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Question #40969

Difficulty: 8/10 | 🟢 Free

The Question

Let $f: \left(0, \frac{\pi}{4}\right) \to (0, \infty)$ be a continuous function such that for all $x \in \left(0, \frac{\pi}{4}\right)$ , $(\sin x)^{3x} \le f(x) \le (\tan x)^{2x}$ Let $S$ be the set of all possible values of the limit $L = \lim_{x \to 0^+} (f(x))^{\frac{1}{x \ln x}}$ assuming the limit exists. Find the supremum of the set $S$ .

Options

A.
$e^2$
B.
$e^3$
C.
$e^{3/2}$
D.
$e^5$

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

L'Hopital's Pitfall|Logarithmic Differentiation Domain|Cross Multiplication Trap

Why This Question Is Tricky

Common Mistakes

This question tests the student's sensitivity to domain-dependent signs when manipulating inequalities within limits.

Students often assume that to maximize the limit of $f(x)^{g(x)}$ , they should maximize the base $f(x)$ , leading them to choose the upper bound $(\tan x)^{2x}$ and incorrectly obtain $e^2$ .

However, because the exponent $g(x) = \frac{1}{x \ln x}$ is strictly negative as $x \to 0^+$ , the operation of raising to the power of $g(x)$ (or dividing by $x \ln x$ after taking logarithms) reverses the inequality.

Thus, the maximum value of the limit is actually obtained by minimizing the base $f(x)$ , exposing the 'cross-multiplication trap' where students fail to account for the negative sign of the logarithmic term.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Limits and Derivatives - Class 11
Concept	L'Hopital's Pitfall
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free