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Question #41327

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = f(x)$ be the unique continuous real-valued function defined implicitly by $y^3 - 3y - x = 0$ for $x \in [-2, 2]$ such that $f(0) = 0$ . Let $g(x) = (x^2-4) f''(x)$ . Consider the applicability of Rolle's Theorem to $g(x)$ on the interval $[-2, 2]$ and the value of the limit $L = \lim_{x \to 0} \frac{g(x)}{x}$ . Which of the following statements is correct?

Options

A.
$\text{Rolle's Theorem is applicable, } L = \frac{8}{27}$
B.
$\text{Rolle's Theorem is NOT applicable, } L = \frac{8}{27}$
C.
$\text{Rolle's Theorem is applicable, } L = -\frac{8}{27}$
D.
$\text{Rolle's Theorem is NOT applicable, } L = -\frac{8}{27}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Second Order Derivative Continuity|Implicit Differentiation Rate|Boundary Value Theorem

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to analyze the boundary behavior of implicitly defined functions and their higher-order derivatives.

The primary trap lies in blindly applying Rolle's Theorem by assuming $g(\pm 2) = 0$ because of the $(x^2-4)$ factor.

Students often calculate $f''(x)$ correctly but fail to realize that at the boundaries $x = \pm 2$ (where vertical tangents occur), $f''(x)$ has a pole of order 3, which overpowers the order 2 zero of $(x^2-4)$ .

Consequently, $g(x)$ is discontinuous at the endpoints, rendering Rolle's Theorem inapplicable.

The limit calculation further tests rigorous substitution using the implicit relation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Second Order Derivative Continuity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free