If `f(x) =Pi_(i=1)^3(x-a_i) + sum_(i=1)_3 a_i - 3x` where `a_i < a_(i+1)` for i = 1,2, then f(x) = 0 have

Options

A. only one distinct real root
B. exactly two distnict real roots
C. exactly 3 distinet real roots
D. 3 equal real roots

Answer

To solve the problem, we need to analyze the function \( f(x) = \prod_{i=1}^{3}(x - a_i) + \sum_{i=1}^{3} a_i - 3x \) and find the values of \( x \) for which \( f(x) = 0 \).

### Step-by-Step Solution:

1. **Write down the function**:
\[ f(x) = (x - a_1)(x - a_2)(x - a_3) + (a_1 + a_2 + a_3) - 3x \]

2. **Expand the product**:
The product \( (x - a_1)(x - a_2)(x - a_3) \) can be expanded using the formula for the product of linear factors:
\[ (x - a_1)(x - a_2)(x - a_3) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_2a_3 + a_3a_1)x - a_1a_2a_3 \]

3. **Substitute the expansion into \( f(x) \)**:
\[ f(x) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_2a_3 + a_3a_1)x - a_1a_2a_3 + (a_1 + a_2 + a_3) - 3x \]

4. **Combine like terms**:
\[ f(x) = x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_2a_3 + a_3a_1 - 3)x - a_1a_2a_3 + (a_1 + a_2 + a_3) \]

5. **Set \( f(x) = 0 \)**:
Now we need to solve the equation:
\[ x^3 - (a_1 + a_2 + a_3)x^2 + (a_1a_2 + a_2a_3 + a_3a_1 - 3)x - (a_1a_2a_3 - (a_1 + a_2 + a_3)) = 0 \]

6. **Analyze the roots**:
Since \( a_1 < a_2 < a_3 \), we can analyze the behavior of \( f(x) \) at specific points:
- As \( x \to -\infty \), \( f(x) \to -\infty \).
- At \( x = a_1 \):
\[ f(a_1) = 0 + (a_1 + a_2 + a_3 - 3a_1) = a_2 + a_3 - 2a_1 \]
Since \( a_2 > a_1 \), \( f(a_1) > 0 \).
- At \( x = a_3 \):
\[ f(a_3) = 0 + (a_1 + a_2 + a_3 - 3a_3) = a_1 + a_2 - 2a_3 \]
Since \( a_1 < a_3 \), \( f(a_3) < 0 \).

7. **Conclusion**:
By the Intermediate Value Theorem, since \( f(x) \) changes from positive to negative between \( a_1 \) and \( a_3 \), and since it is a cubic polynomial, it must have three distinct real roots.

### Final Answer:
The function \( f(x) = 0 \) has **three distinct solutions**.