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Question #40721

Difficulty: 8/10 | 🟢 Free

The Question

Let f:(−13,13)→Rf: \left(-\frac{1}{3}, \frac{1}{3}\right) \to \mathbb{R} be a function defined by the series f(x)=∑n=1∞3nn(n+1)x∣x∣nf(x) = \sum_{n=1}^{\infty} \frac{3^n}{n(n+1)} x |x|^n Let L1=lim⁡x→0+f′′(x)L_1 = \lim_{x \to 0^+} f''(x) and L2=lim⁡x→0−f′′(x)L_2 = \lim_{x \to 0^-} f''(x). Find the value of L1−L2L_1 - L_2.

Options

  • A.

    0

  • B.

    3

  • C.

    6

  • D.

    9

Check Your Answer

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|Second Order Derivative Continuity|Modulus Sign Dependency

Why This Question Is Tricky

Common Mistakes

This question tests the convergence and differentiability of a logarithmic series intertwined with a modulus function.

The core trap is the 'Modulus Sign Dependency' at the critical point x=0x=0.

Students frequently differentiate the series term-by-term without splitting the domain, or they incorrectly apply the chain rule to ∣x∣|x| by omitting the sgn(x)\text{sgn}(x) multiplier.

This leads to the false conclusion that f′′(x)f''(x) is continuous at x=0x=0 (yielding an incorrect difference of 0).

Recognizing the piecewise nature of the expansion is mandatory to identify the jump discontinuity in the second derivative.

Quick Facts About This Question

Question Information

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