The roots of `(x-41)^(49)+(x-49)^(41)+(x-2009)^(2009)=0` are

Options

A. all necessarily real
B. non-real except one positive real root
C. non-real except three positive real roots
D. non-real except for three real roots of which exactly one is positive

Explanation

To solve the equation \((x-41)^{49} + (x-49)^{41} + (x-2009)^{2009} = 0\), we will analyze the function step by step.

### Step 1: Define the Function
Let \( f(x) = (x-41)^{49} + (x-49)^{41} + (x-2009)^{2009} \).

### Step 2: Analyze the Function
We need to determine the behavior of \( f(x) \) as \( x \) varies.

1. **For \( x < 41 \)**:
- \( (x-41)^{49} \) is negative (since it is raised to an odd power).
- \( (x-49)^{41} \) is negative (since it is raised to an odd power).
- \( (x-2009)^{2009} \) is negative (since it is raised to an odd power).
- Therefore, \( f(x) < 0 \).

2. **For \( x = 41 \)**:
- \( f(41) = 0 + (41-49)^{41} + (41-2009)^{2009} \).
- The second term is negative and the third term is also negative, hence \( f(41) < 0 \).

3. **For \( 41 < x < 49 \)**:
- \( (x-41)^{49} > 0 \) (positive).
- \( (x-49)^{41} < 0 \) (negative).
- \( (x-2009)^{2009} < 0 \) (negative).
- The positive term \( (x-41)^{49} \) dominates, so \( f(x) > 0 \).

4. **For \( x = 49 \)**:
- \( f(49) = (49-41)^{49} + 0 + (49-2009)^{2009} \).
- The first term is positive, and the third term is negative, hence \( f(49) > 0 \).

5. **For \( x > 49 \)**:
- \( (x-41)^{49} > 0 \) (positive).
- \( (x-49)^{41} > 0 \) (positive).
- \( (x-2009)^{2009} < 0 \) (negative).
- The positive terms dominate, so \( f(x) > 0 \).

### Step 3: Find the Derivative
Now, we find the derivative \( f'(x) \):
\[ f'(x) = 49(x-41)^{48} + 41(x-49)^{40} + 2009(x-2009)^{2008} \]
Since all terms are raised to even powers and multiplied by positive coefficients, \( f'(x) > 0 \) for all \( x \).

### Step 4: Conclusion on Roots
Since \( f(x) \) is continuous and strictly increasing:
- It starts below zero for \( x < 41 \).
- It becomes positive between \( 41 < x < 49 \).
- It remains positive for \( x > 49 \).

Thus, there is exactly one point where \( f(x) = 0 \), which is a positive real root.

### Final Answer
The roots of the equation \((x-41)^{49} + (x-49)^{41} + (x-2009)^{2009} = 0\) are:
**Non-real except one positive real root.**