Consider a quadratic equation `ax^(2) + bx + c = 0`, where 2a + 3b + 6c = 0 and let `g(x) = a(x^(3))/(3)+b(x^(2))/(2)+cx`.
Statement-1 The quadratic equation has at least one root in the internal (0, 1).
Statement-2 The Rolle's Theorem is applicable to function g(x) on the interval [0, 1].

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 step by step, we will analyze the given quadratic equation and the function \( g(x) \) to determine the validity of the statements.

### Step 1: Understand the Given Conditions
We have a quadratic equation of the form:
\[ ax^2 + bx + c = 0 \]
with the condition:
\[ 2a + 3b + 6c = 0 \]
We also have a function defined as:
\[ g(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx \]

### Step 2: Check Continuity and Differentiability of \( g(x) \)
The function \( g(x) \) is a polynomial, which means it is continuous and differentiable everywhere, including the interval \([0, 1]\).

### Step 3: Evaluate \( g(0) \) and \( g(1) \)
Now, we will evaluate \( g(0) \) and \( g(1) \):
- For \( g(0) \):
\[ g(0) = \frac{a}{3}(0)^3 + \frac{b}{2}(0)^2 + c(0) = 0 \]

- For \( g(1) \):
\[ g(1) = \frac{a}{3}(1)^3 + \frac{b}{2}(1)^2 + c(1) = \frac{a}{3} + \frac{b}{2} + c \]

### Step 4: Substitute the Condition into \( g(1) \)
From the condition \( 2a + 3b + 6c = 0 \), we can express \( g(1) \):
\[ g(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6} = \frac{0}{6} = 0 \]

### Step 5: Apply Rolle's Theorem
Since \( g(0) = 0 \) and \( g(1) = 0 \), and \( g(x) \) is continuous and differentiable on \([0, 1]\), by Rolle's Theorem, there exists at least one \( \alpha \) in the interval \( (0, 1) \) such that:
\[ g'(\alpha) = 0 \]

### Step 6: Find the Derivative and Relate it to the Quadratic Equation
The derivative of \( g(x) \) is:
\[ g'(x) = ax^2 + bx + c \]
Setting \( g'(\alpha) = 0 \) gives us:
\[ a\alpha^2 + b\alpha + c = 0 \]
This means that \( \alpha \) is a root of the quadratic equation \( ax^2 + bx + c = 0 \).

### Conclusion
Since we have shown that there exists at least one root \( \alpha \) in the interval \( (0, 1) \) for the quadratic equation, we conclude that:
- **Statement 1** is true: The quadratic equation has at least one root in the interval \( (0, 1) \).
- **Statement 2** is also true: The conditions for applying Rolle's Theorem are satisfied for the function \( g(x) \) on the interval \([0, 1]\).