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Question #41485

Difficulty: 8/10 | 🟢 Free

The Question

Let $C$ be the curve defined by the equation $\ln x = y \ln y + \ln(1 + \ln y)$ . Let $R$ be the finite region in the first quadrant bounded by the curve $C$ , the $y$ -axis, and the line $y=1$ . The area of the region $R$ is

Options

A.
$1 - e^{-1/e}$
B.
$0$
C.
$e^{-1/e}$
D.
$1 - \frac{1}{e}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Horizontal vs Vertical Strips|Axis of Integration Selection|Logarithmic Differentiation Domain

Why This Question Is Tricky

Common Mistakes

This question tests the selection of the axis of integration and the identification of hidden domain constraints.

The curve is given in a form that makes integrating with respect to $x$ (vertical strips) analytically impossible, forcing the student to integrate with respect to $y$ (horizontal strips).

The critical trap lies in the logarithmic domain: the term $\ln(1 + \ln y)$ restricts the domain to $y > 1/e$ .

Students who ignore this constraint often incorrectly assume the region extends down to $y=0$ (since $\lim_{y \to 0^+} y^y = 1$ ), leading to an incorrect area evaluation of $1 - 1 = 0$ .

Recognizing the domain naturally provides the lower limit of integration where the curve asymptotically meets the $y$ -axis.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Integrals - Class 12
Concept	Horizontal vs Vertical Strips
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free