The function `f(x)=x^(3)-ax` has a local minimum at `x=k`, where `k ge 2,` then a possible value of a is

Options

A. 9
B. 11
C. 13
D. 8

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 - ax \) and determine the conditions under which it has a local minimum at \( x = k \), where \( k \geq 2 \).

### Step-by-step Solution:

1. **Find the derivative of the function**:
\[ f'(x) = \frac{d}{dx}(x^3 - ax) = 3x^2 - a \]

2. **Set the derivative equal to zero to find critical points**:
Since \( f(x) \) has a local minimum at \( x = k \), we set the derivative equal to zero:
\[ f'(k) = 3k^2 - a = 0 \]
Rearranging gives:
\[ a = 3k^2 \]

3. **Analyze the condition for \( k \)**:
We know that \( k \geq 2 \). Therefore, we can substitute the minimum value of \( k \) into the equation for \( a \):
\[ a = 3k^2 \geq 3(2^2) = 3 \times 4 = 12 \]

4. **Conclusion about the value of \( a \)**:
From the inequality \( a \geq 12 \), we can conclude that any value of \( a \) that is greater than or equal to 12 is a possible candidate.

5. **Select a possible value of \( a \)**:
If we are given options for \( a \), we need to choose a value that satisfies \( a \geq 12 \). For example, if one of the options is 13, then that is a valid choice.

### Final Answer:
A possible value of \( a \) is 13.