Consider the equation `x^(3)-nx+1=0, n in N, n ge 3`. Then

Options

A. Equation has atleast one rational root.
B. Equation has exactly one rational root.
C. Equation has all real roots belonging to (0, 1).
D. Equation has no rational root.

Answer

To solve the equation \( x^3 - nx + 1 = 0 \) where \( n \in \mathbb{N} \) and \( n \geq 3 \), we will analyze the behavior of the function and its derivatives to determine the nature of its roots.

### Step 1: Define the function
Let \( f(x) = x^3 - nx + 1 \).

### Step 2: Find the first derivative
To find the critical points, we calculate the first derivative:
\[ f'(x) = 3x^2 - n \]

### Step 3: Set the first derivative to zero
Setting the first derivative to zero to find critical points:
\[ 3x^2 - n = 0 \implies x^2 = \frac{n}{3} \implies x = \pm \sqrt{\frac{n}{3}} \]

### Step 4: Analyze the critical points
Since \( n \geq 3 \), we can evaluate the critical points:
- For \( n = 3 \):
\[ x = \pm \sqrt{1} = \pm 1 \]
- For \( n > 3 \):
\[ x = \pm \sqrt{\frac{n}{3}} \text{ which will be greater than } 1 \text{ or less than } -1. \]

### Step 5: Find the second derivative
Next, we find the second derivative to determine the nature of the critical points:
\[ f''(x) = 6x \]
- At \( x = 1 \):
\[ f''(1) = 6 > 0 \quad (\text{local minimum}) \]
- At \( x = -1 \):
\[ f''(-1) = -6 < 0 \quad (\text{local maximum}) \]

### Step 6: Evaluate the function at critical points
Now we evaluate \( f(x) \) at the critical points:
- At \( x = 1 \):
\[ f(1) = 1^3 - n \cdot 1 + 1 = 2 - n \]
- At \( x = -1 \):
\[ f(-1) = (-1)^3 - n(-1) + 1 = -1 + n + 1 = n \]

### Step 7: Analyze the function values
- For \( n = 3 \):
- \( f(1) = 2 - 3 = -1 \) (negative)
- \( f(-1) = 3 \) (positive)

Thus, by the Intermediate Value Theorem, there is at least one root between \( -1 \) and \( 1 \).

- For \( n > 3 \):
- \( f(1) < 0 \) (since \( 2 - n < 0 \))
- \( f(-1) > 0 \) (since \( n > 3 \))

This indicates that there is at least one root between \( -1 \) and \( 1 \).

### Step 8: Conclusion
From the analysis, we conclude that the equation \( x^3 - nx + 1 = 0 \) has at least one rational root for \( n \geq 3 \).

### Final Answer
The correct option is: **The equation has at least one rational root.**