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Question #41558

Difficulty: 8/10 | 🟢 Free

The Question

Let $R$ be the region in the $xy$ -plane defined by the inequality $|x^2 - 2x| + |y - x| \le 1$ . The area of the region $R$ is

Options

A.
$\frac{16\sqrt{2} - 16}{3}$
B.
$\frac{16\sqrt{2}}{3}$
C.
$\frac{8\sqrt{2}}{3}$
D.
$\frac{8\sqrt{2} - 8}{3}$

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

Bounded vs Unbounded Regions|Horizontal vs Vertical Strips|Modulus Inequality Logic

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to choose vertical strips over horizontal strips for complex modulus inequalities and correctly handle the piecewise nature of non-linear modulus arguments.

The primary trap lies in the integral of the modulus function.

Students often treat the modulus as a single continuous expression without splitting the domain at the roots of the inner function ( $t=1$ ), leading to incorrect integration over the entire interval $[0, \sqrt{2}]$ .

This results in the trap answers.

Furthermore, attempting to use horizontal strips leads to an algebraically intractable inverse relation, punishing poor strategic choices in area evaluation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Bounded vs Unbounded Regions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free