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Question #40537

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all real values of $x$ satisfying the equation $\tan^{-1}\left(\frac{x+1}{x-1}\right) + \tan^{-1}\left(\frac{x-1}{x}\right) = \tan^{-1}(7) - \pi$ Find the sum of the squares of all elements in $S$ .

Options

A.
$\frac{15 - 3\sqrt{21}}{2}$
B.
$15$
C.
$\frac{15 + 3\sqrt{21}}{2}$
D.
$9$

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

Denominator Restriction|The Sum-Difference Formula Constraints|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to navigate the domain restrictions and branch cuts of inverse trigonometric sum formulas.

The common trap is blindly applying the tangent function to both sides of the equation $\tan^{-1}(A) + \tan^{-1}(B) = k$ , which algebraically yields a quadratic with two roots.

However, this algebraic step ignores the piecewise nature of the identity.

By forcing students to evaluate the sum of squares of the valid solutions, they cannot simply guess.

They must explicitly test both roots against the original equation.

The negative root produces positive arguments for the inverse tangents, making the LHS positive while the RHS is strictly negative, acting as a rigorous secondary filter that eliminates the extraneous solution.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Linear Inequalities - Class 11
Concept	Denominator Restriction
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free