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Question #41381

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all values of the parameter $c$ such that the solution $y = f(x)$ of the differential equation $\frac{dy}{dx} - y = \cos x - \sin x$ satisfying $y(0) = 2c + 1$ has exactly two points of inflection in the interval $(0, \pi)$ . Then $S$ is equal to:

Options

A.
$\left( -\frac{1}{2}, \frac{e^{-\pi/4}}{2\sqrt{2}} - \frac{1}{2} \right)$
B.
$\left( \frac{1}{2}, \frac{e^{\pi/4}}{2\sqrt{2}} \right)$
C.
$\left( -\frac{1}{2}, \frac{e^{-3\pi/4}}{2\sqrt{2}} - \frac{1}{2} \right)$
D.
$\left( 0, \frac{e^{-\pi/4}}{\sqrt{2}} \right)$

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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

Point of Inflection Conditions|Arbitrary Constants vs Parameters|Integrating Factor (IF) Sign Sensitivity|General vs Particular Solutions

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to link the solution of a linear differential equation to second-order derivative conditions for points of inflection.

The primary trap is Integrating Factor (IF) sign sensitivity.

A common mistake is taking the IF as $e^x$ instead of $e^{-x}$ , which completely changes the family of curves to $y = \cos x + K e^{-x}$ , leading to the incorrect interval in Option B.

Another trap is forgetting to substitute the initial condition $y(0) = 2c+1$ and directly finding the range for the integration constant $K$ , leading to Option D.

The problem requires careful analysis of the transcendental function $g(x) = e^{-x} \sin x$ to ensure exactly two intersections, verifying the change of sign of $y''$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Application of Derivatives - Class 12
Concept	Point of Inflection Conditions
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free