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Question #72125

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Rf: \mathbb{R} \to \mathbb{R} be a continuous function such that f(1)=2e−1f(1) = 2e^{-1} and satisfies the differential equation (x2−3x)∣x−2∣f′(x)=(x2−3x)(x−2)(f(x)−x)(x^2 - 3x) |x-2| f'(x) = (x^2 - 3x) (x-2) (f(x) - x) for all x∈R∖{0,2,3}x \in \mathbb{R} \setminus \{0, 2, 3\}. Let SS be the set of all points in R\mathbb{R} where f(x)f(x) is not differentiable. If k=∑x∈Sxk = \sum_{x \in S} x, then the value of ∫0kf(x)dx\int_0^k f(x) dx is:

Options

  • A.

    2−2e−22 - 2e^{-2}

  • B.

    2+2e−22 + 2e^{-2}

  • C.

    4−2e−24 - 2e^{-2}

  • D.

    2−4e−22 - 4e^{-2}

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

Implicit to Explicit Conversion|Right-Hand vs Left-Hand Derivatives|The Zero-Factor Cancellation Trap

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

The question tests the 'Zero-Factor Cancellation Trap' by embedding roots x=0 and x=3 into the differential equation.

Students often blindly cancel the algebraic factor and incorrectly assume that points of undefined DE behavior automatically translate to points of non-differentiability for the continuous solution.

This leads to an incorrect set S={0,2,3} and a wrong upper limit k=5 for the integral.

The rigorous approach requires recognizing that continuity at x=0 and x=3 forces the integration constants to match, ensuring smooth transitions, leaving x=2 as the sole point of non-differentiability due to the absolute value term.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterDifferential Equations - Class 12
ConceptImplicit to Explicit Conversion
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree