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Question #40723

Difficulty: 8/10 | 🟢 Free

The Question

Let $f(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n}$ be defined for all $x$ where the series converges. Let $H(x) = \int_0^x (f(t) + t f'(t)) dt$ . Consider the limit $L = \lim_{x \to 1^-} (1-x) H'(x)$ and the applicability of the Mean Value Theorem (MVT) to $H(x)$ on the closed interval $[0, 1]$ . Which of the following is correct?

Options

A.
$L = 1$ , MVT is applicable on $[0, 1]$
B.
$L = 1$ , MVT is not applicable on $[0, 1]$ as $H(1)$ diverges
C.
$L = 0$ , MVT is applicable on $[0, 1]$
D.
$L = 0$ , MVT is not applicable on $[0, 1]$ as $H(1)$ diverges

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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|Integrating f(x) + xf'(x) forms|Mean Value Theorem Applicability

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize the differential form d/dx[x f(x)] = f(x) + x f'(x) and integrate it to find a logarithmic series.

The primary trap lies in the application of the Mean Value Theorem (MVT).

Students often mechanically apply MVT to functions that are well-behaved and differentiable on the open interval (0, 1), forgetting that MVT strictly requires continuity on the closed interval [0, 1].

Here, the resulting integral H(x) = -ln(1-x) corresponds to the Harmonic series at x=1, which diverges.

Consequently, H(x) is not continuous at the endpoint x=1, invalidating the hypothesis of MVT.

This exposes a common 'convergence-at-boundary' oversight in Advanced calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Logarithmic Series
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free