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Question #41334

Difficulty: 8/10 | 🟢 Free

The Question

Let $F(x) = \int_1^x \frac{[t]}{t^2} dt$ , where $[ \cdot ]$ denotes the greatest integer function. Let $L_1 = \lim_{x \to 2^+} \frac{F(x) - F(2)}{x - 2}$ and $L_2 = \lim_{x \to 2^-} \frac{F(x) - F(2)}{x - 2}$ . If $L = \frac{L_1}{L_2}$ , consider the sequence $I_n = \int_1^n \frac{[t]}{t^L} dt$ for integers $n \ge 2$ . Which of the following inequalities strictly holds true for all $n \ge 2$ ?

Options

A.
$\ln(n) - 1 < I_n < \ln(n)$
B.
$\ln(n) < I_n < \ln(n) + 1$
C.
$n - 1 - \ln(n) < I_n \le n - 1$
D.
$I_n = \ln(n) - \frac{1}{n}$

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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

Step Function Discontinuities|Inequalities in Definite Integrals|L'Hopital's Pitfall|Right-Hand vs Left-Hand Derivatives

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of the Fundamental Theorem of Calculus when applied to integrands with jump discontinuities, specifically step functions.

The 'Lethal Collision' occurs if a student blindly applies L'Hopital's Rule to evaluate the limits $L_1$ and $L_2$ .

Because the derivative of an integral with a discontinuous integrand is undefined at the jump, assuming $F'(2) = [2]/2^2 = 1/2$ leads to the incorrect conclusion that both left and right limits equal $1/2$ , yielding $L=1$ .

This perfectly aligns with the distractor inequality $n - 1 - \ln(n) < I_n \le n - 1$ .

Correctly partitioning the integral reveals the left-hand derivative is $1/4$ , giving $L=2$ , which transforms the sequence into a Riemann sum bounding problem for the harmonic series.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Continuity and Differentiability - Class 12
Concept	Step Function Discontinuities
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free