Statement-1: There is a value of k for which the equation `x^(3) - 3x + k = 0` has a root between 0 and 1.
Statement-2: Between any two real roots of a polynomial there is a root of its derivation.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given question, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
We need to determine if there is a value of \( k \) such that the equation
\[ x^3 - 3x + k = 0 \]
has a root between 0 and 1.

### Step 2: Define the Function
Let's define the function:
\[ f(x) = x^3 - 3x + k \]
We want to find \( k \) such that \( f(x) = 0 \) has a root in the interval \( (0, 1) \).

### Step 3: Evaluate the Function at the Endpoints
We will evaluate \( f(x) \) at the endpoints of the interval:
1. Calculate \( f(0) \):
\[ f(0) = 0^3 - 3(0) + k = k \]
2. Calculate \( f(1) \):
\[ f(1) = 1^3 - 3(1) + k = 1 - 3 + k = k - 2 \]

### Step 4: Apply the Intermediate Value Theorem
For \( f(x) \) to have a root in the interval \( (0, 1) \), \( f(0) \) and \( f(1) \) must have opposite signs. This gives us the conditions:
- \( f(0) = k \)
- \( f(1) = k - 2 \)

We need:
\[ k \cdot (k - 2) < 0 \]

### Step 5: Solve the Inequality
The product \( k(k - 2) < 0 \) implies that \( k \) must be between the roots of the equation \( k(k - 2) = 0 \), which are \( k = 0 \) and \( k = 2 \). Therefore, the solution to the inequality is:
\[ 0 < k < 2 \]

### Step 6: Conclusion for Statement 1
Thus, there exists a value of \( k \) (specifically any \( k \) in the interval \( (0, 2) \)) such that the equation \( x^3 - 3x + k = 0 \) has a root between 0 and 1. Therefore, Statement 1 is **true**.

### Step 7: Analyze Statement 2
Statement 2 states that between any two real roots of a polynomial, there is a root of its derivative. This is a direct consequence of Rolle's Theorem, which is a fundamental theorem in calculus. Therefore, Statement 2 is also **true**.

### Final Conclusion
- Statement 1: True
- Statement 2: True