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Question #40302

Difficulty: 8/10 | 🟢 Free

The Question

Let SS be the set of all real values of α\alpha for which the improper integral ∫−11xcos⁡x⋅∣x∣α−21+exdx\int_{-1}^{1} \frac{x \cos x \cdot |x|^{\alpha-2}}{1 + e^x} dx converges. Which of the following represents the set SS?

Options

  • A.

    (−1,∞)(-1, \infty)

  • B.

    (0,∞)(0, \infty)

  • C.

    [−1,∞)[-1, \infty)

  • D.

    (1,∞)(1, \infty)

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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

Universal Set Relativity|Even and Odd Function Symmetry|The Modulus Trap|Improper Integrals

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of improper integrals and the limitations of definite integral properties (specifically King's Property).

A common trap in JEE Advanced is to blindly apply the identity ∫−aaf(x)1+exdx=∫0af(x)1−ex1+exdx\int_{-a}^{a} \frac{f(x)}{1+e^x} dx = \int_{0}^{a} f(x) \frac{1-e^x}{1+e^x} dx for an odd function f(x)f(x).

While this identity holds for convergent integrals, applying it here artificially increases the degree of the numerator (since 1−ex∼−x1-e^x \sim -x), masking the true singularity at x=0x=0.

Students falling for this 'Modulus/Symmetry Trap' will incorrectly conclude the integral converges for α>−1\alpha > -1 because the Cauchy Principal Value exists.

However, true convergence of an improper integral over [−a,a][-a, a] requires independent convergence on [−a,0)[-a, 0) and (0,a](0, a], which reveals the correct domain α>0\alpha > 0.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSets - Class 11
ConceptUniversal Set Relativity
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree