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Question #40766

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the region in the complex plane consisting of all points $z$ that satisfy $|z - 1| \le 1$ and $\frac{\pi}{4} \le \arg(z - 1) \le \frac{3\pi}{4}$ . Let $L$ be the line containing the ray $\arg(z - 2) = \frac{2\pi}{3}$ . If $S'$ is the image of $S$ under reflection across the line $L$ , find the area of the region consisting of all points $w \in S'$ that satisfy $\text{Re}(w) \ge 2$ .

Options

A.
$\frac{\pi + 6}{24}$
B.
$\frac{\pi + 3}{12}$
C.
$\frac{\pi + 8}{24}$
D.
$\frac{2\pi + 3}{24}$

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

Reflection Principle|Locus Problems via Transformation|Geometry of Complex Inequalities

Why This Question Is Tricky

Common Mistakes

This question tests the Reflection Principle and the Geometry of Complex Inequalities.

The primary trap lies in the transformation from the Cartesian to the Argand plane when the line of reflection is defined by a complex argument constant.

Students often fail to correctly apply the reflection formula $w = z_0 + e^{i2\theta}(\bar{z} - \bar{z}_0)$ and subsequently miscalculate the boundary conditions of the inequalities post-transformation.

Specifically, they ignore the domain restriction imposed by the argument constant on the image point, failing to realize that the conjugation reverses the argument bounds and shifts the center.

Furthermore, intersecting the transformed polar region with a Cartesian inequality ( $\text{Re}(w) \ge 2$ ) requires carefully splitting the area into a circular sector and a triangle, penalizing those who assume the entire region remains a simple sector.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Reflection Principle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free