Question

If \(2a + 3b + 6c = 0\), then at least one root of the equation \(ax^2 + bx + c = 0\) lies in the interval

• A. (0, 1)
• B. (1, 2)
• C. (2, 3)
• D. None of these

Hint:

When coefficients satisfy a linear relation, constructing a function and applying Rolle's theorem is a standard technique to find intervals containing roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Consider the function \(f(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx\). If we can find two points where \(f(x)\) takes the same value, then by Rolle's theorem, there exists a point where \(f'(x) = 0\) which is exactly the given quadratic.

Step 2: Detailed Explanation:
Let \(f(x) = \frac{a}{3}x^3 + \frac{b}{2}x^2 + cx\). Then \(f(0) = 0\). Also, \(f(1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2a + 3b + 6c}{6} = \frac{0}{6} = 0\). Thus, \(f(0) = f(1) = 0\). By Rolle's theorem, there exists at least one \(c \in (0,1)\) such that \(f'(c) = 0\). But \(f'(x) = ax^2 + bx + c\). Hence, the quadratic has a root in (0,1).

Step 3: Final Answer:
The root lies in the interval (0, 1), which corresponds to option (A).