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Question #40356

Difficulty: 8/10 | 🟢 Free

The Question

Let y(x)y(x) be a continuous function on R\mathbb{R} satisfying the differential equation (x2+1)dydx+2xsgn(x)y=2x(x^2+1) \frac{dy}{dx} + 2x \text{sgn}(x) y = 2x for all x≠0x \neq 0, with the boundary condition y(−1)=−1y(-1) = -1. If A=lim⁡x→∞y(x)A = \lim_{x \to \infty} y(x), then the value of 1π∫0∞(A−y(x))dx\frac{1}{\pi} \int_0^\infty (A - y(x)) dx is equal to

Options

  • A.

    0

  • B.

    12\frac{1}{2}

  • C.

    1

  • D.

    2

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Rational Function Asymptotes|Integrating Factor (IF) Sign Sensitivity|Signum Function Discontinuity

Why This Question Is Tricky

Common Mistakes

This question tests the integration of first-order linear ODEs with piecewise continuous coefficients (specifically involving the signum function).

The primary trap lies in the jump discontinuity of the integrating factor's form across x=0x=0.

Students frequently integrate both sides independently and either ignore the constant of integration for x>0x>0 or incorrectly assume the same constant applies globally.

By forcing the evaluation of the piecewise constant through the continuity of y(x)y(x) at x=0x=0, the problem shifts the horizontal asymptote profile.

Failing to match y(0−)=y(0+)y(0^-) = y(0^+) leads to an incorrect numerator in the rational function, directly altering the area computed in the final improper integral.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 11
ConceptRational Function Asymptotes
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree