Before You Begin

Question #41332

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \lim_{t \to 0} \left( \cos t + \sin\left(t \sqrt{x^4 - 2x^2 + 1}\right) \right)^{\csc t}$ for $x \in [-2, 2]$ . Let $N$ be the number of points in $(-2, 2)$ where $f(x)$ is non-differentiable, and $M$ be the number of points in $(-2, 2)$ where $f'(x) = 0$ . Find the value of $N + M$ .

Options

A.
1
B.
2
C.
3
D.
4

Check Your Answer

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

Rolle's Theorem Root Existence|Chain Rule Inner Function Differentiability|The 1^∞ Form Shortcut

Why This Question Is Tricky

Common Mistakes

This question tests the careful application of the $1^\infty$ limit shortcut and the chain rule involving absolute values.

The primary trap lies in the mechanical simplification of $\sqrt{x^4 - 2x^2 + 1}$ to $x^2 - 1$ without the absolute value.

Students who fall for this will incorrectly deduce that $f(x) = e^{x^2 - 1}$ , which is differentiable everywhere on $(-2, 2)$ , leading to $N=0$ and an incorrect final sum of 1.

Recognizing the absolute value reveals the hidden non-differentiable points at $x = \pm 1$ , which would also disqualify Rolle's Theorem on intervals containing them.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Rolle's Theorem Root Existence
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free