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Question #41564

Difficulty: 8/10 | 🟢 Free

The Question

Find the area of the region enclosed by the curve given parametrically by $x(t) = 2|\cos t|$ and $y(t) = 3\sin 2t$ for $t \in [0, 2\pi]$ .

Options

A.
$4$
B.
$8$
C.
$12$
D.
$16$

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

Parametric Curve Area|Symmetry Utilization|Modulus Function Symmetry|Polynomial Function Parity

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to identify hidden symmetry and retracing in parametric curves caused by modulus functions.

A common trap is blindly applying the parametric area formula $\frac{1}{2} \int_0^{2\pi} |x y' - y x'| dt$ , which yields 16 because the curve traces the same region twice.

Recognizing the parity and periodicity forces the student to partition the interval, convert to Cartesian coordinates, and realize the true enclosed area is only half of the blind integral's value.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Parametric Curve Area
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free