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Question #66025

Difficulty: 8/10 | 🟢 Free

The Question

Let $x > 0$ be a real number and $y = x + \frac{2}{x}$ . Consider the infinite series: $S(x) = \sum_{n=1}^{\infty} \tan^{-1}\left( \frac{y}{1 + (x - (n-1)y)(x - ny)} \right)$ Which of the following expressions is equal to $S(x)$ ?

Options

A.
$\tan^{-1}(x) + \frac{\pi}{2}$
B.
$\tan^{-1}(x) - \frac{\pi}{2}$
C.
\cot^{-1}(x)
D.
$-\tan^{-1}(x)$

Check Your Answer

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|The Telescopic Series Trap|Extraneous Solution Rejection|Domain Restrictions

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'Telescopic Series Trap' combined with 'Domain Restrictions' of inverse trigonometric functions.

Students commonly assume $\tan^{-1}\left(\frac{A-B}{1+AB}\right)$ universally expands to $\tan^{-1}(A) - \tan^{-1}(B)$ , leading to the extraneous solution $\tan^{-1}(x) + \frac{\pi}{2}$ .

However, the principal value branch requires $AB > -1$ .

By defining the sequence such that the product of the first two terms is strictly less than $-1$ ( $a_0 a_1 = -2$ ), a hidden $-\pi$ phase shift is introduced at the very first term.

Recognizing this boundary condition violation is a hallmark of JEE Advanced mathematics.

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