Before You Begin

Question #66042

Difficulty: 8/10 | 🟢 Free

The Question

Let $\alpha$ be the minimum integer value of $x$ for which the infinite geometric progression with common ratio $r(x) = \frac{2x+1}{x-2}$ converges.

Let $S = \sum_{n=1}^{\infty} \arctan\left( \frac{ |\alpha| \cdot r^{n-1} (1-r) }{ 1 + \alpha^2 r^{2n-1} } \right)$ , where $r = r(\alpha)$ .

Find the value of $\sec^2(S) + \alpha^2$ .

Options

A.
5
B.
8
C.
9
D.
13

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

Infinite Series involving Inverse Tangent|Infinite GP Convergence|Domain Constraint Intersections|Cross Multiplication Trap

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the 'Cross Multiplication Trap' in rational inequalities and the evaluation of telescoping inverse trigonometric series.

When solving the convergence condition $-1 < \frac{2x+1}{x-2} < 1$ , students frequently multiply by $(x-2)$ assuming it is positive.

This illegal cross-multiplication leads to an empty set ( $x < -3$ and $x > 1/3$ ), causing them to lose the valid negative domain and fail to find the minimum integer.

A rigorous approach requires the wavy curve method.

Furthermore, the problem requires recognizing that the argument of the arctan function perfectly factors into the difference formula $\frac{x-y}{1+xy}$ , creating a telescoping sum that converges to a clean trigonometric value.

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	170 sec
Access	Free