If `x_(1),x_(2),x_(3),…,x_(n)` are the roots of the equation `x^(n)+ax+b=0`, the value of
`(x_(1)-x_(2))(x_(1)-x_(3))(x_(1)-x_(4))…….(x_(1)-x_(n))` is

Options

A. `nx _(1) +b`
B. ` n x _(1)^(n-1) +a`
C. `nx _(1) ^(n-1)`
D. `nx _(1) ^(n-1)`

Answer

To solve the problem, we need to find the value of the expression:

\[ (x_1 - x_2)(x_1 - x_3)(x_1 - x_4) \ldots (x_1 - x_n) \]

where \(x_1, x_2, x_3, \ldots, x_n\) are the roots of the polynomial equation:

\[ x^n + ax + b = 0 \]

### Step-by-Step Solution:

1. **Understand the Polynomial Representation**:
The polynomial can be expressed in terms of its roots as:
\[ x^n + ax + b = (x - x_1)(x - x_2)(x - x_3) \ldots (x - x_n) \]

2. **Divide by \(x - x_1\)**:
We can divide both sides of the equation by \(x - x_1\):
\[ \frac{x^n + ax + b}{x - x_1} = (x - x_2)(x - x_3) \ldots (x - x_n) \]

3. **Take the Limit as \(x\) Approaches \(x_1\)**:
We take the limit as \(x\) approaches \(x_1\):
\[ \lim_{x \to x_1} \frac{x^n + ax + b}{x - x_1} = \lim_{x \to x_1} (x - x_2)(x - x_3) \ldots (x - x_n) \]

4. **Differentiate the Numerator and Denominator**:
Since both the numerator and denominator approach 0 as \(x\) approaches \(x_1\), we apply L'Hôpital's Rule:
- Differentiate the numerator:
\[ \frac{d}{dx}(x^n + ax + b) = nx^{n-1} + a \]
- The derivative of the denominator is:
\[ \frac{d}{dx}(x - x_1) = 1 \]

5. **Evaluate the Limit**:
Now we evaluate the limit:
\[ \lim_{x \to x_1} \frac{nx^{n-1} + a}{1} = n x_1^{n-1} + a \]

6. **Evaluate the Right Side**:
On the right side, we can directly substitute \(x = x_1\):
\[ (x_1 - x_2)(x_1 - x_3) \ldots (x_1 - x_n) \]

7. **Final Equation**:
Setting both sides equal gives us:
\[ n x_1^{n-1} + a = (x_1 - x_2)(x_1 - x_3) \ldots (x_1 - x_n) \]

8. **Conclusion**:
Therefore, the value of the expression \((x_1 - x_2)(x_1 - x_3)(x_1 - x_4) \ldots (x_1 - x_n)\) is equal to \(n x_1^{n-1} + a\).

### Final Answer:
\[ \text{The value is } n x_1^{n-1} + a \]