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Question #41124

Difficulty: 8/10 | 🟢 Free

The Question

Let f:R→Sf: \mathbb{R} \to S be a surjective, twice-differentiable function satisfying the differential equation e2f(x)(f′(x))2−2xef(x)f′(x)+x2(1−e2x)=0e^{2f(x)} (f'(x))^2 - 2x e^{f(x)} f'(x) + x^2 (1 - e^{2x}) = 0 for all x∈Rx \in \mathbb{R}, with f(0)=0f(0) = 0. Which of the following sets represents SS?

Options

  • A.

    [0,∞)[0, \infty)

  • B.

    (−∞,∞)(-\infty, \infty)

  • C.

    [ln⁡2,∞)[\ln 2, \infty)

  • D.

    (−∞,0](-\infty, 0]

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Surjective Mapping (Onto)|Variable Transformation Substitution|Implicit to Explicit Conversion

Why This Question Is Tricky

Common Mistakes

The trap tests the student's ability to handle implicit to explicit conversions involving variable transformations (u=ef(x)u = e^{f(x)}).

The differential equation factors into two branches.

The 'Lethal Collision' occurs when students blindly integrate both branches and fail to check the global domain constraints.

The negative branch yields a function that eventually becomes negative, making the logarithmic back-substitution undefined and thus failing the requirement that ff is defined on all of R\mathbb{R}.

Furthermore, the twice-differentiable condition strictly prohibits piecewise branch switching at the critical point x=0x=0 (as u′′(0)u''(0) differs between branches).

Unwary students will either assume both branches are valid and guess (−∞,∞)(-\infty, \infty), or fail to properly minimize the correct branch, falling into a piecewise logic error.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterRelations and Functions - Class 12
ConceptSurjective Mapping (Onto)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree