Consider the cubic equation `f(x)=x^(3)-nx+1=0` where `n ge3, n in N` then `f(x)=0` has

Options

A. (a) atleast one root in (0, 1)
B. (b) atleast one root in (1, 2)
C. (c) atleast one root in `(-1, 0)`
D. (d) data insufficient

Answer

To solve the cubic equation \( f(x) = x^3 - nx + 1 = 0 \) where \( n \geq 3 \) and \( n \in \mathbb{N} \), we need to determine the intervals in which at least one root exists.

### Step 1: Analyze the function
The function is given by:
\[ f(x) = x^3 - nx + 1 \]
We want to find intervals where \( f(x) = 0 \) has at least one root. According to the Intermediate Value Theorem, if \( f(a) \) and \( f(b) \) have opposite signs, then there exists at least one root in the interval \( (a, b) \).

### Step 2: Evaluate \( f(0) \) and \( f(1) \)
Let's first check the interval \( [0, 1] \):
- Calculate \( f(0) \):
\[ f(0) = 0^3 - n \cdot 0 + 1 = 1 \]
- Calculate \( f(1) \):
\[ f(1) = 1^3 - n \cdot 1 + 1 = 2 - n \]

### Step 3: Determine the sign of \( f(1) \)
For there to be a root in the interval \( [0, 1] \), we need \( f(0) \) and \( f(1) \) to have opposite signs:
\[ f(0) = 1 \quad \text{(positive)} \]
\[ f(1) = 2 - n \]
We require:
\[ 1 \cdot (2 - n) < 0 \implies 2 - n < 0 \implies n > 2 \]
Since \( n \geq 3 \), this condition is satisfied. Therefore, there is at least one root in the interval \( [0, 1] \).

### Step 4: Evaluate \( f(1) \) and \( f(2) \)
Now, let's check the interval \( [1, 2] \):
- Calculate \( f(2) \):
\[ f(2) = 2^3 - n \cdot 2 + 1 = 8 - 2n + 1 = 9 - 2n \]

We need to check the signs of \( f(1) \) and \( f(2) \):
\[ f(1) = 2 - n \quad \text{(already calculated)} \]
\[ f(2) = 9 - 2n \]

### Step 5: Determine the sign of \( f(1) \) and \( f(2) \)
We need:
\[ (2 - n)(9 - 2n) < 0 \]
This product is negative if one of the factors is positive and the other is negative.

1. **Case 1:** \( 2 - n < 0 \) (i.e., \( n > 2 \)) and \( 9 - 2n > 0 \) (i.e., \( n < 4.5 \)).
2. **Case 2:** \( 2 - n > 0 \) (i.e., \( n < 2 \)) and \( 9 - 2n < 0 \) (i.e., \( n > 4.5 \)).

Since \( n \geq 3 \), we can only consider the first case. Thus, \( n \) must be in the interval \( (2, 4.5) \). Since \( n \) is a natural number, valid values for \( n \) are 3 and 4.

### Step 6: Evaluate \( f(-1) \) and \( f(0) \)
Now, let's check the interval \( [-1, 0] \):
- Calculate \( f(-1) \):
\[ f(-1) = (-1)^3 - n(-1) + 1 = -1 + n + 1 = n \]
Since \( n \geq 3 \), \( f(-1) \) is positive.

### Conclusion
From our analysis:
- There is at least one root in the interval \( [0, 1] \).
- The interval \( [1, 2] \) can have roots depending on the value of \( n \).
- The interval \( [-1, 0] \) does not contain roots since \( f(-1) \) is positive.

Thus, the only certain conclusion is that there is at least one root in the interval \( [0, 1] \).