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Statement-1: If a, b, c, A, B, C are rea...

Statement-1: If a, b, c, A, B, C are real numbers such that `a lt b lt c`, then `f(x) = (x-a)(x-b)(x-c) -A^(2)(x-a)-B^(2)(x-b)-C^(2)(x-c)` has exactly one real root.
Statement-2: If f(x) is a real polynomical and `x_(1), x_(2) in R` such that `f(x_(1)) f(x_(2)) lt 0`, then f(x) has at least one real root between `x_(1) and x_(2)`

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.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze both statements provided in the question. ### Step 1: Analyze Statement 2 **Statement 2:** If \( f(x) \) is a real polynomial and \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) f(x_2) < 0 \), then \( f(x) \) has at least one real root between \( x_1 \) and \( x_2 \). **Solution:** 1. Since \( f(x) \) is a polynomial, it is continuous. 2. If \( f(x_1) f(x_2) < 0 \), it means one of the values is positive and the other is negative. 3. By the Intermediate Value Theorem, since \( f(x) \) is continuous, there must be at least one point \( c \) in the interval \( (x_1, x_2) \) where \( f(c) = 0 \). 4. Therefore, Statement 2 is **true**. ### Step 2: Analyze Statement 1 **Statement 1:** If \( a < b < c \), then \( f(x) = (x-a)(x-b)(x-c) - A^2(x-a) - B^2(x-b) - C^2(x-c) \) has exactly one real root. **Solution:** 1. First, observe the function \( f(x) \): \[ f(x) = (x-a)(x-b)(x-c) - A^2(x-a) - B^2(x-b) - C^2(x-c) \] 2. Evaluate \( f(a) \): \[ f(a) = (a-a)(a-b)(a-c) - A^2(a-a) - B^2(a-b) - C^2(a-c) = 0 - 0 - B^2(a-b) - C^2(a-c) \] Since \( a < b < c \), \( f(a) \) is positive. 3. Evaluate \( f(c) \): \[ f(c) = (c-a)(c-b)(c-c) - A^2(c-a) - B^2(c-b) - C^2(c-c) = 0 - A^2(c-a) - B^2(c-b) \] Since \( a < b < c \), \( f(c) \) is negative. 4. Since \( f(a) > 0 \) and \( f(c) < 0 \), by the Intermediate Value Theorem, there is at least one root in the interval \( (a, c) \). 5. Next, we need to check if there are more than one root. The polynomial \( (x-a)(x-b)(x-c) \) is a cubic polynomial, which can have up to three roots. The additional terms \( -A^2(x-a) - B^2(x-b) - C^2(x-c) \) can modify the behavior of the function, but they do not guarantee that there will only be one root. 6. Thus, it is possible for \( f(x) \) to have more than one real root, contradicting the claim of Statement 1. Therefore, Statement 1 is **false**. ### Conclusion - Statement 1 is **false**. - Statement 2 is **true**. Thus, the correct answer is **Option D**: Statement 1 is false, and Statement 2 is true.
To solve the problem, we need to analyze both statements provided in the question. ### Step 1: Analyze Statement 2 **Statement 2:** If \( f(x) \) is a real polynomial and \( x_1, x_2 \in \mathbb{R} \) such that \( f(x_1) f(x_2) < 0 \), then \( f(x) \) has at least one real root between \( x_1 \) and \( x_2 \). **Solution:** 1. Since \( f(x) \) is a polynomial, it is continuous. 2. If \( f(x_1) f(x_2) < 0 \), it means one of the values is positive and the other is negative. ...
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