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Question #41195

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \frac{x}{2x^2+2x+1}$ for $x > 0$ . Let $\alpha$ be the unique point of inflection of the curve $y = f(x)$ . Evaluate the limit: $L = \lim_{n \to \infty} \sum_{k=1}^{\lfloor \alpha n \rfloor} \arctan\left( \frac{k}{n^2 + 2nk + 2k^2} \right)$

Options

A.
$\frac{1}{2} \ln(\sqrt{3}+1) - \frac{\pi}{12}$
B.
$\frac{1}{2} \ln\left(\frac{4}{3}\right)$
C.
$\frac{1}{2} \ln(\sqrt{3}+1) - \frac{5\pi}{24}$
D.
$\frac{1}{2} \ln(\sqrt{3}+1) - \frac{\pi}{24}$

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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

Infinite Series involving Inverse Tangent|Integral as the Limit of a Sum|Definition of Limit|Point of Inflection Conditions

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous definition of the definite integral as the limit of a Riemann sum, combined with differential calculus to identify integration bounds.

The primary trap lies in the algebraic manipulation of the general term: students often rewrite $\frac{k}{n^2+2nk+2k^2}$ as $\frac{1}{n+k} \frac{k/(n+k)}{1+(k/(n+k))^2}$ and incorrectly assume $dx = \frac{1}{n+k}$ with $x = \frac{k}{n+k}$ .

This false substitution leads to the integral $\int_0^{1/\sqrt{3}} \frac{x}{1+x^2} dx = \frac{1}{2}\ln(4/3)$ , perfectly masking the divergence of the non-uniform partition.

The correct approach requires factoring $1/n$ to form a uniform partition $dx = 1/n$ , revealing the true integrand $f(x) = \frac{x}{2x^2+2x+1}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Infinite Series involving Inverse Tangent
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free