Which one of the following is correct in respect of the function `f(x)=(x^(2))/(|x|)` for `xne0`and `f(0)=0`?

Options

A. f(x) is discontinuous everywhere
B. f(x) is continuous everywhere
C. f(x) is continuous at x = 0 only
D. f(x) is discontinuous at x = 0 only

Answer

To solve the problem regarding the function \( f(x) = \frac{x^2}{|x|} \) for \( x \neq 0 \) and \( f(0) = 0 \), we will analyze the function step by step.

### Step 1: Simplify the function for \( x > 0 \) and \( x < 0 \)

1. **For \( x > 0 \)**:
- The absolute value \( |x| \) is equal to \( x \).
- Therefore, \( f(x) = \frac{x^2}{|x|} = \frac{x^2}{x} = x \).

2. **For \( x < 0 \)**:
- The absolute value \( |x| \) is equal to \( -x \).
- Therefore, \( f(x) = \frac{x^2}{|x|} = \frac{x^2}{-x} = -x \).

So, we can summarize the function as:
\[ f(x) = \begin{cases} x & \text{if } x > 0 \\ -x & \text{if } x < 0 \end{cases} \]

### Step 2: Check the continuity of the function at \( x = 0 \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to evaluate the left-hand limit (LHL), right-hand limit (RHL), and the value of the function at that point.

1. **Right-Hand Limit (RHL)**:
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0 \]

2. **Left-Hand Limit (LHL)**:
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0 \]

3. **Value of the function at \( x = 0 \)**:
\[ f(0) = 0 \]

### Step 3: Conclusion on continuity

Since:
- RHL = 0
- LHL = 0
- \( f(0) = 0 \)

We have:
\[ \text{LHL} = \text{RHL} = f(0) \]

Thus, \( f(x) \) is continuous at \( x = 0 \).

### Final Result

The function \( f(x) \) is continuous for all \( x \), including at \( x = 0 \).