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Question #41450

Difficulty: 8/10 | 🟢 Free

The Question

Let $A_n = \left\{ k \in \mathbb{N} : 1 \le k \le 5n \text{ and } \left| \frac{2k}{n} - 3 \right| \le \frac{k}{n} + 1 \right\}$ . Find the value of: $\lim_{n \to \infty} \frac{1}{n} \sum_{k \in A_n} 2^{\frac{3k}{n}}$

Options

A.
$\frac{1364}{\ln 2}$
B.
$\frac{32767}{3 \ln 2}$
C.
$\frac{1365}{\ln 2}$
D.
$\frac{4088}{3 \ln 2}$

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

Integral as the Limit of a Sum|Selection of GP Terms|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to convert a limit of a Riemann sum into a definite integral while simultaneously handling a dynamic domain restriction.

The sum forms a Geometric Progression where the terms are filtered by a modulus inequality.

The primary trap lies in blindly applying the integration limits from the universal set ( $1 \le k \le 5n \implies 0$ to $5$ ) without resolving the modulus inequality $|2x - 3| \le x + 1$ to find the true effective domain $[2/3, 4]$ .

Failing to intersect these conditions leads to incorrect bounds and falling for the distractor options.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Integrals - Class 12
Concept	Integral as the Limit of a Sum
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free