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Question #40421

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \frac{\sqrt{1 - \sin^2 x} + \cos x}{\sin(2x)}$ . If $L_1 = \lim_{x \to \frac{\pi}{2}^-} f(x)$ and $L_2 = \lim_{x \to \frac{\pi}{2}^+} f(x)$ , then the value of $L_1^2 + L_2^2$ is:

Options

A.
1
B.
2
C.
0
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

Square Root and Modulus Logic|Right-Hand vs Left-Hand Derivatives|The L'Hôpital's Rule Misuse|Chain Rule Complexity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly handle the square root of a squared trigonometric function, specifically the transition $\sqrt{1 - \sin^2 x} = |\cos x|$ .

The trap is set for students who blindly apply L'Hôpital's rule to the $0/0$ form or incorrectly simplify the radical as simply $\cos x$ across the entire domain.

Doing so leads to a false continuous limit of 1 for both sides, yielding an incorrect sum of squares (2).

By forcing the evaluation of one-sided limits at a point where the inner function changes sign, it exposes the misuse of algebraic simplification and highlights the non-differentiable sharp corner of the modulus function.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Square Root and Modulus Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free