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Question #41189

Difficulty: 8/10 | 🟢 Free

The Question

Consider the linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ for $x \in (1, \sqrt{2})$ , where $P(x) = \frac{2}{x\sqrt{x^2-1}(\sec^{-1}x - \csc^{-1}x)}$ Let $I(x) = e^{\int P(x) dx}$ be the integrating factor of this differential equation, where the constant of integration is taken to be zero. The value of $\int_{1}^{\sqrt{2}} I(x) dx$ is:

Options

A.
$2\ln(\sqrt{2} + 1) - \frac{\pi}{2}$
B.
$\frac{\pi}{2} - 2\ln(\sqrt{2} + 1)$
C.
$\ln(\sqrt{2} + 1) - \frac{\pi}{4}$
D.
$\frac{\pi\sqrt{2}}{2} - \ln(\sqrt{2} + 1)$

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

Algebraic-Trigonometric Graphical Mapping|Linear Differential Equation Structure|Integrating Factor (IF) Sign Sensitivity

Why This Question Is Tricky

Common Mistakes

This question tests the student's rigorous understanding of the Integrating Factor (IF) formula $I(x) = e^{\int P(x) dx}$ .

A common and lethal mistake is evaluating $\int \frac{f'(x)}{f(x)} dx$ as $\ln(f(x))$ instead of $\ln|f(x)|$ .

In this problem, $f(x) = 2\sec^{-1}x - \frac{\pi}{2}$ is strictly negative in the given domain $x \in (1, \sqrt{2})$ .

Failing to apply the absolute value results in an incorrect IF of $2\sec^{-1}x - \frac{\pi}{2}$ , which completely inverts the sign of the final definite integral.

The problem elegantly punishes pattern memorization by ensuring the incorrect sign leads directly to a plausible distractor option, distinguishing top-tier students who respect domain-range constraints of transcendental functions.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Algebraic-Trigonometric Graphical Mapping
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free