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Question #41894

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \left\{ x \in \mathbb{R} : \frac{x^2 - 6x + 8}{x - 5} > 0 \text{ and } x < \frac{9}{2} \right\}$ . Let $I_A(x)$ be the indicator function of the set $A$ , defined as $I_A(x) = 1$ if $x \in A$ and $0$ otherwise. Let $y(x)$ be a continuous function on $\mathbb{R}$ satisfying the differential equation $\frac{dy}{dx} = \frac{2x - 6}{x^2 - 6x + 5} I_A(x)$ with the initial condition $y(0) = 0$ . Find the sum of all real solutions to the equation $y(x) = \ln\left(\frac{4}{3}\right)$ .

Options

A.
3
B.
6
C.
0
D.
9

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

Indicator Function Logic|Logarithmic Integration Modulus|System of Inequalities

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of domain restrictions from an indicator function and the absolute value requirement when integrating to a natural logarithm.

The system of inequalities restricts the active domain of the differential equation to the interval (2, 4).

Within this interval, the argument of the logarithm is negative, necessitating the modulus |x^2 - 6x + 5| = -x^2 + 6x - 5.

Students who ignore the indicator function's domain or the modulus will incorrectly solve x^2 - 6x + 1 = 0 or 3x^2 - 18x - 5 = 0, both of which yield a sum of roots equal to 6, falling directly into the trap.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Indicator Function Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free