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Question #40942

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = (\cos x + i \sin x) \left( \frac{\sqrt{1+\sin 2x} + i\sqrt{1-\sin 2x}}{\sqrt{1+\sin 2x} - i\sqrt{1-\sin 2x}} \right)^{\sin x}$ for $x \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$ , where $i = \sqrt{-1}$ . Evaluate the limit: $\lim_{h \to 0} \frac{\left| f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right) \right|}{|h|}$

Options

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

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

First Principle of Derivative|Euler's Form and Rotations|De Moivre’s Theorem Applications

Why This Question Is Tricky

Common Mistakes

This question tests the integration of De Moivre's Theorem within a limit-based derivative definition, specifically targeting three classic JEE Advanced traps.

First, students often mistake the limit of the modulus difference for the difference of moduli, incorrectly evaluating the limit as 0 since |f(x)| = 1.

Second, it tests the branch cut and absolute value trap of the radical trigonometric identity; students who blindly write the square root of (1 - sin 2x) as (cos x - sin x) instead of |cos x - sin x| will obtain a phase shift of (pi/2 - 2x), leading to an incorrect derivative modulus of 1.

Finally, the question requires recognizing that the limit definition exactly represents the modulus of the complex derivative |f'(x)|, bridging algebraic complex manipulation with differential calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	First Principle of Derivative
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free