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Question #40719

Difficulty: 8/10 | 🟢 Free

The Question

Consider the parametric curve defined by the equations: x(t)=∑n=1∞tnnx(t) = \sum_{n=1}^{\infty} \frac{t^n}{n} y(t)=∫0tln⁡(1+u)1+u2duy(t) = \int_0^t \frac{\ln(1+u)}{1+u^2} du Let L=y(1)+dydx∣t=1L = y(1) + \left. \frac{dy}{dx} \right|_{t=1}. Which of the following represents the value of LL?

Options

  • A.

    π8ln⁡2\frac{\pi}{8}\ln 2

  • B.

    π4ln⁡2\frac{\pi}{4}\ln 2

  • C.

    0

  • D.

    \text{Does not exist}

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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

Logarithmic Series|Logarithmic Integration via Trig Manipulation|Parametric Derivative Endpoint Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Parametric Derivative Endpoint Trap' combined with logarithmic series convergence and trigonometric integration.

Students typically evaluate the definite integral correctly using the u=tan⁡θu = \tan \theta substitution to get π8ln⁡2\frac{\pi}{8}\ln 2.

For the derivative, the common mistake is to blindly apply the parametric derivative formula dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, which yields an apparent value of 00 at t=1t=1, leading to the ghost solution π8ln⁡2\frac{\pi}{8}\ln 2.

However, this ignores the domain restriction of the logarithmic series x(t)=∑tnnx(t) = \sum \frac{t^n}{n}, which diverges at t=1t=1 (the harmonic series).

Because the function x(t)x(t) is undefined at the endpoint, the derivative at that point does not exist.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptLogarithmic Series
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree