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Question #40784

Difficulty: 8/10 | 🟢 Free

The Question

Find the area of the region in the $xy$ -plane defined by the inequalities $x + y \ge \sqrt{2(x^2 + y^2 - 8)}$ and $x^2 + y^2 \le 32$ .

Options

A.
$8\sqrt{3} + \frac{4\pi}{3}$
B.
$8\sqrt{3} + \frac{16\pi}{3}$
C.
$16\sqrt{3} + \frac{8\pi}{3}$
D.
$16\sqrt{3} + \frac{32\pi}{3}$

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

Normal Form Parameters|Locus of a Moving Point|Geometrical Interpretation of Modulus

Why This Question Is Tricky

Common Mistakes

This question tests two critical traps in coordinate geometry and algebra.

First, the 'Modulus/Square Root Sign Trap': squaring $x+y \ge \sqrt{\dots}$ requires the condition $x+y \ge 0$ , which restricts the locus to a half-plane (bifurcating the full lines into rays).

Second, the 'Implicit Domain Trap': the expression $\sqrt{2(x^2+y^2-8)}$ is only real when $x^2+y^2 \ge 8$ .

Failing to subtract the area of this inner undefined region, or failing to halve the area due to the $x+y \ge 0$ condition, leads directly to the carefully constructed distractor options.

Quick Facts About This Question

Question Information

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