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Question #40674

Difficulty: 8/10 | 🟢 Free

The Question

Let In=∫1n+11n(2xcos⁡−1(cos⁡πx)−πsin⁡(πx)∣sin⁡(πx)∣)dxI_n = \int_{\frac{1}{n+1}}^{\frac{1}{n}} \left( 2x \cos^{-1}\left(\cos \frac{\pi}{x}\right) - \pi \frac{\sin\left(\frac{\pi}{x}\right)}{\left| \sin\left(\frac{\pi}{x}\right) \right|} \right) dx. Find the value of 100π∑n=199In\frac{100}{\pi} \sum_{n=1}^{99} I_n.

Options

  • A.

    100

  • B.

    99

  • C.

    0

  • D.

    101

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

The Telescopic Series Trap|Integrating f(x) + xf'(x) forms|Composition Inverse Identity Traps

Why This Question Is Tricky

Common Mistakes

This question tests the integration of f(x)+xf′(x)f(x) + xf'(x) forms combined with the domain restrictions of inverse trigonometric functions.

The core trap lies in the false assumption that cos⁡−1(cos⁡θ)=θ\cos^{-1}(\cos \theta) = \theta for all θ\theta.

If a student incorrectly simplifies cos⁡−1(cos⁡πx)\cos^{-1}(\cos \frac{\pi}{x}) to πx\frac{\pi}{x}, the antiderivative becomes Ftrap(x)=x2(πx)=πxF_{trap}(x) = x^2(\frac{\pi}{x}) = \pi x.

Evaluating this flawed antiderivative yields Ftrap(1)−Ftrap(1/100)=π(1−1/100)=99π100F_{trap}(1) - F_{trap}(1/100) = \pi(1 - 1/100) = \frac{99\pi}{100}, leading directly to the distractor answer 99.

The true antiderivative x2cos⁡−1(cos⁡πx)x^2 \cos^{-1}(\cos \frac{\pi}{x}) correctly accounts for the principal value branch shifts, evaluating to exactly π\pi.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptThe Telescopic Series Trap
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree