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Question #41884

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \begin{cases} \frac{1 - \cos(2x)}{|x|} & \text{for } x \neq 0 \\ 0 & \text{for } x = 0 \end{cases}$ . Let $L$ and $R$ be the right-hand and left-hand derivatives of $f(x)$ at $x=0$ , respectively. Let $N$ be the number of points $c \in (-\pi, \pi)$ such that $f'(c) = 0$ . Find the value of $L - R + N$ .

Options

A.
4
B.
5
C.
6
D.
8

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

Trigonometric Standard Limits|Mean Value Theorem Applicability|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the applicability of Rolle's Theorem and the Mean Value Theorem by introducing a modulus function that creates a non-differentiable 'kink' at the origin.

Students must use standard trigonometric limits to correctly evaluate the left-hand and right-hand derivatives, avoiding the common trap of missing the sign change caused by $|x|$ for $x < 0$ .

Furthermore, the question lures students into analyzing $f'(c)=0$ (the conclusion of Rolle's Theorem), requiring careful derivative analysis and application of the Intermediate Value Theorem to count the roots, thereby testing both the prerequisites and the mechanical outcomes of MVT.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	Trigonometric Standard Limits
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free