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Question #41863

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: I=∫090(2x−1)ln⁡∣2[x]−2x+1∣ dxI = \int_0^{90} (2x - 1) \ln | 2[x] - 2x + 1 | \, dx where [x][x] denotes the greatest integer less than or equal to xx.

Options

  • A.

    -8010

  • B.

    -8190

  • C.

    -4005

  • D.

    0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Modulus Sign Dependency|Logarithmic Integration Modulus|Integral Part and Fractional Part

Why This Question Is Tricky

Common Mistakes

This problem orchestrates a 'Lethal Collision' between greatest integer functions and modulus-dependent logarithmic integration.

The primary trap lies in the argument of the logarithm, 2[x]−2x+12[x] - 2x + 1, which simplifies to 1−2{x}1 - 2\{x\}.

Students who fail to respect the modulus sign dependency will incorrectly evaluate the improper integral ∫01ln⁡(1−2t)dt\int_0^1 \ln(1 - 2t) dt, often encountering undefined real values for t>1/2t > 1/2 and abandoning the problem.

The modulus forces a split at t=1/2t = 1/2, revealing an absolute value symmetry ∫−11ln⁡∣u∣du\int_{-1}^1 \ln|u| du.

Furthermore, the secondary trap involves the linear term (2x−1)(2x - 1), which must be expanded as (2k+2t−1)(2k + 2t - 1).

Students frequently miss the odd symmetry of (2t−1)ln⁡∣1−2t∣(2t - 1)\ln|1 - 2t|, leading to catastrophic integration by parts errors.

Finally, the summation bounds from k=0k=0 to 8989 (not 9090) test boundary precision.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Inequalities - Class 11
ConceptModulus Sign Dependency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree