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Question #40350

Difficulty: 8/10 | 🟢 Free

The Question

Let $y = f(x)$ be a differentiable function defined on $(-\infty, 0)$ satisfying the differential equation $x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + 2y^2 - x^2 = 0$ If $f(-1) = 0$ and $f'(-1) > 0$ , then the value of $f(-e^{-\pi/6})$ is:

Options

A.
$\frac{1}{2} e^{-\pi/6}$
B.
$-\frac{1}{2} e^{-\pi/6}$
C.
$\frac{\sqrt{3}}{2} e^{-\pi/6}$
D.
$-\frac{\sqrt{3}}{2} e^{-\pi/6}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Polynomial Function Parity|Homogeneous Function Homogeneity|Implicit to Explicit Conversion

Why This Question Is Tricky

Common Mistakes

This question tests the implicit-to-explicit conversion of a homogeneous differential equation in a strictly negative domain.

The lethal trap lies in the algebraic manipulation of $\frac{1}{x}\sqrt{x^2 - y^2}$ .

Students routinely write this as $\sqrt{1 - (y/x)^2}$ , ignoring that for $x < 0$ , $x = -|x| = -\sqrt{x^2}$ .

The correct conversion introduces a parity-induced sign change, yielding $+\sqrt{1 - (y/x)^2}$ .

Failing to account for this absolute value property leads to an inverted sign in the differential equation, resulting in the incorrect solution $y = -x \sin(\ln(-x))$ and ultimately trapping the student in option B.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 11
Concept	Polynomial Function Parity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free