The value of `c` in Rolles theorem for the function `f(x)=x^3-3x` in the interval `[0,\ sqrt(3)]` is (a) 1 (b) `-1` (c) `3//2` (d) `1//3`

Explanation

To find the value of `c` in Rolle's Theorem for the function \( f(x) = x^3 - 3x \) in the interval \([0, \sqrt{3}]\), we will follow these steps:

### Step 1: Verify the conditions of Rolle's Theorem
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one \(c\) in \((a, b)\) such that \(f'(c) = 0\).

1. **Check continuity**: The function \(f(x) = x^3 - 3x\) is a polynomial, which is continuous everywhere.
2. **Check differentiability**: The function is also differentiable everywhere since it is a polynomial.
3. **Check endpoints**: Calculate \(f(0)\) and \(f(\sqrt{3})\):
- \(f(0) = 0^3 - 3 \cdot 0 = 0\)
- \(f(\sqrt{3}) = (\sqrt{3})^3 - 3(\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0\)

Since \(f(0) = f(\sqrt{3})\), the conditions for Rolle's Theorem are satisfied.

### Step 2: Find the derivative of the function
Next, we need to find the derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3 \]

### Step 3: Set the derivative equal to zero
To find the value of \(c\), we set the derivative equal to zero:
\[ 3x^2 - 3 = 0 \]
\[ 3x^2 = 3 \]
\[ x^2 = 1 \]
\[ x = \pm 1 \]

### Step 4: Determine the valid value of \(c\)
Since we are looking for \(c\) in the interval \((0, \sqrt{3})\), we only consider the positive value:
\[ c = 1 \]

### Conclusion
Thus, the value of \(c\) in Rolle's Theorem for the function \(f(x) = x^3 - 3x\) in the interval \([0, \sqrt{3}]\) is:
\[ \boxed{1} \]