Statement-1: `If alpha and beta` are real roots of the quadratic equations `ax^(2) + bx + c = 0 and -ax^(2) + bx + c = 0`, then `(a)/(2) x^(2) + bx + c = 0` has a real root between `alpha and beta`
Statement-2: If f(x) is a real polynomial and `x_(1), x_(2) in R` such that `f(x_(1)) f_(x_(2)) lt 0`, then f(x) = 0 has at leat one real root between `x_(1) and x_(2)`.

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 the two statements given and establish the relationship between them.

### Step 1: Understanding the Quadratic Equations
We have two quadratic equations:
1. \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \).
2. \( -ax^2 + bx + c = 0 \) with roots \( \beta \) and \( \alpha \) (the roots are the same).

### Step 2: Define the New Quadratic Function
We define a new quadratic function:
\[ \phi(x) = \frac{a}{2} x^2 + bx + c \]

### Step 3: Evaluate \( \phi(\alpha) \)
Since \( \alpha \) is a root of \( ax^2 + bx + c = 0 \), we have:
\[ a\alpha^2 + b\alpha + c = 0 \implies b\alpha + c = -a\alpha^2 \]
Substituting this into \( \phi(\alpha) \):
\[ \phi(\alpha) = \frac{a}{2} \alpha^2 + b\alpha + c = \frac{a}{2} \alpha^2 - a\alpha^2 = -\frac{a}{2} \alpha^2 \]

### Step 4: Evaluate \( \phi(\beta) \)
Similarly, since \( \beta \) is a root of \( -ax^2 + bx + c = 0 \), we have:
\[ -a\beta^2 + b\beta + c = 0 \implies b\beta + c = a\beta^2 \]
Substituting this into \( \phi(\beta) \):
\[ \phi(\beta) = \frac{a}{2} \beta^2 + b\beta + c = \frac{a}{2} \beta^2 + a\beta^2 = \frac{3a}{2} \beta^2 \]

### Step 5: Determine the Sign of \( \phi(\alpha) \) and \( \phi(\beta) \)
From the evaluations:
- \( \phi(\alpha) = -\frac{a}{2} \alpha^2 \) (which is negative if \( a > 0 \) and \( \alpha \neq 0 \))
- \( \phi(\beta) = \frac{3a}{2} \beta^2 \) (which is positive if \( a > 0 \) and \( \beta \neq 0 \))

### Step 6: Apply the Intermediate Value Theorem
Since \( \phi(\alpha) < 0 \) and \( \phi(\beta) > 0 \), by the Intermediate Value Theorem, there exists at least one real root of \( \phi(x) = 0 \) between \( \alpha \) and \( \beta \).

### Conclusion
Thus, we conclude that:
- Statement 1 is true: \( \frac{a}{2} x^2 + bx + c = 0 \) has a real root between \( \alpha \) and \( \beta \).
- Statement 2 is also true and correctly explains Statement 1.

### Final Answer
Both statements are true, and Statement 2 is the correct explanation for Statement 1.

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