`f(x) ={{:((x)/(2x^(2)+|x|),xne0),( 1, x=0):}`then f(x) is

Options

A. continuous but non- differentiable at x=0
B. differentiable at x=0
C. discontinuous at x=0
D. none of these

Answer

To determine the nature of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} \frac{x}{2x^2 + |x|} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

we need to analyze the continuity and differentiability of \( f(x) \) at \( x = 0 \).

### Step 1: Find the Right-Hand Limit as \( x \to 0^+ \)

For \( x > 0 \), we have:

\[ f(x) = \frac{x}{2x^2 + x} = \frac{x}{x(2x + 1)} = \frac{1}{2x + 1} \]

Now, we calculate the limit as \( x \) approaches \( 0 \) from the right:

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

### Step 2: Find the Left-Hand Limit as \( x \to 0^- \)

For \( x < 0 \), we have:

\[ f(x) = \frac{x}{2x^2 - x} = \frac{x}{x(2x - 1)} = \frac{1}{2x - 1} \]

Now, we calculate the limit as \( x \) approaches \( 0 \) from the left:

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

### Step 3: Compare the Limits and the Function Value at \( x = 0 \)

Now we compare the limits we found:

- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 1 \)
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = -1 \)
- Function value at \( x = 0 \): \( f(0) = 1 \)

Since the right-hand limit and left-hand limit are not equal:

\[ \lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x) \]

### Conclusion

Thus, \( f(x) \) is not continuous at \( x = 0 \). Since continuity is a prerequisite for differentiability, \( f(x) \) is also not differentiable at \( x = 0 \).

### Final Answer

The function \( f(x) \) is **discontinuous at \( x = 0 \)**.

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