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Question #41606

Difficulty: 8/10 | 🟢 Free

The Question

Let y=f(x)y = f(x) be the solution to the initial value problem dydx+1xln⁡xy=1x2∣ln⁡x∣\frac{dy}{dx} + \frac{1}{x \ln x} y = \frac{1}{x^2 |\ln x|} with f(e)=1−1ef(e) = 1 - \frac{1}{e}, defined on its maximal interval of existence. Let SS be the set of all possible values of f(1/e)f(1/e). Find SS.

Options

  • A.

    { e - 1 }

  • B.

    { 1 - e }

  • C.

    { 0 }

  • D.

    \emptyset

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

Integrating Factor (IF) Sign Sensitivity|Variable Separable Constraints|Extraneous Solution Rejection

Why This Question Is Tricky

Common Mistakes

This question tests Integrating Factor (IF) sign sensitivity and Extraneous Solution Rejection.

StudentsoftenblindlycomputetheIFasln⁡xwithouttheabsolutevalue,ignoringthedomainrestrictionx>1.Students often blindly compute the IF as \ln x without the absolute value, ignoring the domain restriction x > 1.

By forcing a variable separable transformation via the IF method, the DE reveals a singularity at x=1.

Whilethesolutionforx>1hasaremovablesingularityatx=1(limitis1),thevalidsolutionforx<1(whereIF=−ln⁡x)hasalimitof−1.While the solution for x > 1 has a removable singularity at x=1 (limit is 1), the valid solution for x < 1 (where IF = -\ln x) has a limit of -1.

This jump discontinuity means the initial value problem cannot be extended across x=1.

Students who ignore the IF sign will incorrectly match the branches and obtain an extraneous value for f(1/e).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptIntegrating Factor (IF) Sign Sensitivity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree