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Question #61073

Difficulty: 8/10 | 🟢 Free

The Question

Let $p$ and $q$ be distinct positive real numbers. Consider the limits: $L_1 = \lim_{x \to 0^+} \frac{|x|^{p|x|} - |x|^{q|x|}}{x \ln|x|} \quad \text{and} \quad L_2 = \lim_{x \to 0^-} \frac{|x|^{p|x|} - |x|^{q|x|}}{x \ln|x|}$ Which of the following represents the ordered pair $(L_1 + L_2, L_1 - L_2)$ ?

Options

A.
$(0, 2(p-q))$
B.
$(2(p-q), 0)$
C.
$(p-q, p-q)$
D.
$(0, p-q)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Logarithmic Differentiation|L'Hopital's Pitfall|Right-Hand vs Left-Hand Derivatives

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'L'Hopital's Pitfall' and the handling of absolute values in logarithmic differentiation.

Students often blindly apply L'Hopital's Rule by differentiating $x^{px}$ instead of $|x|^{p|x|}$ , ignoring the chain rule on $|x|$ which yields $\text{sgn}(x)$ .

This leads to the incorrect assumption that the left-hand and right-hand limits are identical (yielding $2(p-q)$ for the sum).

By forcing the expansion of $e^{p|x|\ln|x|}$ , the problem highlights the non-differentiability at $x=0$ and the resulting sign flip in the left-hand limit.

Quick Facts About This Question

Question Information

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