Statement-1: `If a, b, c in R and 2a + 3b + 6c = 0`, then the equation `ax^(2) + bx + c = 0` has at least one real root in (0, 1).
Statement-2: If f(x) is a polynomial which assumes both positive and negative values, then it has at least one real root.

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 problem, we need to analyze both statements given in the question.

### Step 1: Analyze Statement-2
Statement-2 states that if \( f(x) \) is a polynomial that assumes both positive and negative values, then it has at least one real root.

**Solution:**
This statement is true because of the Intermediate Value Theorem. If a continuous function (like a polynomial) takes on both positive and negative values, it must cross the x-axis at least once. Therefore, there exists at least one real root.

### Step 2: Analyze Statement-1
Statement-1 states that if \( 2a + 3b + 6c = 0 \), then the quadratic equation \( ax^2 + bx + c = 0 \) has at least one real root in the interval \( (0, 1) \).

**Solution:**
1. **Define the function:** Let \( f(x) = ax^2 + bx + c \).
2. **Evaluate \( f(0) \):**
\[ f(0) = c \]
3. **Evaluate \( f(1) \):**
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
4. **Use the given condition:** We know that \( 2a + 3b + 6c = 0 \). We can manipulate this equation to express \( c \):
\[ c = -\frac{2a + 3b}{6} \]
5. **Substituting \( c \) into \( f(0) \) and \( f(1) \):**
- From the above, we have \( f(0) = c = -\frac{2a + 3b}{6} \).
- For \( f(1) \):
\[ f(1) = a + b + c = a + b - \frac{2a + 3b}{6} \]
- Combine the terms:
\[ f(1) = a + b - \frac{2a + 3b}{6} = \frac{6a + 6b - 2a - 3b}{6} = \frac{4a + 3b}{6} \]
6. **Check the signs of \( f(0) \) and \( f(1) \):**
- If \( f(0) = c \) and \( f(1) = \frac{4a + 3b}{6} \), we need to analyze the signs of these two expressions.
- Since \( 2a + 3b + 6c = 0 \), we can see that if \( c \) is negative (which could happen if \( 2a + 3b > 0 \)), then \( f(0) < 0 \).
- If \( 4a + 3b < 0 \), then \( f(1) < 0 \).
- If \( 4a + 3b > 0 \), then \( f(1) > 0 \).

7. **Conclusion using Rolle's Theorem:**
Since \( f(0) \) and \( f(1) \) can take opposite signs, by the Intermediate Value Theorem (or Rolle's Theorem), there must be at least one root in the interval \( (0, 1) \).

### Final Conclusion:
Both statements are true. Statement-1 is true because the quadratic equation has at least one real root in the interval \( (0, 1) \) given the condition \( 2a + 3b + 6c = 0 \). Statement-2 is also true as it follows from the Intermediate Value Theorem.