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Question #41592

Difficulty: 8/10 | 🟢 Free

The Question

Let f:(0,6)∖{3}→Rf: (0, 6) \setminus \{3\} \to \mathbb{R} be a differentiable function satisfying the differential equation (x−3)dydx+(x2−6x+9)y=0(x-3) \frac{dy}{dx} + (x^2 - 6x + 9) y = 0 with the initial condition f(2)=e4f(2) = e^4. Let AA be the number of points of inflection of f(x)f(x) in its domain. Let B=lim⁡x→3ln⁡(f(x))B = \lim_{x \to 3} \ln(f(x)). Let C=1C = 1 if f(x)f(x) attains a global maximum on the interval [1,5]∖{3}[1, 5] \setminus \{3\}, and C=0C = 0 otherwise. Find the value of 2A+2B+C2A + 2B + C.

Options

  • A.

    12

  • B.

    13

  • C.

    14

  • D.

    15

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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

Variable Separable Constraints|Concavity Analysis|Limit Existence at Discontinuity|Global Extrema on Intervals

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to rigorously analyze a transcendental function derived from a variable separable differential equation with a removable singularity.

The primary trap lies in the evaluation of the global maximum (Trap 266).

Students often blindly find the critical point (x=3x=3), calculate the function's limit, and incorrectly assume this supremum is the attained global maximum, leading them to erroneously set C=1C=1.

However, because x=3x=3 is explicitly excluded from the domain (Trap 243), the maximum does not exist.

Furthermore, the concavity analysis requires correctly applying the product rule to f′(x)f'(x) to find the true inflection points at x=2x=2 and x=4x=4, bypassing the non-existent critical point.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptVariable Separable Constraints
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree