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Question #40448

Difficulty: 8/10 | 🟢 Free

The Question

Find the solution set of the inequality $\frac{\cos^2 x (2\sin x - 1)}{\tan x - \sqrt{3}} \ge 0$ for $x \in [0, 2\pi]$ .

Options

A.
$\left[0, \frac{\pi}{6}\right] \cup \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \cup \left[\frac{5\pi}{6}, \frac{4\pi}{3}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right]$
B.
$\left[0, \frac{\pi}{6}\right] \cup \left(\frac{\pi}{3}, \frac{\pi}{2}\right] \cup \left[\frac{5\pi}{6}, \frac{4\pi}{3}\right) \cup \left[\frac{3\pi}{2}, 2\pi\right]$
C.
$\left[0, \frac{\pi}{6}\right] \cup \left(\frac{\pi}{3}, \frac{5\pi}{6}\right] \cup \left(\frac{4\pi}{3}, 2\pi\right]$
D.
$\left[\frac{\pi}{6}, \frac{\pi}{3}\right) \cup \left(\frac{\pi}{2}, \frac{5\pi}{6}\right] \cup \left(\frac{4\pi}{3}, \frac{3\pi}{2}\right)$

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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

Extraneous Solution Rejection|Even Multiplicity Roots|System of Inequalities

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle extraneous solutions and domain constraints in trigonometric inequalities.

A common trap is mechanically applying the wavy curve method and including the roots of the even-multiplicity term $\cos^2 x = 0$ ( $x = \pi/2, 3\pi/2$ ).

However, these points are asymptotes of $\tan x$ , making the expression undefined.

Another critical trap is failing to recognize that the sign of $\tan x - \sqrt{3}$ changes at its asymptotes, which must be treated as critical points.

Ignoring the asymptotes leads to incorrect sign intervals (Option C), while forgetting to exclude the domain violations leads to Option B.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Extraneous Solution Rejection
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free