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Question #40453

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all complex numbers $c$ such that the function $f(t) = \text{Arg}(e^{it} - c)$ is discontinuous for at least one $t \in \mathbb{R}$ , where $\text{Arg}(z) \in (-\pi, \pi]$ denotes the principal argument. Find the area of the region $S \cap \{c \in \mathbb{C} : |c| \le 2\}$ .

Options

A.
$\frac{7\pi}{6} + \sqrt{3}$
B.
$\frac{2\pi}{3} + \sqrt{3}$
C.
$\frac{4\pi}{3} + 2\sqrt{3}$
D.
$\frac{5\pi}{6} + \frac{\sqrt{3}}{2}$

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

Symmetry and Periodicity|Argument and Principal Value|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

Students often assume that since $e^{it} - c$ is a continuous and periodic function of $t$ , its argument must also be continuous.

They fail to account for the branch cut of the principal argument $\text{Arg}(z) \in (-\pi, \pi]$ , which creates a jump discontinuity whenever the locus crosses the negative real axis.

This question forces students to reconcile local function behavior with global domain constraints by identifying the exact geometric envelope of circles that intersect the branch cut.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Symmetry and Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free