Let us consider an equation `f(x) = x^(3)-3x +k=0`.then the values of k for which the equation has
Exactly one root which is negative, then k belong

Options

A. `(2 , oo)`
B. (0,2)
C. (-2,0)
D. `(-oo,-2) `

Answer

To solve the problem, we need to determine the values of \( k \) for which the equation \( f(x) = x^3 - 3x + k = 0 \) has exactly one negative root.

### Step-by-step Solution:

1. **Identify the function**: We start with the function given by the equation:
\[ f(x) = x^3 - 3x + k \]

2. **Find critical points**: To analyze the behavior of the function, we first find its derivative:
\[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1) \]
Setting the derivative equal to zero gives us the critical points:
\[ f'(x) = 0 \implies x = -1, 1 \]

3. **Evaluate the function at critical points**: Next, we evaluate \( f(x) \) at these critical points:
- For \( x = -1 \):
\[ f(-1) = (-1)^3 - 3(-1) + k = -1 + 3 + k = k + 2 \]
- For \( x = 1 \):
\[ f(1) = (1)^3 - 3(1) + k = 1 - 3 + k = k - 2 \]

4. **Determine conditions for exactly one negative root**:
- For the function to have exactly one negative root, it must cross the x-axis only once in the negative region. This occurs when:
- \( f(-1) > 0 \) (the function is above the x-axis at \( x = -1 \))
- \( f(1) < 0 \) (the function is below the x-axis at \( x = 1 \))

From our evaluations:
- \( f(-1) > 0 \) gives:
\[ k + 2 > 0 \implies k > -2 \]
- \( f(1) < 0 \) gives:
\[ k - 2 < 0 \implies k < 2 \]

5. **Combine the inequalities**: We need to satisfy both conditions:
\[ -2 < k < 2 \]

6. **Conclusion**: The values of \( k \) for which the equation has exactly one negative root are:
\[ k \in (-2, 2) \]

### Final Answer:
The values of \( k \) for which the equation \( f(x) = x^3 - 3x + k = 0 \) has exactly one negative root are:
\[ k \in (-2, 2) \]