Before You Begin

Question #40420

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be the solution set of the inequality $\left( \sum_{j=1}^{8} \cos((2j-1)x) \right) \sin x \ (\cos 6x - 1) \ge 0$ for $x \in (0, 2\pi)$ . Let $I$ be the subset of $A$ consisting of all isolated points (i.e., points in $A$ that do not belong to any open interval contained in $A$ ). Find the sum of all elements in $I$ .

Options

A.
$\frac{7\pi}{3}$
B.
$\frac{5\pi}{3}$
C.
$5\pi$
D.
0

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

Double and Triple Angle Ratios|Sigma Notation Properties|Even Multiplicity Roots

Why This Question Is Tricky

Common Mistakes

This question tests the 'Even Multiplicity Root' trap within the Wavy Curve method, combined with a telescoping trigonometric series.

Students often divide by $(\cos 6x - 1)$ assuming it is strictly negative, thereby missing the isolated solutions where the term equals zero.

Furthermore, they must verify which of these roots are truly isolated by checking the sign of the collapsed series $\sin 16x$ , preventing the blind summation of all roots.

This forces a 'Lethal Collision' between procedural inequality solving and conceptual rigor regarding root multiplicity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Double and Triple Angle Ratios
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free