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Question #40515

Difficulty: 8/10 | 🟢 Free

The Question

Find the complete set of real values of $x$ that satisfy the inequality: $\frac{\cos^{-1}(|x - 1| + |x + 1| - 3)}{\sin^{-1}(|x - 1| + |x + 1| - 3)} < 1$

Options

A.
$\left(-\frac{3}{2}, \frac{3}{2}\right) \cup \left[-2, -\frac{3}{2} - \frac{1}{2\sqrt{2}}\right) \cup \left(\frac{3}{2} + \frac{1}{2\sqrt{2}}, 2\right]$
B.
$\left[-2, -\frac{3}{2} - \frac{1}{2\sqrt{2}}\right) \cup \left(\frac{3}{2} + \frac{1}{2\sqrt{2}}, 2\right]$
C.
$\left(-\frac{3}{2}, \frac{3}{2}\right) \cup \left(-\infty, -\frac{3}{2} - \frac{1}{2\sqrt{2}}\right) \cup \left(\frac{3}{2} + \frac{1}{2\sqrt{2}}, \infty\right)$
D.
$\left[-2, 2\right] \setminus \left\{-\frac{3}{2}, \frac{3}{2}\right\}$

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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

Sign Flipping Rule|Modulus Inequality Logic|Triangle Inequality Mind Games|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

The core trap is the 'Sign Flipping Rule' combined with the 'Principal Value Branch Boundaries'.

Students often blindly cross-multiply the denominator $\sin^{-1}y$ , forgetting that its principal value branch includes negative values for $y < 0$ .

This causes them to miss the entire inner interval $\left(-\frac{3}{2}, \frac{3}{2}\right)$ where the sign flips, making the inequality universally true for that domain.

Additionally, the 'Triangle Inequality' acts as a psychological distractor; students might waste time solving the lower bound of the domain algebraically instead of recognizing $|x-1|+|x+1| \ge 2$ , which naturally satisfies the $-1$ lower bound of the inverse trigonometric domain.

Quick Facts About This Question

Question Information

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