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Question #41198

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following two sets of real numbers: $A = \left\{ x \in \mathbb{R} \setminus \{0\} : \sum_{n=1}^{\infty} \frac{x}{n^2 + x^2} < \int_0^{\infty} \frac{x}{t^2 + x^2} dt \right\}$ $B = \left\{ x \in \mathbb{R} \setminus \{0\} : \int_1^{\infty} \frac{x}{t^2 + x^2} dt = \arctan(x) \right\}$ Which of the following statements is correct?

Options

A.
$A = (0, \infty)$ and $B = (0, \infty)$
B.
$A = \mathbb{R} \setminus \{0\}$ and $B = (0, \infty)$
C.
$A = (0, \infty)$ and $B = \mathbb{R} \setminus \{0\}$
D.
$A = \mathbb{R} \setminus \{0\}$ and $B = \mathbb{R} \setminus \{0\}$

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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

Infinite Series involving Inverse Tangent|Inequalities in Definite Integrals|Region Ambiguity|Conditional Identity Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle region ambiguities in improper integrals and infinite series bounds.

The primary trap lies in Set B: students commonly assume $\lim_{t \to \infty} \arctan(t/x) = \pi/2$ for all $x$ , leading them to incorrectly conclude that the identity only holds for $x > 0$ .

However, for $x < 0$ , the limit is $-\pi/2$ , and the branch cut of the arctan addition formula perfectly compensates, making the identity universally true for all non-zero reals.

Set A requires correctly applying the integral test bound, recognizing that for $x < 0$ , the series sum is strictly greater than the integral, thus restricting $A$ to positive reals.

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