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Question #41522

Difficulty: 8/10 | 🟢 Free

The Question

Let Ar=(1rn(rn)2(rn)21rnrn(rn)21)A_r = \begin{pmatrix} 1 & \frac{r}{n} & \left(\frac{r}{n}\right)^2 \\ \left(\frac{r}{n}\right)^2 & 1 & \frac{r}{n} \\ \frac{r}{n} & \left(\frac{r}{n}\right)^2 & 1 \end{pmatrix}. Evaluate the limit: L=lim⁡n→∞1n∑r=1n−1ln⁡(det⁡(Ar))L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n-1} \ln(\det(A_r))

Options

  • A.

    3ln⁡3−6+π33 \ln 3 - 6 + \frac{\pi}{\sqrt{3}}

  • B.

    3ln⁡3−3+π233 \ln 3 - 3 + \frac{\pi}{2\sqrt{3}}

  • C.

    2ln⁡3−6+π32 \ln 3 - 6 + \frac{\pi}{\sqrt{3}}

  • D.

    3ln⁡3−6−π33 \ln 3 - 6 - \frac{\pi}{\sqrt{3}}

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

Integral as the Limit of a Sum|Factor Theorem in Determinants|Singular vs. Non-Singular Threshold

Why This Question Is Tricky

Common Mistakes

This question tests the 'Singular vs Non-Singular Threshold' trap by embedding an improper integral within a Riemann sum limit.

Students must first use the Factor Theorem to evaluate the circulant determinant, yielding (1−x3)2(1-x^3)^2.

The critical trap lies in the conversion to the integral ∫01ln⁡(1−x3)dx\int_0^1 \ln(1-x^3) dx, which has a singularity at x=1x=1.

If students blindly apply the Fundamental Theorem of Calculus without taking the proper left-hand limit as x→1−x \to 1^-, they will encounter an undefined ln⁡(0)\ln(0) and likely abandon the correct path or make algebraic errors.

The sum explicitly terminates at n−1n-1 to maintain non-singularity in the discrete phase, punishing those who do not rigorously handle boundary limits in improper integrals.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptIntegral as the Limit of a Sum
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree