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Question #40317

Difficulty: 8/10 | 🟢 Free

The Question

Let A={z∈C:Re(z)≥1}A = \{ z \in \mathbb{C} : \text{Re}(z) \ge 1 \}, B={z∈C:log⁡∣z∣/2(∣z−3∣)≤1}B = \left\{ z \in \mathbb{C} : \log_{|z|/2} (|z - 3|) \le 1 \right\}, and C={z∈C:∣z−4∣≤2}C = \{ z \in \mathbb{C} : |z - 4| \le 2 \}. Let Ω=(B∪C)∩(A∪C)∩(B∪Ac)\Omega = (B \cup C) \cap (A \cup C) \cap (B \cup A^c). If the area of the region Ω\Omega in the complex plane is Δ\Delta, then the value of Δ\Delta is:

Options

  • A.

    4π4\pi

  • B.

    16π3−3\frac{16\pi}{3} - \sqrt{3}

  • C.

    8π8\pi

  • D.

    10π3+3\frac{10\pi}{3} + \sqrt{3}

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

Set Identity Laws|Geometry Constraint Modeling|Geometry of Complex Inequalities|Symmetry Utilization

Why This Question Is Tricky

Common Mistakes

The question tests the rigorous application of Set Identities alongside complex logarithmic inequalities.

The primary trap is 'Symmetry Utilization': students often assume the logarithmic inequality simply resolves to the symmetric disk C, ignoring the fractional base case (0 < |z|/2 < 1).

This omission hides a non-symmetric 'hole' (the segment D intersected with A) revealed only through careful Set Identity expansion.

Students who rely on heuristic geometric symmetry will incorrectly simplify the region to just C, yielding 4π, while the true region breaks this symmetry and requires calculating the area of the hidden circular segment.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSets - Class 11
ConceptSet Identity Laws
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree