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Question #41547

Difficulty: 8/10 | 🟢 Free

The Question

Let AA be the area of the region bounded by the curves y=sin⁡−1(sin⁡x)y = \sin^{-1}(\sin x) and y=cos⁡−1(sin⁡x)y = \cos^{-1}(\sin x) for x∈[0,2π]x \in [0, 2\pi]. The value of AA is:

Options

  • A.

    5π24\frac{5\pi^2}{4}

  • B.

    25π28\frac{25\pi^2}{8}

  • C.

    π2\pi^2

  • D.

    5π28\frac{5\pi^2}{8}

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

Intersection Point Accuracy|General Solution Principal Values|The Sum-Difference Formula Constraints

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle the restricted principal domains of inverse trigonometric functions when calculating areas bounded by periodic curves.

The 'Lethal Collision' occurs if a student blindly assumes sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x and cos⁡−1(sin⁡x)=π2−x\cos^{-1}(\sin x) = \frac{\pi}{2} - x over the entire interval [0,2π][0, 2\pi], ignoring the piecewise nature of the principal branches.

This false assumption leads to integrating ∣2x−π2∣|2x - \frac{\pi}{2}| over [0,2π][0, 2\pi], yielding the trap answer 25π28\frac{25\pi^2}{8}.

Furthermore, students might incorrectly solve for intersection points by applying the general solution of sin⁡x=cos⁡x\sin x = \cos x without verifying if those points lie on the correct branch of the inverse functions, leading to double-counting or missing area segments when the integration bounds cross the principal domain thresholds (e.g., at x=π2x = \frac{\pi}{2} and x=3π2x = \frac{3\pi}{2}).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterApplication of Integrals - Class 12
ConceptIntersection Point Accuracy
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree