step 4: Set the limits equal to \( k \)

For \( f(x) \) to be continuous at \( x = 0 \):
\[ \lim_{x \to 0} f(x) = f(0) = k \]
Thus, we have:
\[ k = 0 \]

### Final Answer

The value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:
\[ \boxed{0} \]