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Question #41109

Difficulty: 8/10 | 🟢 Free

The Question

Define a relation $\sim$ on $\mathbb{R}$ by $\alpha \sim \beta \iff \sin \alpha = \sin \beta$ . For $x > 1$ , consider the function $S(x) = \arcsin\left(\frac{2x}{1+x^2}\right) + \arccos\left(\frac{1-x^2}{1+x^2}\right) + \arctan\left(\frac{2x}{1-x^2}\right)$ Let $P(x)$ be the unique value in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ such that $P(x) \sim S(x)$ . Which of the following expressions is identically equal to $P(x)$ for all $x > 1$ ?

Options

A.
$2\arctan x - \pi$
B.
$\pi - 2\arctan x$
C.
$2\arctan x$
D.
$\frac{\pi}{2} - \arctan x$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Transitivity of Identity|The Sum-Difference Formula Constraints|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to navigate the 'Lethal Collision' between general trigonometric relations and Principal Value Branches.

Students commonly make the mistake of blindly applying identities like $\arcsin(\frac{2x}{1+x^2}) = 2\arctan x$ without checking the domain constraints ( $x > 1$ forces the output into a different branch).

Furthermore, the equivalence relation $\alpha \sim \beta$ acts as a trap: even if a student correctly calculates $S(x) = 2\arctan x$ , they must recognize that $2\arctan x \notin [-\pi/2, \pi/2]$ for $x > 1$ , requiring a final modular shift to $\pi - 2\arctan x$ to satisfy the principal value condition.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Transitivity of Identity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free