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Question #41422

Difficulty: 8/10 | 🟢 Free

The Question

Let f(θ)=∫0π/2ln⁡(1−cos⁡2θsin⁡2x) dxf(\theta) = \int_0^{\pi/2} \ln(1 - \cos^2\theta \sin^2 x) \, dx for θ∈(0,2π)\theta \in (0, 2\pi). Find the value of f′′(7π6)f''\left(\frac{7\pi}{6}\right).

Options

  • A.

    −2π3-\frac{2\pi}{3}

  • B.

    −2π-2\pi

  • C.

    2π3\frac{2\pi}{3}

  • D.

    2π2\pi

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Trigonometric Identical Substitutions|Leibniz Rule for Differentiation under Integral|Logarithmic Differentiation Domain|Domain and Range Nuance

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of Leibniz integration and domain/range nuances.

When evaluating the integral ∫0∞dt1+sin⁡2θt2\int_0^\infty \frac{dt}{1 + \sin^2\theta t^2}, students frequently write π2sin⁡θ\frac{\pi}{2\sin\theta} instead of the rigorously correct π2∣sin⁡θ∣\frac{\pi}{2|\sin\theta|}.

Because θ=7π6\theta = \frac{7\pi}{6} lies in the third quadrant where sine is negative, omitting the absolute value leads to a fundamentally incorrect derivative function f′(θ)=π(sec⁡θ−tan⁡θ)f'(\theta) = \pi(\sec\theta - \tan\theta), resulting in the trap answer −2π-2\pi.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterIntegrals - Class 12
ConceptTrigonometric Identical Substitutions
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree