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Question #41176

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: [-1, 1] \to \mathbb{R}$ be defined as $f(x) = \sin^{-1} \left( \frac{2g(x)}{1 + (g(x))^2} \right)$ where $g(x) = (x^2 - 1)\text{sgn}(x)$ and $\text{sgn}(\cdot)$ denotes the signum function. Let $S$ be the set of all points $c \in (-1, 1)$ such that $f'(c) = 0$ . The number of elements in $S$ is:

Options

A.
0
B.
1
C.
2
D.
3

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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

Derivative of Implicit Inverse Forms|Rolle's Theorem Root Existence|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the strict continuity and differentiability prerequisites of Rolle's Theorem, specifically using a hidden jump discontinuity.

Students often observe $f(-1) = f(1) = 0$ and blindly apply Rolle's Theorem to conclude that $f'(c) = 0$ must have at least one solution in $(-1, 1)$ .

Furthermore, formal differentiation ignoring the signum function's domain yields a critical point at $x=0$ , reinforcing the false assumption that $c=0$ is a valid root.

The trap lies in the signum function nested within the inverse trigonometric identity, which creates a non-removable discontinuity at $x=0$ .

Because the function is neither continuous nor differentiable at $x=0$ , Rolle's Theorem is invalidated, and the actual number of points where $f'(c)=0$ is zero.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Derivative of Implicit Inverse Forms
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free