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Question #41199

Difficulty: 8/10 | 🟢 Free

The Question

Let ω(x)=1+ix1+x2\omega(x) = \frac{1 + ix}{\sqrt{1+x^2}} for x∈Rx \in \mathbb{R}. Define the function f(x)f(x) by the infinite series: f(x)=∑n=1∞1nIm(ω(x)2n)f(x) = \sum_{n=1}^{\infty} \frac{1}{n} \text{Im}\left( \omega(x)^{2n} \right) Let g(x)=cos⁡(f(x))g(x) = \cos(f(x)). Find the value of 125⋅g′(−34)125 \cdot g'\left(-\frac{3}{4}\right).

Options

  • A.

    -64

  • B.

    64

  • C.

    -48

  • D.

    48

Check Your Answer

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

Infinite Series involving Inverse Tangent|De Moivre’s Theorem Applications|Chain Rule Complexity|Implicit to Explicit Conversion

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly transition from an implicit algebraic complex number to its explicit trigonometric form using De Moivre's Theorem, while rigorously handling the principal argument and branch cuts.

The core trap lies in the evaluation of the infinite sine series (derived via the Taylor series of the complex logarithm).

Students frequently memorize the standard Fourier series result ∑sin⁡(nθ)n=π−θ2\sum \frac{\sin(n\theta)}{n} = \frac{\pi - \theta}{2} without the domain restriction θ∈(0,2π)\theta \in (0, 2\pi).

For negative xx, the argument θ=2tan⁡−1x\theta = 2\tan^{-1} x falls in (−π,0)(-\pi, 0).

Because the modulus of the complex logarithm must be positive, a phase shift of π\pi occurs, requiring the odd extension of the series which yields −π+θ2-\frac{\pi + \theta}{2}.

Failing to account for this branch cut leads to a sign error in the explicit function g(x)g(x), resulting in the trap answer 6464 instead of −64-64 after applying the Chain Rule.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptInfinite Series involving Inverse Tangent
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree