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Question #40512

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all complex numbers $z \neq 0$ satisfying the inequality $\frac{1}{|z|^2 - 6|z| + 8} \le \frac{1}{|z|^2 + \frac{9}{|z|^2} + 2}$ If the area of the region represented by $S$ in the complex plane is $A\pi$ , find the value of $A$ .

Options

A.
8
B.
10
C.
12
D.
16

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

Geometry of Complex Inequalities|Arithmetic Mean Inequality|Sign Flipping Rule

Why This Question Is Tricky

Common Mistakes

This question tests the interaction between AM-GM and complex moduli, specifically targeting the 'Sign Flipping Rule' trap.

Students mechanically cross-multiply or take reciprocals of the inequality $1/A \le 1/B$ to get $B \le A$ , implicitly assuming $A > 0$ .

By doing so, they arrive at $6|z|^3 - 6|z|^2 + 9 \le 0$ , which has no real positive roots, leading them to incorrectly conclude the solution set is empty.

The trap punishes mechanical manipulation; geometric and algebraic rigor requires checking the sign of the denominator.

Since the RHS denominator is strictly positive (via AM-GM), the inequality is naturally satisfied whenever the LHS denominator is negative, revealing the true bounded annular region.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Complex Numbers and Quadratic Equations - Class 11
Concept	Geometry of Complex Inequalities
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free