Problem 110 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Consider a quadratic equation \(a x^{2}+b x+c=0\), where \(2 a+3 b+6 c=0\) and let \(g(x)=a \frac{x^{3}}{3}+b \frac{x^{2}}{2}+c x\) |Online May 19, 2012] Statement 1: The quadratic equation has at least one root in the interval \((0,1)\). Statement 2: The Rolle's theorem is applicable to function \(g(x)\) on the interval \([0,1]\). (a) Statement 1 is false, Statement 2 is true. (b) Statement 1 is true, Statement 2 is false. (c) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1 . (d) Statement 1 is true, Statement 2 is true,, Statement 2 is a correct explanation for Statement 1 .

Step-by-Step Solution

Answer

Option (c): Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

Step 1: Understand Statement 1

Statement 1 claims that the quadratic equation \(ax^2 + bx + c = 0\) has at least one root in the interval \((0,1)\). To verify this, note the given condition \(2a + 3b + 6c = 0\) plays a crucial role. Let's express the quadratic as \(f(x) = ax^2 + bx + c\). Since there's a special condition \(2a + 3b + 6c = 0\), it will affect the specific roots of the equation, possibly indicating a root in the desired interval.

Step 2: Applying Rolle's Theorem

Statement 2 says Rolle's theorem applies to \(g(x) = a \frac{x^3}{3} + b \frac{x^2}{2} + c x\) over \([0,1]\). For Rolle's theorem to be applicable, \(g(x)\) must be continuous and differentiable on \([0,1]\) with \(g(0) = g(1)\). First, compute \(g(0)\) and \(g(1)\): \(g(0) = 0\) from each term vanishing at \(x = 0\), and \(g(1) = \frac{a}{3} + \frac{b}{2} + c \). If \(g(1) = 0\), then Rolle's theorem is applicable.

Step 3: Verifying Equal Values for Rolle's Theorem

Calculate \(g(1)\) using \(g(1) = \frac{a}{3} + \frac{b}{2} + c \). By rearranging to match Rolle's requirement, substitute \(a\), \(b\), and \(c\) in terms of one another using the given condition \(2a + 3b + 6c = 0\). Solving, we find \(\frac{a}{3} + \frac{b}{2} + c = 0\) indeed follows from the condition, confirming Rolle's theorem is applicable, so Statement 2 is accurate.

Step 4: Determine the Relationship between Statements

Rolle's theorem guarantees at least one point \(x\) in \((0,1)\) such that \(g'(x) = 0\); however, proving \(g'(x)\) is zero doesn't necessarily show \(f(x) = 0\) has a root in \((0,1)\). Consequently, Rolle's theorem indicates a critical point of \(g(x)\) rather than directly confirming \(f(x)\)'s root within the interval \((0,1)\).

Step 5: Final Step: Conclusion

Rolle's theorem applies, making Statement 2 true as it verifies \(g(x)\) reaching equal values at boundaries. Statement 1 remains unsupported directly by Rolle's theorem, because it shows a specific behavior of \(g(x)\), many not translating to the roots of \(f(x)\). Thus, Statement 2, although true, doesn't correctly explain Statement 1's truth value.

Key Concepts

Rolle's Theorem

Rolle's Theorem is a key result in calculus that connects continuity, differentiability, and the behavior of function roots. It states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and if \(f(a) = f(b)\), then there exists at least one point \(c\) in the interval \((a, b)\) such that the derivative \(f'(c) = 0\).
This theorem is quite useful because it ensures that somewhere between \(a\) and \(b\), there's a point where the slope of the tangent to the curve is horizontal (i.e., a stationary point).

• Continuity ensures there are no breaks or jumps in the function over \([a, b]\).
• Differentiability means the function has a defined slope at every point in \((a, b)\).
• The condition \(f(a) = f(b)\) suggests the function returns to its initial value at \(b\).

This theorem can help predict critical points of polynomials, where changes in increasing or decreasing behavior occur, but it doesn't guarantee roots of the function itself. In the exercise given, even if Rolle's theorem applies, it does not imply that the quadratic portion \(f(x)=ax^2+bx+c\) has its root in the interval \((0,1)\).
Instead, it only affirms the existence of some critical point for the function \(g(x)\).

Roots of a Polynomial

The roots of a polynomial equation are values of \(x\) that make the equation equal to zero. For a quadratic equation like \(ax^2 + bx + c = 0\), the roots can be found using various methods:

• Factoring: Expressing the quadratic as a product of two binomials, if possible.
• Completing the Square: Rewriting the quadratic in the form \((x-h)^2 = k\).
• Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), a general solution method for any quadratic.

The discriminant \(b^2 - 4ac\) helps determine the nature of roots:

• Positive discriminant: Two distinct real roots.
• Zero discriminant: Exactly one real root (a repeated or double root).
• Negative discriminant: Two complex roots.

In the exercise, knowing Room's theorem doesn't directly assure these roots, since it connects more to the critical points of \(g(x)\). The roots of the quadratic \(f(x)\) may not coincide with those critical points, illustrating a subtle yet essential distinction from just applying the theorem
on the polynomial itself.

Continuity and Differentiability

Continuity and differentiability are fundamental concepts in calculus that often go hand-in-hand but are not synonymous.
Continuity of a function at a point means that as \(x\) approaches a specific value, the values of the function approach a single finite limit, and the function is defined at that point.

• A function is continuous at a point \(x = c\) if \(\lim_{x \to c} f(x) = f(c)\).
• On an interval, a function is continuous if it is continuous at every point within the interval.

Differentiability, on the other hand, means the function has a derivative at every point in its domain:

• If a function is differentiable at \(x = c\), it means a tangent at that point exists with a well-defined slope \(f'(c)\).
• Differentiability implies continuity, but not the other way around.

In the problem context, applying Rolle's theorem requires both conditions: continuity of \(g(x)\) over \([0, 1]\) and differentiability over \((0, 1)\). These conditions ensure there's no break in the function or irregular slope changes, which is essential for finding a place where a tangent slope can be zero.
Continuity lets \(g(x)\) flow smoothly between values; differentiability lets us find the critical \(x\) where the stationary point exists, crucial for applying the theorem correctly.