Question

Consider the quadratic equation \( ax^2 + bx + c = 0 \), where \( 2a + 3b + 6c = 0 \) and let \[ g(x) = \frac{a x^3}{3} + \frac{b x^2}{2} + c x \] Statement-I: The given quadratic equation \( ax^2 + bx + c = 0 \) has at least one root in \( (0, 1) \). Statement-II: Rolle's theorem is applicable to \( g(x) \) on \( [0, 1] \). Then:

• A. Statement-I is false, Statement-II is true
• B. Statement-I is true, Statement-II is false
• C. Statement-I is true, Statement-II is true but Statement-II is not a correct explanation of Statement-I
• D. Statement-I is true, Statement-II is true and Statement-II is a correct explanation of Statement-I

Hint:

When a quadratic’s derivative is related to a cubic satisfying Rolle’s theorem over an interval, it guarantees at least one real root within the open interval. Always check continuity, differentiability, and endpoint equality for applying Rolle's theorem.

Answer 1

The Correct Option is D

We are given: \[ g(x) = \frac{a x^3}{3} + \frac{b x^2}{2} + c x \] To apply Rolle's theorem on \([0,1]\), the function must be continuous, differentiable on \([0,1]\) and \(g(0) = g(1)\). Check continuity and differentiability: Since \(g(x)\) is a polynomial, it is continuous and differentiable everywhere including \([0, 1]\). Check \( g(0) \) and \( g(1) \): \[ g(0) = 0 \] \[ g(1) = \frac{a}{3} + \frac{b}{2} + c \] Given: \[ 2a + 3b + 6c = 0 \] Dividing both sides by 6: \[ \frac{a}{3} + \frac{b}{2} + c = 0 \] So: \[ g(0) = g(1) \] Therefore, Rolle's theorem is applicable on \([0, 1]\). Now, Statement-II is true. Check Statement-I: By Rolle’s theorem: If a continuous and differentiable function has equal values at the endpoints of a closed interval, then there exists at least one point in \((0,1)\) where the derivative is zero. Compute: \[ g'(x) = a x^2 + b x + c \] Then by Rolle’s theorem, there exists at least one point \( \alpha \in (0, 1) \) such that: \[ g'(\alpha) = a \alpha^2 + b \alpha + c = 0 \] Which means the quadratic equation \( a x^2 + b x + c = 0 \) has at least one real root in \( (0,1) \). Thus, Statement-I is also true. And since Statement-II (Rolle’s theorem applicable) is the reasoning for Statement-I (existence of root in (0,1)), Statement-II is the correct explanation of Statement-I. Therefore, the correct answer is option (4).