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Question #68612

Difficulty: 5/10 | 🟢 Free

The Question

Let SS be the set of all continuously differentiable functions f:[0,π]→Rf: [0, \pi] \to \mathbb{R} satisfying (f′(x))2=sin⁡2(2x)(f'(x))^2 = \sin^2(2x) and f(0)=0f(0)=0. Let RR be the region bounded by all the curves in SS. Let S′S' be the subset of SS consisting of functions that do NOT satisfy the condition ∫0π/2f(x)cos⁡(2x) dx≥0\int_0^{\pi/2} f(x) \cos(2x) \, dx \ge 0, and let R′R' be the region bounded by the curves in S′S'. If AA and A′A' are the areas of regions RR and R′R' respectively, what is the value of A−A′A - A'?

Options

  • A.

    π2\frac{\pi}{2}

  • B.

    π\pi

  • C.

    3π2\frac{3\pi}{2}

  • D.

    2π2\pi

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Geometric Interpretation|Inequalities in Definite Integrals|Extraneous Solution Rejection|The Sign-Consistency Principle

Why This Question Is Tricky

Examiner Levers

Negative Thinking

Uses "NOT", "EXCEPT", or "INCORRECT" to flip the logic gate.

Difficulty Boost : +2

Common Mistakes

This question tests the student's ability to handle absolute value ODEs and extraneous solutions combined with integral inequalities (Sign-Consistency Principle).

The ODE (f′(x))2=sin⁡2(2x)(f'(x))^2 = \sin^2(2x) implies f′(x)=±sin⁡(2x)f'(x) = \pm \sin(2x).

Because f′(x)f'(x) must be continuous, it can only change signs at the roots of sin⁡(2x)\sin(2x), specifically at x=π/2x = \pi/2.

This creates 4 distinct valid branches for f(x)f(x).

Students often fail to identify all 4 branches, assuming f′(x)f'(x) must be uniformly sin⁡(2x)\sin(2x) or −sin⁡(2x)-\sin(2x), which would yield an incorrect total area AA.

Furthermore, the negative phrasing 'do NOT satisfy' forces students to correctly evaluate the integral inequality to isolate the specific subset S′S'.

Failing to apply the sign-consistency check or misinterpreting the geometric boundaries of the piecewise functions leads to incorrect area calculations.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptGeometric Interpretation
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree