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Question #41793

Difficulty: 8/10 | 🟢 Free

The Question

Let RR be the region in the complex plane consisting of all points z=x+iyz = x+iy that satisfy the following system of inequalities: 0≤Re(zeiπ/3)≤60 \le \text{Re}(z e^{i\pi/3}) \le 6 0≤Im(zeiπ/3)≤40 \le \text{Im}(z e^{i\pi/3}) \le 4 Re(z)+3Im(z)≤10\text{Re}(z) + \sqrt{3}\text{Im}(z) \le 10 3Re(z)−Im(z)≤10\sqrt{3}\text{Re}(z) - \text{Im}(z) \le 10 Find the area of the region RR.

Options

  • A.

    3233\frac{32\sqrt{3}}{3}

  • B.

    100−13633100 - \frac{136\sqrt{3}}{3}

  • C.

    2020

  • D.

    24−323324 - \frac{32\sqrt{3}}{3}

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

Word Problem Modeling|Redundant Constraints|Slope Sensitivity|Euler's Form and Rotations

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to handle rotational transformations in the complex plane and identify redundant constraints in linear programming.

The primary trap is 'Slope Sensitivity': if a student ignores the rotation eiπ/3e^{i\pi/3} and treats the initial bounds as Cartesian xx and yy, they will incorrectly apply the constraints to an unrotated rectangle, leading to a completely different area (100−1363/3100 - 136\sqrt{3}/3).

Another trap is applying the rotation in the wrong direction, which yields an area of 20.

The problem elegantly demonstrates how a constraint that appears active in one orientation becomes strictly redundant under rotation.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterLinear Programming - Class 12
ConceptWord Problem Modeling
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree