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Question #41602

Difficulty: 8/10 | 🟢 Free

The Question

Let $y=y(x)$ be the solution to the differential equation $\cos x \ln(\sec x + \tan x) \frac{dy}{dx} + y = \tan x \ln(\sec x + \tan x)$ for $x \in \left(0, \frac{\pi}{2}\right)$ , satisfying the initial condition $y\left(\frac{\pi}{4}\right) = \sqrt{2} - \frac{1}{\ln(\sqrt{2} + 1)}$ . Then $y\left(\frac{\pi}{3}\right)$ is equal to:

Options

A.
$2 - \frac{\sqrt{3}}{\ln(2 + \sqrt{3})}$
B.
$2 + \frac{\sqrt{3}}{\ln(2 + \sqrt{3})}$
C.
$\sqrt{3} - \frac{2}{\ln(2 + \sqrt{3})}$
D.
$1 - \frac{\sqrt{3}}{\ln(2 + \sqrt{3})}$

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

Linear Differential Equation Structure|Exact Differential Recognition|Logarithmic Integration via Trig Manipulation

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to navigate a non-standard linear differential equation that masks an exact differential.

The primary trap lies in the integration of the Integrating Factor (IF), which requires recognizing that the derivative of $\ln(\sec x + \tan x)$ is $\sec x$ , leading to a nested logarithmic IF.

Students often mismanage this integration or fail to apply integration by parts correctly on the resulting product $\sec x \tan x \ln(\sec x + \tan x)$ .

By forcing the equation into an exact differential form $d(y \cdot \text{IF}) = Q(x) \cdot \text{IF} dx$ , the problem filters out those who blindly memorize standard forms without mastering the underlying logarithmic and trigonometric manipulations required in JEE Advanced.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Differential Equations - Class 12
Concept	Linear Differential Equation Structure
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free