If `f'(x)=(x-a)^(2010)(x-b)^(2009) and agtb`, then

Options

A. f(x) has relative maxima at x = b
B. f(x) has relative minima at x = b
C. f(x) has relative maxima at x = a
D. f(x) has neither maxima, nor minima at x = a

Explanation

To solve the problem, we need to analyze the function \( f'(x) = (x-a)^{2010}(x-b)^{2009} \) given that \( a > b \). We will determine the nature of the critical points where \( f'(x) = 0 \).

### Step 1: Find the critical points
To find the critical points, we set \( f'(x) = 0 \):
\[ (x-a)^{2010}(x-b)^{2009} = 0 \]
This equation is satisfied when either \( (x-a)^{2010} = 0 \) or \( (x-b)^{2009} = 0 \).

- From \( (x-a)^{2010} = 0 \), we get \( x = a \).
- From \( (x-b)^{2009} = 0 \), we get \( x = b \).

Thus, the critical points are \( x = a \) and \( x = b \).

### Step 2: Determine the nature of the critical points
Next, we analyze the nature of these critical points using the first derivative test.

1. **At \( x = b \)**:
- For \( x < b \), \( f'(x) \) will be negative because \( (x-b)^{2009} \) is negative and \( (x-a)^{2010} \) is positive (since \( a > b \)).
- For \( x > b \), \( f'(x) \) will be positive because both factors \( (x-a)^{2010} \) and \( (x-b)^{2009} \) are positive.

Therefore, \( f'(x) \) changes from negative to positive at \( x = b \), indicating that \( x = b \) is a point of **local minimum**.

2. **At \( x = a \)**:
- For \( x < a \), \( f'(x) \) is positive (since \( (x-a)^{2010} \) is negative and \( (x-b)^{2009} \) is positive).
- For \( x > a \), \( f'(x) \) remains positive (since both factors are positive).

Therefore, \( f'(x) \) does not change sign at \( x = a \), indicating that \( x = a \) is neither a maximum nor a minimum.

### Conclusion
- \( x = b \) is a point of local minimum.
- \( x = a \) is neither a maximum nor a minimum.

### Final Answer
Thus, the conclusion is:
- \( x = b \) is a point of local minimum.
- \( x = a \) is neither a maximum nor a minimum.