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Question #40387

Difficulty: 8/10 | 🟢 Free

The Question

If $S$ is the complete solution set of the inequality $\tan^{-1}(x+1) + \tan^{-1}(x-1) \le \frac{\pi}{4}$ , then $S$ is equal to:

Options

A.
$(-\infty, \sqrt{3}-1]$
B.
$(-\infty, -\sqrt{3}-1] \cup (-\sqrt{2}, \sqrt{3}-1] \cup (\sqrt{2}, \infty)$
C.
$(-\sqrt{2}, \sqrt{3}-1]$
D.
$(-\infty, \sqrt{3}-1] \cup (\sqrt{2}, \infty)$

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

Codomain vs Range Distinction|Domain Constraint Intersections|The Sum-Difference Formula Constraints|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of domain constraints and the principal value branches of inverse trigonometric functions.

The primary trap lies in blindly applying the identity $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\left(\frac{A+B}{1-AB}\right)$ , which is only valid for $AB < 1$ .

Students who do this will solve the rational inequality $\frac{2x}{2-x^2} \le 1$ , leading to the erroneous solution set $(-\infty, -\sqrt{3}-1] \cup (-\sqrt{2}, \sqrt{3}-1] \cup (\sqrt{2}, \infty)$ .

This false path introduces extraneous intervals (like $x > \sqrt{2}$ , where the true sum is $> \pi/4$ ) and misses valid regions due to sign flips in the denominator.

The optimal bypass relies on recognizing the strict monotonicity of the function over $\mathbb{R}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Codomain vs Range Distinction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free