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Question #41477

Difficulty: 8/10 | 🟢 Free

The Question

Consider the parametric curve defined by x(t)=t2−2tx(t) = t^2 - 2t and y(t)=∣t∣(3−t)y(t) = |t|(3 - t) for t∈Rt \in \mathbb{R}. The curve intersects itself at a single point, forming a closed loop. The area of the region enclosed by this loop is:

Options

  • A.

    73\frac{7}{3}

  • B.

    143\frac{14}{3}

  • C.

    523\frac{5\sqrt{2}}{3}

  • D.

    823\frac{8\sqrt{2}}{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

Axis of Integration Selection|Parametric Curve Area|Modulus Function Breakdown

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle parametric curves involving modulus functions.

The primary trap lies in the 'Modulus Function Breakdown'.

Students often find the intersection parameters t=1±2t = 1 \pm \sqrt{2} and blindly apply the area formula ∫ydx\int y dx or 12∫(xdy−ydx)\frac{1}{2} \int (x dy - y dx) over the entire interval without partitioning the domain at the zero of the modulus (t=0t=0).

Ignoring the modulus sign change leads to integrating a single polynomial, yielding the incorrect trap answers 823\frac{8\sqrt{2}}{3} or 523\frac{5\sqrt{2}}{3}.

The correct approach forces a multi-split integral to account for the piecewise nature of the curve.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptAxis of Integration Selection
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree