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Question #41228

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = f(x)$ be the continuous solution to the differential equation $\frac{dy}{dx} + y \tan(\arcsin(\sin x)) = y^3 \sin x$ for $x \in \left( \frac{\pi}{2}, \pi \right)$ . If $f\left(\frac{2\pi}{3}\right) = \frac{1}{2}$ , then the value of $\left( f\left(\frac{5\pi}{6}\right) \right)^{-2}$ is

Options

A.
$9 + \sqrt{3}$
B.
$\frac{13 - 3\sqrt{3}}{9}$
C.
$15 - \sqrt{3}$
D.
$\frac{11 + 3\sqrt{3}}{9}$

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

Variable Transformation Substitution|Linear Differential Equation Structure|Logarithmic Integration Modulus

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to handle variable transformations in Bernoulli differential equations while navigating a lethal domain trap involving inverse trigonometric functions.

The 'Lethal Collision' occurs when evaluating $\arcsin(\sin x)$ for $x \in (\pi/2, \pi)$ .

Students who mechanically assume $\arcsin(\sin x) = x$ will incorrectly formulate the linear differential equation as $v' - 2v \tan x = -2\sin x$ , leading to an incorrect integrating factor of $\cos^2 x$ instead of $\sec^2 x$ .

This fundamentally alters the integration trajectory, forcing a sign flip in the exponent of the integrating factor, and results in a highly plausible but incorrect trap option.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Variable Transformation Subsitution
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free